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ALL RIGHTS RESERVED
©COPYRIGHT BY THE AUTHOR
The whole or any part of this publication may not be reproduced or transmitted in any
form or by any means without permission in writing from the publisher. This includes
electronic or mechanical, including photocopying, recording, or any information storage
and retrieval system.
Every effort has been made to obtain copyright of all printed aspects of this publication.
However, if material requiring copyright has unwittingly been used, the copyrighter is
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acknowledgement can be made by the author.
Maths Textbook Grade 9 NCAPS
ISBN 13:
978-1-869-21973-4 ( Perpetual book )
978-1-869-21775-4 ( 1 Year License )
Product Code:
MAT 145
Authors:
M.D Phillips
J. Basson
J. Odendaal
First Edition:
August 2014
Second Edition: November 2014
Third Edition:
August 2015 (Minor revisions)
PUBLISHERS
ALLCOPY PUBLISHERS
P.O. Box 963
Sanlamhof, 7532
Tel: (021) 945 4111, Fax: (021) 945 4118
Email: info@allcopypublishers.co.za
Website: www.allcopypublishers.co.za
i
MATHS TEXTBOOK
GRADE 9 NCAPS INFORMATION
The AUTHORS
MARK DAVID PHILLIPS
B.SC, H Dip Ed, B.Ed (cum laude) University of the Witwatersrand.
Mark Phillips has over twenty-five years of teaching experience in both state and private schools and has a
track record of excellent matric results. He has presented teacher training and learner seminars for various
educational institutions including Excelearn, Isabelo, Kutlwanong Centre for Maths, Science and
Technology, Learning Channel, UJ and Sci-Bono. Mark is currently a TV presenter on the national
educational show Geleza Nathi, which is broadcast on SABC 1. He has travelled extensively in the United
Kingdom, Europe and America, gaining invaluable international educational experience. Based on his
experiences abroad, Mark has successfully adapted and included sound international educational
approaches into this South African textbook. The emphasis throughout the textbook is on understanding
the processes of Mathematics.
JURGENS BASSON
B.A (Mathematics and Psychology) HED RAU (UJ)
Jurgens Basson is a Mathematics consultant with more than 25 years experience and expertise in primary,
secondary and tertiary education. His passion is the teaching of Mathematics to teachers, students and
learners. Jurgens founded and started the RAU / Oracle School of Maths in 2002 in partnership with
Oracle SA. Due to the huge success of this program in the community, there are now several similar
programs running in the previously disadvantaged communities that are being sponsored by corporate
companies. During 2012, one of his projects sponsored by Anglo Thermal Coal in Mpumalanga was voted
Best Community Project of the Year. Jurgens had the privilege to train and up skill more than 15 000
teachers countrywide since 2006. He was also part of the CAPS panel that implemented the new syllabus
for Grade 10 - 12 learners. He successfully co-authored the Mind Action Series Mathematics textbooks,
which received one of the highest ratings from the Department of Education.
JACO ODENDAAL
B.Sc (Mathematics and Applied Mathematics)
Jaco Odendaal has been an educator of Mathematics and Advanced Programme Mathematics for the past
ten years. He is currently the Head of Mathematics at Parktown Boys’ High School in Parktown,
Johannesburg. He has worked as a teacher trainer in many districts in Limpopo and Mpumalanga. Jaco
has a passion for Mathematics and the impact it can have in transforming our society. He has been
successful in teaching and tutoring Mathematics at all levels, from primary school to university level. He
has also worked with learners at both ends of the spectrum, from struggling to gifted learners. He is known
for his thorough, yet simple explanation of the foundational truths of Mathematics.
The CONCEPT
The focus in this textbook is on providing learners with crucial background knowledge and skills needed to
cope with Mathematics in the higher grades. The emphasis is on the understanding of concepts which are
reinforced through more than enough quality examples and exercises. Each exercise is based on given
concepts at a “standard grade” level in order to ensure that learners master the basics. At the end of each
chapter, revision exercises consolidate all of the concepts by providing learners with an opportunity to
tackle questions of a mixed nature. These are the typical examination-type questions. The challenge
questions are of a “higher grade” nature and serve to extend the learners and develop problem-solving
skills. The Grade 8 and 9 publications serve to bridge the gap between the Senior Phase and FET phase.
The educator’s guide contains approaches to the teaching of the particular topic, detailed solutions to the
exercises, and various assessment tasks. The publication is also available in e-book/e-pub format.
ENDORSEMENT
This textbook is truly amazing. It is such pleasure to be able to work through a book that has thorough
explanations and excellent examples. I am blown away with the different approaches and methodologies
used and the layout is user friendly. The exercises are lengthy and cater for students of varied abilities.
Each section is well presented and thoroughly researched. It is such a breath of fresh air to work with a
book that is completely CAPS aligned. The challenges at the end of the chapter really extend my learners
and I am able to alleviate the boredom of my bright learners while working with my weaker learners.
This textbook is truly in a league of its own and will help to raise the standard of Maths in this country.
Belinda Pretorius (Curro Inter-schools Head of Mathematics)
ii
MATHEMATICS
TEXTBOOK
GRADE 9 NCAPS
CONTENTS
CHAPTER 1
WHOLE NUMBERS
1
CHAPTER 2
INTEGERS
34
CHAPTER 3
COMMON FRACTIONS
39
CHAPTER 4
DECIMAL FRACTIONS
47
CHAPTER 5
EXPONENTS
56
CHAPTER 6
NUMERIC AND GEOMETRIC PATTERNS
69
CHAPTER 7
FUNCTIONS AND RELATIONSHIPS
80
CHAPTER 8
ALGEBRAIC EXPRESSIONS
94
CHAPTER 9
ALGEBRAIC EQUATIONS
110
CHAPTER 10
CONSTRUCTIONS
127
CHAPTER 11
GEOMETRY OF 2D SHAPES
143
CHAPTER 12
GEOMETRY OF LINES
175
CHAPTER 13
THEOREM OF PYTHAGORAS
181
CHAPTER 14
AREA AND PERIMETER OF 2D SHAPES
191
CHAPTER 15
GRAPHS
212
CHAPTER 16
SURFACE AREA AND VOLUME OF 3D SHAPES
238
CHAPTER 17
TRANSFORMATIONS
254
CHAPTER 18
GEOMETRY OF 3D SHAPES
273
CHAPTER 19
DATA HANDLING
290
CHAPTER 20
PROBABILITY
308
iii
CHAPTER 1: NUMBERS, OPERATIONS AND RELATIONSHIPS
TOPIC: WHOLE NUMBERS
What is a number? This is a very important question in mathematics, but not an easy one to answer.
If you could ask someone living in 500 BC, you would not get the same answer as when you ask a
21st century mathematician. Many numbers we work with today were totally unheard of or
considered very strange in earlier times. Just think of your own journey in Mathematics so far.
Before you started school, you may have thought of numbers merely as tools for counting. Your
entire number system would have been: 0; 1; 2; 3; 4; ... etc. During your primary school career, you
learnt about negative numbers, fractions and decimals. In Grade 8 you came across interesting
numbers like 2 and π . In fact, the way in which humanity discovered new kinds of numbers is not
all too different from the way a typical child finds out about these numbers today. It just happens
much quicker now, because it’s all been done before. We will now see how one type of number leads
to the next by asking the right questions.
THE NUMBER SYSTEM
Some of the most important building blocks of mathematics are numbers, operations and
relationships. Think of the simple sentence: 1  2  3 . Here we have all three of the building blocks.
The ‘1’, ‘2’ and ‘3’ are numbers, the ‘+’ is the operation and the ‘  ’ is the relationship. You will see
that, as we introduce more operations, we need new types of numbers to answer the questions and
this is how our number system grows.
Whole Numbers
Whole numbers are the numbers we use to count: 0; 1; 2; 3; 4; 5; ... You should remember, from
Grade 8:
The set of natural numbers  N  1; 2;3; 4;5;...
The set of whole numbers  N0  0;1; 2;3; 4;5;...
Some mathematicians don’t distinguish between natural numbers and whole numbers and include 0
in the natural numbers. We will keep the distinction. Any two whole numbers can be added or
multiplied and the result will still be a whole number. When we subtract or divide, we may run into
trouble. What is 4 – 7 or 2  5 , for example? To find the answers to these questions, we need new
types of numbers. Let’s start with the subtraction issue.
Integers
In order to get an answer for something like 4 – 7 (where we subtract a greater number from a
smaller number), we need negative numbers. You should know that 4  7  3 . The number –3 does
not belong to the set of whole numbers. If we take all the whole numbers, together with all the
negative numbers, we get a new set of numbers called integers. The symbol for integers is Z . This
comes from the German word ‘zahlen’ which means number. Z  ...; 3; 2; 1;0;1; 2;3;... .
Note that the integers include the whole numbers. All whole numbers are also integers, but careful –
not all integers are whole numbers.
We can add, multiply or subtract any two integers and the result will always be another integer, but
what about division? We have no integer to give as an answer to 2  5 .
Rational Numbers
So what is the answer to 2  5 ? From your knowledge of fractions and decimals, you should know
2
that the answer is or 0,4. If we allow for fractions, we can perform any division of two integers:
5
a
a b 
where a and b are integers. The only exception is division by 0. We cannot divide by 0.
b
This is undefined.
1
If we take all the integers, together with all the fractions, we get a new set of numbers called the
rational numbers. The symbol for rational numbers is Q . This comes from the word ‘quotient’. The
a
rational numbers is the set of all numbers of the form with the following conditions: a must
b
belong to the integers, b must belong to the integers and b may not be 0.’ To summarise this in easyto-understand language, we can say that a rational number is a number of the form:
any integer
3
. Note that this includes integers as well, for example: 3  .
any non-zero integer
1
a

We write Q   : a  Z ; b  Z ; b  0 
b

Note on Equivalent Fractions
Fractions can be equivalent. This means that we can have two fractions that look different, but are
4 2
equal in value, for example:  . To check quickly whether two fractions are equivalent, we use
6 3
the cross-multiplication rule:
a c
4
2

If
then a  d  b  c . If you do this test on
and
you will find: 4  3  12  6  2 ,
b d
6
3
indicating that they are equivalent.
The rational numbers include:
All integers (3; 0; – 2 etc.)

 1 2 4

etc. 

All proper fractions  ; ;
2 3 5




1
 13 2

All improper fractions or mixed numbers  ; 1 ;  11 etc. 
3
2
7

All terminating decimals (decimals that end i.e. 0,4; 2,65; 3,14159265 etc.)
   0,272727...;0,256  0,2565656... etc )
All recurring decimals ( 0,3  0,333333...; 0,27
Note on Terminating and Recurring Decimals




225
9

.
1000 40
All recurring decimals can be written as common fractions too. There is a simple procedure for
doing this, which we will discuss in Grade 10, but it is worth knowing a few tricks here:
4
1
2
3 1
   23 ; 0,36
   36  4 etc.
0, 23
0,1  ; 0, 2  ; 0,3   ; 0, 4  etc.
9
99
99 11
9
9
9 3
12 3 6 1 36  1 37
  

Can you figure out this one: 1, 23  
?
10 90 5 30
30
30
It is also very interesting, and perhaps surprising, to note that 0,9  1 ; 1,9  2 ; 0, 29  0,3 etc.
All terminating decimals can be written as common fractions. e.g. 0, 225 
Note again that the rational numbers include the integers and therefore also the whole numbers and
natural numbers. All integers are also rational numbers, but be careful – not all rational numbers are
integers. We can add, multiply, subtract or divide any two rational numbers and the result will always
be another rational number. The only exception is division by 0, which is undefined.
Rational numbers are dense. This means that, between any two rational numbers, we can always find
1
3
5
1
5
9
more rational numbers. Between
and
we have , between
and
we have
and so we
2
4
8
2
8
16
can carry on forever. So are there still other types of numbers?
2
Irrational Numbers
Mathematicians believed for a long time that rational numbers were sufficient. They thought that any
quantity can be expressed as a fraction of two integers. But certain discoveries, especially in the
world of geometry, suggested otherwise.
Right Angled Triangles
What is the length of the hypotenuse of the given right angled triangle?
x
1
Using the theorem of Pythagoras, we get:
x 2  12  12
1
 x2  2
x  2
This number
2 cannot be written as a terminating or as a recurring decimal. In fact,
2  1, 4142135... No matter how many decimals we work out, we will never be able to represent
the exact value of 2 in decimal form. The digits don’t recur either. This means that it is impossible
to write it as a common fraction (i.e. a ratio between two integers). The number 2 is an example of
what we call an irrational number.
Circles
Circumference
What is the ratio of the circumference of a circle to its diameter?
Tradesmen have known, from the earliest times that the
Diameter
circumference of a circle is a bit more than three times
its diameter, but the exact value of this ratio proved illusive.
22
Approximations include 3,14, 3,1416,
etc. It is, however,
7
impossible to express this ratio as a common fraction using an integer numerator and denominator or
as a terminating or recurring decimal. Mathematicians gave this ratio the name  (pi) and with the aid
of computers, more decimal places are being calculated every day, but it will never stop.  is an
irrational number. Here are the first 100 decimal places:
  3,1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679...
Irrational numbers are non-terminating, non-recurring decimals. They cannot be expressed as a ratio
between integers. Examples include:
 Square roots of numbers that are not perfect squares 2; 3; 5 etc.


3
2; 3 3; 3 4 etc.

Cube roots of numbers that are not perfect cubes

(This can be extended to other types of roots as well)
Transcendental numbers  π ; e etc.


These strange, but important, numbers are called transcendental because they cannot be
produced by performing addition, subtraction, multiplication, division, powers, roots or ordinary
algebra on integers or rational numbers. In this way they ‘transcend’ the rational numbers.
3
Real Numbers
If we think of a number as any possible length or any point on a number line, we come to what we
call the real numbers. They include all the numbers we have already discussed. The real numbers
are all the rational numbers and irrational numbers put together. Consider the following illustration.
-3
-1,5
-2
-1
-1
Rational Number
Real Number
Integer
Rational Number
Real Number
0
1
2
3
0
Whole Number
2
3
2
2
Integer
Rational Number
Rational Number
Real Number
π
Irrational Number
Natural Number
Irrational Number
Real Number
Whole Number
Real Number
Real Number
Integer
Rational Number
Real Number
So any number that has a position on the number line is a real number. Believe it or not, there are still
more numbers that don’t have positions on the number line, but are used by mathematicians. We will
not study any numbers apart from real numbers in school mathematics though.
You have to be able to recognise calculations that do not produce real numbers. There are mainly
two such calculations:

Square roots of negative numbers do not produce real numbers 2; 4 etc



These numbers exist, but don’t have a position on the number line. We call them non-real
numbers.
 6 2 0 π

Division by zero does not produce a real number  ; ; ; etc 
0 0 0 0

These calculations are completely undefined and there are in fact no such numbers.
Summary of the Number System
N  1; 2;3; 4;5;...
1.
Natural Numbers:
2.
Whole Numbers:
3.
Integers:
Z  ...; 3; 2; 1;0;1; 2;3;...
4.
Rational Numbers:
a

Q   : a  Z ; b  Z ; b  0
b

5.
N0  0;1; 2;3; 4;5;...
- Integers
- Proper Fractions
- Improper fractions and mixed numbers
- Terminating decimals
- Recurring decimals
Irrational Numbers: Q|
- Non-terminating, non-recurring decimals.
- Square roots of numbers that are not perfect squares, cube roots of numbers that are not
perfect cubes etc.
-
4
6.
7.
Real Numbers: R
Any number on the number line. All rational and irrational numbers put together.
Calculations that do NOT produce real numbers:
- Square roots of negative numbers
- Division by zero
R
2,15246675082532.....
6
7
2
0,7265
0,4
2
1
Q
Z
7
4

4
3
4
0
N0
55
1
100
3
3

5
100
7 2 3
1
2
52

5
Example 1
State whether the following numbers are rational, irrational or neither:
π 8
(a)
0,25
(b)
–4
(c)
(d)
11
2
5
(h)
(e)
11
(f)
(g)
9
3
0
π
(i)
(j)
(k)
0,3232323232323232...
0, 6820045370518346...
π
π
4
3
3
(o)
(m)
(n)
(l)
8
7
3
9
4
2π
0π
(p)
(q)
(r)
(s)
16
49
Solutions
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l)
(m)
(n)
Rational (Terminating decimal)
Rational (Integer)
Irrational (The sum of an irrational number and a rational number is always irrational)
Irrational (11 is not a perfect square, therefore the square root of 11 is irrational)
Neither (Square root of a negative is non-real)
Rational (9 is a perfect square, therefore the square root of 9 is rational)
Rational (Any common fraction is rational)
Neither (Division by 0 is undefined)
Rational (This equals 1, which is an integer and therefore rational)
Rational (Recurring decimal)
Irrational (Non-terminating, non-recurring decimal)
Rational (This equals –2 since –8 is the cube of –2. –2 is an integer and therefore rational)
Irrational (It is real, because cube roots of negatives are real, but, since 7 is not a perfect
cube, the cube root of –7 is irrational).
Irrational (Any fraction of an irrational number is irrational)
5
(o)
(p)
(q)
(r)
(s)
4 2
 which is a common fraction)
9 3
Irrational (An irrational number multiplied by a rational number is always irrational, except
when the rational number is 0 – see (q))
Rational (This equals 0 which is an integer and therefore rational)
Neither (Just like the square root of a negative the 4th root of a negative is also non-real.
The same applies for 6th roots, 8th roots, 10th roots etc).
Irrational (9 + 4 = 13 which is not a perfect square, therefore its square root is irrational)
Rational (4 and 9 are both perfect squares.
Example 2
Find a rational number between
3
2
and .
5
3
Solution
Start by getting the lowest common denominator of the fractions and rewrite them over this
3 9
2 10

and  . Since we can’t find any fifteenths between 9 fifteenths and 10
denominator:
5 15
3 15
fifteenths, we can try doubling the denominators (and therefore also the numerators):
9 18
10 20
19

and

. Now we have a fraction, namely,
, between the two fractions.
15 30
15 30
30
Example 3
Between which two consecutive integers do the following irrational numbers lie?
(a)
11
(b)
 11
(c)
π
2
(d)
3π
Solution
(a)
9  11  16
 9  11  16
(b)
 3  11  4
11 lies between 3 and 4.
9  11  16
 9  11  16
 3  11  4
4   11  3
 11 lies between 4 and 3 .
(c)
  3,14159.......
 3,14159.......
 
 1,570796......
2
2
1  1,57259.....  2
π
1   2
2
(d)
3π  9, 42477......
10  9, 42477......  9
10  3  9
Example 4
Find a rational numbers between 3,14 and π . You may use your calculator.
Solution
π  3,1415... (and many more decimal places) as your calculator will show you.
Now it’s easy to see that 3,1401; 3,1402; 3,1403; ... ; 3,1414; 3,1415 are all between 3,14 and π .
Any one of them will do.
6
Example 5
Give an example, if possible, of:
(a)
two irrational numbers of which the product is rational
(b)
two irrational numbers of which the sum is rational
(c)
a rational number and an irrational number with a rational product
(d)
a rational number and an irrational number with a rational sum
Solutions
(b)
1
 1 and (2  3)(2  3)  1 *
π
* This last one will make more sense once you have done the algebra section.
Some possibilities are: 2   2  0 , π    π   0 , π  (5  π)  5 and
(c)
(d)
2,3142679508...  1, 2857320491...  3, 6 *
* Test this last one for yourself – it’s quite interesting!
The only way to do this is to use 0 as the rational number, for example: 0  5  0
This is impossible. Can you explain why?
(a)
Some possibilities are:

3  2 3  6, π

You may not use a calculator in this exercise.
EXERCISE 1
(a)
2 2  2,
Complete the following table:
Number
e.g. −3
(a) 0
(b) 7
(c) 
(d)  23
(e) 0,73
Natural Whole Integer Rational Irrational Real
No
No
Yes
Yes
No
Yes
(f) 1, 2357
(g)
(h)
π
3π
3
0
2 14
(i)
(j) 1,9
16
(k)
(l)
(b)
(c)
3
8
3
; 2; 9; 0; 2; 4 , write down all the
4
(1)
natural numbers
(2)
whole numbers
(3)
integers
(4)
rational numbers
(5)
irrational numbers
(6)
real numbers
State whether each of the following numbers are rational, irrational or neither:
π
1
3
4,01345
33
(2)
(4)
(3)
(1)
2
121
2
3
(5)
2
(6)
 2
(7)
27  1
(8)
2
From the list of numbers: 3;
7
(d)
(e)
(f)
(g)
(h)
List the possible values of a for which a is between 6 and 7 and a  N .
Find a rational number between
1
1
2 and 5
π and 4
(1)
(3)
and
(2)
2
3
Find an irrational number between
2 and 3
(3)
π and 2π
(1)
2 and 3
(2)
Say whether the following statements are true or false:
(1)
The sum of two rational numbers is always a rational number.
(2)
The sum of two irrational numbers is always a rational number.
(3)
The product of two rational numbers is always a rational number.
(4)
The product of two irrational numbers is always an irrational number.
(5)
The product of two real numbers is always a real number.
(6)
When a real number is divided by a real number the result is always a real number.
(7)
The sum of a rational number and an irrational number is always an irrational number.
(8)
The product of a rational number and an irrational number is always an irrational
number. (Be careful!)
Between which two consecutive integers do the following irrational numbers lie?
(1)
7
(2)
 3
(3)
97
PRIME NUMBERS AND THEIR USES
In Grade 8, you learnt that a prime number is a natural number with exactly two factors (no more and
no less): the number itself and 1. This means that 1 is not prime because it has only one factor
(namely 1). Also, 4 is not a prime number because it has 3 factors (namely 1, 2 and 4). Prime
numbers are: 2; 3; 5; 7; 11; 13; 17; 19; 23; 29; ...
A number with more than two factors is called a composite number.
You also learnt how to decompose any natural number (except for 1) into prime factors and how to
use this to check whether one number is a factor of another and to find the highest common factor
and the lowest common multiple of any two (or more) numbers. Let’s revise this briefly.
REVISION OF PRIME NUMBERS – GRADE 8
Decomposing a number into prime factors
Example 6
Decompose 144 into prime factors.
Solution
2 144
2 72
2 36
2 18
39
33
1
144  24  32
A rule for checking whether one number is a factor of another
A number a is a factor of number b if:
(1)
all the prime factors of a are also prime factors of b.
(2)
the number of times a specific prime factor is multiplied together in the decomposition of a is
not more than the number of times it is multiplied together in the decomposition of b.
8
Example 7
Example 8
Use prime factors to determine whether 18 is a
factor of 252.
Use prime factors to determine whether 14 is a
factor of 180.
Solution
Solution
18  2  32
14  2  7
252  22  32  7
Notice that:
(1) all prime factors of 18 are prime factors of
252, and
(2) 18 contains 2 whereas 252 contains 22
which means that 252 contains more 2’s
than 18. Also 18 and 252 both contain 32
which is the same number of 3’s.
The number of 2’s and 3’s in 18 are not
more than those in 252.
Therefore we can say that 18 is a factor of 252.
Example 9
180  22  32  5
Notice that:
14 contains the prime factor 7 which 180
doesn’t. This means that not all of the prime
factors of 18 are prime factors of 252.
The rule in this case breaks down and 14 is not a
factor of 252.
Use prime factors to determine whether 60 is a multiple of 18.
Solution
18  2  32
60  22  3  5
Notice that 18 contains 32 whereas 60 contain 3. This means that 18 contains more 3’s than 60. The
rule breaks down in this case. Therefore 18 is not a factor of 60, or, 60 is not a multiple of 18.
HIGHEST COMMON FACTORS
When one number is a factor of two (or more) different numbers, we call it a common factor of those
numbers. Two (or more) numbers may have more than one common factor. The biggest of all these
common factors is called the Highest Common Factor (HCF).
We can determine the HCF of two (or more) numbers by using prime factors.
Example 10
Find the HCF of 225, 315 and 495.
Solution
2
225  3  5
Alternative method:
Factors of 225:
1, 3, 5, 9, 15, 25, 45, 75, 225
Factors of 315:
1, 3, 5, 7, 9, 15, 21, 45, 63, 105, 315
Factors of 495:
1, 3, 5, 9, 11, 15, 33, 45, 55, 99, 165,
The highest common factor is 45
2
315  32  5  7
495  32  5 11
These three numbers have the number 32 and
one 5 in common. Now take the lowest power
of each prime factor:
 HCF  225;315; 495  32  5  45
LOWEST COMMON MULTIPLES
When one number is a multiple of two (or more) different numbers, we call it a common multiple of
those numbers. Two (or more) numbers always have infinitely many common multiples, but the
smallest of all these common multiples is called the Lowest Common Multiple (LCM).
We can also determine the LCM of two (or more) numbers by using prime factors.
9
Example 11
Find the LCM of 24 and 90.
Solution
24  23  3
Alternative method:
Multiples of 24:
24, 48, 72, 96, 120, 144, 168, 192, 216,
240, 264, 288, 312, 336, 360, 384, ….
Multiples of 90:
90, 180, 270, 360, 450, …
The lowest common multiple is 360
90  2  32  5
So we take three 2’s, two 3’s and a 5 :
(Take the highest power of each prime factor.)
 LCM  24;90   23  32  5
 8 9 5
 360
Some other interesting uses of prime numbers
There are many uses for prime numbers in the different branches of Mathematics. You will have to
use prime factors to deal with exponential expressions and equations later so it is important that you
get comfortable with the idea of using them as soon as possible. We will show you a few interesting
things you can do with prime numbers here to help you get used to using them and to get a feel for
how powerful they are.
Using prime factors to simplify fractions
Example 12
Write the fraction
336
in its simplest form, without the use of a calculator.
980
Solution
Start by decomposing the numerator, as well as the denominator into prime factors:
336  24  3  7 and 980  22  5  7 2
336 24  3  7


980 22  5  7 2
22  3

5 7
12
=
35
[divide powers of the same base by subtracting the exponents]
Prime factors and powers of 10
If we decompose the powers of 10 into prime factors, we notice something interesting:
101  10  2  5
102  100  22  52
103  1000  23  53
104  10000  24  54 etc.
Do you see it?
(1)
There are only 2’s and 5’s as prime factors.
(2)
There are just as many 5’s as 2’s.
The following example illustrates how we can use this fact.
10
Example 13
If we could calculate 23  54  7100 and write the answer down, what would the last three digits be?
Solution
This number is far too large to calculate, even with a calculator! What we are given here is really the
decomposition of the number into prime factors. Because there are both 2’s and 5’s in the product
we know that the number has a power of 10 as a factor. Notice that there are more 5’s than 2’s here,
so let’s rewrite the number so as to group the same number of 2’s and 5’s together:
23  54  7100
 23  53  5  7100
 (23  53 )  (5  7100 )
Now the part (23  53 ) is actually a power of 10, namely, 103 , which is 1000. This means that we
are multiplying 5  7100 by 1000 and therefore the result will end on three 0’s. This is all we were
asked for. The last three digits will be 000.
Using prime numbers to determine whether a fraction is a terminating or a recurring decimal
We said earlier that fractions, when written in decimal form, will either be terminating or recurring.
Here are a few examples of fractions that are terminating decimals:
1
3
7
111
 0,5
 0,6
 0, 28
 0,555
2
5
25
200
And some fractions that are recurring decimals:
1
5
3


 0,333...  0,3
 0,8333...  0,83
 0, 428571428571428571...  0, 428571
3
6
7
Do you have an idea of how to tell whether a fraction will be terminating or recurring as a decimal?
The answer is very simple. First decompose the denominator into prime factors.
 If it contains only 2’s and/or 5’s (not necessarily both) then the decimal will be terminating.
 If it contains any other prime factors (3’s, 7’s etc.) then the decimal will be recurring.
The reason for this is that, in order to write a fraction as a terminating decimal, we have to be
able to turn the denominator into a power of 10 by multiplication. For example:
7
74
28


 0, 28 . But remember that powers of 10 contain only 2’s and 5’s when
25 25  4 100
decomposed into prime factors. So, if the denominator contains any other prime factors, it will
be impossible to turn it into a power of 10 by multiplication.
Example 14
If the following fractions are written as decimals, will the result be terminating or recurring? Don’t
use your calculator.
5
3
4
7
(b)
(c)
(d)
(a)
16
22
125
20
1
3
5
3
(f)
(g)
(h)
(e)
100
20
100
60
120
2 7
6  25  4200
Solutions
(a)
(b)
(c)
(d)
(e)
Denominator: 16  24
Denominator: 22  2  11
Denominator: 125  53
Denominator: 20  22  5
Denominator: 60  22  3  5
Terminating
Recurring
Terminating
Terminating
Recurring
11
(f)
(g)
(h)
Careful, simplify the fraction first:
3
1

120 40
Denominator: 40  23  5
Denominator: 2100  7 20
3
3
1


100
200
2 100
2 100
200
6  25  4
2  3  (5 )  (2 )
2  5  2400
Only 2’s and 5’s in denominator
Terminating
Recurring
Terminating
EXERCISE 2
(a)
(b)
(c)
(d)
(e)
(f)
(g)
Consider the numbers 24 and 504.
(1)
Write both numbers as a product of primes.
(2)
Is 24 a factor of 504?
Consider the numbers 72 and 180.
(1)
Write both numbers as a product of primes.
(2)
Is 180 a multiple of 72?
Determine the highest common factor (HCF) of each of the following lists of numbers:
(1)
16; 12
(2)
13; 15
(3)
14; 154; 28 (4)
135; 225; 315
(5)
675; 1125
(6)
756; 432
(7)
176; 1331
(8)
125; 352
(9)
36; 45; 162 (10) 36; 78; 156
Determine the lowest common multiple (LCM) of each of the following lists of numbers:
(1)
9; 6
(2)
8; 14
(3)
2; 9; 54
(4)
54; 36
(5)
16; 12
(6)
10; 92; 115 (7)
56; 308
(8)
55; 20; 154
(9)
21; 153; 119 (10) 108; 396
1176
Simplify the fraction
by using prime factors.
504
What is the sum of the digits of the following numbers?
298  3  599
(1)
(2)
298  3  599  1
When each of the following fractions are written in decimal form, will the result be
terminating or recurring? Motivate your answer.
3
11
20
33
(1)
(2)
(3)
(4)
35
80
21
220
Note: If a number is a perfect square, then all the exponents in the prime factor decomposition of
the number are divisible by 2 (even numbers).
If a number is a perfect cube, then all the exponents in the prime factor decomposition of the
number are divisible by 3.
Use this information to answer the following questions.
(h)
(i)
Consider the number 1728
(1)
Write the number as a product of prime numbers.
(2)
Is the number a perfect square?
(3)
Is the number a perfect cube?
Consider the number 350  530  7 20
(1)
Is 225 a factor of this number?
(2)
Is 321  531  7 20 a multiple of this number?
(3)
Is the number a perfect square?
(4)
Is the number a perfect cube?
(5)
When the number is calculated, what is its last digit?
12
REVISION OF CALCULATION TECHNIQUES, ESTIMATING AND ROUNDING
Although, in this day and age, calculators and computers can do basic and advanced calculations for
us, it is very important that you understand how to do calculations mentally and manually.
You have studied this in detail in primary school and revised it in Grade 8, so we will just show you
some examples here to remind you of the procedures.
Example 15 (Addition)
Example 16 (Subtraction)
233  708  435
Calculate:
Solution
2
7
 4
1 3
535  377
Calculate:
Solution
1
3
0
3
7
3
8
5
6
4

5
3
1
12
3
7
5
1
5
7
8
Example 17 (Multiplication)
63  42
Calculate:
Solution
6 3
4 2
6
1
1
2 4
2 6
2
2
0
4
0
0
0
6
Or even shorter:
1
6 3
4 2
(3  2)
(3  40)
(60  2)
(60  40)
1 2
2 5 2
2 6 4
6
0
6
(63  2)
(63  40)
Example 18
Calculate 441  3 by using long division.
Solution
1 4 7
3
(3 goes into 4 once, into 14 four times and into 21 seven times)
4 4 1
3
(1 3  3)
1 4
(4  3  1 and bring down the 4)
1 2
(3  4  12)
2
1
(14  12  2 and bring down the 1)
2
1
0
(7  3  21)
(21  21  0 and so there is no remainder and the answer is 147)
Now for division by a two digit number.
13
Example 19
Calculate 384350  12 without a calculator.
Solution
 3 2 0 2 9
1 2
3 8 4 3 5 0
3 6
2 4
2 4
 3
0
3 5
2 4
1 1 0
1 0 8
2
The remaining 2 indicates that 384 350 is not exactly divisible by 12. The answer of 32029 is the
whole number part of the actual answer (also called the quotient) and the 2 is called the remainder.
Rounding
Example 20
Example 21
Round 22378 off to the nearest hundred.
Round 3,14159 off to the nearest hundredth (i.e.
to two decimal places).
Solution
The hundreds digit is a 3, so it will either stay a
3 or become a 4 depending on the next digit (i.e.
the tens digit):
Solution
3,14159
Since the thousandths digit is 1 (less than 5),
we leave the hundredths digit 4 and drop the
subsequent digits: 3,14159  3,14
22378
Since the tens digit is 7 (5 or higher), we change
the 3 to a 4 and replace the subsequent digits
with 0’s: 22378  22400
EXERCISE 3
(a)
(b)
For each of the following calculations, estimate the answer first and then calculate the
exact result without the use of a calculator, using any method of your choice:
349  388
614  289
34  92
(1)
(2)
(3)
7084  7
85712  72553
4500  2688
(4)
(5)
(6)
206

356
95623

11
825
 339
(7)
(8)
(9)
(10) 10000  4321
(11) 777  999
(12) 420684  12
(13) 38  55
(14) 98  473
(15) 798  291
(16) 743  566
(17) 2073  22
(18) 42444  16
Round the following off as indicated:
(1)
2196 to the nearest ten
(2)
25527 to the nearest hundred
(3)
23988 to the nearest thousand
(4)
372,555 to the nearest ten
(5)
357,384 to two decimal places
(6)
775,552 to the nearest whole number
(7)
345,678 to two decimal places
(8)
566,2238 to three decimal places
14
(9)
(11)
(13)
78,66 to the nearest whole number
550,0055 to the nearest hundred
897,0074 to three decimal places
(10)
(12)
(14)
699,2525 to the nearest ten
89,8989 to the nearest thousand
791,0001 to three decimal places
RATIOS
A ratio is a comparison between two numbers or two quantities with the same unit. Here follows
some revision of the work you did on ratios in Grade 8.
Revision of ratio calculations (Grade 8)
Example 22
Example 23
Write the ratio of 24 to 36 in simplest form.
Johan walks 2 km to school and Thabang walks
800 m. Express the ratio of Johan’s distance to
Thabang’s distance as a ratio, in its simplest
form.
Solution
Divide both numbers by their HCF, which is 12:
24 : 36
24 36

:
12 12
 2:3
Solution
First, make the units the same:
2 km : 800 m
 2000 m : 800 m
Now, divide by the HCF (400 m)
2000 m 800 m

:
400 m 400 m

5 : 2
Example 25
(Dividing a number in a certain ratio)
Example 24
(Finding the missing number/quantity)
Two numbers are in the ratio 5:7. If the smaller
number is 20, what is the greater number?
Divide 72 in the ratio 3 : 4 : 5.
Solution
Solution
Method:
5 : 7
1. Add all the parts of the ratio together:
3  4  5  12
= 20 : ?
72
2. Divide 72 by this result:
6
Method:
12
Divide the known number by the part of the ratio 3. Now multiply this result by each of the parts
it corresponds to and multiply by the part
of the ratio:
corresponding to the missing number.
6  3  18
6  4  24
6  5  30
20
So, dividing 72 in the ratio 3 : 4 : 5 gives 18 ; 24
7
5
and 30.
 4 7
 28
 The greater number is 28.
Dealing with decimals
When working with a ratio between decimals, we can easily turn it into a ratio between whole
numbers by multiplying all the parts of the ratio by the same power of 10. Powers of 10 are:
10; 100; 1000; 10 000; 100 000; 1 000 000 etc.
15
Example 26
Express 0,2 : 0,15 as a ratio between whole numbers, in its simplest form.
Solution
Look at the number with the most decimal places (in our example, this is 0,15).
To make this a whole number, we need to multiply by 100.
Now multiply both parts by 100:
0, 2 100 : 0,15 100
 20
:
15
Now simplify this by dividing both parts by the HCF:
20 15
:
5 5
 4:3
SOME MORE INTERESTING EXAMPLES
Since the last time you studied ratios, in Grade 8, you have learnt some new interesting mathematical
facts and techniques like equations, geometry etc. In the following examples we will show you how
to make use of these in ratio problems.
A technique that comes in very handy when dealing with ratios is to introduce a variable. If two
numbers are in a ratio a : b we can let the first number equal ak and the second number equal bk .
Example 27
A shop sells two soft drinks: Fizzy Cool and Berry Spark. Their prices are in the ratio:
Price of Fizzy Cool : Price of Berry Spark  3 : 4 .
You buy 5 Fizzy Cools and 7 Berry Sparks and pay a total of R52,89.
What is the price of each drink?
Solution
Let the price of Fizzy Cool be 3k and the price of Berry Spark 4k
5  3k  7  4k  52,89
15k  28k  52,89
 43k  52,89
52,89
k 
43
 1, 23
Berry Spark costs 4  R1,23  R4,92
 Fizzy Cool costs 3  R1,23  R3,69
Example 28
The perimeter of a rectangle is 280 m. The ratio of its length to its breadth is 5 : 2 .
Determine the area of the rectangle.
Solution
Let the length be 5k and the breadth 2k
Perimeter  2  Length  2  Breadth
 280  2  5k  2  2k
 280  10k  4k
 280  14k
 k  20
16
 Length  5  20  100 m
Breadth  2  20  40 m
Area  Length  Breadth
 100 m  40 m
 4000 m 2
Example 29
The diagonal of a rectangular TV-screen is 25 inches long. If the length and breadth is in a ratio
4 : 3 , determine the length and the breadth of the screen.
Solution
Let the length be 4k and the breadth 3k
By the theorem of Pythagoras: (4k ) 2  (3k ) 2  252
16k 2  9k 2  625
 25k 2  625
3k
 k 2  25
k  5
25 inches
 Length  4  5  20 inches
Breadth  3  5  15 inches
4k
Example 30
If two numbers are in a ratio 2:3, what is the ratio between their squares?
Solution
Let the first number be 2k and the second 3k
The ratio between their squares is:
(2k ) 2 : (3k ) 2
 4 k 2 : 9k 2
4k 2 9k 2
:
k2 k2
 4:9

Example 31 (Ratio of fractions)
1 7
Simplify the ratio 2 :1 without using a calculator.
4 8
Solution
9 15
:
4 8

9 8 2 15 8
 : 
 1 8 1
14
[convert to improper fractions]
[multiply both parts by LCD]
 18 :15
18 15
:
3 3
 6:5

[divide both parts by HCF]
17
EXERCISE 4
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l)
(m)
(n)
(o)
(p)
(q)
(r)
(s)
(t)
You may use a calculator in this exercise.
Express the following as a ratio in its simplest form:
(1)
18 to 12
(2)
24 to 32
(3)
15 to 25
(4)
30 to 20
(5)
9 to 15 to 21
(6)
1000 to 10 to 100
(7)
1,6 to 1
(8)
0,36 to 0,06
(9)
0,01 to 0,00001
(10) 2,4 to 0,8 to 0,16
(11) 30 cm to 1,5 m
(12) 6 kg to 600 g to 3000 g
1
11
(15) 2 x 2 y to 6 xy 2
(13) 48 seconds to 1 minute
(14) 3 to
7
14
Calculate the missing number:
(1)
3 : 5 = 30 : ?
(2)
7:3=?:9
(3)
15 : ? = 5 : 3
(4)
20 000 : ? = 1000 : 1 (5)
? : 400 = 4 : 100
(6)
? : 1000 = 400 : 10 000
(7)
? : 0,2 = 7 : 6
(8)
8 cm : 1 m = ? : 250
(9)
20 seconds : 2 minutes = 1 : ?
(10) 30 g : ? mg = 40 : 1
(11) ? hours : 10 seconds = 720 : 1
Divide the number 720 into the following ratios:
(1)
5:4
(2)
1:9
(3)
2:1:3
Divide each of the following numbers in a ratio of 5 : 3:
(1)
80
(2)
176
(3)
100
Divide the following numbers in a ratio 1 : 5 : 5:
(1)
11
(2)
33
(3)
253
The ratio of boys to girls at a party is 4 : 3. If there are 20 boys, how many girls are there?
The ratio of Lions supporters to Sharks supporters in a sports bar is 2:5. If there are 35 people
in the bar, how many are Sharks supporters?
Two numbers are in a ratio 3 : 5 . What is the ratio between their squares?
The ratio between the square of a number and its cube is 1:3. What is the number?
The length of square A is twice the length of the side of square B. What is the ratio between
the area of square A and square B?
The angles of a triangle are in a ratio 1 : 3 : 8. What are the angles?
The ratio between the interior angles of a triangle is 1 : 2 : 6. What is the ratio between the
exterior angles?
The scale on a map is 5 cm = 10 km. If two cities are 8 cm apart on the map, how far are they
apart in reality?
The ratio of the length to the breadth of a rectangle is 7 : 3. Calculate the dimensions of the
rectangle if:
(1)
the perimeter is 10 cm.
(2)
the area is 525 m2.
The ratio between the two shorter sides of a right angled triangle is 5 : 12. If the hypotenuse
measures 65 cm, what is the area of the triangle?
The sketch shows a square and its inscribed circle.
What is the ratio of:
(1)
the circumference of the circle to the perimeter of the square?
(2)
the area of the circle to the area of the square?
The ratio of the length to the breadth of an A4 sheet of paper is 2 :1 .
1 2
The area of an A4 sheet of paper is
m . Calculate its length and breadth.
16
At the start of the week a bookshop had fiction and non-fiction books in the ratio 2 : 5. By
the end of the week, 20% of each type of book were sold and 2240 books (in total) were
unsold. How many of each type were there at the start?
Susan was 20 years old when her daughter, Janet, was born. What will Janet’s age be when
the ratio of their ages is 7 : 2?
The angles of a triangle are in the ratio 1:1:2. What is the ratio between its sides?
18
RATE
A rate is a measure of how rapidly one quantity changes in comparison to another. The two quantities
usually have different units (as opposed to having the same unit in the case of ratios).
When expressing a rate, we use the word per, which means for every and we write this using the
symbol /.
Revision of rate calculations – Grade 8
To do rate calculations, we use the following simple method:
Look at the unit in which the rate is measured. It will usually be a combination of two units with a
per sign (/) in between (e.g. km/h, ml/s, kg/m3 etc).
Each one of these units measures a specific quantity. For example, km measures distance and h
measures time.
So if a rate is measured in unit A / unit B, then unit A measures quantity A and unit B measures
quantity B. The following formulae then apply:
rate 
quantity A
quantity B
quantity A  rate  quantity B
quantity B

quantity A
rate
To illustrate how this works, consider speed as a rate measured in km/h:
km / h
quantity A  distance
speed 
distance
time
rate  speed
quantity B  time
distance  speed  time
time

distance
speed
Example 32
Example 33
If a car drives 200 km in 2 hours, determine the
speed in km/h.
I run at 6 m/s. How far can I run in a minute?
Solution
rate unit: m/s
quantity A  distance (m)
quantity B  time (s)
quantity A  rate  quantity B
 distance  speed  time
Solution
rate unit: km/h
quantity A  distance (km)
quantity B  time (h)
quantity A
rate 
quantity B
distance
 speed 
time
200 km

2h
 100 km/h
Example 34
 (6 m/s)  (60s)
 360 m
My car can travel 10 km on every litre of petrol. How much petrol will I need to drive 400 km?
Solution
rate unit: km/l
quantity A  distance (km)
quantity B  volume of fuel (l)
quantity A
quantity B 
rate
volume of fuel



19
distance
fuel consumption rate
400 km
10 km/l
40 l
SOME MORE INTERESTING EXAMPLES
Example 35
Example 36
If Suzette can run 2 km in 8 minutes, how long
will it take her to run 5 km if she maintains her
speed?
From town A to town B a car travels at an average
speed of 60 km/h. From town B back to town A
the car travels at an average speed of 100 km/h.
What was the average speed for the entire
journey?
Solution
rate unit: km/min
quantity A  distance (km)
quantity B  time (min)
quantity A
rate 
quantity B
distance
 speed 
time
2 km

8 min
 0,25 km/min
 time
distance
speed
5 km

0,25 km/min
 20 min

Solution
Suppose the distance from town A to town B is d.
distance d
 Time from town A to town B 

speed
60
d
distance

Time from town B to town A 
speed
100
d
d
Total time taken 

60 100
3d  5d

300
8d

300
2d

75
total distance
 Average speed 
total time
2d
75

 2d 
 75 km/h
2d
 2d 


 75 
EXERCISE 5
You may use a calculator in this exercise. Assume that the rates remain constant.
(a)
A car travels at 80 km/h.
(1)
How long does it take to travel 320 km?
(2)
How far will the car travel in 10 hours?
(3)
Convert this rate (80 km/h) to a rate measured in m/s.
(b)
If you can consistently run 80 m in 16 s,
(1)
how long will it take you to run 400 m?
(2)
how far can you run in 1 minute and 4 seconds?
(c)
A car consumes fuel at a rate of 8l/100 km.
(1)
How many litres is needed to travel 400 km?
(2)
How far you can travel with 50 l?
(d)
Water is drained from a tank at a rate of 25 litres per minute.
(1)
If there is 3000 litres in the tank, how long will it take till the tank is empty?
(2)
How much water was in the tank if it could be emptied in 10 minutes?
(e)
A large truck uses 25 l of diesel to travel 200 km. How many litres would be needed to travel
150 km?
(f)
The density of a certain liquid is 50 grams per cubic centimetre. What is the mass, in kg, of
5 kilolitres (5m3) of this liquid?
(g)
If pump A can empty a tank in 2 hours and pump B can empty the tank in 4 hours, how long
will they take to empty the tank if they run together?
20
DIRECT PROPORTION
Two variable quantities are said to be directly proportional if they change in such a way that the
ratio between them remains constant and is positive. If one variable increases (or decreases), then the
other variable will increase (or decrease) as well. If the one variable doubles, then so will the other
variable. If one variable is increased five times, so will the other. For example, suppose that 10
pencils are bought for R30, 20 pencils are bought for R60, 30 pencils are bought for R90 and so
forth. The relationship between the number of pencils and the cost can be represented in a table.
Number of pencils (x)
Cost in rands (y)
10
30
20
60
40
120
x 10 20 30 40 1





y 30 60 90 120 3
y
There is a positive constant ratio between the two variables. In this example,  3 or y  3 x
x
y
In general, if two variables x and y are directly proportional, then  k or y  kx where k is the
x
constant ratio (or proportionality constant).
Notice:
y 30 60 90 120




3
x 10 20 30 40
30
90
or
Example 37
If it takes you 15 minutes to staple 90 booklets, how many booklets can you staple in 20 minutes?
Solution
It would be reasonable to assume that the number of booklets you staple is directly proportional to
the time taken. The more booklets you staple, the longer it takes.
Let x be the number of booklets you staple and y the time you take in minutes.
Number of booklets stapled (x)
Time taken in minutes (y)
90
15
a
20
Since the constant ratio is the same for all ratios in this example, we can write:
15 20

90 a
15a  1800
[cross-multiply]
 a  120
You can staple 120 booklets in 20 minutes.
Example 38
The following table shows the voltage (in V for volts) connected to a 5Ω resistor and the current (in
A for amperes) that is measured flowing through it.
Voltage (V)
Current (A)
15
3
30
6
45
a
60
12
b
15
Calculate the value of a and b.
Solution
15 30 60


5
3
6 12
The number 5 is the proportionality constant and in this case it represents the resistance of the
resistor i.e. 5Ω. The fact that voltage is directly proportional to current, when resistance is constant,
is known in Science as Ohm’s Law.
45 15
b 15


a
3
15 3
15a  135
 3b  225
a  9
 b  75
The ratio between the variables remains constant since
21
Example 39 (A different approach to example 35)
Suzette can run 2 km in 8 minutes. Assume she maintains her speed. Let x be the time she takes and
y the distance she covers.
(a)
How long will it take her to run 5 km?
(b)
What is the proportionality constant? What is the meaning of this constant in this case?
Solutions
Since her speed remains constant, the distance covered (y) will be directly proportional to the
time taken (x).
(a)
Time taken in minutes (x)
Distance covered in km (y)
(b)
8
2
a
5
a 8

5 2
 2a  40
[cross-multiply]
 a  20
She will take 20 minutes.
y 2 km
k 
 0, 25 km/min
x 8 min
This is the speed at which Suzette is running.
Graphical representation of two variables in direct proportion
If we represent two variables that are directly proportional by means of a graph, the result will always
be a straight line, with a positive slope (rising from left to right) and passing through the origin (the
point where both variables are 0). Let’s look at the data from Example 38 again.
Example 40
The following table shows the voltage (in V for volt) connected to a 5Ω resistor and the current (in
A for ampere) that is measured flowing through it:
Current (A)
Voltage (V)
3
15
6
30
9
45
12
60
15
75
Sketch the graph of voltage vs. current, placing voltage on the y-axis (vertical axis) and current on
the x-axis (horizontal axis).
Solution
Voltage (V)
As you can see, the graph is a positively-sloped
straight line, passing through the origin.
The graph showing two directly proportional variables is a positively sloped straight line,
passing through the origin.
22
EXERCISE 6
(a)
(b)
(c)
(d)
(e)
(f)
In the following tables, the two variables x and y are directly proportional. In each case:
(i)
Find the values of a and b.
(ii)
Draw a graph of y versus x, using an appropriate scale on each of the axes (not
necessarily the same scale on both.)
(1)
x
y
6
4
9
a
b
12
(2)
x
y
3
5
6
a
b
8
(3)
x
y
0,4
2,5
0,8
a
b
10
(4)
x
y
100
3
400
a
b
6
If 15 sandwiches are sold for R12, what would 12 sandwiches be sold for?
If 6 men can dig 4 ditches in an hour. How many men would be required to dig 6 ditches in an
hour?
If you can drive 300 km on 16 litres of petrol, how far can you drive on 20 litres?
200 ml of a liquid has a mass of 300 g.
(1)
What would the mass of a litre of this liquid be?
(2)
Draw a graph of the mass of the liquid (in g) versus the volume (in ml).
The acceleration (in m/s/s) of a body is directly proportional to the force (in N for newton)
acting on the body. If a certain body accelerates at 10 m/s/s when subjected to a force of 3 N,
what will the acceleration be under a force of 8 N?
INVERSE PROPORTION
Two variable quantities are said to be inversely proportional or indirectly proportional if they
change in such a way that their product remains constant. If the one variable is doubled the other one
will be halved; if the one is multiplied by five, the other will be divided by five. If the one increases
(or decreases), then the other will decrease (or increase). For example, if 5 men take 4 hours to do a
job then 10 men should take 2 hours to do the same job. Notice that the number of men doubled and
the time taken is halved. Also, the product of the number of men and the time taken in hours remains
20. The relationship between the number of men and the time taken can be represented in a table.
Number of men (x)
5
10
15
20
Time taken in hours (y)
4
2
4
3
1
Notice:
5  4  10  2  15  43  20  1  20
There is a positive constant product between the two variables. In this example, xy  20 or y  20x .
In general, if two variables x and y are inversely proportional, then xy  k or y  kx where k is the
constant product.
Example 41
Jabulani has a fixed amount of money to buy sweets. If the sweets cost R2 each, he can buy 50
sweets. How many sweets will he be able to buy if the sweets cost R5 each?
Solution
Clearly the number of sweets Jabulani can afford is inversely proportional to the price of a sweet.
The higher the price, the less sweets you can buy. Let x be the price of a sweet and y the number of
sweets Jabulani can buy.
Price of a sweet in rands (x)
2
Number of sweets (y)
50
5  a  2  50
 5a  100
 a  20
Jabulani can by 20 sweets at R5 each.
5
a
23
Example 42
A fixed mass of gas is kept inside a large container at a fixed temperature. The capacity of the
container, and hence the volume of the gas, is variable. The following table shows the volume (in
m3) of the gas in the container and the pressure (in kPa for kilopascal) inside the container.
Volume (m3)
0,4
0,8
1,2
1,6
b
100
75
40
Pressure (kPa)
300
a
(a)
Do the two variables appear to be directly or inversely proportional? Motivate your answer.
(b)
Assuming that the two variables are in fact inversely proportional, calculate a and b.
Solutions
Inversely proportional. 0, 4  300  1, 2  100  1,6  75  120 . The product of the variables
remains constant.
b  40   0, 4  300 
0,8a   0, 4  300 
 0,8a  120
 40b  120
 a  150
b  3
(a)
(b)
Graphical representation of two variables in inverse proportion
If we represent two variables that are inversely proportional by means of a graph, the result will
always be a special type of curve called a hyperbola. The curve approaches the axes, ever getting
closer to them, but never actually touching or cutting them.
Example 43
The following table shows the number of men working on a job and the time in hours it takes to
complete the job:
Number of men
Time in hours
1
20
2
10
4
5
5
4
10
2
20
1
Sketch the graph of time in hours vs. number of men, placing time in hours on the y-axis (vertical
axis) and number of men on the x-axis (horizontal axis).
Solution
y
20
(1; 20)
The type of curve we have here is called a
hyperbola. When sketching an hyperbola we always
use the same scale on both axes.
(Notice that we don’t draw a solid curve, since
fractions of men are not possible and therefore not
all points on the curve are possible).
The graph showing two inversely proportional
variables is a hyperbola.
Time in hours
15
10
5
4
3
2
1
(2 ;10)
(4 ; 5)
(5 ; 4)
(10 ; 2)
1 2 3 4 5
10
Number of men
(20 ;1)
20
15
24
x
EXERCISE 7
(a)
(b)
(c)
In the following tables, the two variables x and y are inversely proportional. In each case:
(i)
Find the values of a and b.
(ii)
Draw a graph of y versus x, using an appropriate scale.
(Use the same scale on both axes.)
(1)
x
y
6
4
8
a
b
12
(2)
x
y
6
5
5
a
b
10
(3)
x
y
0,4
2,5
0,2
a
4
b
(4)
x
y
1
4
4
a
b
2
Use the concept of indirect proportion to answer the following questions:
(1)
Two mothers each buy the same pack of sweets. The one mother has 3 children and
can give each child 12 sweets. How many sweets can the other mother give to each
child if she has 6 children?
(2)
If 6 men can dig a ditch in 2 hours. How many men would be required to dig the ditch
in 3 hours?
(3)
If you travel at 100 km/h a journey takes 40 minutes. How long does the same journey
take if you travel at 80 km/h?
The current (in A for ampere) flowing through a resistor is inversely proportional to the
resistance (in  for ohm) of the resistor if connected to a constant voltage source. When the
resistance is 5  , the current is 6 A.
(1)
What would the current be when the resistance is 2  ?
(2)
Draw the graph of current versus resistance.
FINANCIAL MATHEMATICS
REVISION OF VARIOUS FINANCIAL CALCULATIONS – GRADE 8
Exchange Rates
An exchange rate is used to convert from one currency to another.
Example 44
Given the exchange rate:
$1 = R10,47
(a)
How many rands will you get for $50?
(b)
How many dollars will you get for R50?
Solutions
(a)
(b)
Since one dollar is worth R10,47, 50 dollars will be worth: 50  R10,47 = R523,50
Since we need R10,47 for one dollar, what we want to know is how many times R10,47 goes
50
 $4,78
into R50. For this we divide R50 by R10,47:
10, 47
Example 45
A calculator costs £12,80. You buy it over the internet and so you have to pay £3,30 for delivery
and other costs. You end up paying R289,32. What is the exchange rate for a pound in rands?
(£1 = R?)
Solution
The total cost in pounds is: £12,80 + £3,30 = £16,10.
So £16,10 is equal to R289,32
289,32
 R17,97
This means £1 is equal to
16,10
Exchange Rate:
£1 = R17,97
25
Example 46
In 2010, the following exchange rates applied: $1 = R7,53 and £1 = R12,52
How many pounds (£) could you get for $50?
Solution
First, for $50 you could get 50  7,53  R376,50
376,50
 £30,07
Then, for R376,50 you could get
12,52
Profit, loss, discount and VAT
The following table summarises all the calculations you will need to solve problems involving
percentage increases/decreases:
Increasing by a Percentage
100 + percentage
number 
100
Reversing an Increase
100
number 
100 + percentage
Decreasing by a Percentage
100  percentage
number 
100
Reversing a Decrease
100
number 
100  percentage
Example 47
Example 48
An item is bought for R80 and sold at a profit of
20%. What price is the item sold for?
Thabo sells stationery, usually at a profit, but a
certain batch of pens has been on his shelves for
too long and he now decides to sell them at a
loss of 40% to get rid of them. He bought the
pens for R8 each. What will he sell them for?
Solution
R80 
120
= R96
100
Solution
60
= R4,80
100
Example 50
R8 
Example 49
All items sold with VAT of 14% in South
Africa. If the price of a calculator is R190
without VAT, what is the final selling price,
including VAT?
Solution
R190 
114
= R216,60
100
Melcia sells T-shirts at R150 each. Louis buys
20 of these T-shirts and agrees to pay cash for
them. Melcia gives him 10% discount for
paying cash. How much will Louis pay, in total,
for the 20 T-shirts?
Solution
R150  20 = R3000
Example 51
90
= R2700
100
Example 52
Petrus sells coffee mugs at 45% profit. If he
sells the mugs for R21,75 each, what did he buy
them for?
Lindiwe bought exam pads at a 20% discount
sale for R9,60. What was the original price of
the exam pads?
Solution
Solution
R3000 
R21,75 
100
= R15
145
R9,60 
26
100
= R12
80
Percentage calculations can also be done using the “ratio method”. When using this method, the
number on which a percentage is based is always associated with 100%.
Example 53
Example 54
The price of an item, including 25% profit, is
R800. Determine the cost price (excluding
profit).
The selling price of an item is R575 after 15%
profit was added. What would the selling price
have been if the profit were 25%?
Solution
Solution
Since the 25% profit is based on the cost price,
we associate the cost price with 100% and the
selling price with 125%. This means that the
ratio of cost price : selling price is 100 : 125.
Cost Price : Selling Price

100 : 125

?
: R800
R800
 Cost Price 
 100  R640
125
115 : 125
 R575 : ?
R575
125  R625
115
EXERCISE 8
(a)
You may use a calculator in this exercise.
The following shows the value of $1 (one United States Dollar) in some other
currencies:
US Dollar ($)
1
(b)
Euro (€)
0,75611
British Pound (£)
0,65988
(1)
How many Pounds can you get for $25?
(2)
How many Dollars can you get for €100?
(3)
If R10,2147 = $1, how many Pounds can you get for R500?
(4)
Complete the exchange rate for Pounds to Rupees: £1 = ? Indian Rupees
The following shows the value of R1 in some other currencies:
SA Rand (R)
1
US Dollar ($)
0,08995
British Pound (£)
0,05225
(1)
(2)
(3)
(c)
(d)
(e)
(f)
(g)
(h)
Indian Rupee
61,23117
Australian Dollar ($)
0,09996
How many Australian Dollars can you get for R400?
How many Rands will you need to buy 600 US dollars?
You buy the following items over the internet:
 A fountain pen for $40,51.
 A graphical calculator for £65,42.
What is the combined total cost in Rands for these purchases?
Thabiso bought an item in Brazil for R$400 (the symbol R$ is for Brazilian Real).
The Rand-value of this is R1980.
Complete:
R$1 = R ?
Express the following as percentages:
(1)
R25 of R150
(2)
R180 of R120
(3)
50c of R8
Calculate the following:
(1)
120% of R50
(2)
2% of R2 500 000
(3)
15% of R12
Complete:
(1)
R21 is 70% of ...
(2)
R36 is 30% of ...
(3)
R46 is 115% of ...
(1)
Increase R180 by 50%
(2)
Increase 16 by 12,5%
(3)
Decrease R480 by 87,5%
Jaydi buys and sells flash disks. He buys the disks at a cost of R100 and sells them at a
profit of 25%. What is his selling price?
27
(i)
(j)
(k)
(l)
(m)
(n)
(o)
(p)
A flat screen TV is worth R12 000, excluding 14% VAT. What is the selling price of the TV,
including VAT?
Jabu spends R20 to prepare a take-away meal for his Spaza-shop. He wants to sell the takeaways at 60% profit. What should he sell them for?
A shop sells packets of chips at R5,70 each, including 14% VAT. What is the price of a
packet of chips, excluding VAT?
Johnno buys shorts at R40 each. He adds a profit of 66 23 % and then he has to add 14%
VAT on this increased price.
(1)
What is the selling price of the shorts, excluding VAT?
(2)
What will the buyer pay for the shorts, including VAT?
Thulani sells chocolates at R12 each, after he has added 40% profit. What was the cost price
of these chocolates?
Mymoena buys pens for R3,20 each and then sells them for R5,60 each. What percentage
profit does she make?
A supermarket decides to give a 33 13 % discount on all food items. A bag of tomatoes
originally sold for R18. What will they now sell it for?
Complete the following:
(1)
An increase of 30%, followed by an increase of 20%, has the same effect as a
single increase of ...
(2)
A decrease of 20%, followed by an increase of 20%, has the same effect as a
single increase/decrease (choose) of ...
SIMPLE INTEREST
Simple interest is interest based on the original amount invested or borrowed.
In Grade 8, we used the following formula to work out simple interest:
percentage interest per period
Simple interest  original amount 
 number of periods
100
If we use the symbols:
S for simple interest,
P for the original amount (principle amount),
r for the interest rate as a percentage and
n for the number of periods (usually years),
then this can be written as: S 
Pnr
100
To work out the final amount saved/owed: Final Amount  Original Amount + Simple Interest
If we use the symbol A for final amount (accumulated amount), this can be written as: A  P  S
If we wish to get the final amount directly, with one formula, we can easily condense these two
formulae into one:
A  PS
Pnr
A  P 
100
nr 

 A  P 1 

 100 
28
Example 55
Example 56
Ingrid invests R45 000 for 12 years at an interest
rate of 13% per annum simple interest. How
much will she have saved at the end of the 12
years?
Johannes started to save money six years ago.
The current value of his investment is R38 000.
The interest rate for the investment was 7% per
annum simple interest. How much did he invest
six years ago?
Solution
Solution
nr 

A  P 1 

 100 
 67 
38 000  P 1 

 100 
 38 000  P(1, 42)
 12  13 
A  45 000 1 

100 

A  R115 200
38 000
P
1, 42
 P  R26 760,56

Example 57
Nkosinathi invested R6000 and it accumulated to R10 000 after 3 years. Find the annual interest rate
as a percentage if the investment earned simple interest. Round off your answer to one decimal
place.
Solution
 3 r 
10 000  6000 1 

 100 
10 000
3r

 1
6000
100
5
3r
  1
3
100
5
3r
 1 
3
100
2 3r
 
3 100
 9r  200
 r  22, 2
The interest rate was 22,2% p.a.
HIRE PURCHASE
A Hire Purchase Agreement (HP) is a short-term loan. Household appliances and furniture are often
bought on HP. The buyer signs an agreement with the seller to pay a specified amount per month.
The interest paid on a hire purchase loan is simple interest and it is calculated on the full value of the
loan over the repayment period. Normally a deposit is paid initially and the balance is paid over a
short time period. The buyer will be required to pay the total interest charged on the loan even if the
loan can be paid off in a shorter time period.
Example 58
Agnes buys a tumble dryer for R4000. She takes out a hire-purchase loan involving equal monthly
payments over three years. The interest rate charged is 14% per annum simple interest. She also
takes out an insurance premium of R12,40 per month to cover the cost of damage or theft. Calculate:
the actual amount paid for the tumble dryer.
(a)
29
Solutions
(a)
nr 

A  P 1 

 100 
 14  3 
 A  4000 1 
100 

 A  R5680
(b)
The monthly loan repayments:
5680
 R157,78
36
With insurance:  R157,78  R12,40  R170,18
Example 59
Ruhan buys a computer costing R12 000. He pays a 20% deposit and then takes out a 24 month hire
purchase loan on the balance. The interest rate charged on the loan is 12% per annum simple
interest. Calculate his monthly payments and what he will actually pay for the computer.
Solution
20
 R2 400
100
Balance on HP  R12000  R2400  R9600 over a period of 24 months (2 years)
 2 12 
A  9600 1 

100 

 A  R11904
11904
 R496
His monthly repayments will be:
24
Ruhan will actually pay R2400  R11904  R14 304 for the computer.
Deposit  12 000 
EXERCISE 9
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
Jacob invests R6 000 for 4 years at 7,5% p.a. simple interest.
(1)
What is the interest earned over the 4 years?
(2)
What is the total value of Jacob’s investment after 4 years?
You borrow R10 000 for 3 years, on the condition that you pay it back with 12% p.a.
simple interest. How much will you have to pay back?
Nelson invests R12 000 at 11,5% p.a. simple interest for 18 months. What amount will he
have in the end?
Mmusi borrows an amount of money from Aggie. After two years, Mmusi pays back the
money, plus 18% simple interest per year. He has to pay back an amount of R4080. How
much money did he borrow from Aggie?
Nelis invests R3 000 for 5 years. He ends up with R4500. What was the simple interest rate?
Ettiene wants to know for how long he needs to invest his R2 000 at 15% p.a. simple interest
in order to have R11 000. Answer Ettiene’s question.
Jurgen wants to buy a tumble dryer for R1 500. He cannot afford to pay all at once, so he
buys it on hire purchase. According to the agreement, he will pay back the money, plus 12%
simple interest per year, by means of monthly payments, over a two-year period.
(1)
Calculate the total amount of money he’ll have to pay.
(2)
Calculate his monthly payment.
Janice buys a car costing R100 000 on hire purchase. She makes a R25 000 deposit and
pays back the balance, with 20% p.a. simple interest, over a period of 54 months.
Calculate her monthly payment.
Adelaide needs a printer for her new office. The cost of the printer is R15 000. This is
much too expensive for Adelaide to pay at once, so she opts for a hire purchase agreement,
whereby she will have to pay a deposit of 15% of the cost and pay back the balance, plus 18%
p.a. simple interest, by means of monthly payments over a period of 5 years. Adelaide works
out that she can afford the deposit and can manage to pay a maximum of R500 per month.
Will she be able to afford the deal?
30
(j)
(k)
A play station 2 costs R5500. Andries buys a play station on HP and agrees to pay a deposit
of R700 and 36 monthly payments of R200. Calculate the total simple interest paid and the
rate of simple interest.
Alexander wants to buy a laptop. He can only afford to pay R500 per month. He wants to take
out a hire purchase loan over 24 months at an interest rate of 20% per annum. Calculate the
price of the computer that he can afford to buy.
COMPOUND INTEREST
If interest is calculated on the original sum plus interest already earned, then it is called compound
interest. In practice, financial institutions use compound interest. The interest or growth on your
money is reinvested back into your investment. In this sense, money makes money. By keeping your
debts to a minimum and rather concentrating on investing your money wisely, you can harness the
power of compound interest and become financially independent or even wealthy. Einstein called
compound interest the most powerful principle in the universe.
Let’s compare what happens to R10 000 invested at 10% p.a. simple interest for 3 years and the same
amount invested for 3 years at 10% p.a. compounded yearly:
10% p.a. simple interest:
Beginning of year total
R10 000
R11 000
R12 000
First year
Second year
Third year
Interest for the year
10% of R10 000 = R1 000
10% of R10 000 = R1 000
10% of R10 000 = R1 000
End of year total
R11 000
R12 000
R13 000
Interest for the year
10% of R10 000 = R1 000
10% of R11 000 = R1 100
10% of R12 100 = R1 210
End of year total
R11 000
R12 100
R13 310
10% p.a. compound interest:
Beginning of year total
R10 000
R11 000
R12 100
First year
Second year
Third year
If you look at these tables, you will notice that in the case of simple interest, the interest is R1 000
every year (10% of the original R10 000 invested). In the case of compound interest, the interest is
more and more each year (in each case 10% of the new total).
Fortunately, there is a formula to calculate the accumulated amount when compound interest is used:
r 

A  P 1 

 100 
n
where
P  present value of the investment (original amount at the beginning)
A  accumulated amount (future value) of the investment after n periods
n  time period
r = the compound interest rate as a percentage
Example 60
Diego invests R18 000 for 6 years at 15% p.a. compounded annually. Find the future value of his
investment after 6 years and the interest he receives.
Solution
n
r 

A  P 1 

 100 
6
15 

A  18 000  1 

 100 
 A  R41 635,09
Interest received
 R41 635,09  R18 000
 R23 635,09
31
Example 61
Example 62
Tiana has just opened a small coffee shop
and takes out a loan to provide the initial
capital to start the business. She agrees to
repay the loan four years later by means of
a payment of R800 000. The bank charges
her an interest rate of 18% per annum
compounded annually. What was the amount
of money she originally borrowed?
R6 800 is invested for 6 years and grows in
value to R12 500. Find the interest rate if interest
is compounded annually.
Solution
r 

A  P 1 

 100 
r 

12 500  6800  1 

 100 
Solution
r 

A  P 1 

 100 
n
18 

 800 000  P 1 

 100 
n
r 
12 500 

 1 

6800  100 
4
6
6
r
 12 500 
6 
  1
100
 6800 
 800 000  P(1,18) 4
r
 12 500 
6 
 1 
100
 6800 
r

 0,1067941374
100
 r  10, 7% p.a.
800 000

P
(1,18) 4
 P  R412 631,10
EXERCISE 10
(a)
(b)
(c)
(d)
(e)
(f)
(g)
Find the final amount that R1 200 will grow to invested for 4 years at
(1)
13,5% p.a. compound interest.
(2)
17,8% p.a. compound interest.
Find the final amount that R20 000 will grow to invested for 5 years at
(1)
8% p.a. compound interest?
(2)
2,3% p.a. compound interest?
Phila invests R200 000 in an account paying 12% p.a. compounded annually. Calculate the
future value of his investment after 15 years.
Bradley borrows money from a bank in order to finance his business. The bank charges him
an interest rate of 12% p.a. compounded annually. Calculate the amount he originally
borrowed, if he repays the loan in 6 years’ time with a payment of R500 000.
R6 000 is invested for 5 years and grows to R7 500. Find the interest rate if interest is
compounded annually.
Find the annual compound interest rate that makes R3 000 double in 6 years.
R20 000 is invested at 12% p.a. simple interest for 3 years. Thereafter, the total amount is reinvested in a different financial institution at 20% p.a. compound interest for 2 more years.
What is the future value of the investment after the five-year period?
REVISION EXERCISE
(a)
(b)
Consider the following list of numbers:
1
0  
22
3 ;  ; 3 8 ; 9 ; ; ; ; 4 ; 3,6 ; 0,9 ;
 0 8
3
7
Write down, from this list, all the
(1)
whole numbers
(2)
integers
(3)
rational numbers
(4)
irrational numbers
(5)
real numbers
Consider the numbers 126, 108 and 36.
(1)
Write each number as the product of prime numbers.
(2)
Determine the lowest common multiple of these numbers.
(3)
Determine the highest common factor of these numbers.
32
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l)
(m)
(n)
(o)
(p)
(q)
(r)
Find the missing number:
(1)
80 : 16 = ? : 3
(2)
300 ml : 1,5 l = 1 : ?
Share R96 in a ratio 1 : 2 : 3.
There are R5-coins and R2-coins in a jar. The ratio of the number of R5-coins to the number
of R2-coins is 10:7. There is a total amount of R256 in the jar. How many of each type of
coin are there?
Two co-interior angles between parallel lines have sizes in a ratio 4:5. Find the angles.
The ratio of a father’s age to his son’s age is 8 : 3. In 5 years’ time, the sum of their ages will
be 54 years. How old are they?
A car is travelling at a speed of 84 km/h.
(1)
How far can the car travel in 45 minutes?
(2)
How long will it take for the car to travel a distance of 378 km?
The table shows the number of dogs an animal shelter has to feed and the mass of dog food
they need.
Number of dogs
20 30 b
Mass of food (kg)
5
20
a
(1)
Are these two variables directly or indirectly proportional?
(2)
Find the values of a and b.
(3)
Sketch the graph of the mass of food vs. the number of dogs.
The table shows the number of rats feeding on a pile of food and the time they take to finish
it.
Number of rats
10 15 b
Time taken (minutes) 3
5
a
(1)
Are these two variables directly or indirectly proportional?
(2)
Find the values of a and b.
(3)
Sketch the graph of the number of rats vs. the time taken to finish the food.
A box of highlighters is bought for R40 and sold at a profit of 22%. What is the box sold for?
If an item is sold at a price of R60, after 20% discount, what was the original price of the
item?
A book is bought for R50 and sold for R80. What is the percentage profit?
Assume the rand-dollar exchange rate is $1 = R10,56.
An item is bought for $ x and sold, with 70% profit added, for R359,04. Find the value of x.
R8 000 is invested for 6 years. What is the accumulated amount if the interest rate is
(1)
12% p.a. simple interest?
(2)
7% p.a. compounded annually?
Calculate the original amount you have to invest in order to have R10 000 in 5 years’ time if
interest is calculated at
(1)
8% p.a. simple interest.
(2)
7,2% p.a. compound interest.
At what interest rate must R1 000 be invested to grow to R3 000 in 10 years’ time if
(1)
simple interest is used?
(2)
compound interest is used (compounded annually)?
A couch worth R9 500 is bought on HP. A deposit of 30% is paid first and then the balance,
plus 15% p.a. simple interest is repaid over 3 years by means of monthly payments. What is
the value of the monthly payment?
SOME CHALLENGES
(a)
(b)
The pressure (P) on a diver under water is directly proportional to the square of her depth (d)
below the surface of the water. The diver has reached a depth of 10 metres. How much further
must she descend for the pressure to double?
The light intensity (I) at a distance (d) from a light source is inversely proportional to the
square of the distance from the light source. At a distance 2 cm from the light source, the light
intensity is 128 units. Calculate the light intensity at a distance of 8 cm from the light source.
33
CHAPTER 2: NUMBERS, OPERATIONS AND RELATIONSHIPS
TOPIC: INTEGERS
REVISION OF GRADE 8 WORK
Here is a summary of the rules from Grade 8.
ADDITION
The following rules can help you to add integers:

When adding two positive numbers the result is positive. Simply add the numbers: 4 + 5 = 9

When adding two negative numbers the result is negative. Add the numbers (ignoring signs)
and then write a negative sign (−) in front to indicate the negative result: (−4) + (−5) = −9

When adding a positive number and a negative number (regardless of the order), the result
will have the sign of the “biggest” number (ignoring signs). Subtract the numbers (ignoring
signs) and then write the sign of the “biggest” number (ignoring signs) in front:
4 + (−5) = −1 (5 is bigger than 4: take the sign in front of 5, which is – and then subtract the
numbers 5 – 4 = 1)
(−4) + 5 = +1 (5 is bigger than 4: take the sign in front of 5, which is + and then subtract the
numbers 5 – 4 = 1)
Identity element of addition
0 is the identity element of addition, which means that when you add 0 to a number, the number
remains unchanged, so −5 + 0 = −5, for example.
Additive inverses
Additive inverses are two numbers that add up to 0. Examples are 2 and −2 (since 2 + (−2) = 0),
3 and −3 (since 3 + (−3) = 0) etc. They are numbers with the same “size”, but opposite signs.
When two additive inverses appear in a sum of integers, they cancel each other:
If we calculate, for example, 3 + 2 + (−8) + (−2), we may cancel the 2 and the (−2), since they are
additive inverses:
3 + 2 + (−8) + (−2)
= 3 + (−8)
= −5
SUBTRACTION
To subtract an integer is the same as to add its additive inverse.
In symbols:
a  (  b )  a  ( b )
and a  ( b )  a  (  b )
For example: 4  ( 3)  4  ( 3)  1 and 4  ( 3)  4  ( 3)  7.
The negative of a negative:
For example: (5)  5
(a )   a
MULTIPLICATION
  
  
  
  









DIVISION



SQUARES AND CUBES OF INTEGERS
() 2  
( ) 2  
(  )3  
(  )3  
Note: 32 does not mean the same as (3) 2 :
32  9 (only the 3 is squared, not the −)
34
(3)2  9 (both the – and the 3 are squared)
Example 1
Calculate the following:
(a)
45
(d)
(4)  (5)
(g)
(+7) × (+5)
(j)
(−4) × (−6)
12
(m)
4
4
(p)
3 
2
2
3
(s)
3
(v)
64
4  (5)
2  ( 3)
(−5) × (+3)
−5 × 5
100
10
(c)
(f)
(i)
(l)
(q)
(5)2
(r)
(5)3
(t)
(3)2
(u)
25
(w)
3
(x)
4
(b)
(e)
(h)
(k)
(n)
(o)
1
(4)  5
3  5  ( 8)  ( 4)
(+8) × (−2)
(−2)(−1)
64
4  4
Solutions
(c)
(f)
(h)
(k)
4  (5)  1
2  ( 3)
 2  3
1
−15
−25
−3
(n)
10
(o)
 3  2  6
9
4
(q)
(t)
(w)
25
9
−1
(r)
(u)
(x)
(g)
(j)
459
(4)  (5)
 4  5
 9
35
24
(m)
(p)
(s)
(v)
(a)
(d)
(b)
(e)
(i)
(l)
(4)  5  1
3  5  ( 8)  ( 4)
 3  5  8  4
 12
−16
2
64
 4
16
−125
5
not possible to calculate
MIXED OPERATIONS
Order of operations
When we perform a calculation containing more than one operation, there is a strict order we need to
follow:
FIRST:
SECOND:
THIRD:
FOURTH:
BRACKETS OVERRULE THE ORDER
Any calculation in brackets is always done first.
POWERS AND ROOTS
MULTIPLICATION AND DIVISION
Multiplication and division are equal in preference and if more than one
multiplication and/or division occur in a calculation they are simply done from left
to right.
ADDITION AND SUBTRACTION
Addition and subtraction are equal in preference and if more than one addition
and/or subtraction occur in an expression they are simply done from left to right.
Caution:
In calculations like
brackets:
28
and
23
5  20 the division line and the root line work like
28
 (2  8)  (2  3)  10  5  2 and
23
5  20  (5  20)  25  5
35
Example 2
Calculate:
(a)
2
2  3 3
6  4  (4  9)  (5)
(b)
(c)
Solutions
(a)
3  23
2  6
3  8
  9  25 
8
24
  95 
8
  45  3
(b)  6  4  (4  9)  (5)
2  3  32
 2  3 9
 2  27
(c)  32  9  16 
 6  4  (5)  (5)
  6  2  (5)  (5)
  6  (10)  (5)
  6  (10)  5
  25
3  23
3  9  16 
2  6
2
  16  5
  11
  42
EXERCISE 1 (Revision)
(a)
(b)
(c)
(d)
(e)
(f)
Determine the values of K, L and M in the following pattern:
−10; K; −4; −1; 2; L; M
Determine the values of X, Y and Z on the following number line:
X -20 Y 10 Z
Replace the  in each of the following with  ,  or :
(2)
−11  9
(1)
−35  −37
(5) 2   52
(5)
(3) 2  (  3)3
(4)
Arrange the following list of integers in descending order:
500; 50; 5 000; −5 000; −50; −500
Arrange the following list of integers in ascending order:
−8; −11; 17; −50; 23; −100; 0
Calculate, if possible:
(17)  (7)
(20)(5)
(1)
(2)
30  (6)
11  (12)
(4)
(5)
55
(16)  8
(8)
(7)
11
(11) (11)  (6)
(10) 15  (6)
(13) (27)  (9)  (3)
(14) (17)  (4)
(3)
( 2) 2   22
(6)
3  ( 5)  3  5
(3)
(6)
17  14
9  20
(9)
1000  (  300)  ( 700)
(12)
(15)
( 2)( 15)
2  10
(16)
21  5  (3)
(17)
(9)
3  3
(18)
5 
(19)
(7)2
(20)
81
(21)
( 4)3
(23)
(26)
112
3
125
(24)
(27)
 121
(92 )
(30)
 (  53 )
(22)
(25)
121
3
5
9
3  1
 ( 9) 2
(28)
3
125
(29)
(31)
 3 125
(32)
16  144
(33)
(9) 2  3 8
(34)
 16  4
(35)
25  16
(36)
25  16
36
(g)
Calculate:
(1)
(3)
(5)
(2)2
 (4)  49
4
42  (5)2  (2)3  (3)
(12)(3)
 2(3)2
(2)(3)
(7)
(1  2)  (2)2  (3  32 )
(9)
3 3

27   25

(2)
(4)(2) (2)3

8
(1)2
(4)
(2)3  3  3  22
(6)
(1  2)  (2)2
(8)
(3  2  12 )  10
(10)
64
4
INTEGERS IN ALGEBRA
Integers play a very important role as coefficients in algebraic expressions. You have already come
across integers in this context when studying Algebra and as you continue your journey through
Algebra this year you will continue to see integers featuring a great deal. The following examples
illustrate how the principles of dealing with integers feature in Algebra.
Addition and subtraction
Example 3
Simplify the following:
(a)
2 x  (3x)
(b)
5 x  9 x
(c)
(7 x)  (9 x)
(d)
2 x 2  14 y  7 x 2  3 y  8 y 2  11y 2
(e)
2 x 2  3x  (5 x 2  2 x  1)  8
Solutions
(a)
2 x  3x
(b)
(c)
=  14 x
= 5x
(d)
 5x  9 x
 (7 x)  (9 x )
 7x  9x
  2x
(  2 x 2  7 x 2 )  (14 y  3 y )  (8 y 2  11y 2 )
(e)
  9 x 2  11 y  3 y 2
2 x 2  3x  5 x 2  2 x  1  8
  3x 2  5 x  9
Multiplication and division
Example 4
Simplify the following:
(a)
7 x  2 xy
8a 2b8
2a 3b 6
Solutions
(d)
(a)
(d)
(f)
 14x 2 y
(b)
(3 xy )  (2 xy 2 )
(c)
(e)
3m 2 n(2m  3n  5)
(f)
(b)
6x 2 y 3
(c)
4b 2
(e)
a
10 pq 5 p 2 q 15 pq 2


5 pq 5 pq 5 pq
= 2  p  3q
6m3n  9m 2 n  15m 2 n
37
14a 2b
7ab 2
10 pq  5 p 2 q  15 pq 2
5 pq
2 a
b
Powers
  even number  
Remember the following rules:
  odd number  
Example 5
Simplify the following:
(a)
( 3 x 2 y 3 ) 2
Solutions
(a)
9x 4 y 6
(b)
(2 x3 y 2 )3
(c)
(2 p10 q 4 ) 4
(d)
3(2m 2 n3 )5
(b)
8x9 y 6
(c)
 (16 p 40 q16 )
(d)
 3(32m10 n15 )
  16 p 40 q16
 96m10 n15
Mixed operations
Example 6
Simplify the following:
12 x 2  5 x  3 x
(a)
(c)
3a  5ab  (2a) 2  b
(b)
5a  4ab  2a (b  2)
(d)
3a 3  (2b3  6b3 )  (3ab)3
3a 2b
Solutions
(a)
 12 x 2  15 x 2
(b)
 3x 2
(c)
 15a 2b  (4a 2 )  b
2
(d)
2
  15a b  4a b
  19a 2b
 5a  4ab  2ab  4a
  a  2ab
3a3  (4b3 )  (27a3b3 )
3a 2b
12a3b3  27a3b3

3a 2b
39a3b3

3a 2b
  13ab 2
EXERCISE 2 (Revision)
Simplify the following:
(a)
5a  7 a
(b)
3 y  10 y
(c)
8 p  5 p
(d)
9 p  11 p
(e)
2 x  8 y  10 x  9 y
(f)
7 x2  2 x  8x2  5x
(g)
2 p  3 p  5 p
(h)
18 p 2 q
9 pq
(i)
7 p  6 p 2 q
14 p3q 2
(j)
2 xy (3x  2 xy  1)
(k)
(5ab) 2
(l)
(3c 2 d 3 )3
(m)
(3mn) 2 (2m 2 n)3
(n)
(2 x) 2  (3x) 2
(o)
3 x 2  2 x 2  x  2 x 2
(p)
6 xy   2 xy  x  3 y 
(q)
(r)
2t  (3t  5t  t )  2t 2
(s)
(t)
6m 2 n 2  12mn2  6n 2
6 n 2
x2
 x  (3x)  3x 
x
2
4
3
(5 x) y  16 x y  2( xy ) 2 
2
2
38
2
CHAPTER 3: NUMBERS, OPERATIONS AND RELATIONSHIPS
TOPIC: COMMON FRACTIONS
REVISION OF BASIC CONCEPTS (GRADE 8)
a
Common fractions are numbers of the form
where a and b are integers and b  0 .
b
a

The fraction
represents the division calculation a  b .
b
a

In the fraction , a is called the numerator and b is called the denominator.
b
Equivalent fractions
Equivalent fractions look different but have the same value (i.e. the same position on the number
2
4
and :
line). An example is
3
6
2
2
0
1
3
0
1
2
4
6
A quick way to check whether two fractions are equivalent, is to use the cross-multiplication rule:
a c
2
4
If  then a  d  b  c . For example in and
you will find: 6  2  12 and 3  4  12.
b d
3
6
We can produce equivalent fractions of any given fraction by multiplying or dividing the numerator
and the denominator by the same number, for example:
5 5  2 10
5 5  3 15
20
20  20 1






6 6  2 12
6 6  3 18
100 100  20 5
Integers as common fractions
Integers may be regarded as special cases of common fractions. Simply “write them over 1”.
3
5
.
For example: 3  and 5 
1
1
The simplest form of a fraction
Any fraction can be simplified to its simplest form by dividing the numerator and denominator by
their HCF (highest common factor).
Example 1
Simplify the following fractions:
21
(a)
49
54
36
(b)
Solutions
(a)
The HCF of 21 and 49 is 7:
21 21  7 3


49 49  7 7
21
3
is
The simplest form of
49
7
(b)
The HCF of 54 and 36 is 18:
54 54  18 3
3



36
36  18
2
2
54
3
The simplest form of
is 
36
2
3 3 3
Note:  

2 2 2
39
Ordering and comparing common fractions
To compare fractions, it is advisable to first write them with the same denominator. We use the LCM
(lowest common multiple) of the denominators (also called the lowest common denominator or
LCD).
Example 2
5 12

6 15
Replace the  in the following with >, < or  :
Solution
The LCM of 6 and 15 is 30.
5 5  5 25
12 12  2 24




6 6  5 30
15 15  2 30
25 24
5 12
Clearly

 
30 30
6 15
Example 3
7
17
 
9
21
Replace the  in the following with >, < or  :
Solution
The LCM of 9 and 21 is 63.
17
17  3
51
7
77
49
 

 

9
9 7
63
21
21  3
63
49 51
49
51
Careful now, although
 , the given fractions are negative and so   
63 63
63
63
7
17
  
9
21
REVISION OF OPERATIONS WITH FRACTIONS
The following table summarises how to deal with the different operations:
Addition
a b ab
 
d d
d
Subtraction
a b a b
 
d d
d
Multiplication
a c ac
 
b d bd
Division
a c a d ad
   
b d b c bc

Denominators must be the same.
If not, use the LCD.

Add only the numerators.

Keep the denominator the same.

Denominators must be the same.
If not, use the LCD.

Subtract only the numerators.

Keep the denominator the same.

Multiply numerator by numerator.

Multiply denominator by denominator.

Cancel common factors between
numerator-denominator pairs before
multiplying (not essential, but easier!).
“Tip and times”:

Change division to multiplication.

Change the fraction after the division
sign to its reciprocal.

Multiply.
40
Squares & Cubes
Square Roots &
Cube Roots
2
3
a2
a3
a
a
and


 
 
b2
b3
b
b
a
a

and
b
b
3
a 3a

b 3b


Square/Cube the numerator.
Square/Cube the denominator.

Square Root/Cube Root of the
numerator.
Square Root/Cube Root of the
denominator.

Note: Always rewrite mixed numbers into improper fractions before doing fraction calculations.
Example 4
Calculate:
2 5 1
(a)
 
3 2 8
(b)
1
2
3 4
3
5
(c)
2
1 4
1   
3
2 7
Solutions
(a)
2 5 1
[LCD = 24]
 
3 2 8
2 5 1
  
3 2 8
2  8 5 12 1 3



3  8 2 12 8  3
16 60 3

 
24 24 24
73

24
1
3
24
(b)
10 22
[LCD = 15]

3
5
50 66


15 15
16

15
1
 1
15
(c)
2
1 4
1   
3
2 7
5
1 7
  
3
2 4
35

24
11
1
24
EXERCISE 1
You may not use a calculator in this exercise.
(a)
Fill in >, < or  :
1 1
1
1


(1)
(2)
(3)
3 9
5
8
3
7
2
2
1

(4)
(5)
(6)
4
4
3
3
(b)
Arrange the following list of numbers in ascending order:
1 2 1 6
9
1 ;  ;  ; ; 0;
3 3 6 5
7
(c)
Calculate:
2 1
1 1
3 1

(1)
(2)
(3)


5 3
2 4
5 6
1 2
2 5 1
1
2
(6)
(7)
(5)

 
2 3
9 3
3 2 8
2
3
1 2
2 4
2
3  1 

5
(9)
(10)
(11)
3 5
3
2 3
8 3
1
1
12 5


(13)
(14)   
(15)
9 2
5
2
10 3
41
5
6
23
71
(4)
(8)
(12)
(16)
9
11
23
73
4 1

3 2
1
2
3 4
3
5
2 9

3 4
9
11
 
22
18
5
1 7
 
3
2 4
1
2
3  2  3
2 3
3
9
4
1
2
2
5
(17)
(21)
(25)
(29)
(18)
(22)
(26)
(30)
5 9 7
 
(19)
6 14 5
2
1
4 3
(23)
3
2
100

 1000 (27)
11
 23  24  25 
   
 24  25  26 
1
1
3 4
(20)
3
5
7
9
4
(24)
100
10
2
1
 (4)  (28)
5
3
4 2

5 25
7 49

4 2
1
1
1  7  11
6
3
Squares, cubes, square roots and cube roots
Example 5
Calculate:
(a)
 1
 1 
 2
(d)
3

3
1 1
  
 2 3
(b)
64
125
2
4
1  
5
(e)
(c)
6
1
4
(c)
6
1
4
2
Solutions
(a)
 1
 1 
 2
3
 3 
 
 2 
1 1
  
3 2
(b)
3
2 3
  
6 6
 1 
 
 6 
1

36
27

8
3
 3
8
(d)
3
64

125

2
4
5

25
4

25
4

5
1
2
2
2
2
4
1  
5
(e)
2
2
25 16

25 25
3
9


25 5

EXERCISE 2
You may not use a calculator in this exercise.
(a)
Calculate the following:
2
(1)
1
 
2
3
(4)
2
 
5
(7)
 1
2 
 2
3
(2)
1
 
2
(5)
 2
 
 3
(8)
 1
1 
 3
2
(3)
(6)
 1
 
 4
(9)
 1
 1 
 2
2
3
42
2
2
 
5
3
2
(b)
2
(10)
 2
3 
 3
(13)
1

 2 1 
2

2 1
  
3 2
(14)
  1 3 
1     
  2 


2
Calculate the following (if possible):
1
(2)
(1)
16
8
3
(5)
(4)
27
7
8
2
(11)
(8)
3
1 1
  
 4 3
(15)
  1 3 
  
 2  


2
1
8

1
4
(6)
2
10
27
(9)
1
11 1

16 8
2
(10)
16 9

25 25
(11)
16
9

25
25
(12)
(13)
1 3
1 
2 4
(14)
3
8 5

27 9
(15)
2
4
9
(3)
(7)
3
2
(12)
3

1
8
7
9
3
1
64
Mixed Calculations
Remember the order of operations:
(1) Brackets, (2) Powers/Roots,
(3) Multiplication/Division, (4) Addition/Subtraction.
Example 6
Calculate:
(a)
1 1 4 1 
   
4 3  5 10 
3
(b)
 1  1
      2 
 2  2
2
(c)
1 1

2 3
1
5 3
3
Solutions
(a)
1 1 4 1 
   
4 3  5 10 
1 1 8 1
    
4 3  10 10 
1 1 9
   
4 3  10 
1 3
 
4 10
5
6


20 20
11

20
3
(b)
 1  1
      2 
 2  2
2
 1  5
     
 8  2
1 25
 
8 4
1 50
 
8 8
49

8
1
 6
8
43
(c)
2
3 2

6 6
16 9

3 3
5
 
6
  
 25 
 
 3 
1
5
31


25 5
2 6

1
10
EXERCISE 3
You may not use a calculator in this exercise.
Calculate the following:
(a)
1 4  2
1 1    
3 5  3
2
3
(e)
113  111   112 

 

111  112   113 
(h)
1 1
1
12     1
4 2
8
3
1
 1 
  1 1
3
 3 
(d)
3
(g)
1
 1
1  2 
3
 2
1 1

2 3

27  25
 
 1
8
 16

(b)
(c)
1
 1
1  2 
2
 3
1 2

2 3
(f)
3
2
2

125  4
 
 1
27
 9

2
3
Percentages as common fractions
Example 7
Express the following percentages as fractions:
(a)
30%
(b)
75%
Solutions
(a)
30
100
30  10

100  10
3

10
(b)
75
100
75  25

100  25
3

4
(c)
33 13 %
(d)
(c)
33 13
100
(d)
115%
115
100
115  5

100  5
23

20
3
1
20
1
 33  100
3
100 100


3
1
100
1


3
100
1

3
Common fractions as percentages
Example 8
Express the following fractions as percentages:
3
2
(b)
(c)
(a)
5
3
7
2
(d)
1
Solutions
3 100

(a)
5 1
7 100

2 1
(d)
5 100

4 1
3 100


1
15
 60
60%
(b)
20
2 100

3 1
(c)
200

3
2
 66
3
2
66 %
3
7 100


1
12
44
50
1
4
5 100


1
14
 350
 125
350%
125%
25
EXERCISE 4
(a)
(b)
(c)
Express the following percentages as fractions:
(1)
25%
(2)
60%
(3)
12 12 %
Express the following fractions as percentages:
3
7
6
(1)
(2)
(3)
5
4
10
Complete
(1)
20 is …% of 50?
(2)
24 is …% of 30?
(3)
2
5
is ...% of
(6)
(4)
30 is …% of 20?
(5)
3
6
(4)
125%
(4)
4
1
2
115 is …% of 50?
11
1
is ...% of 2
25
5
COMMON FRACTIONS IN ALGEBRAIC EXPRESSIONS
Common fractions play a very important role as coefficients in algebraic expressions.
Example 9
Simplify:
(a)
2
1
1
x x2
3
6
2
(b)
x 1 x  3

1
5
2
Solutions
(a)
(b)
2
1
1
x x2
3
6
2
2x x 5

 
3 6 2
2x 2
x
5 3



 
3 2
6
2 3
4 x x 15

 
6 6 6
4 x  x  15

6
5 x  15

6
x 1 x  3

1
5
2
( x  1) ( x  3) 1



5
2
1
( x  1) 2
( x  3) 5
1 10




 
5
2
2
5
1 10
2( x  1)
5( x  3)
10



10
10
10
2( x  1)  5( x  3)  10

10
2 x  2  5 x  15  10

10
3 x  7

10
[write each fraction as
numerator
]
denominator
[LCD  6 ]
[convert to equivalent fractions]
[write as a single fraction]
[simplify]
[put brackets around terms in the numerator]
[LCD  10 ]
[convert to equivalent fractions]
[write as a single fraction]
[distributive property]
[simplify]
45
EXERCISE 5
(a)
Simplify to a single fraction:
(b)
1
2
x x
2
3
2
m
(4)
m
5
10
3
x
1
(7)
x  3
4
7
2
Simplify to a single fraction:
2
x 1
x
(1)
3
3
4x x  2
(4)

3
2
2
x
2 x 3x

 1
(7)
2
7
2
(1)
(2)
(5)
(8)
(2)
(5)
(8)
y 1
 y
8 2
1
3
2 p
4
5
2
14
k 2
3
k
2x 1 x  2

4
2
1
a 1
a
6
8
x4 1
2
 ( x  1)
4
8
(3)
(6)
(9)
(3)
(6)
(9)
1
1
x 1
4
2
1
1
a  a 1
4
6
1
2 20
1 x  
3
7 21x
2x  3
2
2 x 3x x  1
 
3
2
6
x3
 x 3
3
2
SOME CHALLENGES
(a)
(b)
(c)
(d)
(e)
(f)
(g)
John asks Thabo how many learners there are in his class. Thabo answers: “If you take twice
our number, plus half of our number, plus a quarter of our number, plus 1, the answer is 100.”
How many learners are there in Thabo’s class?
5
You spend of your money and have R42 left. How much money did you have initially?
7
In a certain school, a test counts 20% of the year mark. The year mark counts 25% of the
final mark. What fraction does the test count of the final mark?
A certain concentrate contains 40% orange juice. 50 ml of the concentrate is diluted with
200 ml of water to make 250 ml of cool drink. What fraction of the drink is orange juice?
2
The length of a square is 3 cm . What is the area of the square?
3
4 22
The volume of a sphere is approximately   (radius)3 . What is the volume of a sphere
3 7
1
with a radius of 1 m?
2
Simplify:
x x
2

2 xy 2
 2  1
2 3



xy
x
(1)
(2)


x x
2
3
 3 

3 6
3
1
1 2 3 
x x

3
2x  x  x   x 
(4)
(3)
  3 x  1
42
4
2

2 
46
CHAPTER 4: NUMBERS, OPERATIONS AND RELATIONSHIPS
TOPIC: DECIMAL FRACTIONS
REVISION OF GRADE 8 WORK
Converting from decimal fractions to common fractions
A decimal fraction can be converted to a common fraction easily by using the appropriate power of
10 as denominator. After this, it is merely a matter of simplification.
Example 1
Write 0,25 as a common fraction.
Solution
Look at the place-value of the right-most digit. Can you see that it is hundredths? This means that
we will use 100 as the denominator and all the digits together as the numerator:
25 1
  0, 25
0, 25 
100 4
Example 2
Write 2, 4 as a common fraction.
Solution
24 12

10 5
12
2
 2, 4 
(or 2 )
5
5
2, 4 
[The right-most digit is a tenths digit]
Converting from common fractions to decimal fractions
To write a common fraction as a decimal fraction requires re-writing the common fraction into an
equivalent form with a power of 10 as denominator.
Example 3
Example 4
3
as a decimal fraction.
5
Solution
21
as a decimal fraction.
20
Solution
Write
Write
3 2 6

(6 tenths)
5  2 10
3
  0,6
5
21 5 105

(105 hundredths)
20  5 100
21
  1, 05
20
21
1
1 5
5
Alternatively:
1 1
1
 1,05
20
20
20  5 100
Example 5
Write
7
as a decimal fraction.
200
Solution
7
75
35


 0, 035
200 200  5 1000
There are cases where it is impossible to convert the denominator to a power of 10! In such cases,
we have to use long division.
47
Example 6
Write
1
as a decimal fraction.
3
Solution
There is nothing we can multiply 3 by to produce a power of 10. Now, remember that
1  3 . If we write 1 as 1,000000… and divide this, digit by digit, by 3 we get:
0 , 3 3 3 …
3 1 , 0 0 0 0
0
1
0
9
1 0
9
1 0
1
Can you see that this process will repeat indefinitely?  0,333333...
3

This is an example of what is called a recurring decimal and can be written in short as 0,3.
1
means
3
Example 7
Write
5
as a decimal fraction.
6
Solution
There is nothing we can multiply 6 by to produce a power of 10. Now, remember that
5  6.
If we write 5 as 5,000000… and divide this, digit by digit, by 6 we get:
0 , 8 3 3 …
6 5 , 0 0 0 0
0
5
0
4
8
2 0
1 8
2 0
Can you see that this process will repeat indefinitely? This means that
5
means
6
5
 0,83 .
6
Common forms
Fraction
Decimal
%
Fraction
Decimal
%
Fraction
Decimal
%
1
2
1
3
2
3
1
4
0,5
50%
0,4
40%
37 12 %
33 13 %
0,6
60%
0,625
62 12 %
0,6
66 23 %
0,8
80%
0,875
87 12 %
0,25
25%
0,16
16 23 %
3
8
5
8
7
8
1
9
0,375
0,3
2
5
3
5
4
5
1
6
0,1
11 19 %
48
0,83
83 13 %
0,125
12 12 %
3
4
1
5
0,75
75%
0,2
20%
Fraction
Decimal
%
Fraction
Decimal
%
5
9
7
9
8
9
1
10
3
10
7
10
0,5
55 %
0,9
90%
0, 7
77 79 %
9
10
1
1
100%
0,8
88 89 %
0,1
10%
0,3
30%
0,7
70%
5
9
5
6
1
8
2
9
4
9
0, 2
22 92 %
0,4
44 94 %
Ordering and comparing decimal fractions
To compare decimal fractions, it is advisable to first write them with the same number of decimal
places after the comma. Add 0’s onto the decimal fraction with the least number of decimal places
and then compare the “numbers” ignoring the comma.
Example 8
Replace the  in the following with >, < or  :
5, 7  5, 07
Solution
5,7 = 5,70
Now comparing 5,70 to 5,07 is comparing 570 hundredths to 507 hundredths.
Clearly 570 is more than 507 and so 5,7 > 5,07.
Example 9
Replace the  in the following with >, < or  :
2,15  1, 338
Solution
−2,15 = −2,150 which is −2150 thousandths and −1,338 is −1338 thousandths.
Can you see that −2150 < −1338? So −2,15 < −1,338.
Example 10
Arrange 0,3;
1
and 0,28 in descending order.
3
Solution
0,3 = 0,300000….
1
3
 0,333333...
Comparing the first 3 places, we can already see that
49
0,28 = 0,280000…
1
 0,3  0, 28 .
3
Rounding
In practice, we are often interested in approximate, rather than exact answers. To round off to a
specified number of decimal places, go to the digit in the position to which you want to round off.
Look at the next digit. If it is 5 or higher we round up and if it is 4 or lower we round down.
Example 11
Round 0,2378 off to the nearest hundredth.
Solution
The hundredths digit is a 3, so it will either stay a 3 or become a 4 depending on the next digit (i.e.
the thousandths digit):
0,2 3 7 8
Since the thousandths digit is 7 (5 or higher), we change the 3 to a 4 and drop the subsequent
digits: 0, 2378  0, 24
Example 12
Round 2,14149 off to the nearest thousandth (i.e. to three decimal places).
Solution
2,14149
Since the next digit is 4 (less than 5), we leave the thousandths digit 1 and drop the subsequent
digits: 2,14149  2,141
Example 13
Round 3,9521 off to the nearest tenth (i.e. to one decimal place).
Solution
3, 9 5 2 1
Since the hundredths digit is 5, we have to change the 9 to 10, which means we get an overflow of 1
into the units: 3,9821  4, 0.
Notice how we wrote 4,0 instead of just 4 to indicate that we have rounded off to 1 decimal.
Writing simply 4 is not incorrect, but it doesn’t give the same information (namely the degree of
accuracy).
ADDITION AND SUBTRACTION
Example 14
Example 15
Calculate 2,314 + 3,72
Calculate 3,18 – 2,713
Solution
Solution
1
+
2
2 , 3 1 4
3 , 7 2 0
6 , 0 3 4
+
3 ,
2 ,
0 ,
1
1
7
4
7
8
1
6
1
0
3
7
MULTIPLICATION
Multiplying decimal fractions by powers of 10
When decimal fractions are multiplied by powers of 10 (10, 100, 1000 etc.), we move the comma to
the right by the same number of places as the numbers of zeros in the power of 10 we are multiplying
by.
50
Example 16
Calculate:
(a)
0,0234 × 100
(b)
2,708 × 100 000
Solutions
(a)
(b)
Move the comma 2 places to the right: 0,0234 × 100 = 002,34 = 2,34
(Remove excess 0’s to the left of 2)
Move the comma 5 places to the right: 2,708 × 100 000 = 270800, = 270800
(After we move the comma 3 places we need to add in two extra 0’s to complete the 5-place
move)
Multiplying decimal fractions by other decimal numbers and integers
Example 17
Calculate 0,6 × 0,3.
Solution
Method 1 (Using Common Fractions)
0, 6  0,3
6 3

10 10
18

100
 0,18

Method 2 (Direct Approach)
To multiply two decimal fractions:
 We add the number of decimal places of the two numbers. This gives us the number of decimal
places we must allow for in our answer.
 We then multiply the numbers, ignoring the comma.
 Finally, we add in the comma, allowing for the number of decimal places we’ve worked out
initially.
6 × 3 = 18
Number of decimal places:
 0, 6  0,3  0,18
1 (from 0,6) + 1 (from 0,3) = 2
Example 18
Calculate 2,5 × −1,34.
Solution
Let’s determine the sign first:     
Next, the number of decimal places: 1 (from 2,5) + 2 (from 1,34) = 3
Now, multiply the numbers, ignoring the signs:
1 3 4
×
2 5
6 7 0
2 6 8 0
3 3 5 0
Insert the comma to allow for 3 decimal places: 3,350 (Use the full answer, including the 0.)
∴ 2,5 × −1,34 = −3,350 = −3,35
51
DIVISION
Dividing decimals by powers of 10
When decimal fractions are divided by powers of 10 (10, 100, 1000 etc.), we move the comma to the
left by the same number of places as the numbers of zeros in the power of 10 we are dividing by.
Example 19
Calculate:
5,302 ÷ 1000
Solution
Move the comma 3 places to the left: 5,302 ÷ 1000 = 0,005302 (Insert 0’s to make this possible).
Dividing decimals by integers
Example 20
Calculate: 903,807 ÷ 5
Solution
1
5 9
5
4
4
8 0 , 7 6 1 4
0 3 , 8 0 7 0
0
0
· 3
0
3
3
8
5
3 0
3 0
· 7
5
2 0
2 0
·
903,807 ÷ 5 = 180,7614
Example 21
Calculate: −21,23 ÷ −9
Solution
2 , 3
9 2 1 , 2
1 8
3
2
2
7
5
4
5 8 8 8 …
3 0 0 0 …
3
5
8 0
7 2
8 0
7 2
8 0
7 2
8 …
It is clear that this will continue indefinitely,
so the 8 recurs.
Signs: − ÷ − = +
So −21,23 ÷ −9 = 2,358
52
Dividing decimals (or integers) by decimals
When dividing by a decimal, we can convert the divisor to a whole number by writing the division
calculation as a fraction and then multiplying by an appropriate power of 10 (both numerator and
denominator) to create an equivalent fraction.
Example 22
Calculate:
(a)
0,6 ÷ 0,03
(b)
14,245 ÷ 0,7
Solutions
(a)
(b)
0,6
0,6  100 60


 20
0,03 0,03  100 3
14, 245 14, 245  10 142, 45


0,7
0,7  10
7
· 2 0 ,
7 1 4 2 ,
3
4
2
5
5
3
14,245 ÷ 0,7 = 20,35
[We have used short division here. It saves time!]
SQUARES AND CUBES
Squares and cubes of decimal fractions can easily be calculated by converting them to common
fractions first, but there is a method that can be used to calculate them directly. In the following
examples we will explore this.
Example 23
Calculate:
(a)
0, 22
(b)
(0,9)2
(c)
1,12
(b)
81
 9 
( 0,9) 2     
 0,81
 10  100
(d)
25
 5 
0, 052  
 0, 0025
 
 100  10000
(d)
0,052
Solutions
2
(a)
4
 2
0, 22    
 0, 04
 10  100
(c)
121
 11 
1,12      
 1, 21
100
 10 
2
2
2
We can square a decimal by squaring the number formed by the digits of the base, ignoring the
comma, and then placing the comma in the correct position, so that the result has double the number
of decimal places as the original base.
Example 24
Calculate:
0, 23
(a)
(b)
(0,1)3
(b)
1
 1
( 0,1)3      
 0, 001
1000
 10 
(c)
0,033
Solutions
3
3
(a)
8
 2
0, 23    
 0, 008
 10  1000
(c)
27
 3 
0, 033   
 0, 000027
 
1000000
 100 
3
53
We can cube a decimal by cubing the number formed by the digits of the base, ignoring the comma,
and then placing the comma in the correct position, so that the result has three times the number of
decimal places as the original base.
SQUARE ROOTS AND CUBE ROOTS
Example 25
Calculate:
0,09
(a)
(b)
0,25
(b)
0, 25 
0, 0064
(c)
Solutions
9
3
  0, 3
100 10
64
8
0, 0064 

 0, 08
10000 100
0, 09 
(a)
(c)
25
5
  0, 5
100 10
We can take the square root of a decimal by taking the square root of the number formed by the digits
of the base, ignoring the comma, and then placing the comma in the correct position, so that the result
has half the number of decimal places as the original base.
Example 26
Calculate:
3
0, 027
(b)
3
0, 000001
27
3

 0,3
1000 10
(b)
3
0, 000001  3
(a)
(c)
3
0,125
Solutions
(a)
3
0, 027  3
(c)
3
0,125  3 
1
1

 0, 01
1000000 100
125
5
   0,5
1000
10
We can take the cube root of a decimal by taking the cube root of the number formed by the digits of
the base, ignoring the comma, and then placing the comma in the correct position, so that the result
has a third as decimal places as the original base.
(Grade 8 Revision)
EXERCISE 1
You may not use a calculator for this exercise:
(a)
(b)
(c)
(d)
(e)
Write 3,24 as a common fraction.
Write the following in decimal form:
22
3
(2)
(1)

40
9
Arrange in ascending order: 3,305 ; 3,15 ; 0 ;  3; 3,4 ;
Round 453,4778 off to the nearest hundredth.
Calculate:
(1)
0,256 + 2,78
(2)
3,2 – 2,907
(4)
−4,387 – 2,995
(5)
0,06 × 0,4
(7)
2,32 × 6,7
(8)
−3,6 × 1,7
(10) −3,03 × −2,4
(11) 27 ÷ 0,09
(13) −1,25 ÷ 0,04
(14) 11,4 ÷ 0,04
(16) 2,1 – 0,06 × 0,5
(17) 0,05 × (2,2 – 1,8)
(19)
3
(0, 2) 2  (0,013)
(20)
3
 0,000001
54
 3,35
(3)
(6)
(9)
(12)
(15)
(18)
−3,8 + 5,54
−4 × 0,07
4,6 × −2,55
−0,36 ÷ 0,4
−2,15 ÷ 0,3
3 0,125  1, 21  ( 0, 2) 2
DECIMAL FRACTIONS IN ALGEBRA
Decimal fractions are sometimes encountered as coefficients in algebraic expressions. The following
example illustrates this.
Example 27
Simplify the following:
(a)
0,1x  2,5 y  0, 2 x  0,3 y
(c)
0,2 x 2  3 xy  1,5 x
0,1x
(b)
0,3x(0,2 x 2  0,1x  5)
(d)
(0, 2 xy ) 2  2 x  0,3 y
(b)
0,06 x3  0,03x 2  1,5 x
(d)
0,04 x 2 y 2  0,6 xy
Solutions
(a)
(c)
(0,1x  0, 2 x)  (2,5 y  0,3 y )
 0,3x  2,8 y
0,2 x 2
3 xy
1,5 x


0,1x 0,1x 0,1x

0, 2 x 2  10 3xy  10
1,5 x  10


0,1x  10 0,1x  10 0,1x  10
2 x 2 30 xy 15 x


1x 1x
1x
 2 x  30 y  15

EXERCISE 2
Simplify the following expressions:
(a)
0,3x  0, 4 y  0, 23 x  0,01 y
2,52 x  0,71 y  0,02 x  0,5 y
0, 2 xy (0,1x  0,3xy  3 y )
(d)
0,01a(2a 2  3a  1,5)
(e)
0,3 p (0,02 p 2  0,01 p  0,04)
(f)
0,32m3  0,02m 2  0,18m
0,04m
(h)
0,01x  20 y  0,1x  0,3 y
(j)
0,8 xy  0, 2  (0,1 y  0,5 y )
(l)
(0, 2 x) 2  x  (0,5 x)3
(i)
(k)
2
(b)
(c)
(g)
2
0, 24 x 2  0,03 x  1, 2 xy
4 x
2
0,3x y  0, 25 x  1,2 xy
2
2
3
(0,3 x y )  0,4 x  0,1xy
2
55
CHAPTER 5: NUMBERS, OPERATIONS AND RELATIONSHIPS
TOPIC: EXPONENTS
REVISION OF GRADE 8
Recall, from Grade 8, the exponential notation:
The meaning of this is:
a n  a  a  a  a  ... (n factors)
a
n
Exponent
Power
Base
We also learnt some important exponential laws.
LAW 1: MULTIPLYING POWERS OF THE SAME BASE
When multiplying two powers of the same base, we leave the base the same and add the exponents:
a m  a n  a m n
Example 1
Simplify the following:
x  (  x) 2
(a)
x 2  x3
(e)
24  x  35  y 5  23  x5  y 2
(b)
(c)
a  b  a 2  b3
(d)
6m2  2m3
(c)
a  b  a 2  b3
(d)
6m2  2m3
Solutions
(a)
x 2  x3
(b)
x  (  x) 2
 x5
(e)
 x  x2
 x3
24  x  35  y 5  23  x5  y 2
 a 3b 4
 12m5
 (24  23 )  35  ( x  x5 )  ( y 5  y 2 )
 27  35  x6  y 7
LAW 2: DIVIDING POWERS OF THE SAME BASE
When dividing two powers of the same base, we leave the base the same and subtract the exponents.
am
 a mn
n
a
Example 2
Simplify the following and leave your answers in exponential form:
x8
28
(b)
(a)
x2
23
= 25
 x6
When the exponent in the denominator is greater than the exponent in the numerator, we write the
power in the denominator:
am
1
 nm
n
a
a
56
Example 3
Simplify the following and leave your answers in exponential form:
(a)
(d)
23
28
x4 y 2
xy 3
(b)
(e)
Solutions
23 1

(a)
28 25
(d)
x 4 y 2 x3

y
xy 3
(f)
9( y )3 x 2
12 yx3

9   y3  x2
12 yx 3
3

 9 y3 x2
4
12 yx3
x2
x8
6 x3 y 2
2 xy 3
(c)
(f)
(b)
x2
1
 6
8
x
x
(e)
6 x3 y 2 3x 2

y
2 xy 3
(c)
23  312
28  38
9( y )3 x 2
12 yx3
23  312 34

28  38 25
Notice that we still divide the −6 and −2
normally. They are not in exponential
form.
Notice that we deal with the 9 and the 12
as we would normally do in fractions, i.e.
divide them by both their HCF. They are
not in exponential form.
3 y 2

4x
Combining multiplication and division (Rules 1 and 2)
Example 4
Simplify the following and leave your answer in exponential form:
(a)
57  33  36  58
37  520  31
(b)
3 x 2  4 xy 7
2 x2 y  6 x2  y
(b)
3x 2  4 xy 7
2 x2 y  6 x2  y
7 x 2 y 5 18 x

14 x3 y 6 y
(c)
2x2 y 
(c)
7 x 2 y 5 18 x

2x y 
14 x3 y 6 y
Solutions
(a)
57  33  36  58
37  520  31
515  39
 20 8
5 3


2 x 2 y y 4 3x
 
1
2x y
6 x3 y 5

 3x 2 y 4
2 xy
y5

x
3
 5
5
The zero exponent rule:
12 x3 y 7
12 x 4 y 2
2
a0  1
Example 5
Simplify:
(a)
x0
(b)
(4 x)0
(c)
4x0
(b)
(4 x)0  1
(c)
4 x0  4 1  4
Solutions
(a)
x0  1
57
LAW 3: RAISING A POWER TO A POWER
m n
mn
When raising a power to a power, we multiply the exponents: (a )  a
Example 6
Simplify the following and leave your answer in exponential form:
(a) (23 )5
(b)
( x2 )4
(c)
[( x 2 )4 ]3
 215
 x8
 ( x8 )3
 x 24
LAW 4: RAISING A PRODUCT TO A POWER
When raising a product to a power, we apply the exponent to each factor of the product, and multiply
the results. This is also known as the distributive rule of powers over multiplication.
(a  b)n  a n  bn
Example 7
Simplify the following:
(a)
(ab)5
(b)
(2 x)3
(c)
(b)
(2 x)3
(c)
(3abc) 2
Solutions
(a)
(ab)5
 a 5b 5
(3abc ) 2
 23 x3
 ( 3) 2 a 2b 2c 2
 8 x3
 9 a 2b 2 c 2
Combining rules 3 and 4
We can combine rules 3 and 4 as follows: (a mbn ) p  (a m ) p (bn ) p  a m p  bn p
Example 8
Simplify the following:
(a)
(a 2b3 )5
(b)
(2 xy 2 )3
Solutions
(a)
(a 2b3 )5
(b)
 a10b15
( 2 xy 2 )3
 ( 2)3 x 3 y 6
  8 x3 y 6
SUMMARY OF THE DEFINITIONS AND EXPONENTIAL LAWS (GRADE 8)
Definitions
a n  a  a  a  ... (to n factors)
a0  1
Negative bases
(1)even number  1
( 1)odd number  1
(a)even number   a even number
( a )odd number   a odd number
58
Laws of Exponents
Law 3:
a m  a n  a m n
am
 a m  n when m  n
n
a
(a m )n  a mn
Law 4:
(a  b)n  a n . bn
Law 1:
Law 2:
am
1
 n  m when m  n
n
a
a
MIXED EXAMPLES
Example 9
Simplify the following:
(a)
2 x5  (3x3 y)2
4 x 2  x  (5)2
(c)
(2 x3 )2
Solutions
(a)
2 x5  ( 3 x3 y ) 2
(b)
(2 xy3 )3
4 xy 4  2 x 2 y 5
(d)
2(3x)2  xy 2  32
(b)
(2 xy 3 )3
4 xy 4  2 x 2 y 5
 2 x5  9 x 6 y 2
8 x 3 y 9
 1

8 x3 y 9
11 2
 18 x y
(c)
4 x 2  x  (5) 2
(2 x3 )2
2(3 x) 2  xy 2  32
(d)
 2(9 x 2 )  xy 2  9
3
4 x  25
4 x6
100 x3

4 x6
25
 3
x

 162 x3 y 2
EXERCISE 1 (Grade 8 Revision)
(a)
(b)
Simplify the following and leave your answer in exponential form:
374  373
239  113
(312  19)3
(2)
(3)
(1)
315  ( 193 )2
375
2317  114
Simplify:
a17
x10
x 2  x8
(2)
(3)
(1)
a 21
x4
( x 6 )3
( xy )5
4 x3 y 9  5xy 2
(4)
(5)
(6)
(7)
( 7 pq5 )2
(8)
3x 2 y 0  (3x3 )0
(9)
b2  (ab3 )2  ( b3 )2
(10)
10 xy 3
8 x 3 y
(11)
( 2 x 5 y 6 ) 2
(4 xy 4 )3
(12)
2a 2bc3  9abc 2
( 6a )2 b0c7
(13)
(  m 2 )3  (  n 2 p ) 3
( mnp )2  ( 2mn )0
[a{a( a )2 }2 ]3
(14)
59
(15)
(  x 3 ) 2
3
(c)
Determine the values of A, B and C, in each of the following:
( m 2n A )3
3 2 A
6 C
 m3n 9t C
( 5 x y )  Bx y
(2)
(1)
B
m
SOME MORE EXPONENTIAL LAWS AND DEFINITIONS
Raising a fraction to a power
When raising a fraction to a power, we apply the exponent to both the numerator and the
denominator. This is also known as the distributive rule of powers over division.
n
an
a

 
bn
b
Example 10
Simplify the following:
a
 
b
(a)
5
2
(b)
 3 
 3
x 
2
(b)
 3 
 3
x 
3
(c)
 3x 2 y 
 3 
 z 
3
(c)
 3x 2 y 
 3 
 z 
Solutions
5
a5
a

 
b5
b
(a)


(3)2
( x3 ) 2
9

27x 6 y 3
z9
x6
Negative exponents
Consider the expression:
x3
x5
xm
x3
mn
here, we get 5  x35  x 2 . But we also know that
If we apply the law n  x
x
x
3
x
x. x. x.
1
1

 2 . So we have reason to assume that x 2  2 . (They both result from the
5
x . x . x .x.x x
x
x
same calculation, approached in different ways.)
1
1
x 1 
In general, we may say that: x  n  n Take special note of the case:
x
x
A few more rules concerning negative exponents are:
1. ax
4.
n
a
 n
x
a
ax n

b
bx  n
ax  n
a
2.
 n
b
bx
5.
x
 
 y
n
 y
 
x
3.
a
 ax n
n
x
n
In a nutshell: As long as an expression is made up entirely of multiplication and division, any
power with a negative exponent in the numerator may be moved to the denominator and the
exponent changed to a positive. Similarly any power with a negative exponent in the denominator
may be moved to the numerator and the exponent changed to a positive.
60
Example 11
Simplify the following, leaving your answer with positive exponents:
(a)
2
3 5
2
3 5
3 x y  6 x y
(b)
 4 x 3 y 3
6 x  5 y 2
(b)
 4 x 3 y 3
6 x  5 y 2
3
(c)
 3 x 2 
 3 
 2y 
3
(c)
 3 x 2 
 3 
 2y 
Solutions
(a)
3 x y  6 x y
 18 xy 4
18 x
 4
y
2 x5
 3 2 3
3x y y
 3 y3 
  2 
 2x 
3
2 x2
 5
3y
 2 x2 
  3 
 3y 
3

8 x6
27 y 9
When there is addition and subtraction involved, powers cannot simply be moved across the fraction
line. In such cases each power has to be dealt with individually.
Example 12
Simplify the following:
x 1  y 1
x y
Solution
1
x y
x y
1
1 1  x y



1
1
x y  xy  x  y





xy
x y
x y
x  y xy
SUMMARY OF ALL THE EXPONENTIAL DEFINITIONS AND LAWS
Definitions
a n  a  a  a  ... (to n factors)
1
an  n
a
Negative bases
a0  1
(1)even number  1
( 1)odd number  1
( a)even number   a even number
( a )odd number   a odd number
Laws of exponents
Law 1:
a m  a n  a m n
m n
mn
Law 3:
(a )  a
Law 5:
an
a

 
bn
b
Law 2:
Law 4:
n
61
am
n
 a mn
a
(a  b)n  a n . bn
When can exponent laws not be applied?
The first two exponent laws only apply when powers with the same base are multiplied or divided.
Expressions like x 2  x3 and x3  x 2 are outside the scope of these laws. An expression like x3  x3
is simplified by adding like terms and so x3  x3  2 x3 . A common mistake is to confuse this with
the law for multiplying powers: x3  x3  x 6 .
The fourth and fifth laws are about the distributivity of exponents over multiplication and division.
Exponents are not distributive over addition and subtraction, so ( x  y ) n  x n  y n and
( x  y )n  x n  y n .
A special note on distributivity
 Multiplication is distributive over addition and subtraction:
a( x  y)  ax  ay and a( x  y)  ax  ay
 Division is distributive over addition and subtraction:
x y x y
x y x y
a
a a
a
a a
 
 
and
*Careful:
and
 
 
a
a a
a
a a
x y x y
x y x y
 Exponents are distributive over multiplication and division:
n
x
xn
( xy )  x y and    n
y
 y
Roots are distributive over multiplication and division:
a na
n
a  b  n a  n b and n  n
b
b
Exponents and roots are not distributive over addition and subtraction:
(a  b) n  a n  b n , (a  b) n  a n  b n , n a  b  n a  n b and n a  b  n a  n b
n


n n
EXERCISE 2
(a)
Simplify the following:
2 x3  2 x3
2 x3  2 x 3
(1)
(2)
3 x3  2 x3  6 x 2  2 x 4
(3)
(3 x3  2 x3 )  (3 x3  2 x3 )
(4)
(b)
Simplify the following, leaving your answer with positive exponents:
x 12
x 7
a 6 b8 c
(1)
(2)
(3)
(4)
x 7
x 12
a 8b6 c 3
(c)
Simplify the following, leaving your answers with positive exponents:
( x3 )4
(2)
( x2 )5
(3)
( x 2 ) 2
(1)
(d)
(4)
(7)
(4 x 3 )3
(10)
(2 x3 )4
2( x 4 ) 2  4( x 2 ) 4
(5)
(8)
(3 x 4 ) 2  (2 x 2 ) 4
(11)
3( 2a 3 ) 2  2( 3a 2 )3
(  x 3 ) 2 . (  x 2 )3
(6)
(9)
4 p4
4 p 4
( x 4 )2  ( x 2 )4
2( x 4 ) 2  4( x 2 ) 4
(12)
(3x 4 ) 2 . (2 x 2 ) 4
(13) 3(2a 3 ) 2  2(3a 2 )3
Simplify the following, leaving your answers with positive exponents:
2
(1)
 a4 
 6
b 
(4)
 3 x 5 
 7 
 9x 
(7)
x 3 y 6
(2)
(5)
 2a3 .4a 2 

3 2 
 8(a ) 
2
( x 3 y 5 ) 2
3
 10 x3 y 

2 
 5 xy 
(8)
(3)
2
3( a 1b 2 ) 3  ( ab) 5
( 3b 5 ) 2
62
(6)
(9)
  2 p 4 q 2 


6
 4p

 x2  x2 
 2 2
 x x 
x y
x 1  y 1
2
3
EXPONENTIAL EQUATIONS
In an exponential equation, the exponent is the unknown.
For example, the equation 3x  9 is called an exponential equation because the unknown variable is
the exponent. Clearly, the solution to this equation is x  2 because 32  9 .
Getting the bases on both sides of the equation to be the same and then equating the exponents will
solve the equation as follows:
3x  9
3x  32
[write 9 to base 3]
x  2
[equate the exponents]
Example 13
Solve for x:
32 x  9
(a)
1
(d)
5 
25
Solutions
x
(a)
(d)
(e)
1
  2
4
5x 
1
25
1
5
(c)
2
 5 x  52
 x  2
x
1
  2
4
x
5.5x  125
x
(b)
5.5x  125
Method 1
 5x  25
 5 x  52
x  2
5x 
(e)
2 x 4  1
32 x  9
 32 x  32
 2x  2
x 1
(c)
(b)
[write 25 to base 5]
[equate exponents]
[divide by 2]
[divide both sides by 5]
[write 25 to base 5]
[equate exponents]
2 x 4  1
 2 x 4  20
x  4  0
x  4
Method 2
51.5x  125
 51 x  125
 51 x  53
1  x  3
x  2
[ 1  20 ]
[equate exponents]
[solve]
[add exponents of like bases]
[add exponents of like bases]
[write 125 to base 5]
[equate exponents]
[express 25 as 52 ]
1
1
as 52 by using the rule a  n  n in reverse]
2
a
5
[equate exponents]
[rewrite
 1 
  2   2 [write 4 to base 2]
2 
1
1
 (22 ) x  2 [rewrite 2 as 2 2 by using the rule a  n  n in reverse]
a
2
2 x
[multiply exponents]
2
2
2 x
1
[write 2 as 21 ]
2
2
2 x  1
[equate exponents]
1
[solve]
x  
2
63
EXERCISE 3
(a)
(b)
Solve the following equations:
2x  2
2x  4
(1)
(2)
2x  1
(5)
(6)
3x  3
(9)
3x  81
(10) 4 x  16
(13) 5 x  25
(14) 5 x  125
(17) 6 x1  36
(18) 23 x1  64
(21) 32( x1)  81 (22) 2.2 x  32
(25) 2.3x  162
(26) 7 2 x. 7 x  7
Solve the following equations:
1
1
(1)
(2)
2x 
3x 
2
3
1
1
(6)
(5)
2x 
3x 
4
9
(9)
1
5 
125
(13)
1
  9
3
x
(10)
7
(14)
1
  1
6
x
(5)
(9)
4  16
1
49
x
(7)
(11)
(15)
(19)
(23)
(27)
3x  1
4 x  64
10 x  100
7 4 x  49
3.3x  9
8 x. 2  128
(8)
(12)
(16)
(20)
(24)
(28)
3x  27
5x  1
11x  121
8x2  1
5.5 x5  5
9.3x  92
(4)
6x 
1
4
1
3x 
27
4x 
x1
 343
(2)
x
(11)
(15)
1
   16
4
(19)
3
(3)
1 . 32 x
3
(6)
(10)
9
 27
x
x1
3 9
x 1
(7)
 81 (11)
 13 
(3
x1
x 1 3
x
(12)
(16)
1
  9
9
(20)
1 . 22 x
8
(4)
1
1
  
27
3
x 1
(8)
1
 
2
x
2
 
3
x 2
x
 13
1
x
1
) 9
1
6
1
4x 
16
1
  4
2
x
144 x  12
x 1
(8)
1
  2
2
x
4 x  16 x1
x
2 x  16
(7)
1
1
1
(17)   
(18) 3    81
27
3
3
Solve the following equations:
(1)
(4)
x
x
(c)
2x  8
(3)
1

49
x1
(3)
x 3
0, 4  0,064 (12)
 4 x

8
27
SCIENTIFIC NOTATION
In Grade 8 we learnt about writing large numbers in scientific notation. Let’s revise this briefly.
Writing large numbers in scientific notation
When writing a large number in scientific notation, the number before the comma must be greater or
equal to 1, but less than 10. This means that you must make sure there is only one digit (but not 0)
before the comma. So, to write a large number in scientific notation:
Move the comma from the left to right after the first non-zero digit.

Count the number of digits after between the new position of the comma and the original

position of the comma.
Write 10number of digits counted

Example 14
(a)
(b)
(c)
The distance between the earth and the sun (also known as an astronomical unit), is
149 597 870,691 km. Write this in scientific notation.
The radius of the planet Saturn is approximately 60 268 000 m.
Write this in scientific notation.
The mass of the moon is approximately 7,35 1022 kg . Write this as a normal number.
64
Solutions
1, 49597870691108 km
(a)
73 500 000 000 000 000 000 000 kg
(c)
6,0268 107 m
(b)
When large numbers (numbers greater than 10) are written in scientific notation, the exponent of
the 10 will always be positive.
Writing small numbers in scientific notation
Negative exponents allow for using scientific notation to represent very small numbers. For example:
1,23×1014  1,23  100 000 000 000 000  0,000 000 000 000 0123
0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,23
Shift the comma 14 places to the left.
When writing a small number in scientific notation, the number before the comma must be greater or
equal to 1, but less than 10. This means that you must make sure there is exactly one digit (but not 0)
before the comma. So, to write a small number in scientific notation:
Move the comma to the right, till right after the first non-zero digit

Count the number of digits after between the new position of the comma and the original

position of the comma
Write 10 (number of digits counted)

Example 15
(a)
(b)
(c)
The mass of an electron is 0,000 000 000 000 000 000 000 000 000 911 g . Write this
number in scientific notation.
The probability of winning the lottery is approximately 0,000 000 071 511. Write this
number in scientific notation.
The charge of an electron is 1,6  1019 (C is the symbol for coulomb, the unit for charge.)
Write this as a normal number.
Solutions
(a)
9,111028 g
(b)
7,1511108
(c)
0,000 000 000 000 000 000 16 C
When small numbers (numbers between 0 and 1) are written in scientific notation, the exponent
of the 10 will always be negative.
Multiplication and division of numbers in scientific notation
When multiplying or dividing numbers in scientific notation, we rely on the exponent laws to deal
with the powers of 10: 10m 10n  10m n and 10m  10n  10mn
Example 16
Calculate the following, without using a calculator, and leave your answer in scientific notation.
6 1010
(2,5  104 )  (3  108 )
(a)
(b)
3 104
2 1012
5
11
(9  10 )  (5  10 )
(d)
(c)
8 1015
3,6  1010
(3, 2  1010 )  (2,5  1011 )
(f)
(e)
4  105
65
Solutions
(a)
(b)
(c)
(2,5  3)  (104  108 )
[commutative and associative properties]
 7,5 1012
[add exponents of 10]
(6  3)  (1010  104 )
[divide numbers and divide powers of 10]
 2  106
[subtract exponents of 10]
(9  5)  (105 1011 )
 45 106
 4,5 101 106
 4,5  107
(d)
(e)
[write in standard scientific notation]
12
15
(2  8)  (10  10 )
(f)
(3,6  4)  (1010  105 )
 0,25  103
 0,9  1010( 5)
 2,5 101  103
 0,9 105
 2,5  104
 9 101 105
(3,2  2,5)  (1010  1011 )
 9 106
 8  1021
Addition and subtraction of numbers in scientific notation
When adding or subtracting numbers in scientific notation, we need to rewrite the numbers so they all
have the same power of 10 and then add or subtract them as we do like terms in Algebra.
Remember the following when doing this:
 When increasing the exponent of the 10, we move the comma of the number to the left to
compensate for this. Move the comma the same number of places as the number by which you
increased the exponent of the 10.
 When decreasing the exponent of the 10, we move the comma of the number to the right to
compensate for this. Move the comma the same number of places as the number by which you
decreased the exponent of the 10.
Example 17
Calculate the following, without using a calculator and leave your answer in scientific notation:
3, 2  108  2  108
(b) 2,5  1014  2  1013
(c) 3  105  2  108
(a)
2,1 1011  3  1013
(d)
Solutions
(a)
(b)
3, 2  108  2  108
(d)
(3, 2  1010 )  (2,5  1011 )
 2,11011  0, 03 102 1013
 5, 2 108
 2,11011  0, 03 1011
2,5 1014  0, 2 1014
  2, 07 1011
 2,3 1014
(c)
(e)
(e)
(3, 2  1010 )  (2,5  1011 )
3  105  2  108
 3,2 1010  0, 25  101 1011
 0,003  103 105  2 108
 3,2 1010  0, 25  1010
 0,003  108  2 108
 3, 45  1010
 2,003  108
66
EXERCISE 4
(a)
(b)
(c)
(d)
(e)
Write the following numbers out in full:
1, 218  104
(1)
(2)
3, 254  104
(3)
1, 2345  107
(4)
3,6  108
Write the following numbers in scientific notation:
(1)
0,000 013 5
(2)
1 846 000 000
(3)
(4)
(6)
738 million
(5)
23 millionths
−0,000 000 000 000 2
333
1 000 000 000 000
The mass of a single atom of oxygen is approximately
0,000 000 000 000 000 000 000 000 026 6 kg. Write this in scientific notation.
The mass of the planet Pluto is 1, 27  1022 kg .
(1)
Write the mass out in full.
(2)
If it is further given that Pluto has an average density of 2  103 kg/m3 , find the
Mass
)
volume of the planet. (Hint: Volume =
Density
Calculate the following, leaving your answers in scientific notation:
9  1018
(2,5  108 )  (4  1012 )
(2)
(1)
3  109
(3)
(1,3  1014 )  (2  108 )
(4)
(5  108 )(4  109 )
(5)
(7)
(f)
This exercise is to be done without the use of a calculator.
2  103
5  1015
4, 2  1012  3  1013
12
 5,6  10
(8)
(1,5  107 )(4  1020 )
3  106
2,05  1014  3  1012
(10)
1,3  1019  2,3  1020
(6)
13
(9)
4,3  10
(11)
3,57632  1013  2,32  1014
(12)
3 108  5 106
(13)
3  109  2  1010
7,9  1020  1 1021
(14)
2,03  1017  3  1015
4  108
2,5  1014  5  1015
(15)
(16) 8,19  1011  8,1 1010
4
6
2  10  4  10
4,35  1023  1,5  1022
(17)
(18) (3  103 ) 2
15
14
2  10  5  10
(20) 2,5  1013  (5, 2  106  2  105 )
(19) 4 1031  2  1015  3 1016
The gravitational force F, between two bodies, measured in N (for newton) is given by the
G  m1  m2
formula: F 
. In this formula:
r2
 G is the gravitational constant, which has a value of 6,7  1011
 r is the distance between the centres of the bodies
 m1 and m2 are the masses of the bodies in kg
Determine the gravitational force between two bodies with masses 2  108 kg and 1 1014 kg ,
if their centres are 1 105 m apart.
67
REVISION EXERCISE
(a)
Simplify the following expressions, leaving your answers with positive exponents:
14 xy 2
2 x 1  3x
4 x3 y 5  5 xy 2
(2)
(3)
(1)
3
21x y
(4)
3x 2  x3 y  3 x 3
(7)
9a 3
18a
(8)
(10)
( 7 x 7 ) 2
(11)
a 5
b 3
(13)
 3x 1 y 

2 3 
 9 x y 
(14)
 2x 
 2 1 
 3x y 
(16)
3x0
y 3
 2 y 1  2
y
y
(17)
(2 x 3 ) 2  (3x 2 )3 (18)
2
(3x) 2
3mn 2
27m 1n
(6)
2 x 0 y 7 z   3 y 2 z
(9)
 2x 2 
 3 2 
x y 
3 p 3 q 4
2
(15)
(3x 3 )0  3x0 


(20)
3x 2  2 x3
1  xy
x 1  y
3
3x1  3
33n1  81
2  3x1  18
(3)
(6)
(9)
(10)
2  72 n  98  0
(11)
125n  25n1
(12)
(13)
62 x 
1
36
(14)
2(2 x )2  0,5
(15)
x
3
4 p  4 q 3
(12)
2
1
(19) (m  n)     33
3
Solve the following equations:
3x1  1
(2)
(1)
2n
3  81
(5)
(4)
1n
5  3  45
(8)
(7)
0
(b)
(5)
x
4n  64
2k 12  16k
1  5n  0
1
3x2 
9
x
1
  8
2
11
 1 
(17)
(18) 2 11x1  241  1

  1000
   27
3 3
 100 
The distance between the earth and the moon is 3,82 108 m. Write this number out in full.
One cubic centimetre is equal to 0,000 035 31 cubic feet. Write this number in scientific
notation.
The Compton wavelength of an electron is 2, 43  1012 m . Write this number out in full.
Calculate the following and leave your answer in scientific notation:
3  107  2  108
3,1 108  2,3  106
(2)
(1)
(16)
(c)
(d)
(e)
(f)
(3)
(2,5  109 )(4  1011 )
(4)
(5)
5,1 107  2,3  106
1,066  1012
(6)
2  104
8  107
2,8  1017  3  1018
2  106  3  106
SOME CHALLENGES
(a)
(b)
Write each of the following expressions as a single power of 2:
1
2 x 16
(2)
(3)
(1)
8
Simplify the following:
2 x  4 x2
(1) 27 x1  9 x1  315 x
(3)
(2)
8 x 3
68
4 x 1
4
12 x. 3 x
22 x
CHAPTER 6: PATTERNS, FUNCTIONS AND ALGEBRA
TOPIC: NUMERIC AND GEOMETRIC PATTERNS
In this chapter, we will revise the number patterns that you studied in Grade 8. These patterns
included those with a constant difference between consecutive terms, a constant ratio
between consecutive terms and other patterns which did not have a constant difference or
ratio. The general rule for a number pattern with a constant difference was discussed as well.
In Grade 9, we will explore general rules for the other types of number patterns.
However, before going any further, let’s revise some important concepts from last year.
Consider the number pattern (or sequence) 3 ; 5 ; 7 ; 9 ; ……………….
We can describe this pattern as follows:
Start with the number 3; add 2 to give 5; then add 2 to 5 to give 7.
Continue in this way by adding 2 to each previous number.
Each number in the sequence is called a term. The first term is 3, the second term is 5, the
third term is 7 and so forth.
We can name the terms as T1  3 , T2  5 , T3  7 and so forth.
Note:
T2 5
This is the actual
second term in the
sequence. It lies in
the second position.
This number represents
the position of the term in
the sequence. The position of
the term is the second position.
T4  9 means that the number 9 lies in the 4th position in the sequence. It is the 4th term in
the sequence.
The rule in words for this sequence is:
Start with 3. Add 2 to 3 to get 5. Continue adding 2 to each previous term. By adding 2 each
time, we obtain more terms in the sequence called consecutive terms.
T1
T2
T3
T4
Now think about this!
What is the 6th term?
Which term is 15?
T5
T6
T7
The 6th term is 13. The position of 13 is 6th in the sequence.
15 lies in the 7th position in the sequence. It is the 7th term.
Let’s now revise number patterns with a constant difference and then look at the other types
in more detail. The following types of number patterns will be the focus of this chapter:
 Number patterns with a constant difference between terms;
 Number patterns with a constant ratio between terms;
 Other types of number patterns.
NUMBER PATTERNS WITH A CONSTANT DIFFERENCE
With these patterns, there is a constant difference between consecutive terms.
Consider the sequence 5 ; 7 ; 9 ;11; ...................
T2  T1  7  5  2
T3  T2  9  7  2
T4  T3  11  9  2
69
There is a constant difference of 2 between consecutive terms. We call this constant
difference d.
The sequence is generated by adding d to each term. In this example, d  2 .
T1
T2
T3
T4
T5
T6
T7
It is easy to determine consecutive terms since you are adding 2 to each previous term.
Finding the 5th, 6th, 10th or even the 20th term is easy to do. However, if you are required to
determine the 100th term, this would be extremely time-consuming. Therefore we need to
find a rule which helps us do this. This rule is called the general rule for the sequence.
One way to find this general rule is to link the position of the term to the constant difference
and work from there.
Notice that in the sequence above:
T1  5  2(1)  3
where 2 is the constant difference, 1 is the position of 5 (first term) and
3 is added to keep the actual term 5.
T2  7  2(2)  3
where 2 is the constant difference, the 2 in brackets is the position of 7
(second term) and 3 is added to keep the actual term 7.
T3  9  2(3)  3
where 2 is the constant difference, the 3 in brackets is the position of 9
(third term) and 3 is added to keep the actual term 9.
We can continue to generate terms of the sequence in this way. The general rule in words
for the sequence is therefore:
Multiply the constant difference by the position of the term and add 3
Let’s continue to find terms of the sequence using this general rule:
T4  2(4)  3  11
T5  2(5)  3  13
T6  2(6)  3  15
This rule can now help us to find the 10th term or the 100th term:
T10  2(10)  3  23
T100  2(100)  3  203
Notice:
5 ; 7 ; 9 ; 11 ; 13 ; 15 ; 17 ; 19 ; 21 ; 23 ; ................. ; 203 ; ......
T1 T2 T3 T4 T5 T6 T7 T8 T9 T10
T100
A table is useful for determining the general rule of a sequence with a constant difference.
Using the previous example, draw a table as follows:
The position of the term
The constant difference
multiplied by the position of
term
What to do to get the actual
term
The actual term in the
sequence
T1
T2
T3
T4
T5
T10
Tn
2(1)
2(2)
2(3)
2(4)
2(5)
2(10)
2(n)
3
3
3
3
3
3
3
5
7
9
11
13
23
2(n)  3
The general rule using the letter n is Tn  2n  3 where 2 represents the constant difference
and n the position of the term in the sequence. It is important to note that the value of n is
always a natural number. We say that the nth term of the sequence is Tn  2n  3 .
This general rule can now be used to determine any term of the sequence.
70
For example: T9  2(9)  3  21
T30  2(30)  3  63
Let’s discuss a few more examples to make sure that you fully understand these concepts.
Example 1
6 ;10 ;14 ;18 ; ......... is a given sequence.
(a)
State the rule in words.
(b)
What is the constant difference?
(c)
Write down the next three terms.
(d)
Determine the general rule (nth term).
(e)
Calculate the 10th term.
(f)
Calculate the 100th term.
(g)
Which term of the sequence is equal to 242?
Solutions
(a)
Start with 6. Add 4 to 6 to get 10. Continue adding 4 to each previous term to obtain
consecutive terms.
(b)
T2  T1  10  6  4
d  4
(c)
6 ;10 ;14 ;18 ; 22 ; 26 ; 30 ;.........
(d)
Draw a table. The constant difference is 4.
The position of the term
The constant difference
multiplied by the position of
term
What to do to get the actual
term
The actual term in the
sequence
(e)
(f)
(g)
T3  T2  14  10  4
T4  T3  18  14  4
(Add 4 to each previous term)
T1
T2
T3
T4
Tn
4(1)
4(2)
4(3)
4(4)
4(n)
2
2
2
2
2
6
10
14
18
4(n)  2
The nth term is Tn  4n  2
Tn  4n  2
 T10  4(10)  2  42
Tn  4n  2
 T100  4(100)  2  402
The actual term in the sequence is 242 and we want to find its position.
Let Tn  242 where n represents the position to be determined.
Tn  4n  2
 242  4n  2
 242  2  4n
 240  4n
 60  n
 T60  242 [242 is the 60th term in the sequence]
Example 2
3 ; 8 ;13 ;18 ; ......... is a given sequence.
(a)
Determine the general rule (nth term).
(b)
Calculate the 24th term.
(c)
Calculate the 700th term.
(d)
Which term is equal to 748?
71
Solutions
(a)
Draw a table. The constant difference is 5.
The position of the term
The constant difference 
by the position of term
What to do to get the actual
term
The actual term in the
sequence
(b)
(d)
The nth term is Tn  5n  2
Tn  5n  2
 T5  5(24)  2  118
Let Tn  748
Tn  5n  2
(c)
T1
T2
T3
T4
Tn
5(1)
5(2)
5(3)
5(4)
5(n)
2
2
2
2
2
3
8
13
18
5(n)  2
Tn  5n  2
 T700  5(700)  2  3 498
 748  5n  2
 750  5n
150  n
 T150  748
EXERCISE 1
For each of the following number patterns:
(1)
State the rule in words.
(2)
(3)
Write down the next three terms.
(4)
(5)
Calculate the 15th term.
(6)
Write down the constant difference.
Determine the general rule (nth term).
Calculate the 120th term.
(a)
(d)
(g)
(j)
(m)
(p)
5 ; 7 ; 9 ;11; ........
6 ; 9 ;12 ;15 ; ........
2 ; 8 ;14 ; 20 ; .......
2 ; 0 ;  2 ;  4 ; .......
9 ; 4 ;  1;  6 ; ........
12 ;  6 ; 0 ; 6 ; ........
(b)
(e)
(h)
(k)
(n)
(q)
6 ;11;16 ; 21; ........
15 ; 25 ; 35 ; .......
6 ;10 ;14 ;18 ; .......
10 ; 7 ; 4 ;1; ........
7 ;  11;  15 ; ........
2 ;  8 ;  18 ; .......
(c)
(f)
(i)
(l)
(o)
(r)
10 ;12 ;14 ;16 ; ........
2 ; 6 ;10 ;14 ; .......
6 ;15 ; 24 ; 33 ; .......
18 ;15 ;12 ; ........
6 ;  11;  16 ; ........
10 ; 9 ; 8 ; 7 ; ........
(s)
1 12 ; 3 12 ; 5 12 ; ....
(t)
(u)
1
2
(v)
1
4
; 12 ; 34 ; 1; ....
(w)
1 ;1 1 ; 2 1 ; 3 1
2
2
2
2
3 ; 2 ; 1 ; 0 ; ....
5 5 5
(x)
0, 25 ; 0,39 ; 0,53 ; ....
(2)
Calculate the 40th term.
(2)
Calculate the 56th term.
(2)
Calculate the 68th term.
(2)
Calculate the 80th term.
; ....
; 2 ; 3 12 ; 5 ; ....
EXERCISE 2
(a)
(b)
(c)
(d)
8 ;13 ;18 ; 23 ; ......... is a given sequence.
(1)
Determine the general rule (nth term).
(3)
Which term is equal to 603?
7 ;16 ; 25 ; 34 ; ......... is a given sequence.
(1)
Determine the general rule (nth term).
(3)
Which term is equal to 1 798?
13 ;19 ; 25 ; 31; ......... is a given sequence.
(1)
Determine the general rule (nth term).
(3)
Which term is equal to 337?
12 ;14 ;16 ;18 ; ......... is a given sequence.
(1)
Determine the general rule (nth term).
(3)
Which term is equal to 810?
72
(e)
1 12 ; 2 ; 2 12 ; .... is a given sequence.
(1)
(3)
(5)
Determine the general rule (nth term).
Calculate the 85th term.
Which term is equal to 45 12 ?
(2)
(4)
Calculate the 70th term.
Which term equals 151?
NUMBER PATTERNS WITH A CONSTANT RATIO BETWEEN TERMS
Consider the sequence 2 ; 6 ;18 ; 54 ; ...................
T3 18
T2 6
T4 54
 3
 3

3
T1 2
T3 18
T2 6
There is a constant ratio of 3 between consecutive terms. We call this constant ratio r.
The sequence is generated by multiplying each term by r. In this example, r  3 .
T1
T2
T3
T4
T5
T6
T7
If you consider the sequence 2 ; 6 ;18 ; 54 ; ......... , it is easy to determine consecutive terms
since you are multiplying each previous term by 3. Finding the 5th, 6th, 10th or even the 12th
term is easy to do. However, if you are required to determine the 100th term, this would be
extremely time-consuming. Therefore we need to find a rule which helps us do this. This rule
is called the general rule for the sequence.
One way to find this general rule is to link the position of the term to the constant ratio and
work from there.
Notice that in the sequence above:
T1  2  2 1  2  30
[ 1  30 ]
T2  6  2  31
T3  18  2  32
T4  54  2  33
Notice that 2 is the first term and the constant ratio is 3. The exponent is the position of the
term minus 1.
The general rule in words for the sequence is therefore:
Multiply the first term 2 by the constant ratio 3 which has been raised to the exponent
which is the position of the term minus 1.
We can write this general rule as follows: Tn  2  3n1
where n represents the position of the term. Remember that n is a natural number.
This general rule or nth term can be used to generate any term of the sequence.
For example
T5  2  34  162
T6  2  35  486
T9  2  38  13 122
Example 3
3 ; 6 ;12 ; 24 ; ......... is a given sequence.
(a)
Determine the general term (nth term).
(c)
Calculate the 20th term.
Solutions
(a)
The first term is 3 and the constant ratio is 2.
T1  3  3 1  3  20
[ 1  20 ]
T2  6  3  21
T3  12  3  22
73
(b)
Calculate the 7th term.
T4  24  3  23
The nth term is Tn  3  2n1
(b)
Tn  3  2n1
Tn  3  2n1
(c)
 T7  3  26  192
 T20  3  219  1 572 864
EXERCISE 3
For each of the following number patterns:
(1)
Write down the constant ratio.
(3)
Determine the nth term.
(2)
(4)
Write down the next three terms.
Determine the 10th term.
(a)
(d)
(g)
2 ; 4 ; 8 ;16 ; ........
(b)
5 ;10 ; 20 ; 40 ; ........ (e)
2 ;10 ; 50 ; 250 ; ........ (h)
1; 3 ; 9 ; 27 ; ........
2 ; 8 ; 32 ; ........
8 ;16 ; 32 ; ........
(c)
(f)
(i)
4 ;12 ; 36 ; ........
3 ; 9 ; 27 ; ........
4 ;16 ; 64 ; ........
(j)
32 ;16 ; 8 ; 4 ; .......
(k)
2 ;  6 ;  18 ; ........
(l)
1
2
(m)
16 ; 4 ;1; 14 ........
(n)
1
2
(o)
7 ........
28 ; 7 ; 74 ; 16
1 ; ........
; 14 ; 81 ; 16
;1; 2 ; 4 ; ........
OTHER TYPES OF NUMBER PATTERNS
These number patterns don’t have a constant difference or constant ratio.
Example 4
Consider the sequence 1; 4 ; 9 ;16 ; ...................
(a)
State the rule in words.
(b)
Write down the next three terms.
Solutions
(a)
(b)
The sequence doesn’t have a constant difference since:
T2  T1  4  1  3
T3  T2  9  4  5
T4  T3  16  9  7
The sequence doesn’t have a constant ratio since:
T3 9
T2 4
T4 16
 4


T1 1
T3 9
T2 4
However, a different pattern can be found:
T1  1  (1) 2
T2  4  (2) 2
T3  9  (3) 2
T4  16  (4) 2
The rule in words is therefore:
Consecutive terms are obtained by squaring each term’s position in the sequence.
T5  (5)2  25
T6  (6)2  36
T7  (7)2  49
You can now insert the next three terms into the sequence as follows:
1; 4 ; 9 ;16 ; 25 ; 36 ; 49 ; ..................
Alternative method
Notice that the difference between
consecutive terms increases by 2 each
time.
T2
T1
T3
T4
T5
T6
T7
The rule in words is: Start by adding 3 to 1 to get 4. Then add 5 to 4 to get 9. Increase the
number added each time by 2 to get consecutive terms.
You can now insert the next three terms into the sequence as follows:
1; 4 ; 9 ;16 ; 25 ; 36 ; 49 ; ..................
74
Example 5
Consider the sequence 2 ; 6 ;12 ; 20 ; ...................
(a)
State the rule in words.
(b)
Write down the next three terms.
Solutions
(a)
The sequence doesn’t have a constant difference since:
T2  T1  6  2  4
T3  T2  12  6  6 T4  T3  20  12  8
The sequence doesn’t have a constant ratio since:
T3 12
T2 6
T4 20 5
 3
 2


T1 2
T3 12 3
T2 6
However, a different pattern can be found:
T3
T2
T1
T4
What is happening is that the difference
between consecutive terms is increased by
2 each time.
Add 4 to T1 to get 6. Then add 6 to T2 to get 12.
Then add 8 to T3 to get 20.
The rule in words is therefore:
Start by adding 4 to 2 to get 6. Then add 6 to 6 to get 12. Increase the number
added each time by 2 to get consecutive terms.
T1  2
T2  2  4  6
T3  6  6  12
T4  12  8  20
(b)
T1
You can now insert the next three
terms into the sequence as follows:
2 ; 6 ;12 ; 20 ; 30 ; 42 ; 56 ; ..................
T2
T3
T4
T5
T6
T7
General rules for number patterns with no constant difference or ratio
Number patterns with no constant difference or constant ratio can often be written in terms of
the position of the term squared. These types of number patterns are called quadratic patterns
and are studied in detail in Grade 11. However, they are interesting patterns to consider at a
basic level in Grade 9.
Example 6
3 ; 6 ;11;18 ; ......... is a given sequence.
(a)
Determine the general term (nth term).
(b)
Calculate the 150th term.
Solutions
(a)
The rule in words for this sequence is:
T2
T1
Start by adding 3 to 3 to get 6. Then add 5
to 6 to get 11. Increase the number added
each time by 2 to get consecutive terms.
To get the nth term of this type of sequence,
work with the position squared and take it from there.
T1  3  (1) 2  2
[square position 1 and add 2 to get 3]
T2  6  (2) 2  2
[square position 2 and add 2 to get 6]
T3  11  (3) 2  2
[square position 3 and add 2 to get 11]
T4  18  (4) 2  2
[square position 4 and add 2 to get 18]
The nth term is Tn  n 2  2
75
T3
T4
(b)
Tn  n 2  2
 T150  (150) 2  2  22 502
EXERCISE 4
For each of the following sequences, determine the general rule (nth term) and hence the
100th term.
1; 4 ; 9 ;16 ;.........
2 ; 5 ;10 ;17 ; .....
4 ; 7 ;12 ;19 ; .....
(b)
(c)
(a)
5 ; 8 ;13 ; 20 ; ....
0 ; 3 ; 8 ;15 ; ....
1; 2 ; 7 ;14 ; ....
(d)
(e)
(f)
2 ; 8 ;18 ; 32 ; ....
3 ;12 ; 27 ; 48 ; ....
4 ; 9 ;16 ; 25 ; ....
(g)
(h)
(i)
(j)
0 ;1; 4 ; 9 ;.........
(k)
1
2
; 2 ; 92 ; 8 ;.........
(l)
1
4
;1; 94 ; 4 ;.........
GEOMETRIC PATTERNS
Geometric patterns are number patterns represented diagrammatically.
Example 7
Consider the diagram made up of black dots.
(a)
How many dots are there in figure 1, 2 and 3?
(b)
How many dots are there in figure 4?
(Draw a picture)
(c)
Determine the general rule to find the number
of dots in the nth figure.
(d)
How many dots are there in the 100th figure?
(e)
Which figure will contain 121 dots?
Solutions
(a)
(b)
(c)
There are 5 dots in figure 1.
There are 9 dots in figure 2.
There are 13 dots in figure 3.
There are 17 dots in figure 4.
The number pattern is:
5 ; 9 ; 13 ; 17 ; ……
There is a constant difference of 4. Draw a table.
The position of the term
(d)
(e)
T1
The constant difference 
4(1)
by the position of term
What to do to get the actual
1
term
The actual term in the
5
sequence
The nth term is Tn  4n  1
Tn  4n  1
 T100  4(100)  1  401
There are 401 dots in the 100th figure.
Tn  4n  1
121  4n  1
120  4n
 30  n
The 30th figure will contain 121 dots.
76
T2
T3
T4
Tn
4(2)
4(3)
4(4)
4(n)
1
1
1
1
9
13
17
4(n)  1
EXERCISE 5
(a)
The goal post nets in the 2014 FIFA World Cup in Brazil were
designed using hexagonal shapes. Consider the following designs
made up of hexagons.
(b)
1
2
3
(1)
How many hexagons are there in design 4, 5 and 6?
(2)
Determine the number of hexagons in design 200.
(3)
Which design will have 450 hexagons?
Designs using matchsticks are shown below.
(c)
Determine:
(1)
the number of matches in design 4, 5 and 6.
(2)
the number of matches in design n.
(3)
the number of matches in design 200.
(4)
which design will contain 601 matches.
Ethnic designs using circles and black oval discs are shown below.
(d)
Determine:
(1)
the number of circles in design 4, 5 and 6.
(2)
the number of discs in design 4, 5 and 6.
(3)
the number of circles in design n.
(4)
the number of discs in design n.
(5)
the number of circles in design 500.
(6)
the number of discs in design 500.
(7)
which design will contain 139 circles .
(8)
which design will contain 438 discs .
Consider the diagram made up of black dots joined by thin black lines.
(1)
How many dots are there in figure 4?
(2)
How many lines are there in figure 4?
(3)
How many dots are there in figure 8?
(4)
How many lines are there in figure 8?
(5)
Determine the general rule to find the
number of dots in the nth figure.
(6)
How many dots are there in the 186th figure?
(7)
Which figure will contain 272 dots?
(8)
Determine the general rule to find the number of lines in the nth figure.
(9)
How many lines are there in the 900th figure?
(10) Which figure will contain 650 lines?
77
(e)
Consider the following designs.
(f)
(1)
How many shaded rectangles are there in design 4, 5 and 6?
(2)
Determine the number of shaded rectangles in design n.
(3)
How many shaded rectangles are there in design 10?
Consider the diagram made up of black dots.
(1)
(2)
(3)
(4)
(5)
How many dots are there in figure 5?
How many dots are there in figure 6?
Describe the pattern in words.
Determine the general rule to find the number of dots in the nth figure.
How many dots are there in the 50th figure?
REVISION EXERCISE
(a)
For each of the following sequences, determine the general rule (nth term) and hence
calculate the 100th term.
(1)
(4)
(7)
5 ; 8 ;11;14 ; ........
3 ; 7 ;11;15 ; ........
4 ; 0 ;  4 ;  8 ; .......
(2)
(5)
(8)
7 ;11;15 ;19 ; ........
10 ;16 ; 22 ; 28 ; .....
0 ;  3 ;  6 ; .......
(10)
5 ;1;  3 ;  7 ; ....
(11)
5 ;  11;  17 ; ........ (12)
(13)
2 12 ; 4 12 ; 6 12 ; .....
(14)
1
4
(16)
(b)
(c)
(d)
(e)
13 ;  7 ;  1; .....
(17)
;1; 74 ; ........
1;  9 ;  19 ; .......
(3)
(6)
(9)
3 ; 8 ;13 ;18; ........
4 ;11;18 ; 25 ; .......
6 ;  11;  16 ; .......
3 12 ; 4 ; 4 12 ; ........
(15)
0,5 ; 0, 7 ; 0,9 ; ....
(18)
12 ;11;10 ; 9 ; .....
For each of the following sequences, determine the general rule (nth term) and hence
calculate the 10th term.
6 ;12 ; 24 ; 48 ; ........ (2)
3 ;15 ; 75 ; 375 ; ....
(3)
1; 7 ; 49 ; ........
(1)
(4)
3 ; 9 ; 27 ; 81; ........ (5)
8 ; 4 ; 2 ;1; ........
(6)
12 ; 6 ; 3 ; ........
(7)
6 ; 9 ;14 ; 21; ....
(8)
4 ;  1; 4 ;11; ....
(9)
16 ; 25 ; 36 ;.........
3 ;10 ;17 ; 24 ; ......... is a given sequence.
(1)
Determine the 40th term.
(2)
Which term of the sequence is 1 403?
19 ;16 ;13 ;10 ; ......... is a given sequence.
(1)
Determine the 65th term.
(2)
Which term of the sequence is 113 ?
Tn  8n  5 is the nth term of a number pattern (sequence).
(1)
Determine the first four terms of the sequence.
(2)
Determine the 30th term.
(3)
Which term is equal to 315?
78
(f)
Below is a sequence of ethnic designs constructed out of short wooden sticks. The
first design is made up of two octagons and a square. The first design consists of 17
short wooden sticks.
(1)
(2)
(3)
(4)
(5)
(6)
How many octagons are there in design 20?
How many squares are there in design 50?
How many short wooden sticks are there in design 100?
Which design contains 50 octagons?
Which design contains 1000 squares?
Which design contains 784 short wooden sticks?
SOME CHALLENGES
(a)
(b)
(c)
(d)
(e)
2 5 8
14
; ; ;1; ; ............
5 7 9
13
(1)
Determine the nth term.
(2)
Calculate the 10th term.
5 14
22 26 30
Consider the sequence: 2 ; ; ;1;
;
;
;............
4 13
23 28 33
(1)
Determine the nth term.
(2)
Calculate the 20th term.
Consider the sequence:
Consider the number pattern: 4  6 ; 7 14 ;10  22 ;13  30 ; ......
(1)
Determine the nth term.
(2)
Determine the 40th term.
1 3 9 27
Determine the general term of the sequence:
; ; ;
; ............
5 8 13 20
Consider the following designs.
(1)
(2)
(3)
Write down the number of circles in design 1, 2, 3, 4, 5 and 6.
Determine the number of circles in design n.
How many circles are there in design 40?
79
CHAPTER 7: PATTERNS, FUNCTIONS AND ALGEBRA
TOPIC: FUNCTIONS AND RELATIONSHIPS
PART 1
(Studied in Term 1 according to CAPS)
In this chapter, we will revise input-output relationships studied in your Grade 8 year.
The relationship between input and output values under a given rule can be represented in
any of the following ways:
(a)
verbally
(b)
in flow diagrams
(c)
in tables
(d)
by means of equations
If we consider the relationship where input values 4 ; 5 ; 6 produce output values
11;13 ;15
using the rule “multiply each input value by 2 and add 3”. Let’s represent this
relationship in the different ways mentioned.
Verbally
Multiply each input value by 2 and add 3:
Input value
Rule
4 2  3
4
5
5 2  3
6
6 2  3
Input values:
Output values:
4 ; 5 ; 6
11;13 ;15
Output value
11
13
15
Note:
If we reverse the operations we can determine the input value when given the output value.
The reverse rule is as follows:
Subtract 3 from each output value and then divide by 2.
Therefore, if the output value is 11, the input value is (11  3)  2  4 .
For the output value 13, the corresponding input value is (13  3)  2  5
For the output value 15, the corresponding input value is (15  3)  2  6
In the examples that follow, you will be required to determine output values when given the
input values. You will also be required to determine the input values when given the output
values. The use of equations will be much easier to do than using inverse operations.
Flow diagram
input value  2  3
Table
Input
Output
n
n2  3
4
11
5
13
6
15
Equation
The relationship can be represented in the form of an equation involving x and y. If we let the
input value be x and the output value be y, then the equation connecting x and y is given by
y  x2  3.
Using algebraic notation, we can write this equation as y  2 x  3 .
80
Let’s now use this equation to determine the output values corresponding to the input values
4 ; 5 ; 6 . The process we use is called substitution.
y  2(4)  3  11
For x  4
y  2(5)  3  13
For x  5
y  2(6)  3  15
For x  6
The set of output values is 11;13 ;15 .
We call x the independent variable since the input values are chosen at random.
We call y the dependent variable since the output values depend on what the input values
are. For example, if the input value is 9, which is chosen as we like, then the output value is
dependent on 9 for its value. The output value for x  9 is y  2(9)  3  21 .
FINDING OUTPUT VALUES WHEN GIVEN INPUT VALUES AND A RULE
Example 1
Determine the output values in the given flow diagram.
Input value (x)
Output value (y)
Rule
2
1
0
1
7x  4
Solution
Input values
Rule
Output values
x  2
7(2)  4
y  10
x  1
7(1)  4
y  3
x0
7(0)  4
y4
x 1
7(1)  4
y  11
The output values can now be represented in the flow diagram.
Input value (x)
Output value (y)
Rule
2
1
0
1
10
3
4
11
7x  4
Example 2
1
If the rule for finding y in the table below is y   x  3 , determine the output values (y)
2
for the given input values (x).
x
y
2
0
2
4
6
8
10
18
Solution
For x  2
For x  2
1
y   (2)  3  4
2
1
y   (2)  3  2
2
For x  0
For x  4
81
1
y   (0)  3  3
2
1
y   (4)  3  1
2
1
y   (6)  3  0
For x  8
2
1
For x  10
y   (10)  3  2
For x  18
2
The output values can now be filled in the table:
For x  6
x
y
2
4
0
3
2
2
4
1
6
0
8
1
10
2
1
y   (8)  3  1
2
1
y   (18)  3  6
2
18
6
EXERCISE 1
(a)
Determine the output values in the following flow diagrams:
(1)
(2)
1
0
1
2
(3)
4x  4
(4)
1
9
13
15
x5
2
(5)
1
2
x7
6
12
18
24
 x 8
1
4
 16 x  6
y  4x
y  4 x  1
y  2x  3
y  4x  5
(2)
(5)
(3)
(6)
y  3 x
y  8 x  6
If the rule for finding y in the table below is y  9 x  6 , determine the output values
(y) for the given input values (x).
x
y
(d)
2
0
2
4
For each of the following equations, determine the output values corresponding to the
input values x  2 ;  1; 0 ;1; 2 ; 3 :
(1)
(4)
(c)
3x  2
(6)
4
0
12
20
(b)
1
1
3
5
1
0
1
2
3
4
5
7
10
1
If the rule for finding y in the table below is y  x  2 , determine the output values
3
(y) for the given input values (x).
x
y
3
0
3
6
9
12
82
24
30
72
FINDING INPUT VALUES WHEN GIVEN OUTPUT VALUES AND A RULE
Example 3
Input value (x)
Rule
Determine the input values in the
given flow diagram.
Output value (y)
9
14
19
24
5x 1
Solution
Using the inverse operations discussed in Grade 8, we can determine the input values.
The rule for determining output values when given input values is ( x  5)  1 .
The reverse rule will help us to determine the input values for the corresponding output
values (y). This rule is ( y  1)  5 .
Rule
Input values (x)
Output values (y)
y 9
(9  1)  5
x2
y  14
(14  1)  5
x3
y  19
(19  1)  5
x4
y  24
(24  1)  5
x5
The output values can now be
represented in the flow diagram.
Input value (x)
2
3
4
5
Rule
5x  1
Output value (y)
9
14
19
24
Example 4
The equation y  3 x  1 is given. If the output values are y  5 ; 8 ;11;14 , determine the
input values.
Solution
Instead of using the method of inverse operations, you may decide to use an algebraic
method (Chapter 9) to determine the input values (x) when given the output values (y). In
this method, the equation is used and the value for y is substituted into the equation and the
value of x is obtained by solving the equation. It is exactly the same as the previous method
but just with the use of some Algebra.
For y  5
For y  8
For y  11
For y  14
5  3x  1
8  3x  1
11  3x  1
14  3x  1
 5  1  3x
 6  3x
6 3x
 
3 3
2  x
x  2
8  1  3x
 9  3x
9 3x
 
3 3
3  x
x  3
11  1  3 x
12  3 x
12 3 x
 
3
3
4  x
x  4
Input values are x  2 ; 3 ; 4 ; 5
83
14  1  3 x
15  3 x
15 3 x
 
3
3
5  x
x  5
EXERCISE 2
(a)
Determine the input values in the following flow diagrams:
(1)
(2)
5 x  10
(3)
1
3
5
10
15
20
x2
5
4
3
2
x4
6
12
18
24
4x  3
7
1
9
17
(4)
1
3
(5)
x3
2
4
6
8
(6)
2
3
x2  1
5
10
17
26
(b)
The equation y  x  3 is given. If the output values are y  4 ; 7 ;10 ;13 ,
(c)
determine the input values.
The equation y  10 x is given. If the output values are y  10 ; 0 ;10 ; 20 ,
(d)
determine the input values.
The equation y  7 x  8 is given. If the output values are y  1; 6 ;13 ; 20 ,
(e)
determine the input values.
The equation y  2 x  6 is given. If the output values are y  10 ;12 ;14 ,
(f)
(g)
determine the input values.
1
The equation y  x  2 is given. If the output values are y  3 ; 5 ; 7 , determine
2
the input values.
1
The equation y   x  1 is given. If the output values are y  0 ;  1;  2 ;  3 ,
5
determine the input values.
FINDING THE RULE WHEN GIVEN INPUT AND OUTPUT VALUES
Example 5
Determine the rule in the given flow diagram.
Input value (x)
2
1
0
1
2
3
Rule
Output value (y)
1
1
3
5
7
9
Solution
Here is a useful method for determining the rule connecting the input and output values.
Start with the output values: 1;1; 3 ; 5 ; 7 ; 9
There is a constant difference of 2 between the terms.
84
Notice the pattern that exists between the output values, constant difference and input
values:
1  2  2  3
1  1 2  3
3  0 2  3
5  1 2  3
7  2 2  3
9  3 2  3
The pattern is as follows:
output value  input value  constant difference 3
The following table is useful for finding the rule.
Input value
The constant difference
between output values
multiplied by input value
What to do to get output
value
Output value
2
1
0
1
2
3
x
2(2)
2(1)
2(0)
2(1)
2(2)
2(3)
2( x)
3
3
3
3
3
3
3
1
1
3
5
7
9
2x  3
The rule for this relationship can be expressed as an equation:
y  2 x  3 where x represents the input values and y represents the output values.
Example 6
Determine the rule in the following table.
Input
Output
1
4
0
1
1
2
2
5
3
8
4
11
x
y
Solution
Start with the output values: 4 ;  1; 2 ; 5 ; 8 ;11
There is a constant difference of 3 between the terms.
Make use of a table.
Input value
The constant difference
between output values
multiplied by input value
What to do to get output
value
Output value
1
0
1
2
3
4
x
3(1)
3(0)
3(1)
3(2)
3(3)
3(4)
3( x)
1
1
1
1
1
1
1
4
1
2
5
8
11
3x  1
The rule for this relationship can be expressed as an equation:
y  3x  1 where x represents the input values and y represents the output values.
Let’s check whether the rule works by substituting the input values into the equation and
checking whether the correct output values are obtained.
For x  1
y  3(1)  1  4
For x  0
y  3(0)  1  1
For x  1
y  3(1)  1  2
For x  2
y  3(2)  1  5
For x  3
y  3(3)  1  8
For x  4
y  3(4)  1  11
Clearly, for the given input values, the output values are correct. Therefore the rule is
correct.
85
Example 7
(a)
(b)
x
y
Determine the rule in the following table.
Calculate the value of a and b.
1
3
0
2
1
7
2
12
3
17
4
22
14
a
b
102
Solution
Start with the output values: 3 ; 2 ; 7 ;12 ;17 ; 22
There is a constant difference of 5 between the terms.
Make use of a table.
0
Input value
1
2
1
The constant difference
5(1)
5(0)
5(1)
5(2)
between output values
multiplied by input value
What to do to get output
2
2
2
2
value
3
Output value
2
7
12
(a)
3
4
x
5(3)
5(4)
5( x)
2
2
2
17
22
5x  2
The rule for this relationship can be expressed as an equation:
y  5 x  2 where x represents the input values and y represents the output values.
(b)
The value of a is an output value (y).
The corresponding input value (x) is 14.
y  5x  2
 y  5(14)  2  72
 a  72
The value of b is an input value (x).
The corresponding output value (y) is 102.
Method 1
The rule for finding an output value given an input value (x) is:
( x  5)  2
The reverse operation for finding an input value given an output value (y) is:
( y  2)  5
The input value b is obtained as follows:
(102  2)  5  20
 b  20
Method 2
Substitute y  102 into the equation y  5 x  2 and solve for x.
102  5 x  2
102  2  5 x
100  5 x
 20  x
 x  20
 b  20
86
EXERCISE 3
(a)
Determine the rule in the following flow diagrams.
(1)
(2)
4
9
14
19
1
2
3
4
(3)
3
6
9
12
(5)
(6)
3
2
1
3
2
3
4
5
10
19
28
37
16
14
12
4
Input (x)
2
3
4
10
Output (y)
5
8
11
29
Rule
Determine the rule in each of the following tables.
(1)
Input
Output
1
4
0
2
1
0
2
2
3
4
4
6
x
y
Input
Output
2
3
1
0
0
3
1
6
2
9
3
12
x
y
Input
Output
1
3
0
3
1
9
2
15
11
69
14
87
x
y
Input
Output
4
9
3
8
2
7
1
6
0
5
1
4
x
y
Input
Output
1
1
0
0
1
1
2
4
3
9
4
16
x
y
Input
Output
1
1
0
0
1
1
2
8
3
27
4
64
x
y
Calculate the value of a and b in each of the following tables:
(1)
0
1
2
5
13
Input
1
Output
3
4
5
6
9
a
b
24
(2)
(3)
(4)
(5)
(6)
(c)
9
11
13
15
(4)
3
4
5
6
(b)
2
3
4
5
(2)
(3)
(4)
(5)
Input
Output
1
9
0
11
1
13
2
15
7
25
15
a
b
91
Input
Output
2
10
1
7
0
4
1
1
2
2
18
a
b
35
Input
Output
1
3
0
7
1
17
a
57
14
b
17
177
32
327
Input
Output
a
20
0
10
1
b
2
0
3
5
4
10
18
80
87
THE GRAPHICAL REPRESENTATION OF RELATIONSHIPS
Another way of representing input-output relationships is graphically. In this chapter, we will
focus on input values that are natural numbers. In Chapter 15 on Graphs, we will explore
other rational input values. Relationships will be represented on the Cartesian plane.
Example 8
A number pattern has a general term (nth term) of Tn  2n  1 where n represents the position
of the term.
(a)
Determine the first four terms.
(b)
Represent the relationship graphically.
Solutions
(a)
(b)
T1  2(1)  1  3
T2  2(2)  1  5
T3  2(3)  1  7
The input values n  1; 2 ; 3 ; 4 are represented
T4  2(4)  1  9
Tn
on the horizontal axis. These values represent the
position of a given term. The output values
Tn  3 ; 5 ; 7 ; 9 are represented on the vertical
axis. These values are the actual terms of the
number pattern.
The graph is made up of dots which each represent
an output value (term) corresponding to a given
input value (term position).
The input values are natural numbers since we
are dealing with the positions of terms in a sequence.
EXERCISE 4
(a)
For each of the following general rules for number patterns, determine the first four
terms and hence represent the relationship graphically. The value of n is a natural
number.
(1)
Tn  n  1
(2)
Tn  2n  1
(3)
Tn  2n  4 (4)
Tn  4n  3
(5)
(b)
6n4
2
(6)
Tn  n 2
(7)
Tn  2n
(8)
Tn  2n1
For each of the following flow diagrams, determine the missing information and
hence represent the relationship graphically. Assume that the input and output values
are natural numbers.
(1)
(3)
(c)
Tn 
1
3
5
7
1
2
3
6
(2)
x 1
2x  2
(4)
6
x
2
x
2
6
10
14
2
4
8
16
For each of the following equations, determine the missing values and hence represent
the relationship graphically. Assume that the input and output values are natural
numbers.
y  x  2 where x  2 ; 3 ; 4 ; 5
(2)
y  2 x  1 where x  1; 3 ;5 ; 7
(1)
(3)
y  x  2 where y  4 ; 5 ; 6 ; 7
88
(4)
y  2 x  4 where y  6 ; 8 ;10
PART 2
(Studied in Term 3 according to CAPS)
RELATIONSHIPS INVOLVING FORMULAE
Example 9
The length of a rectangle is L and the area is A. The perimeter P can be calculated using the
2A
.
formula P  2L 
L
(a)
If L  6 and A  24 , calculate the perimeter.
(b)
If the perimeter is 28 and the length is 8, calculate the area.
Solutions
(a)
P  2L 
2A
L
 P  2(6) 
 P  12  8
 P  20
(b)
2(24)
6
2A
L
2A
 28  2(8) 
8
A
 28  16 
4
A
12 
4
 A  48
P  2L 
Example 10
Given the formulae s  vt and v  12 gt , calculate the positive value of t if it is given that
s  32 and g  4 .
Solution
Substitute the given information into the equations: 32  vt and v  12 (4)t
The value of t cannot be found using either of the equations since each equation has two
unknown quantities.
However, notice how combining the two equations helps:
v  12 (4)t
 v  2t
But 32  vt
Change v in this equation to 2t and you will be able to solve the equation.
32  vt
 32  (2t )t
 32  2t 2
16  t 2
 16  t
t  4
Note: In Chapter 9 (page 118), you will learn that an equation such as t 2  16 has two
solutions which are t  4 or t  4 . If you square both of these solutions, the result
is 16. In this example, you are required to determine only the positive value of t.
If the variable you are solving for represents a length, then only the positive solution
is valid since a length cannot be negative. This applies to calculating lengths using
Pythagoras or calculating the length of the radius of a circle.
89
EXERCISE 5
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
A formula in Physics is v  u  at . Calculate:
(1)
v if u  8, a  13 and t  4 (2)
a if u  16, v  46 and t  10 .
Americans use a different unit for measuring temperature to us. They use the
Fahrenheit scale (  F) whereas we use the Celsius scale (  C). The formula which can
9
be used to convert a temperature in  F to  C or  C to  F is F  C  32 .
5
(2)
Convert 20C to  F
(1)
Convert 35C to  F
(3)
Convert 104F to  C
The circumference of a circle is given by the formula C  2r .
(1)
If r  28 and we take the value of  to be 22
, calculate the value of C.
7
(2)
If C  60 , calculate the value of r.
The perimeter of a rectangle is given by the formula P  2(L  B) where L  length
and B  breadth.
(1)
Calculate the perimeter if the length is 10 cm and the breadth is 6 cm.
(2)
Calculate the breadth if the perimeter is 16 cm and the length is 5 cm.
The air temperature TC outside an aircraft flying at a height of h metres is given by
h
the formula T  26  152
.
(1)
If the height of the aircraft is 5 000 metres, calculate the temperature outside.
(2)
If the temperature outside is 40C , what is the height of the aircraft?
The surface area of a sphere is given by the formula S  4r 2 .
(1)
If r  7 and   3,14 , calculate S rounded off to two decimal places.
(2)
If S  100 , calculate the value of r.
The surface area of a cylinder is given by the formula S  2r (r  h) .
(1)
Calculate S if r  6 and h  10 . Use the value of  on your calculator and
round your answer off to one decimal place.
(2)
Calculate h if S  120 and r  6
Pam invests R5000 (P) at an interest rate (r) of 12% per year for 10 years (n). Use the

r
formula A  P 1  100
(i)
(j)

n
to calculate how much money Pam will have saved (A).
Engineers use two formulae to design electrical circuits: P  I.V and V  I.R
If P  100 and R  4 , calculate the positive value of I.
p  3x  2
Consider the following equations:
y  5 x  2 m  5n 2
(1)
Calculate y if x  6
(2)
Calculate m if n  5
(3)
Calculate x if y  12
(4)
Calculate n if m  45
(5)
Calculate p if x  7
(6)
Calculate y if p  13
(k)
The formulae for the surface area and volume of a cylinder are S  2r 2  2rh and
V  πr 2 h . Calculate the surface area (S) if r  3 and V  45 .
(l)
Given the formula v  u 2  2as .
(1)
Calculate the value of v if u  8 , a  6 and s  3 .
(2)
Calculate the positive value of u if v  6 , a  0,5 and s  11 .
A car towing company provides you with a quotation as follows:
A once-off administrative fee of R300 and a cost of R20 per kilometre towed.
(1)
Write down a formula relating the total cost (C) of towing your car for a
distance of n kilometres.
(2)
Use your formula to calculate the total cost of towing your car 100 km.
(3)
Calculate how many kilometres your car can be towed if you only have R800
available.
(m)
90
EQUIVALENT FORMS
Input-output relationships can have different representations that produce the same outputs.
Consider, for example, the following table.
x
y
0
1
1
2
2
5
3
8
4
11
5
14
The rule for this relationship can be determined in the way discussed in Part 1.
There is a constant difference (d) of 3 between the terms.
Make use of a table.
Input value (x)
0
1
2
3
4
5
dx
3(0)
3(1)
3(2)
3(3)
3(4)
3(5)
What to do to get y
1
1
1
1
1
1
Output value (y)
14
2
5
8
11
1
x
3( x)
1
3x  1
The rule for this relationship can be expressed as an equation:
y  3x  1 where x represents the input values and y represents the output values.
There is another rule for this relationship. It is not necessary for you to know how to
determine this rule. However, you will be required to show that this other rule also works for
the relationship.
The rule is as follows: y  2  ( x  1)(3)
Let’s check this rule to see if the input values (x) produce the output values (y).
For x  0
y  2  (0  1)(3)  1
For x  1
y  2  (1  1)(3)  2
For x  2
y  2  (2  1)(3)  5
For x  3
y  2  (3  1)(3)  8
For x  4
y  2  (4  1)(3)  11
Clearly, for the given input values, the output values are correct. Therefore the rule is correct.
We can use Algebra to show that the two rules are equivalent:
y  2  ( x  1)(3)
 y  2  3x  3
[Use the distributive law]
 y  3x  1
EXERCISE 6
(a)
Consider the following table:
Input (x)
Output (y)
(b)
(c)
(d)
1
5
0
2
1
1
2
4
3
7
4
10
(1)
Determine a rule connecting x and y.
(2)
Show that your rule in (1) and the rule y  4  ( x  2)(3) are equivalent.
Consider the rules y  2 x  2 and y  4  ( x  1)(2) .
Using input values of your choice, draw a flow diagram for each rule to show that
they are equivalent. Hence show that they are equivalent using Algebra.
Consider the rules a  b 2  1 and a  (b  1) 2  2b .
By using the input values b  2 ;  1; 0 ;1; 2 ; 3 , show that the rules are equivalent
by calculating the corresponding output values for each rule.
Show that the following rules are equivalent.
1
3
5
7
1
3
5
7
x2  x
91
x( x  1)
(e)
Show that the following rules are equivalent.
4
5
6
(f)
(g)
3( x  2)  6
12
15
18
3x
In the given square, the length (L) of each side is 4 units.
The diagonals AC and BD are both equal to 32 units.
Show that the following two area formulae are equivalent
when the calculating the area of the square:
A  L2 and A  12 (AC)(BD)
Consider the following equations:
A
y  ( x  2) 2  3
C.
(1)
(2)
B.
4
y  x2  2 x  8
y  ( x  4)( x  2)
y  x2  4 x  1
D.
Use the input values x  1; 0 ;1; 2 to determine which of these equations
are equivalent.
Once you have completed Chapter 8 Part 1, you might like to show
algebraically how the equations are equivalent.
REVISION EXERCISE
(a)
Determine the missing information in the following flow diagrams:
(1)
1
0
1
2
(2)
7 x  9
18 x  14
(3)
(4)
2
1
0
6
(b)
32
4
40
76
23
15
7
41
3
2
1
12
12
8
4
48
Determine the rule connecting x and y in each of the following tables.
(1)
Input
Output
1
9
0
7
1
5
2
3
3
1
4
1
x
y
(2)
Input
Output
1
7
0
4
1
1
2
2
3
5
4
8
x
y
(3)
Input
Output
1
2
0
2
1
6
2
10
3
14
4
18
x
y
92
(c)
Consider the following table.
Input (x)
Output (y)
(1)
(2)
(3)
(4)
(5)
(d)
(f)
(g)
(h)
0
7
1
4
2
1
3
2
4
5
15
a
b
113
Determine an equation connecting x and y.
Show that your rule in (1) and the rule y  ( x  3) 2  x( x  9)  2 are
equivalent by determining the output values.
Calculate y if x  12 .
Calculate x if y  83 .
Calculate the value of a and b.
A formula in Physics is s  ut  12 at 2 .
(1)
(2)
(e)
1
10
Calculate s if u  3, t  5 and a  6 .
Calculate u if s  120, t  5 and a  8 .
The formula for the volume of a cone is V  13 πr 2 h . Assume that π  3,141 .
Calculate the following rounded off to two decimal places where appropriate:
(2)
h if r  3 and V  20
(1)
V if r  3 and h  10
(3)
r if h  3 and V  9π
(4)
r if h  9 and V  5
L
Consider the following Engineering formula: T  2π
.
g
Calculate the value of L rounded off to three decimal places if T  9, 7 and g  9,8 .
b
If M  ab and  ac , determine the positive value of a if M  27 and c  1 .
3
2V
and V  πr 2 h , calculate A if h  10 and r  3 .
If A  2πr 2 
r
Round off your answer to two decimal places.
SOME CHALLENGES
(a)
On a highway to Pretoria, the speed limit is 120 km/h. The fine for exceeding this
speed limit is R50 for each km/h over this limit.
(1)
Write down a formula for F, the fine in rands for a motorist caught driving at a
speed of s km/h. Assume that s  120 .
(2)
Calculate the amount of the fine if the motorist is caught travelling at 160
km/h on the highway.
(3)
If the fine received is R1000, at what speed was the motorist driving?
(b)
Which of the following equations will apply if all the inputs (x) are even numbers and
all the outputs (y) are odd numbers?
(c)
y  ( x  1) 2
1
y  x2  1
y  x( x  2)
y  x7
F.
E.
D.
2
2
2
2
3 ; 4 ; 5 is called a Pythagorean triple since 3  4  5 . This triple can be written as:
a  3  (2) 2  (1) 2
b  4  2(2)(1)
c  5  (2) 2  (1) 2
A.
y  x2
(1)
(2)
Write 8 ; 6 ; 10 in the form a  m 2  n 2
b  2(m)(n)
Why is it not possible to write 9 ; 12 ; 15 in this form?
B.
y  3x
93
C.
c  m2  n2
CHAPTER 8: PATTERNS, FUNCTIONS AND ALGEBRA
TOPIC: ALGEBRAIC EXPRESSIONS
PART 1
(Studied in Term 1 according to CAPS)
REVISION OF GRADE 8 ALGEBRAIC EXPRESSIONS
Let’s revise some important concepts from Grade 8. The revision exercise that follows will
help you to revise the work done last year. It is extremely important for you to attempt all of
the questions in this exercise since these concepts are required for this topic in Grade 9.
A term in algebra is a combination of numbers and letters involving multiplication and
division. Terms are separated by addition and subtraction signs. Whenever terms are placed
inside brackets, the entire expression is treated as one term.
For example, a  2b  c  32c  (3a  6b) consists of four terms: a ; 2bc ; 32c and (3a  6b)
A polynomial is an algebraic expression consisting of one or more terms for which the
exponents are natural numbers 1; 2 ; 3 ; ..... . For example, 3x 4  3x3  2 x  9 .
A monomial is a polynomial consisting of one term. For example, 2x 4 is a monomial.
A binomial is a polynomial consisting of two terms. For example, 3 x  4 is a binomial.
A trinomial is a polynomial made up of three terms. For example, x2  x  6 is a trinomial.
A quadnomial is a polynomial made up of four terms. For example, x 4  3x3  2 x  9 is a
quadnomial.
The three main parts of an algebraic expression:

variable
(the letter in the expression which can vary in value)

coefficient
(the number multiplied by the variable)

constant term (the number that doesn’t change)
For example, in the expression 4 xy  m  4 p 2  3 , the coefficient of x is 4 y , the coefficient
of m is 1, the coefficient of p 2 is 4 and the constant term is 3 .
Like terms have the exact same letters of the alphabet and the same exponents. The
numerical coefficients of the terms may differ. We can only add or subtract like terms and
not unlike terms.
2 x, 5 x and 6 x are like terms since the variables are the same. The coefficients are different
but we can add the terms.
4xy and 3yx are like terms since 3yx is the same as 3xy
3x 2 and 4x2 are like terms since the variable is x 2 (the exponent is the same for each term)
9 x 2 y and 3 x 2 y are like terms since x 2 y is the same for both terms.
x 2 y and xy 2 are unlike terms since the exponents of the letters don’t correlate.
5 x 2 and 6x 4 are unlike terms since the exponents of the letters don’t correlate.
Whenever two monomials are multiplied together, it is important to work through the
following steps:
Step 1:
Multiply the coefficients of the monomials using the integer rules
Step 2:
Multiply the variables by using the laws of exponents
Whenever a monomial is raised to a further power, it is important to work through the
following steps:
Step 1: Evaluate the signs using the fact that ()even number   and ()odd number  
For example:
Step 2:
(1)4  (1)(1)(1)(1)  ()4 (1)4  1
(1)5  (1)(1)(1)(1)(1)  ()5 (1)5  1
Raise the coefficient to the given power by using the laws of exponents.
94
Step 3:
Raise the variable expression to the given power using the laws of exponents.
For example: (2 x3 )4  24. x34  16 x12
Note that  x2 is different from ( x)2 since ( x)2  ( x)( x)  x 2
For example, when simplifying 2(4 x) 2 , it is essential to square first and then
multiply. Always place brackets around the monomial power and work from
there: 2(4 x)2  2[(4 x)2 ]  2[16 x 2 ]  32 x 2
Whenever two monomials are divided, it is important to work through the following steps:
Step 1: Divide the coefficients of the monomials using the integer rules.
Step 2: Divide the variables by using the laws of exponents.
Step 3: Always state the restrictions in the denominator. These are the value(s) of x
for which the denominator becomes 0. Division by 0 is undefined.
Whenever a polynomial is divided by a monomial, it is important to work through the
following steps:
Step 1: Divide each term in the numerator by the monomial in the denominator using
abc a b c
the following rule:
  
d
d d d
Step 2: Simplify using the previous rules of dividing a monomial by a monomial.
Here are some useful rules to help you determine the square or cube root of an algebraic
expression:
n
xn  x 2
and
3
n
xn  x 3
For example:
10
x10  x 2  x5 and
3
18
x18  x 3  x 6
REVISION EXERCISE
(a)
Consider the expression 2 x 2  8 xy  9 y 2  7
(1)
How many terms are there?
(2)
What is the constant term?
2
2
(3)
State the coefficient of the term in x , y , x and y
(4)
Determine the value of the expression if x  2 and y  1
(b)
If x   12 and y  4 , determine the value of the following expressions:
(1)
(c)
(2)
xy
1  3x 3
(7)
( x  y)2
y  2x
(4)
(2)
4m 2 n  7 mn 2
(3)
(m 2 n)(7 mn 2 )(3m 2 )
(5)
(8x3 )(2x3 )
(6)
(3d )(5d 2 )(6d 4 )
(3d )(5d 3 )  6d 4
(8)
(2 x7 )2 . ( x2 )7
(9)
(2 x7 )2  ( x2 )7
(11)
3(a2 )3  (3a2 )3
(12)
(2 x2 )3  (2 x)4
(2)
3(a  b)  2(2a  b)
(3)
( x 2  5 x  6)(2 x)
(5)
(2 p  4 p)(2 p2 1)
(6)
2 p  4 p(2 p2 1)
4a 3b 2  8ab 2
2 ab
2 3
2 3
(10) 3(a ) . (3a )
Simplify:
(m2  3mn)
(1)
(4)
(e)
(3)
y  2 x2
(6)
(5)
Simplify:
(1)
m3  3m3  4m 2
8 x3  2 x3
(4)
(7)
(d)
x y
( x2  5x  6)  2 x
2 xy  5 y
(8)
3
5 x3  2 y  7
3x( x3  2 x2  4 x)  x2 (2 x2  3x  1)
(7)
Simplify:
(1)
12a3b 2 c
8ab7 cd
(2)
(5m3n4 )2
5(mn2 )2
(3)
(4)
3 x 2 12 x
3x
(5)
9 a 2b 18ab 2
3ab
(6)
95
(a 4  a 4 )4
64 a 8 . a 8
(f)
Determine:
(1)
4x 2
(2)
(4)
3
8 y3
(5)
(7)
3
125y 30
(8)
(10)
3
8 x 6  19 x 6
(11)
3
3
25x10
(3)
64 y12
(6)
81x 6  3 27 x9
(9)
25m9  100m9
(12)
9x8
3
8 y15
9 x16  16 x16
3
(4 x3  4 x3 )(2 x5 )3
MULTIPLICATION OF BINOMIALS
Consider the product (a  b)(c  d ) . We can use the distributive, commutative and associative
laws to multiply the two binomials.
(a  b)(c  d )
OUTERS
 ( a  b )c  ( a  b ) d
[distributive law]
FIRSTS
 c ( a  b)  d ( a  b)
[commutative law]
 ca  cb  da  db
[distributive law]
( a  b )( c  d )  ac  ad  bc  bd
 ac  bc  ad  bd
[commutative law]
INNERS
 ac  ad  bc  bd
[associative law]
LASTS
This is done using what we can call the FOIL method.
Here you must first multiply the first terms in each bracket. Then you multiply the outer
terms, then the inner terms and finally the last terms.
Example 1
Expand and simplify the following:
(a)
( x  3)( x  4)
(b)
(3 y  2)(3 y  1)
Solutions
(a)
( x  3)( x  4)
(b)
(3 y  2)(3 y  1)
(c)
(2x3  7 y)( x3  2 y)
(c)
(2x3  7 y)( x3  2 y)
Firsts: ( x )( x )
Firsts: (3 y )(3 y )
Firsts: (2 x3 )( x3 )
Outers: ( x )(4)
Outers: (3 y )(1)
Outers: (2 x3 )(2 y)
Inners: (3)( x )
Lasts: (3)(4)
Inners: (  2)(3 y )
Lasts: ( 2)(1)
Inners: (7 y)( x3 )
Lasts: ( 7 y )( 2 y )
 ( x  3)( x  4)
 (3 y  2)(3 y  1)
 (2 x3  7 y)( x3  2 y)
 x 2  4 x  3x  12
 9 y2  3y  6 y  2
 2 x6  4 x3 y  7 x3 y  14 y 2
 x 2  7 x  12
 9 y2  3y  2
 2 x6  11x3 y  14 y 2
EXERCISE 1
(a)
(b)
Expand and simplify:
( x  2)( x  1)
(1)
(b  1)(b  5)
(4)
(2 x  1)( x  3)
(7)
(10) (5n  2)(n  5)
(13) (m  2n)(m  4n)
(16) (3x  3)(4 x  3)
Expand and simplify:
( x  1)( x  1)
(1)
(m  6)(m  6)
(4)
(3x  2)(3 x  2)
(7)
(2)
(5)
(8)
(11)
(14)
( x  3)( x  2)
( y  3)( y  2)
(3 y  2)( y  2)
(8m  3)(2m  7)
(4  6h)(3  4h)
(3)
(6)
(9)
(12)
(15)
(a  4)(a  6)
( p  7)( p  4)
(4w  1)(2w  3)
( x  2 y )( x  3 y )
(6  2 p)(3  4 p )
(17)
(3  7 x)(9 x  1)
(18)
(a3  2b)(a3  3b)
(2)
(5)
(8)
( x  2)( x  2)
( p  5)( p  5)
(7 p  6)(7 p  6)
(3)
(6)
(9)
(a  4)(a  4)
(2 x  1)(2 x  1)
(12  8 p)(12  8 p)
96
In the next part of the exercise, you need to be aware that (a  b)2  (a  b)(a  b)
(c)
Expand and simplify:
( x 1)2
( x  2)2
( x  3)2
( x  4)2
(2)
(3)
(4)
(1)
(5)
(2 y  3)2
(6)
(4m 7)2
(7)
(8 p  6)2
(8)
(9n  4)2
(9)
( x 1)2
(10)
( x  2)2
(11)
( x  3)2
(12)
( x  4)2
(13)
(2 y  3)2
(14)
(4m 7)2
(15)
(8 p  6)2
(16)
(17)
(12 x  5)2
(18)
(12 x  5)2
(19)
(3x3  4 y 2 )2 (20)
(9n  4)2
(3x3  4 y 2 )2
Investigation
(a)
(1)
(2)
(b)
(1)
(2)
Refer to Exercise 1(b). Can you find a shorter method of determining these
products without going through the steps of FOIL?
Check if your method works with the following products:
(6 x  9 y )(6 x  9 y ) and (12 x  7 y )(12 x  7 y )
Refer to Exercise 1(c). Can you find a shorter method of determining these
products without going through the steps of FOIL?
Check if your method works with the following products:
(3x  4 y)2 and (3x  4 y)2
Conclusion
 When multiplying two binomials which differ in sign only, square the first and last
terms and separate them with a minus sign.
(a  b)(a  b)  a2  b2
or (a  b)(a  b)  a2  b2
 Whenever a binomial is squared, the following direct method is useful:

Square the first term

Multiply the two expressions inside the brackets and then multiply by 2

Square the last term which will always be positive
2
(a  b)  a2  2ab  b2 or
(a  b)2  a2  2ab  b2
Example 2
Use the shorter methods to expand the following:
(a)
(4 x  7 y )(4 x  7 y )
(b)
(3m  12n)(3m  12n)
(c)
2
(d)
(5x3  6 y2 )2
(b)
(3m  12n)(3m  12n)
(4x  3 y)
Solutions
(a)
(4 x  7 y )(4 x  7 y )
2
 (4 x)  (7 y)
2
 (3m)2  (12n)2
 16x2  49 y2
(c)
(4x  3 y)2
 9m 2  144n 2
(d)
(5x3  6 y2 )2
 (4x)2  (4 x)(3 y)  2  (3 y)2
 (5x3 )2  (5x3 )(6 y 2 )  2  (6 y 2 )2
 16 x2  24 xy  9 y 2
 25x6  60 x3 y 2  36 y 4
Note: It is not necessary to use these shorter methods if they seem to be a little complicated
to you. However, with time, you will find that you will gradually start using them in
order to save time. It is quite alright to expand and simplify in the usual way.
In the following exercise, let’s see if you can use the shorter methods. Otherwise,
stick to the usual way of expanding and simplifying.
97
EXERCISE 2
Expand and simplify:
(a)
(d)
( x  6)( x  6)
(8a  7)(8a  7)
(b)
(e)
( x  7)( x  7)
(2a  3b)(2a  3b)
(c)
(f)
(2 x  1)(2 x  1)
(3 p  5q)(3 p  5q )
(g)
(11k  12)(11k  12)
(h)
(4a  3b)(4a  3b)
(i)
(2x  4)2
(j)
(2 x  3)2
(k)
(7m  4n)2
(l)
(7m  4n)2
(m)
(8m  12n)2
(n)
(8m 12n)2
(o)
(x  6 y)2
(p)
(x  3 y)2
(q)
(a4b2  4)(a4b2  4) (r)
(a4b2  4)2
More advanced products
Example 3
Expand and simplify:
(a)
3(3x  2)2
(b)
1 (2 x  4)(2 x  4)  ( x  3) 2
2
(c)
 12 x  1  3x  3
Solutions
(a)
(b)
3(3x  2)2
Shorter way:
 3(3 x  2)(3 x  2)
[write out two brackets]
3(3x  2)2
 3(9 x2  6 x  6x  4)
[expand]
 3(9x2  12x  4)
 3(9x2  12x  4)
[add like terms]
 27 x 2  36 x  12
 27 x 2  36 x  12
[multiply]
Always keep the brackets around the
1 (2 x  4)(2 x  4)  ( x  3)2
binomials while expanding. Then
2
multiply by the number on the left.
 12 (2 x  4)(2 x  4)  ( x  3)( x  3)
 12 (4 x 2  8 x  8 x  16)  1( x 2  3x  3x  9)
 12 (4 x 2  16)  1( x 2  6 x  9) [simplify terms in the brackets]
 2 x2  8  x2  6 x  9
[multiply using the distributive law]
2
 x  6 x  17
[simplify further]
You may also use the shorter ways when expanding and simplifying:
1 (2 x  4)(2 x  4)  ( x  3) 2
2
 12 (4 x 2  16)  ( x 2  6 x  9)
 2 x2  8  x2  6 x  9
(c)
 x 2  6 x  17
1 x 1 x  3
2
3
 
  2x  1  3x  3 
  2x   3x    2x  ( 3)  ( 1)  3x   ( 1)( 3)

[write 12 x as 2x ]
[expand]
2
 x6  32x  3x  3
[simplify]
2
 x6  32x  3x  13
[write 3 as a fraction]
2
 x6  32x  33  3x  22  13  66
[LCD  6 ]
98
2
 x6  96x  26x  18
6
[simplify]
2
 x 9 x62 x 18
[write terms over the LCD]
2
 x 116 x 18
[simplify]
EXERCISE 3
(a)
(b)
Expand and simplify:
(1)
2( x  1)( x  1)
(2)
3(2 x  3)( x  2)
(3)
2( y  3)2
(4)
3(2x 1)2
(5)
5(a  2b)2
(6)
7(3x  1)(3x  1)
(7)
2( p  4)( p  4)
(8)
3(b  2)(b  3)
(9)
5(2m 1)2
(10)
4(2w  3)2
(11)
[4(2 p  q)]2
(12)
4(2 p  q)2
(13)
[2(m  4n)]2
(14)
2(m  4n)2
(15)
(2 y  12 )(2 y  12 )
(16)
(3 y  12 )2
(17)
(2 y  13 )2
(18)
( 3x  3)( 2x  2)
(19)
( 12 a  2)2
(20)
( m3  3)2
(21)
 23 (3m  6)2
(22)
( x4  2 y3 )2
(23)
( x5  3 y 3 ) 2
(24)
(3m3n  2mn3 )2
(25) (8ab  9)(8ab  9)
(26)
Expand and simplify:
(1)
( x  2)( x  3)  x ( x  4)
2
(4b4  3c3 )(4b4  3c3 )
2
(2)
(2 x  5)( x  3)  ( x  1)(3 x  2)
(4)
(2a  7)(3a  2)  ( a  4)(a  2)
(3)
( y  4)  (2 y  3)
(5)
(3m  4)(2m  1)  (m  2)(m  2)
(6)
( x  5 y)2  (3x  y)2
(7)
2(3 x  2)( x  4)  (2 x  1)(3 x  3)
(8)
3n2 (n  3)  (2n  5)2
(9)
( x  2)( x  2)( x2  4)
(10)
(3a  2b)(3a  2b)(9a2  4b2 )
(11)
(13)
(5 y  1)2  (3 y  4)(2  3 y)
(12)
2x6  ( x3  4 y)( x3  4 y)
PART 2
(8m  3n)(4m  n)  (n  3m)(n  3m)
(Studied in Term 3 according to CAPS)
FACTORISATION
When we expand a(b  c) we are multiplying by using the distributive law.
 a (b  c)  ab  ac
If we now reverse this process we get ab  ac  a(b  c) . This reverse process is called
factorisation. The highest common factor (a) has been “taken out” of the expression ab  ac
and the expression is said to be factorised into the product of two factors, namely (a) and
(b  c ) . The two terms have been expressed as one term (or the product of factors).
Factorisation by taking out the highest common factor
Example 4
Factorise the following expressions:
12 x2  8x5
15a 4b6  3ab2
(a)
(b)
Solutions
(a)
12 x2  8x5
The factors of 12 are 1; 2 ; 3 ; 4 ; 6 ; and 12
The factors of 8 are 1 ; 2 ; 4 ; and 8
99
(c)
x ( a  b )  y ( a  b)
Therefore the highest common factor between 12 and 8 is 4.
The expression x 2 can be written as x  x
The expression x5 can be written as x  x  x  x  x
Therefore the highest number of common x’s between the two expressions is x  x .
In other words, the highest common factor between x 2 and x5 is x 2 . Therefore, the
highest common factor between 12 x2 and 8x5 is 4x2 .
Now rewrite the original expression so as to “bring out” the common factor in each
term. Then “take out” the common factor.
12 x 2  8 x5  4 x 2 . 3  4 x 2 . 2 x3
 4 x 2 (3  2 x3 )
The original expression is now expressed as the product of two factors, namely,
4x2 and (3  2x3 ) .
(b)
15a 4b6  3ab2
The HCF  3ab2
15a 4b6  3ab2
 3ab2  5a3b4  3ab2 1
(c)
 3ab2 (5a3b4  1)
x ( a  b )  y ( a  b)
There is a common expression (a  b) in both terms of the expression.
 x(a  b)  y(a  b)
 (a  b)( x  y)
EXERCISE 4
(a)
Factorise the following expressions:
3x  6
4x  8
(1)
(2)
5ab  5ac
9ab  18a
(5)
(6)
(9)
(b)
(c)
7 p  14 pq
2
(10)
2 xy  3 y
2
(11)
2a  12
(4)
15ab  3a
(8)
2
2
15x  10x y (12)
8b  16
6mn  12n
5 x 2  10 x 2 y
(3)
(7)
(13)
x  2x
(14)
2y  4y
(15)
2a 2  2a
(16)
6x  x2
(17)
(18)
(23)
mn 2  m2 n
16 y3  12 y
(20)
(22)
9m2  15m
3a3  12a2
(19)
(21)
5d 2  15d
x3  2 x
πr 2  2πr
6n2  18n3
(25)
x4  2 x
(26)
4 p4  8 p
(27)
9m4  9m3
(28)
(29) 27 g 6  18g 3 (30) 26c7  13c6d (31) 3gh3  33h5
Factorise the following expressions:
18x6 y4  12x4 y7
3a 2b  9ab3
(1)
(2)
(3)
(24)
19 x5  38 x 2
5
2
(32) 17a b  34b
36m5n3  15mn6
(4)
16 pq7  32 p2qr
(5)
45x3 y10  5x2 y8
(6)
27 a 3b 2c  81a 4b3c 2
(7)
25x9 y10  35x7 y5
(8)
3ab2  6a 2b3  9a3b4 (9)
56r 3  8r 4  16r
(10) 7m3n7  14m4 n8  63m5n9
Factorise the following expressions:
(1)
a (b  c)  d (b  c)
(2)
2 x( x  y )  y ( x  y ) (3)
(4)
m(a  2b)  n(a  2b) (5)
3 x( x  1)  2( x  1)
(6)
(7)
3a(b  4c)  6(b  4c) (8)
4 x( x  2)  8( x  2)
(9)
(10)
( x  1)2  3( x  1)
( y  5)3  ( y  5)2
(12)
(11)
100
p (q  r )  q(q  r )
m(n  2)  2n(n  2)
(a  b)2  2(a  b)
( x  2 y) 4  3( x  2 y)3
Factorisation involving the sign-change rule
Consider the following two expressions:
 (b  a) and (b  a)
Let’s remove the brackets from each expression by multiplying.
 (b  a )  b  a
 ( b  a )  b  a
We can therefore conclude that:  (b  a )  (b  a )
Notice that if the sign on the outside of the brackets changes, then the signs inside the
brackets also change. This is called the sign-change rule.
We can also write this as  (b  a )  (a  b)
For example:
(3  x)   (3  x)  ( x  3)
 (5  6 y )  ( 5  6 y )  (6 y  5)
(2  x)  ( 2  x)  (2  x)
(7 n  8m)   ( 7 n  8m)  (8m  7 n)
Note that for the expression (b  a ) , there is no need to use the sign-change rule since
(b  a )  (a  b) from the commutative law.
Example 5
Factorise the following expressions:
(a)
(c)
x( x  3)  y (3  x)
m(m  5n)  (5n  m)
(b)
(d)
4( x  6)  3 x( x  6)
3a(a  4)  6(4  a )
(b)
4( x  6)  3 x( x  6)
 4( x  6)  3 x( x  6)
 ( x  6)(4  3 x)
3a(a  4)  6(4  a )
 3a (a  4)  6(a  4)
 (a  4)(3a  6)
 (a  4)3(a  2)
 3(a  4)(a  2)
Solutions
(a)
(c)
x( x  3)  y (3  x)
 x ( x  3)  y ( x  3)
 ( x  3)( x  y )
m(m  5n)  (5n  m)
 m ( m  5 n )  ( m  5n )
 (m  5n)(m  1)
(d)
EXERCISE 5
Factorise the following expressions:
(a) x(a  b)  y (b  a )
(b)
(d) m(3m  2)  n(2  3m)
(e)
(g) x(a  b)  y (b  a)
(h)
(j) 4 p (3 p  q )  3q (q  3 p ) (k)
(m) 7 a (2  3b)  8b(3b  2)
(n)
(p) 2 x( x  1)  8( x  1)
(s)
2 3
3
a b (c  2)  a b(2  c)
x(a  b)  y (b  a )
m(3m  2)  n(2  3m)
x(a  b)  y (b  a )
4 p (3 p  q )  3q(q  3 p)
7 a (2  3b)  8b(3b  2)
(c) x(a  b)  y (b  a )
(f) m(3m  2)  n(2  3m)
(i) x( a  b)  (b  a )
(l) 4 p(3 p  q )  3q(q  3 p )
(o) 7 a (2  3b)  8b(3b  2)
(q) 8x2 (m  n)  12 x(n  m) (r)
(t)
2
2( x  y)  3( y  x)
3a (4a  1)  15ab(1  4a )
(u) 3(a  4)3  6(4  a)2
Factorisation by grouping in pairs
Whenever there are four terms in an algebraic expression, it is useful to group the terms in
pairs and then proceed to factorise each pair.
Example 6
Factorise the following expressions:
(a)
ax  bx  ay  by
(b)
(c)
ax  3ay  bx  3by
(d)
3ax  6bx  6ay  12by
3x3  6 x  x 2  2
101
Solutions
ax  bx  ay  by
 ( ax  bx )  (  ay  by )
 ( ax  bx )  ( ay  by )
 x(a  b)  y (a  b)
 ( a  b )( x  y )
(a)
[put brackets around the pairs separated by a  sign]
[ ay  ay ]
[factorise each pair]
[take out the common bracket]
(b)
3ax  6bx  6 ay  12by
 (3ax  6bx )  ( 6 ay  12by ) [put brackets around the pairs separated by a  sign]
 (3ax  6bx )  (6ay  12by ) [ 6 ay  6ay ]
[factorise both pairs]
 3 x ( a  2b)  6 y ( a  2b )
[take out the common bracket]
 ( a  2b )(3 x  6 y )
[factorise the expression further]
 ( a  2b )3( x  2 y )
[commutative law]
 3( a  2b )( x  2 y )
(c)
ax  3ay  bx  3by
 ( ax  3ay )  ( bx  3by )
 ( ax  3ay )  (bx  3by )
 a ( x  3 y )  b( x  3 y )
 ( x  3 y )( a  b)
[put brackets around the pairs separated by a  sign]
[apply sign-change rule]
[factorise both pairs]
[take out the common bracket]
3x3  6 x  x 2  2
(d)
There are two methods that you can use.
Method 1
3x3  6 x  x 2  2
 (3x3  6 x)  ( x2  2)
[put brackets around the pairs separated by a  sign]
 (3x3  6 x)  ( x 2  2)
2
[apply the sign-change rule to second pair]
2
 3x( x  2)  ( x  2)
[factorise the first pair]
 ( x 2  2)(3x  1)
Method 2
3x3  6 x  x 2  2
 3 x3  x 2  6 x  2
 (3x3  x 2 )  (6 x  2)
[take out the common bracket]
[group the pairs differently]
[put brackets around the pairs separated by a  sign]
 x 2 (3x  1)  2(3x  1)
[factorise both pairs]
 (3x  1)( x 2  2)
[take out the common bracket]
EXERCISE 6
Factorise the following expressions:
(a) ab  ac  bd  cd
(b) ax  bx  2a  2b
(d) 4 px  8 p  3qx  6 q
(e) 15mx  5my  6nx  2ny
(g)
x 2  2 xy  xy  2 y 2
p3  2 pq 2  4 p 2 q  8q3
(j)
(m) ax  4 x  a  4
(p) ax  2bx  4a  8b
(s) 3mx  9my  2nx  6ny
(v)
3
2
x  3x  x  3
(h) 7 x3  21x  2 x 2 y  6 y
(k)
(n)
(q)
(t)
2m3  4mn  6m2n  12n2
ax  bx  a  b
2ax  3bx  8a  12b
10 ax  15bx  4ay  6by
(w) x3  5x  2 x 2  10
102
(c) 3mx  3nx  m  n
(f) 4ax  4bx  6ay  6by
(i)
4ay 2  8a  12by 2  24b
(l)
(o)
(r)
(u)
ax  bx  a  b
ax  bx  3a  3b
ax  bx  3a  3b
x3  2 x  x 2  2
(x) a3  3ab2  3a 2b  9b3
Factorising the difference of two squares
Consider the product ( a  b)( a  b) .
( a  b)(a  b)
[expand using FOIL]
 a 2  ab  ab  b 2
2
2
[the outer and inner terms cancel each other]
 a b
This only happens if the terms inside the brackets are the same but just differ in sign.
If we reverse this process, we get: a 2  b 2  (a  b)(a  b)
Notice that a 2 and b 2 are square expressions: a 2  a  a and b 2  b  b
When you subtract the two square expressions, namely, a 2  b 2 , you can form two factors of
the form ( a  b)(a  b) . This type of factorisation is called factorising the difference of two
squares (DOTS).
Example 7
Factorise the following expressions:
(a)
4 x2  9
(b)
25a 2  16b 2
(c)
y4 1
(d)
p4  4
(e)
2 m 2  72
(f)
n 2  16
Solutions
(a)
4 x2  9
 2x  2x  3 3
[the two terms are square expressions]
 (2 x  3)(2 x  3)
[DOTS]
You may also write the factors as (2 x  3)(2 x  3)
(b)
25a 2  16b 2
(c)
 5a  5a  4b  4b
 (5a  4b)(5a  4b)
[the two terms are square expressions]
[DOTS]
y4 1
 y 2  y 2  1 1
[the two terms are square expressions]
 ( y 2  1)( y 2  1)
[DOTS]
In the expression ( y  1) , even though y 2 and 1 are square expressions, they are
not separated by a minus sign. The expression is not a difference of two squares and
cannot be factorised further. The expression ( y 2  1) is a difference of two squares
and can be factorised as ( y  1)( y  1) .
2
 y4 1
 ( y 2  1)( y 2  1)
 ( y 2  1)( y  1)( y  1)
(d)
[factorise ( y 2  1) using DOTS]
p4  4
 ( p 2  2)( p 2  2)
[we used DOTS directly]
2
The expression ( p  2) is not an expression involving the difference of two
squares. In the expression ( p 2  2) , the constant 2 cannot be written as the product
of two identical rational numbers. Therefore, the expression p 4  4 is not factorised
further. However, in Advanced Programme Maths, we can factorise further as
follows:
p 4  4  ( p 2  2)( p 2  2)  ( p 2  2)( p 2  ( 2) 2 )  ( p 2  2)( p  2)( p  2)
103
(e)
(f)
2 m 2  72
 2(m 2  36)
 2(m  6)(m  6)
[take out the HCF]
[factorise further using DOTS]
n 2  16
 ( n 2  16)
[put brackets around the terms]
2
 ( n  16)
 (n  4)(n  4)
[apply the sign-change rule]
[factorise using DOTS]
Important note:
When factorising expressions, always stick to the following steps:
Step 1: Apply the sign-change rule if necessary
Step 2: Take out the HCF if it exists
Step 3: If there is a difference of two squares, factorise using DOTS
Here is an example using the above steps:
9 x3  81x
 (9 x3  81x)
[put brackets around the terms]
 (9 x3  81x)
[apply the sign-change rule]
2
 9 x( x  9)
[take out the HCF]
[factorise further using DOTS]
 9x( x  3)( x  3)
EXERCISE 7
(a)
Factorise the following:
(1)
(2)
x2  1
(b)
x2  4
(3)
a 2  16
(4)
p 2  25
(5)
m 2  36
(6)
4 x2  1
(7)
16 x 2  9
(8)
81 y 2  16
(9)
49 n 2  121
(10)
100 d 2  169
(11)
x2  4 y2
(12)
x 4  16
(15)
a 4  81
(16)
16b 2  25c 2
a 2 b 2  4c 4
8 y2  2
(3)
2 x 2  50
(4)
3 p 2  27
(8)
(12)
4 x 2  64
x2  4 x4
(13)
(14)
x4  9
Factorise the following:
(1)
(2)
4 x2  4
(5)
(9)
(6)
5n 2  20
2
3
12 x y  27 y (10)
100m 2  25
75a 2  12b 2
(7)
(11)
4 x 2  16
x 3  36 x
(13)
3 p 2 q  48q 3
(14)
2ax 4  50a
(15)
3 x 2 y  243 y (16)
 x2  4
(17)
 x 2  16
(18)
9 x 2  1
(19)
25 x 2  4
2 p 2  98
(21)
(25)
3n 3  3n
( x  y )2  1
(22)
(26)
8b5  32b
(a  b) 2  9
(23)
(27)
7 a 2b 2  28 (24)
4( m  2 n ) 2  25
(20)
Factorising trinomial expressions
Consider the product ( x  a)( x  b)
By multiplying out, it is clear that this product will become:
( x  a)( x  b)
 x 2  ax  bx  a  b
 x 2  ( a  b) x  ( a  b)
So the expression x 2  (a  b) x  (a  b) can be factorised as ( x  a )( x  b) .
For example, the trinomial x 2  6 x  8 can be factorised as follows:
Write the last term, 8, as the product of two numbers ( a  b ).
1 8 and
4 2
The options are:
104
2a 5b 2  18a
The middle term (a  b) is now obtained by adding the numbers of one of the above options.
The obvious choice will be the option 4  2 because the sum of the numbers 4 and 2 gives 6.
Therefore:
x2  6x  8
 x 2  (4  2) x  (4  2)
 ( x  4)( x  2)
So the trick to factorising trinomials is as follows:
 Write down the last term as the product of two numbers.
 Find the two numbers (using the appropriate numbers from one of the products)
which gets the middle term by adding or subtracting.
 Check that when you multiply these numbers you get the last term.
Example 8
Factorise:
(a)
x 2  7 x  60
The last term can be written as the following products:
1 60, 30  2, 15  4, 10  6, 12  5, 20  3
We now need to get 7 from one of the options above.
Using 12  5 will enable us to get 7 since 12  5  7 which is the middle term
and (12)(5)  60 which is the last term.
Therefore:
Notice that the sign of the last term is
x 2  7 x  60
negative and that the signs in the
brackets are different.
 ( x  12)( x  5)
(b)
a 2  9a  20
The last term can be written as the following products: 1 20, 2 10, 5  4
We now need to get 9 from one of the options above.
Using 5  4 will enable us to get 9 since 5  4  9 which is the middle term
and 5  4  20 which is the last term.
Therefore:
Notice that the sign of the last term is
a 2  9a  20
positive and that the signs in the
 (a  5)(a  4)
brackets are the same (both negative).
(c)
x2  5x  6
The last term can be written as the following products: 3  2, 1 6
We now need to get the middle term 5 from one of the options above.
Try the option 3  2 . Clearly 3  2  5 which is the middle term
and 3  2  6 which is the last term.
Notice that the option 1 6 will not work because even though 6  1  5 is the
middle term, 6  1  6 is not the last term.
Therefore:
Notice that the sign of the last term is
x2  5x  6
positive and that the signs in the
 ( x  3)( x  2)
brackets are the same (both positive).
In summary then, apply the following procedure when factorising trinomials:
 Take out the highest common factor if necessary.
 Write down the last term as the product of two numbers.
 Find the two numbers (using the appropriate numbers from one of the products) which
gets the middle term by adding or subtracting.
105



Check that when you multiply these numbers you get the last term.
If the sign of the last term of a trinomial is positive, the signs in the brackets are the
same (both positive or both negative).
If the sign of the last term of a trinomial is negative, the signs in the brackets are
different.
Example 9
Factorise 2 x 2  20 x  48
Solution
Here it is necessary to first take out the highest common factor:
2 x 2  20 x  48
 2( x 2  10 x  24)
The last term can be written as the following products: 1 24, 12  2, 8  3, 6  4
The signs in the brackets must be the same because the sign of the last term, 24 , is positive.
The option 6  4 will work because:
6  4  10 , which is the middle term, and
( 6)(4)  24 , which is the last term.
Notice that the option 12  2 will not work because even though 12  2  10 is the
middle term, 12 2  24 , is not the last term.
 2 x 2  20 x  48
 2( x 2  10 x  24)
 2( x  6)( x  4)
EXERCISE 8
(a)
(b)
Factorise:
x 2  3x  2
(1)
(2)
a 2  6a  5
(3)
p 2  7 p  12
(4)
y 2  7 y  12
(5)
x 2  11x  18
(6)
k 2  8k  15
(7)
m 2  14m  24
(8)
d 2  14d  40
(9)
y 2  13 y  40
(10)
(13)
(16)
(19)
(22)
(25)
x 2  12 x  35
w2  8w  15
x 2  10 x  16
t 2  4t  60
x2  x  6
x2  2 x  1
(11)
(14)
(17)
(20)
(23)
(26)
x2  5x  6
x 2  11x  28
n 2  n  20
x 2  3 x  88
x 2  5 x  24
a 2  6a  9
(12)
(15)
(18)
(21)
(24)
(27)
x 2  7 x  12
x 2  10 x  9
x 2  7 x  18
r 2  5r  50
k 2  9k  36
y 2  10 y  25
(29)
(32)
(35)
x 2  3 x  54
x 2  22 x  72
64  30 y  y 2
(30)
(33)
(36)
x 2  17 x  38
x 2  25 x  100
k (k  9)  52
(2)
2 y 2  10 y  8
(3)
3a 2  15a  18
(28)
x 2  13 x  30
(31)
a 2  20a  64
27 x  x 2  90
(34)
Factorise fully:
2 x 2  12 x  16
(1)
(4)
5d 2  45d  100
(5)
x3  12 x 2  27 x
(6)
4 p 2  48 p  80
(7)
(10)
 x2  2 x  8
2k 2  22k  52
(8)
(11)
 a 2  22a  75
a 2b  19ab  84b
(9)
(12)
4 x 2  20 x  24
126  36 x  2 x 2
106
The Golden Rules of Factorisation
Two terms
Three terms
Step 1:
Step 1:
Apply the sign-change rule if
Apply the sign-change rule if
necessary.
necessary.
Step 2:
Step 2:
Take out the HCF if it exists.
Take out the HCF if it exists.
Step 3:
Step 3:
Apply DOTS if possible.
Factorise the trinomial.
Four terms
Step 1:
Group in pairs and put brackets
around each pair separated by
the  sign.
Step 2:
Apply the sign-change rule if
necessary.
Step 3:
Factorise the pairs.
Step 4:
Take out the common bracket.
Step 5:
Factorise further if needs be.
Using factorisation to simplify algebraic expressions
Example 10
Simplify:
3a 2b4c 2
(a)
6a5b3c 2
(b)
2x  8
2x
(c)
6 x 2  18 x  60 x  2

8x
8 x 2  40 x
Solutions
3a 2b 4c 2 3 a 2 b 4 c 2 1 3 1
b

 5  3  2   a  b 1  3
(a)
5 3 2
6 a b c
2
6a b c
2a
(b)
(c)
2x  8
2x
There are two methods of simplifying this algebraic fraction:
Method 1 (Grade 8 method)
Method 2 (factorisation)
2x  8
2x  8
2x
2x
2x 8
2( x  4)



2x 2x
2x
4
( x  4)
 1

x
x
x4

x
2
6 x  18 x  60 x  2

8x
8 x 2  40 x
6 x 2  18 x  60 8 x


x2
8 x 2  40 x
2
6( x  3x  10)
8x


8 x( x  5)
( x  2)
6( x  5)( x  2)
8x


8x( x  5)
( x  2)
3( x  2)
8x


4x
( x  2)
6
[change to  ]
[take out the HCF]
[factorise]
[simplify]
107
EXERCISE 9
Simplify the following expressions:
x2  2 x
4 x2  4 x
(a)
(b)
x
2x
(c)
10a 2  15a
10a
(d)
8 p3  16 p 2
12 p
(e)
2 xh2  4 x 2 h
(f)
hx
6 x 2  24
15 x 2  60
(g)
8b8  16b9
8b9  4b8
(h)
6 p 2 q  9 pq 2
12 p 3q 2  18 p 2 q 3
(i)
k2  4
k 2  2k
4m 2  9
4m 2  6m
(k)
5 y 2  15 y
5 y 2  45
(l)
h4  4
3h2  6
(m)
(q)
(j)
6n2  18n
(n)
12n2
3 x( x  1)  2( x  1)
x2  1
x2  5x  6
a 2  9a  20
t 2  81
(o)
(p)
x 2  7 x  12
a 2  a  20
t 2  7t  18
x3  2 x 2  16 x  32
x 2  3 x  18 6 x 2  18 x
(s)
(r)

x2  2 x  8
x2  6 x
(2 x)2
REVISION EXERCISE
(a)
(b)
(c)
Expand and simplify:
4 x(6 x  3 y )
(1)
(3x  5 x)(2 x  1)
(4)
(2)
(5)
4 x(2 x  6 x)
3k  5k (2k  1)
(3)
(6)
4a  (2a  3b)
(3 p  5)(2 p  1)
(7)
(4 y  3)(4 y  3)
(8)
4 y  3(4 y  3)
(9)
(4 x  3)2
(10)
(4n  3)2
(11)
(2b  3c)
(12)
(2b  3c)2
(13)
(2b  3c)(2b  3c)
(14)
(2a 2  3)(a  5)
(15)
(3x3  2 y )(2 x 2  3 y )
(16)
(4n4  4)(4n4  4)
(17)
(4n4  4)(4n4  3)
(18)
(4n4  4)2
(19)
( 13 x  3)( 13 x  3)
(20)
( 13 x  3)( 13 x  3)
(21)
( 13 x  3)2
(22)
(24)
2( x  3)2  [2( x  2)]2
(23) 5(2 y 1)(3 y  2)  (4 y  5)2
(3 p  2q)(4 p  8q)  12 (2 p  16q)(2 p  16q)
The length of a rectangle is (3 x  2) and the width is (2 x  1) . Determine the
perimeter and area in terms of x in simplest form.
Factorise fully:
7 x  14
7 x  28
3x  12
(2)
(3)
(1)
2
3x  12
(5)
(6)
(4)
4a  a
4a 2  2a
(8)
(9)
(7)
4a 2  3a
4 a 2  8a
x2  4
(11) 4n 2  4
(12) 4k 2  16
(10) 4m 2  1
(14) 4d 2  64
(15) 4 x 2  32
(13) 4b 2  32
(17) 8 x 2  8 x
(18) 8 x 2  16 x
(16) 4t 2  64
(20) 3x 2  243 y 2
(21) 8 x 8  8
(19) 2  8 x 2
(22)
(25)
x 2  12 x  27
y 2  6 y  27
(23)
(26)
k 2  12k  27
n 2  4n  77
(24)
(27)
a 2  6a  27
x 2  4 x  77
(28)
(31)
(34)
x 2  18 x  77
2 x 2  4 x  30
4 x (a  b)  3( a  b)
(29)
(32)
(35)
 x2  5x  6
3 x 2  24 x  45
2 x (a  b)  6(b  a )
(30)
(33)
(36)
6  5x  x2
81  18a  a 2
x 2 ( x  1)  4(1  x)
(37)
x2 ( x  4)  9(4  x) (38)
a 3  a 2b  ab 2  b 3
(39)
ax  3a  bx  3b
(40)
ax  3a  bx  3b
(41)
2ax2  4ax  xy  2 y (42)
(43)
2 y3  8 y2  8 y  32
(44)
16b2 x  40b2  8bx  20b
108
x3  3 x 2  9 x  27
(c)
Simplify:
x2  9 x
(1)
x2
(4)
(7)
(10)
(2)
x 2  14 x  24
x 2  11x  12
2 x 2  32
4 x 2  12 x  16
x 4  ( x)2
 x2
(5)
(8)
(11)
x3 y  4 xy 2
x2 y 2
(3)
x 2  14 x
x 2  16 x  28
6 x3  8 x 2  6 x  8
3x  4
16 p 4  1
1
 2
2
4 p 1 4 p 1
(6)
(9)
(12)
x2  9
x 2  3x
x 2  121
x 2  22 x  121
6 x3  8 x 2  6 x  3
2 x2
8 x 6  32 x 2
12 x 6  24 x 4
SOME CHALLENGES
(a)
(b)
(c)
(d)
(e)
(f)
Consider the expression x2 ( x  3)  4( x  3)
(1) Expand and simplify the expression
(2) Factorise the expression fully.
2
2
Consider the expression (2a  3b)  (4a  b)
(1) Expand and simplify the expression
(2) Factorise the expression fully.
Factorise fully:
x2  xy 12 y 2
(2)
(1)
x 4  17 x 2  16
(3)
x2 ( x  2)  x( x  2)  6( x  2)
(5)
(1)
( x  1)2  2( x  1)  8
x2  16 x  64
(6)
From the product of (3 x  4) and (7 x  4) subtract (2 x  4)(2 x  4)
(2)
Subtract (3x  7 y)2 from the quotient when ( x4 16 y 4 ) is divided by
(4)
( x2  5x)2  36
( x2  4 y2 ) . Then add (9x2  40 xy  53 y 2 ) and then square the result.
If a  b  3 and ab  3 , determine the value of a 2  b 2 .
If a 2  b 2  6 and a  b  4 , determine the value of ab .
If 3a  3b  9 and a  b  a  b , determine the value of a 2  b 2 .
(1)
(2)
(3)
Consider the given diagram made up of four congruent rectangles and a shaded
square. Determine:
(1)
the area of one white rectangle.
x
(2)
the area of the large square.
(3)
the area of the shaded square in simplest form
in terms of x.
x6
(g)
In the given diagram, three rectangles are placed next to each other.
Show that the shaded rectangle has an area of A  ( x  7)2
x4
x7
x 3
x
109
CHAPTER 9: PATTERNS, FUNCTIONS AND ALGEBRA
TOPIC: ALGEBRAIC EQUATIONS
PART 1
(Studied in Term 1 according to CAPS)
Solving equations by using additive and multiplicative inverses (Revision)
Example 1
Solve the following equations (find the value for the variable which satisfies the equation)
and then check your answer to see whether your answer actually satisfies the equation.
(a)
(c)
7  5 x  22
6 x  3  7  5x
(b)
(d)
5  4x  7
4x  8  6x  3
Solutions
(a)
7  5 x  22
 7  5 x  7  22  7
5 x  15
5 x 15


5 5
 x  3
Check the answer:
LHS  7  5 x  7  5(3)  22
[subtract 7 from both sides]
[simplify both sides]
[divide both sides by 5 ]
RHS  22
 LHS  RHS
 x  3 is the solution
(b)
5  4x  7
5  7  4x  7  7
12  4x
12 4 x
 
4
4
3  x
[add 7 to both sides]
[simplify both sides]
[divide both sides by 4]
x  3
LHS  5
RHS  4(3)  7  5
 LHS  RHS
 x  3 is the solution
(c)
Here we have variable terms on either side of the equation. If we first add 3 to both
sides and then subtract 5x from both sides, it will be possible to solve for x.
6 x  3  7  5x
 6 x  3  3  7  5x  3
[add 3 to both sides]
 6 x  10  5 x
[simplify both sides]
 6 x  5 x  10  5 x  5 x
[subtract 5 x from both sides]
 x  10
LHS  6(10)  3  57
RHS  7  5(10)  57
 LHS  RHS
 x  10 is the solution
110
(d)
4x  8  6x  3
Method 1
Add 8 to both sides:
 4x  8  8  6x  3  8
 4 x  6 x  11
Subtract 6 x from both sides:
 4 x  6 x  6 x  11  6 x
Simplify both sides and solve:
2 x  11
2 x 11


2 2
11
x  
2
1
 x  5
2
Method 2
Subtract 3 from both sides:
 4x  8  3  6x  3  3
 4 x  11  6 x
Subtract 4 x from both sides:
 4 x  11  4 x  6 x  4 x
Simplify both sides and solve:
11  2 x
11 2 x


2
2
11
  x
2
1
 x  5
2
Shortcut for solving equations
The previous method is quite tedious. We do however have a shortcut for solving equations.
Consider the following equations.
x 3  7
(a)
If we add 3 to both sides we get:
x  73
The constant  3 “moved across” the equal sign and became 3.
There was a sign change from 3 to  3 .
 x  10
(b)
x5  7
If we subtract 5 from both sides we get:
x  75
The constant 5 “moved across” the equal sign and became  5 .
There was a sign change from  5 to  5 .
x  2
So, in conclusion, whenever terms are taken across the equal sign in an equation,
they will change sign on the other side.
4  2 x  12
(c)
 2 x  12  4
[Bring 4 to the right and change its sign]
 2x  8
x  4
5 x  2  6  4 x
(d)
Method 1
5 x  2  6  4 x
5 x  4 x  6  2
[Bring 4 x to the left side and 2 to the right side]
 x  8
 x  8
Method 2
5 x  2  6  4 x
2  6  4 x  5 x [Bring 6 to the left side and 5x to the right side]
8  x
 x  8
111
EXERCISE 1
(a)
Solve the following equations by adding, subtracting, multiplying or dividing both
sides of the equation by a constant.
(1)
2 x  8  12
(2)
4 x  6  18
(3)
3 x  7  14
(4)
3 x  4  20
(5)
3 x  5  10
(6)
4 x  5  13
4x  8  8
8x  4  2
3x  4  0
(7)
(8)
(9)
(10) 3 x  6  6
(11) 7  x  3
(12) 14  2 x  12
(13) 27  3  2 p
(14) 13  16  3x
(15) 0  14m  42
(17) 6  6 x  18
(18) 10  10 x  4
(16) 12  12 x  12
1
1
x25
(20) 5   2 x  2
(21) 2  53 x  1
(19)
2
(22)
(b)
(c)
6  6x  2
8 2 x
4
 8
Solve the following equations and check your answers:
7y 3  4y  6
(1)
5x  5  4 x  9
(2)
Solve the following equations:
8p  7p 3
5x  4 x  2
(2)
(1)
(4)
5x  6  6
(5)
5 x  5  3 x  11
(7)
4t  2  5t  6
(8)
8k  5  9 k  2
(11) 7  3 x  4 x  14
(10) 2  3n  12  5n
(14) 4  4 x  7  7 x
(13) 2 x  5  3 x  10
(17)
x  3  13 x  13
(16) 6 x  5  3 x  10
Example 2
(Examples involving the distributive law)
Solve for x:
6  3( x  2)  4( x  1)
(a)
(c)
(23)
(b)
3x  ( x  4)  2  5 x
(b)
3x  ( x  4)  2  5 x
 3x  x  4  2  5 x
 2 x  4  2  5x
 2 x  5x  2  4
 7 x  2
2
x  
7
2
2( x  1)  (2 x  3)( x  1)
Solutions
(a)
6  3( x  2)  4( x  1)
 6  3x  6  4 x  4
 3x  4 x  4
4  4 x  3 x
4  x
 x  4
(c)
2( x  1) 2  (2 x  3)( x  1)
 2( x 2  2 x  1)  2 x 2  2 x  3x  3
 2 x2  4 x  2  2 x2  x  3
 2 x 2  2 x 2  4 x  x  2  3
3 x  5
3x 5


3 3
5
x 
3
2
x 1
3
112
(24)
12
x
(3)
7  3m  5  2m
(3)
(6)
(9)
(12)
(15)
(18)
6m  4m  14
9 p  7  4 p  12
5 x  11  7 x  3
8 x  1  3x  1  2 x
7  2x  2x  7
3 4 x  3 x
2
 3
EXERCISE 2
(a)
(b)
Solve the following equations and check your answers.
(1)
5( x  1)  4( x  1)
(2)
3( x  2)  2( x  3)
6(3  p )  4(2  p )
(4)
3(m  2)  (3  m)  5
(3)
Solve the following equations.
4(2 x  1)  x  3
(2)
4(2 x  1)  x  3
(1)
4(2 x  1)   x  3
(4)
5(1  3 x)  4(3  x)  2
(3)
3  ( y  3)  6  y
(6)
2  2( x  3)  2( x  4)
(5)
3k  2(k  1)  14  5k
(8)
4( x  2)  1  2( x  4)  0
(7)
(9)
(c)
7  ( x  1)  9  (2 x  1)
(10)
(11) 4( x  1)  2(2 x  3)  2  15  x  3(3  x)
Solve the following equations.
x 2  x( x  2)  4
(2)
(1)
(3)
4 x( x  2)  2 x(2 x  1)  4  0
(4)
1 (2  4 x)  7( x  1)  2
2
2 x 2  7  x(2 x  1)  0
3 x( x  1)  x(3 x  1)  x  6
(5)
8 x(2 x  3)  4 x(4 x  1)  40
(6)
4m  3m(m  2)  3m 2  10
(7)
3t (2t  1)  3t (1  2t )  3
(8)
x2  2 x  1  x2  2 x  3
(9)
2 p2  5 p  2 p2  3 p  4
(10)
 3 y 2  4 y  1  3 y 2  2 y  5
(11)
( x  3)( x  3)  x 2  3 x
(12)
( x  1) 2  x( x  1)  10
(13)
( x  3)( x  2)  x( x  1)  0
(14)
( x  2) 2  x( x  2)
(15)
( x  4) 2  ( x  3) 2
(16)
x( x  3)  ( x  2) 2  0
(17)
(3x  4)(3x  4)  (3x  2) 2  0
(18)
(2 x  3)2  (2 x  1)(2 x  3)  1
Solving equations with fractions
It is extremely important to understand the difference between simplifying an algebraic
expression with fractions and solving an equation with fractions.
In this section, we will work with both concepts together to make sure that you understand
the difference. Remember that simplifying an expression containing more than one fraction
involves writing the expression as a single fraction. Solving an equation involves determining
the value of the variable that solves the equation.
Example 3
(a)
Simplify the expression:
x
x

 3
2
5
(b)
Solve the equation:
x
x

 3
2
5
Solutions
(a)
The lowest common denominator (LCD) is 10. Convert all terms to equivalent
fractions with denominators equal to 10.
x
x

 3
2
5
x
x
3
 

[write 3 in fraction form]
2
5
1
x 5
x 2
3 10
[change to equivalent fractions]
     
2 5
5 2
1 10
5x
2 x 30



10
10 10
113
5 x  2 x  30
10
3x  30

10

(b)
[write expression as one fraction]
[simplify]
There are two ways of solving this equation:
Method 1
The lowest common denominator (LCD) is 10. Convert all terms to equivalent
fractions with denominators equal to 10.
x
x

 3
2
5
x
x
3
 

[write 3 in fraction form]
2
5
1
x 5
x 2
3 10
[change to equivalent fractions]
     
2 5
5 2
1 10
5 x 2 x 30



10 10 10
5 x 2 x  30
[write expressions on both sides as single fractions]


10
10
5x
2 x  30
[multiply both sides by 10]
 10 
 10
10
10
 5 x  2 x  30
[simplify]
 3 x  30
 x  10
[solve for x]
Method 2
A much easier method is to multiply all terms by the LCD right from the start.
x
x

 3
2
5
x
x
3
 10   10  10 [multiply all terms by the LCD of 10]
2
5
1
10 x 10 x 30



2
5
1
 5 x  2 x  30
[simplify]
 3 x  30
 x  10
[solve for x]
EXERCISE 3
(a)
(1)
Simplify:
(b)
(1)
Simplify:
(c)
(1)
Simplify:
(d)
(1)
Simplify:
(e)
(1)
Simplify:
x 2x

5
4 3
3a 2a a

 3
10 5 2
p
3 p   11
4
5 x 3x 1
 1
2
4
2
3x 2
5
 x
2 3
6
114
(2)
Solve:
(2)
Solve:
(2)
Solve:
(2)
Solve:
(2)
Solve:
x 2x

5
4 3
3a 2a a

 3
10 5 2
p
3 p   11
4
5 x 3x 1

1
2
4
2
3x 2
5
  x
2 3
6
Example 4
(a)
Simplify:
x3
x2

 x2
2
3
(b)
Solve for x:
x3
x2

 x2
2
3
Solutions
(a)
(b)
The lowest common denominator (LCD) is 6.
Always put brackets around the expressions in the numerators.
x3
x2

 x2
2
3
( x  3) 3 ( x  2) 2
x 6 2 6
[change to equivalent fractions]

 
    
2
3
3
2
1 6 1 6
3( x  3) 2( x  2)
6x
12




6
6
6
6
3( x  3)  2( x  2)  6 x  12
[write as a single fraction]

6
3 x  9  2 x  4  6 x  12 7 x  1


6
6
Method 1
( x  3)
( x  2)

 x2
2
3
( x  3) 3
( x  2) 2
x 6 2 6

 
    
2
3
3
2
1 6 1 6
3( x  3) 2( x  2)
6x
12
[change to equivalent fractions]




6
6
6
6
3( x  3)  2( x  2) 6 x  12
[write as a single fractions]


6
6
3( x  3)  2( x  2)
6 x  12
[multiply both sides by 6]

6
6
6
6
 3( x  3)  2( x  2)  6 x  12
 3 x  9  2 x  4  6 x  12
[expand and solve for x]
 x  13  6 x  12
5 x  25
x  5
Method 2
( x  3)
( x  2)

 x2
2
3
( x  3)
( x  2)
x
2
[multiply all terms by the LCD]

6 
6  6  6
2
3
1
1
 3( x  3)  2( x  2)  6 x  12
 3 x  9  2 x  4  6 x  12
[expand and solve for x]
 x  13  6 x  12
5 x  25
x  5
115
EXERCISE 4
(a)
(b)
(c)
Solve the following:
x4
(1)
7
(2)
8
3x  1 4 x
(4)

(5)
2
3
x5
(7)
 11  x
(8)
3
Solve the following:
y2
(1)
3
4
2
4x 1
3x  2
3x
(3)


4
2
2
3k  2
2k  3
(5)

 1
7
2
2( x  3)
x7
3x  8
(7)


5
3
2
(1)
Simplify:
(2)
Solve:
Example 5
Solve for x:
3
x7
7
5
2(1  2m)
 6
3
4x  7
3(2 x  10) 
2
(2)
(4)
(6)
(8)
(3)
(6)
(9)
x4
0
4
3
(a  8)  6
4
w
w 4
2
4
6
2a  1
a 50
5
2x  1
3x  2
x


3
6
2
m2 m6 1


4
3
2
1
1
1
(3 x  2)  (2 x  1)  1
3
4
2
3x  8 2 x  1
x2

 x 2 
10
5
2
3x  8 2 x  1
x2

 x 2 
10
5
2
10 2( x  5)

x
x
Solution
In this equation the LCD is x.
10
2( x  5)
x
[multiply all terms by the LCD]
3 x   x 
x
x
 3 x  10  2( x  5)
[simplify]
 3 x  10  2 x  10
[expand brackets]
 3 x  2 x  10  10
x  0
However, if we check the solution, this value of x will not be valid since division by 0 is
undefined.
10
2(0  5)
which is undefined RHS 
which is undefined
LHS  3 
0
0
We say that the equation has no solution since x  0
EXERCISE 5
By solving the following equations, determine whether or not each equation has a solution.
Explain your conclusion in each case.
5 5( x  1)
3 1 2 5
x2
(b)
(a)
(c)
  
 1
7 
x x x 2
2
x
x
4( x  2) 5 x  16
6 3 1
2 1 5
(d)


(e)
 
1
(f)

 0,5
x 2x 4
3x x 3x
x
2x
116
Equations involving the product of factors
Whenever an equation has the form a . b  0 then the solutions to the equation can be
determine by solving the equations as follows:
If a . b  0 then a  0 or b  0 .
Example 6
Solve the following equations:
(a)
( x  2)( x  4)  0
(b)
(2 x  3)(4 x  5)  0
(c)
3x(6 x  8)  0
(d)
3(2 x  4) 2  0
(2 x  3)(4 x  5)  0
 2 x  3  0 or 4 x  5  0
 2 x  3 or 4 x  5
3
5
x 
or x  
2
4
1
1
x 1
or x  1
2
4
2
3(2 x  4)  0
 3(2 x  4)(2 x  4)  0
 3(2 x  4)  0 or 2 x  4  0
Solutions
(a)
( x  2)( x  4)  0
 x  2  0 or x  4  0
 x  2 or x  4
(b)
(c)
3x(6 x  8)  0
 3 x  0 or 6 x  8  0
 x  0 or 6 x  8
8
x 
6
4
1
x  1
3
3
(d)
 6 x  12  0 or 2 x  4  0
 6 x  12 or 2 x  4
 x  2 or x  2
[These are repeated solutions]
EXERCISE 6
(a)
Solve the following equations:
( x  1)( x  2)  0
(1)
( x  8)( x  6)  0
(4)
x( x  5)  0
(7)
(10)  x( x  6)  0
(13) 5( x  5)( x  5)  0
(16) (4 x  8)(3x  6)  0
(19) 2 x(4 x  1)  0
(22) 3(3 x  1)(9 x  12)  0
(25) (2 x  9)(4  3x)  0
( x  3)( x  2)  0
( x  10)( x  9)  0
2 x( x  2)  0
3x( x  8)  0
7( x  7)( x  2)  0
(2 x  3)(3 x  2)  0
(9 x  3)(12 x  8)  0
3x(3  x)  0
(7  9 x)(1  5 x)  0
(3)
(6)
(9)
(12)
(15)
(18)
(21)
(24)
(27)
(29)
x( x  3)( x  7)  0
(30)
( x  4)( x  1)  0
x( x  7)  0
2 x( x  4)  0
2( x  7)( x  1)  0
(3x  6)( x  1)  0
4 x(2 x  5)  0
4(2 x  6)(2 x  1)  0
(3x  4)(4 x  3)  0
2( x  10)(7 x  2)  0
( x  9) 2  0
(28)
3  ( x  3)( x  3)  3
(31)
2
x(2 x  6)  0
(32)
5(2 x  11)  0
(33)
x 2 (2 x  14)  0
(34)
(4 x  9)3  0
(35)
(2 x  12 )(3 x  13 )  0
(36)
1 ( x  4) 2
4
(37)
( x  12 )( x  14 )  0
(38)
( 15 x  1)( 13 x  3)  0
(39)
(40)
(b)
(2)
(5)
(8)
(11)
(14)
(17)
(20)
(23)
(26)
 14 x  3  2x  25x  1  0
2
(41)
0
 2x  2 3x  3  0
(2a  6)(12b  9)(4c  1)  0
Solve the following equations:
(1)
2( x  2)(2 x 1)  1
(2)
3(2 x 3)(3 x  2)  2  9 (3)
117
2
7 2 x 1  7
PART 2
(Studied in Term 3 according to CAPS)
Solving equations using factorisation
Whenever an equation has the form a . b  0 then the solutions to the equation can be
determined by solving the equations as follows:
If a . b  0 then a  0 or b  0 . This is called the zero-factor rule.
Equations written in standard form: expression  0 can be solved by factorising and applying
the principle above.
Example 7
Solve the following equations:
x2  4 x
(b)
(a)
2
(d)
x  2x  8  0
(e)
(g)
x2  4
x( x  5)  6
(c)
(f)
3 x3  27 x
2 x 2  14 x  24  0
4 x 2 ( x  3)  9(3  x)  0
[Note: (f) and (g) are challenges]
Solutions
(a)
(b)
(c)
(d)
(e)
x2  4 x
x2  4 x  0
 x( x  4)  0
 x  0 or x  4  0
 x  0 or x  4
[write in standard form]
[factorise]
[zero-factor rule]
[solve]
x2  4
x2  4  0
 ( x  2)( x  2)  0
 x  2  0 or x  2  0
 x  2 or x  2
[write in standard form]
[factorise]
[zero-factor rule]
[solve]
3 x3  27 x
 x3  9 x
 x3  9 x  0
[divide both sides by 3 to simplify]
[write in the form a . b  0 ]
 x( x 2  9)  0
 x( x  3)( x  3)  0
 x  0 or x  3  0 or x  3  0
 x  0 or x  3 or x  3
[take out HCF]
[factorise binomial]
[zero-factor rule]
[solve]
x2  2 x  8  0
 ( x  4)( x  2)  0
 x  4  0 or x  2  0
 x  4 or x  2
[factorise]
[zero-factor rule]
[solve]
x( x  5)  6
 x2  5x  6
 x2  5x  6  0
 ( x  6)( x  1)  0
 x  6  0 or x  1  0
 x  6 or x  1
[expand]
[standard form]
[factorise]
[zero-factor rule]
[solve]
118
(f)
(g)
2 x 2  14 x  24  0
 0  2 x 2  14 x  24
0 2 x 2 14 x 24
 


2
2
2
2
2
 0  x  7 x  12
 0  ( x  4)( x  3)
 x  4  0 or x  3  0
 x  4 or x  3
[take terms to the right]
[divide all terms by 2]
[factorise]
[zero-factor rule]
4 x 2 ( x  3)  9(3  x)  0
 4 x 2 ( x  3)  9( x  3)  0
[sign change]
 ( x  3)(4 x 2  9)  0
 ( x  3)(2 x  3)(2 x  3)  0
 x  3  0 or 2x  3  0 or 2x  3  0
 x  3 or 2x  3 or 2x  3
3
3
 x  3 or x   or x 
2
2
1
1
 x  3 or x  1 or x  1
2
2
[take out common bracket]
[factorise]
[zero-factor rule]
[solve]
EXERCISE 7
(a)
Solve the following equations:
x2  1
(2)
(1)
2
x  36
(5)
(4)
2
x x
(8)
(7)
2
x  36 x
(11)
(10)
2
x  100 x  0
(14)
(13)
(16)
(19)
(22)
(25)
(28)
(31)
(34)
(37)
(40)
(43)
(b)
2 x 2  32 x
4 x2  9
x 2  7 x  12  0
x 2  4 x  12  0
a 2  12a  35  0
m 2  5m  6  0
x 2  11x  28
0  x 2  13 x  22
3 x 2  3 x  18
x3  2 x 2  8 x  0
(17)
(20)
(23)
(26)
(29)
(32)
(35)
(38)
(41)
(44)
(46) ( x  3) 2  9
(47)
Solve the following equations:
(1)
x( x  4)  2( x  4)  0
x2  9
x 2  49
x2  4 x
x 2  64 x  0
x 2  100  0
(3)
(6)
(9)
(12)
(15)
x 2  25
x 2  64
x 2  25 x
 x 2  49
2 x 2  32
3x 2  27
4 x2  9 x
x 2  7 x  12  0
a 2  9a  20  0
a 2  2a  48  0
m 2  5m  6  0
x2  6 x  9  0
 x 2  12 x  28  0
4a 2  12a  40
( x  2)( x  3)  14
(18)
(21)
(24)
(27)
(30)
(33)
(36)
(39)
(42)
(45)
3x 2  9  9
2 x3  8 x
x 2  4 x  12  0
a 2  11a  12  0
m 2  5m  6  0
x 2  5 x  84
0  x 2  8 x  16
 x 2  10 x  16  0
6a 2  24a  30
( x  6)( x  4)  24
( x  4) 2  25
(48) 2( x  2) 2  72  0
[These are challenges]
(2)
m(2m  4)  7(2m  4)  0
(3)
y (3 y  2)  8(3 y  2)  0
(4)
p 2 (3 p  4)  4(3 p  4)  0
(5)
(7)
n 2 (n  7)  9n(7  n)  0
x( x  5)  2(5  x)  0
(6)
(8)
n 2 (n  7)  9n(7  n)  0
m(2m  8)  2(8  2m)  0
(9)
x 2 (4 x  1)  16(1  4 x)  0
(10)
5 x 2 (3x  4)  50(4  3 x)  0
(11)
8 x 2 (2 x  9)  18 x(2 x  9)  0
(12)
x 2 (5 x  4)  x(5 x  4)  2(5 x  4)  0
119
Setting up equations to describe given situations
In Grade 8, you were required to set up an equation by translating words into mathematical
statements. The Grade 8 textbook contains important words and concepts for this topic. It
might be worthwhile for you to refer back to the Grade 8 textbook (page 103) for this
information.
Example 8
The sum of three consecutive even numbers is equal to 42. Determine the three numbers.
Solution
Let the numbers be 2 x ; 2 x  2 and 2 x  4
 2 x  (2 x  2)  (2 x  4)  42
 6 x  6  42
 6 x  36
x  6
The three consecutive even numbers are:
2 x  2(6)  12
2 x  2  2(6)  2  14
2 x  4  2(6)  4  16
Check: 12  14  16  42 which is true.
Example 9
A father is 16 years older than his son. Six years ago he was three times as old as his son.
Find both their ages now.
Solution
Let the son be x years old now. The father is therefore ( x  16) years old now.
6 years ago the son was ( x  6) years old and the father was ( x  10) years old.
Let’s summarise this information in a table.
Father
Son
x
x  16
Age now:
x  10
x6
Age 6 years ago
But since the father was three times older than the son six years ago:
x  10  3( x  6)
 x  10  3x  18
 28  2 x
 x  14
The son is 14 years old now and the father is 30 years old.
Example 10
A man travels in his car from home, realises that he forgot to bring his laptop, makes a u-turn
at a traffic light and drives back home. His average speed from home to the traffic light is 60
km/h and his average speed back home is 100 km/h. The total time taken for him to travel
from home to the traffic light and back home again is 30 minutes. Calculate the distance
between his home and the traffic light.
In this example we will make use of the following formulae relating to speed, distance and
time:
Distance
Distance
Distance  Speed  Time
Time 
Speed 
Speed
Time
120
Solution
Let x be the distance between his home and the traffic light.
Distance
x

The time taken to the traffic light is: Time 
Speed
60 km/h
Distance
x

The time taken to return home is: Time 
Speed
100 km/h
x
x
30
The total time taken to the traffic light and back home is:
h


60 km/h 100 km/h 60
Let’s now solve this equation:
x
x
30
  300 
 300   300
[ignore units and multiply by the LCD]
60
100
60
 5 x  3 x  150
 8 x  150
 x  150  18, 75
8
The distance between his home and the traffic light is 18,75 km.
Example 11
A rectangle has an area of 8 metres. Its breadth is two
metres less than its length. Find the dimensions of the
rectangle.
x2
Solution
Area  length  breadth
 Area  x( x  2)
But the area is 8 metres
 x( x  2)  8
(equation relating to area)
 x2  2 x  8
 x2  2 x  8  0
 ( x  4)( x  2)  0
x  4 or x  2
But the length of a side can never be negative.
 x  2
x  4 is the solution to the problem.
The length is 4 metres and the breadth is 2 metres.
EXERCISE 8
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
Five times a certain number increased by six is equal to 26. Find the number.
Seven times a certain number decreased by eight is equal to 20. Find the number.
When twice a certain number is subtracted from 15, the result is 11. Find the number.
If you treble a certain number and subtract 8 you get the same answer as when you
add 10 to the number. Determine the number.
If the result of the sum of a certain number and 3 is multiplied by 5, then the answer is
the same as multiplying the number by 8. Determine the number.
The sum of three consecutive even natural numbers is 60. Determine the numbers.
The sum of three consecutive odd natural numbers is 123. Determine the numbers.
The sum of three numbers is 123. The second number is five times the first number
and the third number is 2 more than the second number. Determine the numbers.
A school sold all but 5 of their old computers for R1 200 each. The school received
R18 000 from the sale. How many computers did the school originally have?
121
(j)
(k)
(l)
(m)
(n)
(o)
(p)
(q)
(r)
(s)
(t)
(u)
(v)
(w)
A mother is now nine times as old as her son. In four years’ time she will be five
times as old as her son. What are their present ages?
A father is twice as old as his son. Twelve years ago, the father was three times the
son’s age. How old is the son?
A mother is thirty two years older than her daughter. Ten years ago she was three
times as old as her daughter was then. Determine the present age of both of them.
Mark has three times as many marbles as Sipho. If he gives Sipho 17 marbles he will
have twice as many marbles as Sipho. How many did Sipho have to start with?
A car travels at an average speed of 120 km/h and covers a distance of 400 km. How
many hours has the car travelled?
A car travels for 7 hours and 50 minutes at an average speed of 130 km/h. How far
has the car travelled (in km)?
What is a car’s average speed in km/h if it travels for 80 minutes and covers a
distance of 120 km?
A car completes a trip in 20 minutes. For the first half of the trip the speed of the car
was 100 km/h and the speed for the second half of the trip was 90 km/h. How far did
the car travel (in km) if the distance for the first part is the same at the second part?
The length of a rectangle is 3 metres longer than the breadth.
(1)
If the perimeter is 26 metres, what is the length and breadth?
(2)
If the area is 70 m 2 , what is the length and breadth?
The length of a rectangle is twice its breadth. If the area is 18 m 2 , what is the length?
The length of a rectangle is ( x  3) metres and the breadth is ( x  2) metres. If the
area is 66 m 2 , calculate the value of x.
The area of a rectangle exceeds the area of a square by 2 cm 2 . The length of the
rectangle is ( x  2) and its breadth is ( x  1). The square has a side equal to x.
Calculate the dimensions of the rectangle.
In a right-angled triangle, the length of the hypotenuse in terms of x is ( x  3) . The
other two sides are 8 and ( x  1) . Calculate the numerical value of the perimeter of the
triangle.
In a kite, the one diagonal is 2 cm more than the length of the other diagonal. The area
of the kite is equal to the area of a rectangle with dimensions 6 cm by 4cm. Calculate
the lengths of the diagonals of the kite. (See page 203 for the formula for the area of a
kite)
Substitution in equations and formulae
Example 12
(a)
(b)
If y  x 2  2 x , calculate:
(1)
the value of y if x  1
(2)
the value of x if y  8
The formula for calculating the surface area of a cylinder is S  2πr 2  2πrh .
(1)
State the variables and constants in the formula.
(2)
Calculate the surface area if r  3 cm , h  10 cm and π  3,141 .
(3)
Calculate the value of r if V  (48π) cm3 and h  5 cm .
Solutions
(a)
(1)
x  1
(2)
2
y 8
 y  (1)  2(1)
8  x2  2 x
 y  1 2
y 3
 0  x2  2 x  8
 0  ( x  4)( x  2)
 x  4 or x  2
122
(b)
(2)
π is the constant in the formula since π  3,1415......
The variables are r, h and S.
S  2(3,141)(3) 2  2(3,141)(3)(10)  245 cm3 (to the nearest whole number)
(3)
48π  2πr 2  2πr (5)
(1)
 48π  2πr 2  10πr
48π 2πr 2 10r



2
2
2
2
 24  r  5r
 0  r 2  5r  24
 0  (r  8)(r  3)
 r  8 or r  3
But r  8
r  3
EXERCISE 9
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
If y  2 x  3 , calculate the value of:
y if x  6
(2)
x if y  9
(1)
If y  3 x  6 , calculate the value of:
y if x  2
(2)
x if y  12
(1)
(3)
x if y  11
(3)
x if y  15
If y  x 2 , calculate the value of:
y if x  8
(2)
(1)
x if y  100
(3)
x if y  36
If y  2 x 2 , calculate the value of:
y if x  3
(2)
(1)
x if y  2
(3)
x if y  50
If y  x 2  2 , calculate the value of:
y if x  4
(2)
x if y  14
(1)
(3)
x if y  34
If y  4  x 2 , calculate the value of:
y if x  2
(2)
x if y  12
(1)
(3)
x if y  77
(3)
x if y  60
(3)
x if y  28
2
If y  x  4 x , calculate the value of:
y if x  6
(2)
x if y  5
(1)
2
If y  ( x  1)  2 , calculate the value of:
y if x  3
(2)
x if y  6
(1)
The formula for calculating the surface area of a cylinder is S  2πr 2  2πrh .
(1)
State the variables and constants in the formula.
(2)
Calculate the surface area if r  7 cm , h  12 cm and π  3,141 .
(3)
Calculate the value of r if S  (32π) cm 2 and h  6 cm .
(4)
Calculate the value of r if S  (300π) cm 2 and h  5 cm .
(5)
Calculate the value of r if S  (320π) cm 2 and h  12 cm .
In trapezium ABCD, AD||BC and the length of BC is three times the length of AD.
The height is equal to the length of AD. Let x be the length of AD. If the area of
ABCD is 200 cm 2 , calculate the length of AD. (See page 201 for area formula)
A flea jumps up from the ground, reaches a maximum height and then falls back to
the ground. The equation of its movement is given by the formula D  20t  t 2 where
t is the time in seconds and D is the distance of the flea above the ground in
centimetres.
(1)
How far above the ground is the flea after 2 seconds?
123
(2)
After how many seconds will the flea be 36 cm above the ground?
Explain why there are two valid values for t.
(3)
After how many seconds will the flea be back on the ground?
A dolphin at the surface of the sea dives under water and then swims back to the
surface. The equation of its movement is given by D  3t  t 2 where t is the time in
seconds and D is the depth of the dolphin under water in metres.
(1)
How far below the surface will the dolphin be after 1 second?
(2)
After how many seconds will the dolphin be at a depth of 2 m?
(3)
After how many seconds will the dolphin be back at the surface?
(l)
Using substitution to generate tables of ordered pairs
In this section we will use substitution in equations to generate tables of ordered pairs. An
ordered pair of numbers is ( x ; y ) where x represents the input value in an equation and y the
corresponding output value. This was explored in detail in Chapter 7.
Example 13
The equation y  x 2  3 is given. Complete the following table and then write down the
ordered pairs.
x
y
1
0
1
6
Solution
For x  1
y  (1) 2  3  1  3  2
For x  0
y  (0) 2  3  0  3  3
For x  1
y  (1) 2  3  1  3  2
For y  6
6  x2  3
 0  x2  3  6
 0  x2  9
 0  ( x  3)( x  3)
 x  3 or x  3
The table can now be completed. Notice that for y  6 , there are two x-values that are not in
the table and need to be included.
x
y
3
6
1
2
0
3
1
2
3
6
The set of ordered pairs are:
(3 ; 6) ; (  1;  2) ; (0 ;  3) ; (1;  2) ; (3 ; 6)
In Chapter 15 you will use ordered pairs to represent equations graphically.
EXERCISE 10
For each of the following equations, complete the table and write down the ordered pairs.
(a) y  3x  6
(b) y  4 x  8
0
5
2
x
x
2
1
9
8
12
3
y
y 16
2
2
(c) y  x
(d) y   x
3
0
2
0
2
x
x 3
9
4
9
y
y
4
2
2
(e) y  2 x
(f) y  2 x
0
2
0
2
x
x 2
2
18
2
18
y
y
2
124
(g) y  x 2  1
3
x
y
(i) y  2 x 2  3
x
1
y
(k) y  ( x  2) 2
x
2
y
y  x2  6
x
1
y 2
y   x2  5
x
1
1
y
y  x( x  1)
x 3
0
y
(h)
0
6
2
10
(j)
0
3
3
11
(l)
0
4
9
0
0
4
3
0
4
11
0
2
20
REVISION EXERCISE
(a)
Solve the following equations:
(1)
3x  4  0
(2)
2
(3x  4)( x  2)  3x (5)
3x  4 x  2

1
(8)
(7)
3
2
Solve the following equations:
2x  2
(2)
(1)
2x  2  2
(5)
(4)
2
2x  2x  0
(8)
(7)
(4)
(b)
(c)
3 x  4( x  2)  0
(3)
(3 x  4)( x  2)  0
(3 x  4)( x  2)  8
3x  4 2 x  2
 
x
x
x
(6)
1 (3 x  4)  1 ( x  2)
4
2
(9)
3x  4 2 x  2
 
x
x
x
2 x2  2
2 x2  2  2
2 x2  2 x  4
(3)
(6)
(9)
2 x 2  32
2 x 2  2  74
(2 x  1) 2  1
(10)
(2 x  1) 2  0
(11)
2 x(2 x  1)  x 2
(12)
x 2  2 x  35  0
(13)
(16)
0  x 2  15 x  36
( x  5)( x  4)  70
(14)
(17)
 x 2  4 x  32  0
9  ( x  4) 2
(15)
(18)
0  2 x 2  2 x  84
1 (2 x  4) 2  32
2
Solve the following equations:
(1)
(2 x  1)(2 x  1)  3x 2
(3)
(3 x  4)( x  2)  2 x( x  5)  9
(2 x  1) 2  3x( x  1)  2 x  5
(3 x  4) 2  ( x  2) 2  8( x  1)( x  1)
(2)
(4)
4(2 x  7)  3x 2  ( x  4)( x  2)
(6)
x 2 ( x  3)  81(3  x)  0
 2 x 3  x  1 x  2 

(7)
  
0
3 
 3 4  2
Solve the following equations by using inspection.
(5)
(d)
(1)
(3)
(f)
(2)
x x  27
(3)
x2  2 x
(5)
2 x  2 x  24
x 1 x 1 1


Consider the expression A 
3
5
15
(1)
Write A as a single fraction by simplifying.
(2)
Hence determine the value of A if x  10 1 .
(4)
(e)
x3  27
Solve the equation 15A  x( x  6)  12 .
(6)
x 1  7
(2) x  256
2
Jason argues that the equations 2 x 2  8 x and 2 x 2  8 only each have one solution.
His reasoning is as follows:
2 x2  8
2 x2  8x
2 x2 8x


2x 2x
x  4
 x2  4
x  4
x  2
Explain why Jason is incorrect.
125
SOME CHALLENGES
(a)
(b)
(c)
(d)
(e)
(f)
( x  2) , x and (x  2) are three natural numbers. If the product of the three numbers is
four times their sum, determine the numbers.
The length of a rectangle exceeds the width by 2 cm. If the diagonal is 10cm long,
determine the width.
The perimeter of a rectangle is 14 cm. If the diagonal is 5cm, find the width (x).
A man travels 180 km at a speed of x km/h. If he doubles his speed then the time
taken for him to travel 180 km is one hour less. Calculate his original speed.
A rectangular pool with dimensions 8 m by 6 m,
is surrounded by a cement pathway. The area of
2x
2x
the pool is equal to the area of the cement pathway.
Calculate the value of x and hence the outer perimeter
2x
2x
of the cement pathway.
The solutions of an equation of the form ax 2  bx  c  0 can be found using the
following formulae:
(g)
b  b 2  4ac
b  b 2  4ac
or x 
.
x
2a
2a
(1)
Solve the equation x 2  5 x  6  0 using the zero-factor rule.
(2)
Now use the given formulae to determine the solutions of the equation
x2  5x  6  0 .
(3)
Use these formulae to determine the solutions of the equation 2 x 2  x  1  0 .
Show that the two solutions obtained using these formulae are in fact solutions
of the equation.
A girl throws a ball upwards from the roof of a building (A).
It reaches a maximum height (B) above the ground and then
falls to the ground (C). The equation of the movement of the
ball is given by d  t 2  4t  12 where t is the time in
seconds and d is the height above the ground in metres.
(1)
What does the constant 12 in the equation tell us?
(2)
How high is the roof above the ground?
(3)
The ball reaches a maximum height of 16 m above
the ground. After how many seconds did this happen?
(4)
After how many seconds did the ball reach the ground?
(h)
It is given that x 
(1)
(2)
(3)
(4)
6
 1.
x
Show that x 2  x  6
Show that x3  7 x  6
Show that x 6  49 x 2  84 x  36
Determine the numerical values of x.
126
CHAPTER 10: SHAPE AND SPACE (GEOMETRY)
TOPIC: CONSTRUCTIONS
In this chapter, we will revise the constructions you studied in Grade 8 and take the topic a
little further. Remember that the word “construction” means to draw angles or lines
accurately. We use only a ruler, compass and pencil for constructions. Here are the diagrams
from the Grade 8 textbook that show you the tools for constructing lines, angles triangles and
quadrilaterals.
pencil
protractor
compass
radius
ruler
arc
In Grade 8, you learnt how to:

construct parallel and perpendicular lines

construct angle bisectors

construct angles of 30, 45 and 60 and multiples thereof

construct scalene, isosceles, equilateral and right-angled triangles

construct parallelograms, rhombuses, rectangles, squares and kites
In Grade 9, the focus will be on:

bisecting the angles of a triangle

the sum of the interior angles of a triangle

the relationship between the exterior angle of a triangle and the interior opposite angles

the minimum conditions for two triangles to be congruent

the construction of other polygons (hexagons, octagons)
Example 1
(Revision of the construction of an angle bisector)
D
ˆ without using a protractor.
Bisect acute angle DEF
Solution
Step 1
ˆ with your ruler.
Draw any acute angle DEF
Step 2
Place your compass on point E and draw an arc
that intersects (cuts) ED and EF. Name the points
of intersection P and Q respectively.
Step 3
Place your compass on P and Q and draw arcs
(same radii) that intersect at R.
Step 4
ˆ .
Join ER. Line ER will bisect DEF
ˆ  REQ
ˆ .
We say that PER
F
E
D
P
Q
E
D
F
P
R
E
127
Q
F
Example 2
(Revision of constructing angles)
Construct an angle of 30 .
Solution
Step 1
Draw line BC of any length. Place your compass
at B and draw an arc that intersects BC at Q.
B
Step 2
Place your compass on Q and draw an arc to
intersect the first arc at P. Keep the radius the
same.
C
Q
C
P
B
Step 3
Place your compass on P and draw an arc to
intersect the second arc at A. Keep the radius
the same.
A
P
Step 4
Join B and A.
ˆ  30
ABC
Example 3
Q
30
C
Q
B
(Revision of constructing angles)
Construct an angle of 45 .
Solution
Step 1
Draw line AB any length and mark off point
E between A and B.
Step 2
Place your compass on E and draw an arc close
to A and also close to B. The distance from E
to the arcs (radii) must be the same.
A
A
Step 3
Place your compass on the arc between A and E
and draw an arc above AB and do the same for
the arc between E and B.
A
128
E
B
.
B
E
.
E
B
Step 4
Draw a line from D, the point where the two arcs
intersect, to E on line AB. The two lines DE and
AB are perpendicular. We say that DE  AB at E.
D
.
E
A
.
Step 5
Place your compass on point D and draw an arc
that intersects ED and DB. Name the points of
intersection P and Q respectively.
E
P
B
.
Step 6
Place your compass on P and Q and draw arcs
(same radii) that intersect at F.
E
P
F
.
.
Q
D
A
B
E
P
45
45
D
A
Example 4
Q
D
A
Step 7
ˆ .
Join DF. Line DF will bisect EDB
ˆ  FDQ
ˆ  45 .
EDF
B
F
Q
B
(Revision of constructing angles)
Construct an angle of 60 .
Solution
Step 1
Draw line EF of any length. Place your
compass on E and draw a long arc to intersect
EF at G.
Step 2
Place your compass on G and draw a second arc
(same radius) to intersect the first arc at D.
129
G
E
F
D
E
G
F
Step 3
Join ED.
ˆ  60
DEF
D
60
Example 5
F
G
E
(Revision of constructing parallel lines)
H
In the given drawing, construct line CD through F parallel to line AB.
Solution
G
A
Step 1
Redraw the sketch using a ruler.
The lines may be any length.
B
.
H
F
Step 2
Place your compass on G. Draw an arc to
intersect GB at Q and GH at P.
P
G
A
Step 3
Draw the same arc by placing your compass
on F. The arc cuts FG at R.
Step 4
Place your compass on P, make the radius equal
to the length of PQ and draw an arc. Do the same
at R.
Q
R
.
F
B
H
P
G
A
Step 5
Draw a line through F and the point of intersection
of the two arcs. This line will be parallel to line AB.
Q
R
.
B
F
H
P
G
A
R
Q
F
Bisecting the angles of a triangle and angle relationships in a triangle
In the next example, you will construct angle bisectors for the angles of a triangle.
Angle bisectors intersect at the same point. We say that these lines are concurrent.
Example 6
(a)
(b)
(c)
(d)
(e)
Construct scalene DEF with DE  7 cm , DF  5 cm and EF  8 cm .
Measure D̂ , Ê and F̂ using a protractor.
ˆ  Eˆ  Fˆ . What do you notice?
Calculate D
ˆ D
ˆ and Fˆ . What do you notice?
Extend DE to H and then measure FEH,
Bisect the angles of DEF. What do you notice about the angle bisectors?
130
B
Solutions
(a)
Step 1
Draw a line more than 7 cm in length and then
draw DE  7 cm on this line as shown.
F
Step 2
Set your compass to 5 cm, place it on D and
draw an arc.
Step 3
Set your compass to 8 cm, place it on E and
draw an arc to intersect the first arc at F.
Join DF and FE.
(b)
ˆ  82, Eˆ  38 and Fˆ  60
D
(c)
ˆ  Eˆ  Fˆ  82  38  60  180
D
(d)
ˆ  142
FEH
D̂  82
F̂  60
ˆ  Fˆ  82  60  142
D
ˆ D
ˆ  Fˆ
FEH
(e)
8 cm
5 cm
7 cm
E
D
H
F
60
82
38
142
E
D
H
F
Construct angle bisectors for D̂ , Ê and F̂
using the method in Example 1.
Notice that the angle bisectors intersect at
the same point G.
G
D
E
Conclusion

The sum of the angles of a triangle is equal to 180 .

The exterior angle of triangle is equal to the sum of the interior opposite angles.
EXERCISE 1
(a)
(b)
(1)
Construct scalene ABC with AB  9,8 cm , AC  6 cm and BC  9 cm.
(2)
Measure  , B̂ and Ĉ using a protractor.
ˆ B
ˆ . What do you notice?
ˆ C
(3)
Calculate A
ˆ and A
ˆ . What do you notice?
ˆ C
(4)
Extend AB to D and then measure CBD,
(5)
Bisect the angles of ABC. What do you notice about the angle bisectors?
In PQR , PQ  12 cm , PR  9 cm and QR  15 cm .
(1)
Construct PQR .
(2)
Using your protractor, measure the angles of PQR .
ˆ  Rˆ . What do you notice?
(3)
Calculate Pˆ  Q
(4)
(5)
ˆ and the interior opposite angles. What do you
Extend PQ to S. Measure PQS
notice?
Bisect the angles of PQR. What do you notice about the angle bisectors?
131
(c)
(d)
(e)
(f)
(g)
(1)
Construct isosceles ABC with AB  7 cm , AC  9 cm and BC  9 cm.
(2)
Measure  , B̂ and Ĉ using a protractor.
ˆ B
ˆ . What do you notice?
ˆ C
(3)
Calculate A
ˆ and A
ˆ . What do you notice?
ˆ C
(4)
Extend AB to D and then measure CBD,
(5)
Bisect the angles of ABC. What do you notice about the angle bisectors?
(1)
Construct equilateral ABC with AB  9 cm , AC  9 cm and BC  9 cm.
(2)
Measure  , B̂ and Ĉ using a protractor.
ˆ B
ˆ . What do you notice?
ˆ C
(3)
Calculate A
ˆ and A
ˆ . What do you notice?
ˆ C
(4)
Extend AB to D and then measure CBD,
(5)
Bisect the angles of ABC. What do you notice about the angle bisectors?
In DEF , DE  15 cm , D̂  15 and Ê  60 .
(1)
Construct DEF .
(2)
Using your protractor, measure the size of F̂ .
ˆ  Eˆ  Fˆ . What do you notice?
(3)
Calculate D
ˆ . What do you notice?
(4)
Extend DE to G. Measure FEG
ˆ  75 by first constructing 45 and then 30 .
(1)
Construct ABC
ˆ .
(2)
Construct the bisector of ABC
ˆ with a protractor.
(3)
Measure ABC
ˆ  105 by first constructing 60 and then 45 .
(1)
Construct PQR
ˆ .
(2)
Construct the bisector of PQR
(3)
(h)
(1)
(2)
(3)
ˆ with a protractor.
Measure PQR
ˆ  135 by using 90 and 45 .
Construct LMN
ˆ  135 by using only 45 angles.
Construct LMN
ˆ .
Now construct the bisector of LMN
We can use constructions to prove that the sum of the angles of a triangle add up to 180 .
Example 7
Use constructions to prove that the sum of the angles of a triangle is equal to 180 .
F
Solution
Step 1
Refer to the triangle in Example 6.
Place your compass on D and draw an arc to
cut DF and DE at A and B respectively.
Let the radius of the arc be 2 cm.
8 cm
5 cm A
D 2 cm B
F
Step 2
Now place your compass on E and draw an arc
to cut DE and EF at M and N respectively.
Keep the radius of the arc the same as the arc
in Step 1.
132
A
D 2 cm B
E
7 cm
N
M 2 cm E
Step 3
Now place your compass on F and draw an arc
to cut FD and FE at P and Q respectively.
Keep the radius of the arc the same as the arc
in Step 1.
F2
cm
P
A
Q
N
D 2 cm B
Step 4
Now draw any line and place your compass on a middle
point on the line (X). Draw an arc with the same radius
as the arc in Step 1, namely, 2 cm.
Step 5
Go back to the triangle. Place your compass on A and
draw an arc through B.
M 2 cm E
2 cm X
F
Q
P
A
N
D 2 cm B
E
M
Step 6
Go to the line drawn. Place your compass on the point where
the arc cuts the line on the left (A). Draw an arc with radius
the same length as AB in the triangle to cut the long arc at B.
B
.
A 2 cm X
Step 7
Go back to the triangle. Place your compass on M and
draw an arc through N.
F
Q
P
A
N
D 2 cm B
Step 8
Go to the line drawn. Place your compass on the point
M (where B was). Draw an arc with radius the length
of MN to cut the long arc at N.
Step 9
Go back to the triangle. Place your compass on P and
draw an arc through Q.
M
A
F
P
A
N
X
Q
D 2 cm B
133
E
M
N
M
E
Step 10
Go to the line drawn. Place your compass on the point
P (where N was). Draw an arc with radius the length of
PQ to cut the long arc at Q. Notice that this arc also cuts
the line at Q.
P
.
A
Q
X
B
Step 11
Draw radii BX and XN.
A
N
Q
X
Step 12
Now use your protractor to measure the three angles
on the line. Add them up and you get 180 .
.
X
D̂
A
Ê
F̂
Q
We can use constructions to prove that the angle of triangle is equal to the sum of the interior
opposite angles.
Example 8
Use constructions to prove that the exterior angle of triangle is equal to the sum of the
interior opposite angles.
F
Solution
Step 1
Refer to the triangle in Example 6.
Extend DE to G.
Q
P
A
D 2 cm B
E
G
F
Step 2
Place your compass on E and draw an arc
of radius 2 cm to cut the line DEG at V
and W. Suppose that this arc cut FE at T.
P
A
Q
T
D 2 cm B
F
Step 3
Place your compass on T. Draw an arc
with a radius the length of AB to cut
the long arc at M.
P
A
W
G
W
G
Q
D 2 cm B
134
V 2 cm E
T
V 2 cm E
M
Step 4
Place your compass on M. Draw an arc
with a radius of length PQ to cut the long
arc at W.
Notice that this arc also cuts the line at W.
F
Q
P
A
D 2 cm B
F
Step 5
Draw radius EM.
M
T
V 2 cm E
W
T
M
G
Q
P
A
V 2 cm E
D 2 cm B
F
Step 6
Now use your protractor to measure the angles
ˆ , MEW
ˆ and TEW
ˆ .
TEM
P
ˆ  MEW
ˆ  TEW
ˆ
Notice that TEM
A
However, by construction, it is clear that
ˆ  TEM
ˆ and Fˆ  MEW.
ˆ
D
ˆ  Fˆ  TEW
ˆ
D
W
G
W
G
Q
T
V 2 cm E
D 2 cm B
M
Congruency of triangles
Two triangles are congruent if they are identical in size and shape. This means that their
corresponding sides and angles of the two congruent triangles are equal.
Whenever two triangles are congruent, we can state the following: ABC  DEF
The symbol “  ” means “is congruent to”.
In the investigation that follows, we will focus on the minimum conditions for two triangles
to be congruent.
Investigation on congruent triangles
Activity 1
The first case of congruency (SSS):
If the corresponding sides of two triangles are equal, then the triangles are congruent.
(a)
Construct ABC and DEF exactly as they appear below. The corresponding sides
of the triangles are equal.
A
5 cm
B
(b)
(c)
7 cm
F
E
8 cm
5 cm
8 cm
7 cm
C
D
Now use your protractor to measure the angles of both triangles.
What can you conclude about the corresponding angles of the triangles?
What can you conclude about ABC and DEF ? Give a reason.
135
Activity 2
The second case of congruency (SAA):
If, in two triangles, one pair of corresponding sides are equal and two pairs of corresponding
angles are equal, then the triangles are congruent.
(a)
Construct ABC and DEF exactly as they appear below. The corresponding sides
of the triangles are equal.
C
30
A
(b)
(c)
(d)
(e)
E
45
B
8 cm
8 cm
D
30
45
F
Use your protractor to measure Ĉ and F̂ .
Use your ruler to measure the length of AC, EF, BC and DF.
What can you conclude about the corresponding sides and angles of the triangles?
What can you conclude about ABC and DEF ? Give a reason.
Before doing Activity 3, let’s briefly discuss included angles in triangles. An included angle
in a triangle the angles formed by two sides of a triangle. Consider the following triangle.
The angle  is included between sides AC and AB.
The angle B̂ and Ĉ are angles that are not included between sides AC and AB.
C
The angle B̂ is included between sides AB and BC.
The angle  and Ĉ are angles that are not included between sides AB and BC.
The angle Ĉ is included between sides AC and BC.
B
A
The angle  and B̂ are angles that are not included between sides AC and BC.
Activity 3
The third case of congruency (SAS):
If, in two triangles, two pairs of corresponding sides are equal and the corresponding pair of
included angles are equal, then the triangles are congruent.
(a)
Construct ABC and DEF exactly as they appear below. The corresponding sides
of the triangles are equal.
C
E
9 cm
60
7 cm
7 cm
60
A
(b)
(c)
(d)
(e)
D
9 cm
B
F
Use your protractor to measure B̂ , Ĉ , Ê and F̂ .
Use your ruler to measure the length of BC and EF.
What can you conclude about the corresponding sides and angles of the triangles?
What can you conclude about ABC and DEF ? Give a reason.
136
Activity 4
The fourth case of congruency (RHS):
If, in two right-angled triangles, the hypotenuse of each triangle is equal and one pair of
corresponding sides are equal, then the triangles are congruent.
(a)
Construct ABC and DEF exactly as they appear below. The corresponding sides
of the triangles are equal.
6 cm
E
A
F
10 cm
10 cm
B
(b)
(c)
(d)
(e)
6 cm
D
C
Use your protractor to measure  , Ĉ , D̂ and F̂ .
Use your ruler to measure the length of AB and DE.
What can you conclude about the corresponding sides and angles of the triangles?
What can you conclude about ABC and DEF ? Give a reason.
Activity 5
In this activity, we will consider two triangles in which two pairs of corresponding sides are
equal and one pair of corresponding non-included angles are equal.
7 cm
A
(b)
(c)
(d)
(e)
(f)
m
4c
Construct ABC and ABD exactly as
they appear in the diagram on the right.
Note that  is a non-included angle in
both triangles.
4c
m
(a)
B
30
D
C
ˆ .
In ABC , measure the length of AC and the size of Ĉ and ABC
ˆ and ABD
ˆ .
In ABD , measure the length of AD and the size of ADB
What sides and angles are equal in both triangles?
Are the triangles congruent?
What can you conclude if, in two triangles, two pairs of corresponding sides are
equal but the pair of corresponding equal angles are non-included angles?
Conclusion
Two triangles are congruent if the following minimum conditions are met:
Case 1 (SSS)
The corresponding sides of the two triangles are equal.
Case 2 (SAA)
One pair of corresponding sides are equal and two pairs of corresponding angles are equal.
Case 3 (SAS)
Two pairs of corresponding sides are equal and the corresponding pair of included angles
are equal.
Case 4 (RHS)
The hypotenuse of each triangle is equal, one pair of corresponding sides are equal and one
pair of corresponding angles are right angles.
137
Revision of the construction of quadrilaterals
In the following investigation, you will revise the construction of quadrilaterals and the
properties of quadrilaterals in terms of their sides and angles studied in Grade 8. You will
also investigate the properties of quadrilaterals in terms of their diagonals.
Activity 1
(Parallelograms)
(a)
Construct parallelogram PQRS with Q̂  30 ,
4 cm
QR  6 cm and PQ  4 cm.
(b)
Measure all the other angles and sides.
What can you conclude?
(c)
On your constructed parallelogram, draw
diagonals PR and QS and let them intersect
at T.
(d)
Measure the length of PT, TR, QT and
TS. What can you conclude?
Activity 2
P
S
30
Q
R
6 cm
P
S
T
Q
R
(Rhombuses)
(a)
Construct rhombus ABCD with B̂  60 ,
AB  5 cm and BC  5 cm.
(b)
Measure all the other angles and sides.
What can you conclude?
A
D
5 cm
60
B
(c)
On your constructed rhombus, draw diagonals
AC and BD and let them intersect at E.
(d)
Measure the length of AE, EC, BE and ED.
What can you conclude?
(e)
Measure Eˆ 1 , Eˆ 2 , Eˆ 3 and Ê 4 .
What can you conclude?
Activity 3
A
(a)
Construct rectangle ABCD with B̂  90 ,
AB  5 cm and BC  7 cm.
(b)
Measure all the other angles and sides.
What can you conclude?
C
D
A
5 cm
B
138
D
2 14
E
3
B
(Rectangles)
C
5 cm
7 cm
C
(c)
On your constructed rectangle, draw diagonals
AC and BD and let them intersect at E.
(d)
Measure the length of AC, BD, AE, EC, BE
and ED.
What can you conclude?
D
A
E
B
Activity 4
(Squares)
C
P
(a)
Construct square PQRS with Q̂  90 and all
sides equal to 6 cm.
(b)
Measure all the other angles and sides.
What can you conclude?
Q
(c)
On your constructed square, draw
diagonals PR and QS and let them intersect
at T.
(d)
Measure the length of PR, QS, PT, TR, QT and
TS. What can you conclude?
(e)
S
R
6 cm
P
S
2
Measure Tˆ1 , Tˆ 2 , Tˆ 3 and T̂4 .
What can you conclude?
1
3
4
T
R
Q
Activity 5
(Kites)
(a)
Construct kite ABCD with B̂  30 ,
BC  6 cm and CD  3 cm.
(b)
Measure the length of AB and AD.
What can you conclude?
(c)
Measure the size of  and Ĉ .
What can you conclude?
(d)
On your constructed kite, draw
diagonals AC and BD and let them
intersect at E.
A
D
B
3 cm
30
6 cm
C
A
E1
2
4 3
(e)
Measure the length of AE, EC, BE
and ED. What can you conclude?
(f)
Measure Eˆ 1 , Eˆ 2 , Eˆ 3 and Ê 4 . What can you conclude?
B
139
C
D
The construction of other polygons
Activity 1
(Construction of a regular hexagon)
Construct a regular hexagon with sides equal to 4 cm.
Solution
Step 1
Draw a circle with radius 4 cm. Keep the same
radius, place your protractor on A and start
drawing arcs of radius 4 cm to cut the circle as
shown.
Step 2
Draw the sides of the hexagon constructed.
All six sides will be equal to 4 cm.
Investigation on hexagons
(a)
(b)
(c)
(d)
(e)
Measure the size of the interior angles of the constructed hexagon in Activity 1.
What can you conclude?
Calculate the sum of the interior angles of the hexagon.
How many triangles can be drawn inside the hexagon by joining one vertex to the
other vertices to form diagonals?
Redraw the hexagon to help you.
Multiply the number of triangles by the sum of the angles of each triangle.
What do you notice?
Using the number of sides in a hexagon, the number of triangles that can be drawn
inside the hexagon by joining the diagonals and the fact that the sum of the angles of
a triangle is equal to 180 , try to formulate a rule that will help you calculate the
sum of the angles of a hexagon.
140
Activity 2
(Construction of a regular octagon)
Construct a regular octagon with sides equal to 4 cm.
Solution
Step 1
Draw a line AB of length 4 cm.
Step 2
Construct perpendicular lines at
A and B.
Step 3
Construct angle bisectors at A
and B and mark off 4 cm on each
angle bisector.
45
45
Step 4
At H and C construct perpendicular
lines to meet the line through A and B.
141
45
45
Step 5
Place your compass on H and C
and draw arcs of radius 4 cm to
cut the perpendicular lines.
Step 6
Place your compass on G and D
and draw arc of radius 4 cm to cut
the perpendiculars drawn at A and B.
Join AB, BC, CD, DE, EF, FG , GH
and HA to form the octagon.
Investigation on octagons
(a)
(b)
(c)
(d)
(e)
(f)
(g)
Measure the size of the interior angles of the constructed octagon in Activity 2.
What can you conclude?
Calculate the sum of the interior angles of the octagon.
How many triangles can be drawn inside the octagon by joining one vertex to the
other vertices to form diagonals?
Redraw the octagon to help you.
Multiply the number of triangles by the sum of the angles of each triangle.
What do you notice?
Using the number of sides in a octagon, the number of triangles that can be drawn
inside the octagon by joining the diagonals and the fact that the sum of the angles of
a triangle is equal to 180 , try to formulate a rule that will help you calculate the
sum of the angles of a octagon.
What rule can be used to determine the sum of the angles of any polygon of n sides?
Calculate the sum of the angles of the following polygons using your rule:
(1)
Triangle
(2)
Parallelogram
(3)
Rectangle
(4)
Square
(5)
Rhombus
(6)
Kite
(7)
Trapezium
(8)
Pentagon
(9)
Hexagon
(10) Octagon
142
CHAPTER 11: SHAPE AND SPACE (GEOMETRY)
TOPIC: GEOMETRY OF 2D SHAPES
CLASSIFYING 2D SHAPES
REVISION OF TRIANGLES, QUADRILATERALS AND LINES
Types of triangles
There are three types of triangles that can be classified according to the size of their largest
angle:
Acute-angled triangles
All three interior angles are smaller
than 90 (acute).
Right-angled triangles
The largest interior angle is equal to 90 .
The other two angles are acute.
75
55
70
35
Obtuse-angled triangles
The largest interior angle is greater
than 90 .
The other two angles are acute.
90
35
From Pythagoras:
AB2  AC2  BC2
AC2  AB2  BC2
BC 2  AB2  AC 2
20
30
130
There are three types of triangles that can be classified according to the number of equal
sides and equal angles:
Scalene triangle
No sides or angles are equal.
75
35
70
Isosceles triangle
Two sides are equal and the angles opposite
the equal sides are equal.
We can say that:
ˆ
AB  AC if B̂  C
40
(sides opp equal  's )
ˆ if AB  AC
B̂  C
(  's opp equal sides)
70
Equilateral triangle
Three sides are equal and the interior
angles are equal to 60 .
60
60
143
60
70
Properties of triangles
Property 2
The exterior angle of a triangle is equal
to the sum of the two interior opposite
angles.
Property 1
The sum of the interior angles of a
triangle is 180 .
For any ABC :
ˆ B
ˆ  180
ˆ C
A
2
ˆC  A
ˆ B
ˆ
1
(sum of the  's of  )
(ext   sum of int opp  's )
Properties of straight lines
Vertically opposite angles
Vertically opposite angles are
equal.
Corresponding angles
If AB||CD, then the corresponding
angles are equal.
Alternate angles
If AB||CD, then the alternate angles
are equal.
Co-interior angles
If AB||CD, then the co-interior angles
add up to 180 , i.e. x  y  180
Complementary angles
ˆ B
ˆ  90
In the diagram, B
1
2
Adjacent supplementary angles
ˆ B
ˆ  180
In the diagram, B
1
2
Angles round a point
In the diagram, a  b  c  360
Bisectors of angles
ˆ since B
ˆ B
ˆ
BD bisects ABC
1
2
144
Summary of the definitions and properties of quadrilaterals (sides and angles)
Quadrilateral
Definition
A parallelogram
is a quadrilateral
with two pairs of
opposite sides
parallel.
A rhombus is a
parallelogram
with equal
adjacent sides.
A rectangle is a
parallelogram
with interior
angles equal to
90 .
A square is a
parallelogram
with equal
adjacent sides
and interior
angles equal to
90 .
A trapezium is a
quadrilateral with
one pair of
opposite sides
parallel.
A kite is a
quadrilateral with
two pairs of
adjacent sides
equal.
145
Angles
The opposite
angles of a
parallelogram
are equal.
The interior
angles add up to
360 .
The opposite
angles of a
rhombus are
equal.
The interior
angles add up to
360 .
The interior
angles of a
rectangle are
equal to 90 .
The interior
angles add up to
360 .
The interior
angles of a
square are equal
to 90 .
The interior
angles add up to
360 .
The interior
angles add up to
360 .
Sides
The opposite
sides of a
parallelogram
are parallel and
equal.
One pair of
opposite angles
are equal.
The interior
angles add up to
360 .
Two pairs of
adjacent sides
are equal.
The opposite
sides of a
rhombus are
parallel and all
sides are equal.
The opposite
sides of a
rectangle are
parallel and
equal.
The opposite
sides of a square
are parallel and
all sides are
equal.
One pair of
opposite sides
are parallel.
The following examples and exercise revise the work covered in Grade 8.
Example 1
In ACF , AF||BE. AC is produced to D.
(a)
Calculate x and hence show that EC  BC .
(b)
Calculate, with reasons, the value of y, p, z and q.
80
4x
8x
Statement
(a) 8 x  4 x  80
 4 x  80
 x  20
ˆ  4(20)  80
 EBC
ˆ  BEC
ˆ
 EBC
 EC  BC
(b) 80  80  y  180
160  y  180
80
100
80
80
y  20
p  80
20 160
80
z  80
q  100
Reason
ext   sum of int opp  's
sides opp equal  's
sum of the  's of 
corr  's equal ; AF||BE
corr  's equal ; AF||BE
co-int  's suppl; AF||BE
(or adjacent 's on a line)
Example 2
ˆ ,A
ˆ ,C
ˆ and
ABCE is a parallelogram with Ĉ1  90 . Calculate, with reasons, the size of A
1
2
2
Ĉ3 .
Statement
ˆ B
ˆ  180
ˆ C
A
1
1
Reason
sum of the  's of 
 Â1  74  90  180
 Â1  164  180
 Â1  180  164
74
 Â1  16
74
16
16
74
 2  90
alt  's equal ; BC||AE
Ĉ2  16
alt  's equal ; AB||EC
Ĉ3  74
corr  's equal ; AB||EC
Ê  74
(or adjacent 's on a line )
alt  's equal ; BC||AE
(or opp  's of a parm)
74
146
Example 3
PQRS is a parallelogram. PQ  QT  PT  TR and PT  6cm.
Calculate, with reasons:
(a)
the lengths of the sides of PQRS.
(b)
the size of R̂ and Ŝ .
(a)
Statement
PQ  6cm
Reason
PQ  PT ; given
SR  6cm
QT  6cm
opp sides of a parm
QT  PT ; given
QT  TR ; given
QR  QT  TR
TR  6cm
 QR  12cm
 PS  12cm
60
(b)
Q̂  60
co-int  's suppl; PQ||SR
opp  's of a parm
R̂  120
120
60
opp sides of a parm
PQT is equilateral
Ŝ  60
EXERCISE 1
(a)
Calculate the size of θ in each case:
(1)
(2)
(3)
73
70
θ
43
64
(4)
x
θ
(5)
θ
(6)
160
52

θ
(b)
Calculate the size of x in each of the following.
(1)
ABCD is a parallelogram
30
(2)
ABCD is a parallelogram
120
x
142
147
x
2x
(3)
ABCD is a parallelogram
(4)
ABCD is a rhombus
80
140  2 y
x
y
(5)
(6)
3x
6 x  15
(c)
(d)
x
50
5y
2y
2 x  12
In ABC , calculate, with reasons:
(1)
the size of x and hence the size
of the interior angles.
(2)
the length of AC.
x
5 x  20
2 x  10
x  10
PQST is a rectangle. T̂1  x , T̂2  50 ,
ˆ  y.
and TRS
x
50
Calculate, with reasons, the size of the
angles x and y.
y
(e)
ACDE is a kite and ABGF is a square.
ˆ C
ˆ  96 and AC  12cm .
AB  BC , C
1
2
Calculate, with reasons:
(1)
the length of AE.
(2)
the length of GF.
(3)
the size of Ĉ1 and Ĉ2 .
(4)
the value of x.
96
x
(f)
In trapezium ABCE, AE||BC with Ê  70 .
(1) Calculate the value of x.
(2) Show that ABCD is a parallelogram.
(3) Show that AB  CE .
2 x  10
4 x  10
Quadrilaterals and their diagonals
A diagonal of a polygon is a line segment that joins two opposite
vertices of the polygon. In the diagram alongside, diagonal AC joins
vertices A and C and diagonal BD joins vertices B and D.
The diagonals intersect at E.
148
5 x  40
3x  20
70
Investigation
Note to educator:
The investigation in Chapter 10 on page 138 is useful for allowing the learners to investigate
the properties of quadrilaterals in terms of their diagonals.
Summary of the properties of quadrilaterals (diagonals, angles and sides)
Quadrilateral
Diagonals
The diagonals of a
parallelogram
bisect each other.
The diagonals of a
rhombus bisect
each other at right
angles. The
diagonals bisect
the vertex angles.
The diagonals of a
rectangle bisect
each other and are
equal in length.
45
45
45
45
45
45
45
45
The diagonals of a
square bisect each
other at right
angles and are
equal in length.
The diagonals
bisect the vertex
angles.
The diagonals of a
trapezium
intersect but don’t
bisect each other.
They lie between
parallel lines and
therefore the
alternate angles
are equal.
The diagonals are
perpendicular and
one diagonal
bisects the other.
One of the
diagonals bisects
the vertex angles.
149
Angles
The opposite
angles of a
parallelogram are
equal. The interior
angles add up to
360 .
The opposite
angles of a
rhombus are
equal. The interior
angles add up to
360 .
The interior angles
of a rectangle are
equal to 90 . The
interior angles add
up to 360 .
Sides
The opposite sides
of a parallelogram
are parallel and
equal.
The interior angles
of a square are
equal to 90 . The
interior angles add
up to 360 .
The opposite sides
of a square are
parallel and all
sides are equal.
The opposite sides
of a rhombus are
parallel and all
sides are equal.
The opposite sides
of a rectangle are
parallel and equal.
The interior angles One pair of
opposite sides are
add up to 360 .
parallel.
One pair of
opposite angles
are equal. The
interior angles add
up to 360 .
Two pairs of
adjacent sides are
equal.
Example 4
Diagonals AC and BD intersect at E. ABCD is a rectangle with AC  50 cm and
AD  40 cm . Â1  65 . Calculate the following:
 2 , B̂1 , B̂2 , Ĉ1 , Ĉ2 , D̂1 , D̂2 , BC, AE, DC and AB.
Statement
Reason
ˆ A
ˆ  90
A
1
2
ˆ  90
 65  A
65
interior  of rectangle
2
 Â 2  25
BE  ED
AE  EC
But AC  BD
 AE  EC  BE  ED
25
65
25
65
65
65
25
25
diags of rectangle bisect
diags of rectangle bisect
diags of rectangle equal
 B̂1  65
 's opp equal sides
 B̂2  25
interior  of rectangle
Ĉ1  25
alt  's equal ; AD||BC
Ĉ2  65
alt  's equal ; AB||DC
D̂1  65
alt  's equal ; AB||DC
D̂ 2  25
alt  's equal ; AD||BC
BC  40 cm
opp sides of parm
given
diags of parm bisect
AC  50 cm
AE  EC
 AE  25 cm
DC2  AC2  AD 2
2
Pythagoras; D̂  90
2
 DC  (50)  (40)
2
 DC2  2500  1600
 DC2  900
 DC  30 cm
length DC is positive
opp sides of parm
 AB  30 cm
EXERCISE 2
(a)
A
ABCD is a trapezium with AD||BC.
AB  AD and BD  BC . Ĉ  80 .
Determine the unknown angles.
d
B
150
e
b
c
D
a
80
C
(b)
P
PQRS is a rhombus with Ŝ2  35 .
Calculate the size of all other interior
angles.
35
1
Q
(c)
1
ABCD is a parallelogram.
Â1  20 , B̂1  90 , B̂2  30 ,
DE  2 cm and AE  3 cm . Calculate:
(1)
the length of AC and BD
(2)
the size of θ
2
S
2
R
θ
20
30
(d)
In rhombus PQRS, PQ  26 cm and QS  48 cm .
Calculate the length of PR.
(e)
ABCD is a kite. The diagonals intersect
at E. BD  30 cm , AD  17 cm and
DC  25 cm . Determine:
(1)
AE
(2)
AC
B̂1 if Â1  20
(3)
(f)
ˆ  54 .
ABCD is a square. AEB
Calculate F̂1 .
(g)
ABCD is a rectangle and
A
2
ˆ
DECF is a rhombus. DEC  60
1
(1)
Calculate the size of:
Ê1 , F̂1 , B̂2 and B̂1
(2)
Calculate the length of AC
1
if EC  4 cm
2
B
151
1
F
D
2 3
1
1
2
1
G
2 3
C
60 1
2
E
CONGRUENT SHAPES
Revision of the congruency of polygons
Example 5
Consider the following polygons.
(a)
Polygon ABCDE is identical in size and shape to polygon FGHIJ.
83
102
83
143
102
105 107
143
105 107
The following observations can be made:
ˆ  Fˆ  83
ˆ  102
ˆ H
ˆ  ˆI  107
ˆ  105
D
Eˆ  Jˆ  143
A
B̂  G
C
AB  FG  70 mm BC  GH  56 mm CD  HI  45 mm
DE  IJ  54 mm
EA  JF  37 mm
The conclusion is as follows:
The corresponding angles and sides of two congruent polygons are equal.
(b)
Triangles ABC, DEF and IHG are identical in size and shape.
30
45
45
30
105
30
105
The following observations can be made:
ˆ D
ˆ  Eˆ  H
ˆ  105
ˆ  ˆI  45
B
A
105
ˆ  Fˆ  G
ˆ  30
C
AB  DE  IH  66 mm
BC  EF  HG  97 mm
AC  DF  IG  130 mm
The conclusion is as follows:
The corresponding angles and sides of congruent triangles are equal.
Some congruency notation with regard to triangles
Whenever two triangles are congruent, we can state the following: ABC  DEF
The symbol “  ” means “is congruent to”.
45
45
105
30
105
152
30
45
Note:
The naming of the triangles in terms of their corresponding equal angles is important when
using congruency notation. Consider the above two triangles.
The corresponding equal angles are:
The corresponding equal sides are:
ˆ D
ˆ  45
AB  DE  66mm
A
ABC  DEF
ˆB  Eˆ  105
BC  EF  97mm ABC  DEF
ˆC  Fˆ  30
AC  DF  130mm
The triangles can be named in different ways. However, the order of the corresponding equal
angles is essential. For example, we can name the triangles as follows:
ABC  DEF or ACB  DFE or BAC  EDF or CAB  FDE
Indicating equal sides and angles in two triangles

Whenever the numerical values of the corresponding sides or angles of two triangles are not
given, the sides or angles can be marked as equal by using different symbols. Consider the
two congruent triangles below. In the two triangles, the angles and sides that are equal are

indicated by means of the symbols: 
|
|| |||
This means that:
ˆ D
ˆ
AB  DE
A

ˆ  Eˆ
ABC  DEC
BC  EC ABC  DEC
B
ˆ C
ˆ
C
AC  DC
1
2

One correct possible naming of the triangles is: BCA  ECD
ˆ C
ˆ and A
ˆ D
ˆ .
ˆ  Eˆ , C
since B
1
2
An incorrect naming of the triangles is: ABC  CDE since
ˆ C
ˆ ; Bˆ  D
ˆ  Eˆ
ˆ and C
A
2
1

Example 6
Show that the following triangles are congruent. Make use of congruency notation.
Match the corresponding equal angles:
P̂  Sˆ  20
ˆ  Tˆ  30
Q
ˆ  130 (sum of the  's of  )
Rˆ  U
P̂
Ŝ
Q̂
T̂
R̂
Û
20
20
30
30
Let’s name the triangles as follows: PQR and STU
The corresponding sides are also equal:
 PQR
PQ  ST  6,8 cm
PR  SU  4,8 cm
QR  TU  3, 7 cm
STU
Therefore we can state that the two triangles are congruent using congruency notation:
PQR  STU
(Notice that the order of angles is correct)
153
The four cases of congruency
In Grade 8, we demonstrated that two triangles are congruent by showing that all the
corresponding angles and sides are equal. In Grade 9, we don’t need to do all of this. There
are four minimal conditions that can be used to prove that triangles are congruent.
Investigation on congruency
Note to educator:
An investigation is available on page 135 of the Chapter 10. This investigation is essential
for the learners to fully understand the four cases of congruency. The summary of the four
cases is provided below.
Summary of the four cases of congruency
Case of
What is given or can be
congruency deduced?
SSS
Three pairs of corresponding
sides are equal.
AB  DE
BC  EF
AC  DF
Conclusion about
the two triangles
ABC  DEF
SAA
Two pairs of corresponding
angles and one pair of
corresponding sides are
equal.
AB  DE
ˆ D
ˆ
A
ˆ  Eˆ
B
ABC  DEF
SAS
Two pairs of corresponding
sides and the pair of
corresponding included
angles are equal.
AC  DF
AB  DE
ˆ D
ˆ
A
ABC  DEF
154
What else can now be
assumed?
The three pairs of
corresponding angles
are equal.
ˆ D
ˆ
A
ˆ  Eˆ
B
ˆ  Fˆ
C
The other two pairs of
corresponding sides and
one pair of
corresponding angles
are equal.
BC  EF
AC  DF
ˆ  Fˆ
C
The other two pairs of
corresponding angles
and the other pair of
corresponding sides are
equal.
ˆ  Eˆ
B
ˆ  Fˆ
C
BC  EF
Case of
What is given or can be
congruency deduced?
RHS
One pair of corresponding
angles are right angles, the
pair of corresponding
hypotenuses are equal as well
as one other pair of
corresponding sides.
ˆ D
ˆ  90
A
BC  EF
AC  DF
Conclusion about
the two triangles
ABC  DEF
What else can now
be assumed?
The other two pairs of
corresponding angles
and the other pair of
corresponding sides
are equal.
AB  DE
ˆ  Eˆ
B
ˆ  Fˆ
C
Example 7
For each pair of triangles, state the case of congruency that can be used to prove that the two
triangles are congruent.
Triangles
Corresponding sides and angles
AB  DE (pair of sides)
BC  EF (pair of sides)
AC  DF (pair of sides)
The case of congruency is:
SSS
34
34
85
34
85
AC  DF (pair of sides)
ˆ D
ˆ (pair of equal angles)
A
ˆ  Eˆ (pair of equal angles)
B
The case of congruency is:
SAA
AC  DF (pair of sides)
AB  DE (pair of sides)
ˆ D
ˆ (pair of included equal
A
angles)
The case of congruency is:
SAS
34
155
ˆ D
ˆ
A
BC  EF
(pair of right angles)
(hypotenuse of each triangle
equal)
AC  DF
(pair of sides)
The case of congruency is:
RHS
EXERCISE 3
For each pair of triangles, state the case of congruency that can be used to prove that the two
triangles are congruent.
(a)
(b)
45
45
(c)
(d)
(e)
*(f)
63
Be careful here!
60
46
46
Hint: Fill in angles of ABC that are
equal to the angles of DEC
156
Proving triangles congruent using the four cases of congruency
Example 8
Prove that the following pairs of triangles are congruent:
Statement
In ABC and DEF :
Reason
(1) AB  DE
ˆ D
ˆ
(2) A
44
44
Â
D̂
B̂
Ê
Ĉ
F̂
given
given
given
SAS
(3) AC  DF
ABC  DEF
ABC  DEF
Example 9
Consider ABC and DEF .
(a) Which corresponding angles are equal?
(b) Which corresponding sides are equal?
(c) Prove that ABC  DEF .
(d) Show that AC  DF and AB  DE .
(e) State two reasons which can be used
ˆ D
ˆ .
to prove that A
(a)
(b)
(c)
Statement
ˆ  Fˆ
ˆ  Eˆ and C
B
BC  EF
In ABC and DEF :
(1) BC  EF
ˆ  Eˆ
(2) B
(3) Cˆ  Fˆ
Reason
given
given
given
given
given
SAA
ABC  DEF
ABC  DEF or sum of the  's of 
ABC  DEF
(d)
(e)
AC  DF and AB  DE
ˆ D
ˆ
A
Why is the second reason true in (e)?
Consider the following triangles:
ˆ  Eˆ  130
B
ˆ  Fˆ  20
C
Therefore the only possible size for
 and D̂ is 30 due to the sum of the
angles of a triangle.
ˆ D
ˆ  30
A


20
30
157
130

20
130

30
Notice that the letters in ABC and DEF must correspond. The following diagram can be
used as a check that the corresponding angles and sides of the two triangles correspond:


A B CD E F
ˆ D
ˆ
A
ˆ  Eˆ
B
ˆ  Fˆ
C
20
AB  DE
BC  EF
AC  DF
130

20
130

Example 10
Consider PQS and RSQ .
(a)
Can you assume that P̂ and R̂ are right angles?
(b)
Can you assume that Q̂1 and Ŝ1 are equal?
(c)
Can you assume that Q̂ 2 and Ŝ2 are equal?
(d)
(e)
(f)
Which side is common to both triangles?
Which sides are equal?
Prove that PQS  RSQ .
(g)
Prove that Pˆ  Rˆ
Statement
(a)
(b)
(c)
(d)
(e)
(f)
No
No
Yes
SQ
PS  RQ
In PQS and RSQ :
(1) PS  RQ
ˆ
(2) Sˆ  Q
2
(g)
Reason
These angles are not indicated as right angles.
These angles are not indicated as equal.
These angles are indicated as equal.
These sides are indicated as equal.
given
given
2
(3) QS  SQ
PQS  RSQ
Pˆ  Rˆ
common side
SAS
PQS  RSQ
It might be helpful to redraw the two triangles
so that the corresponding sides and angles
are aligned.
It is now easy to see which angles and sides
correspond.
P Q SR S Q
158
Example 11
Consider ABD and CBD .
(a)
What is the size of B̂2 ? Give a reason.
(b)
Prove that ABD  CBD in two different ways.
(c)
Prove that BD bisects D̂ .
Statement
Reason
(a)
B̂2  90
(b)
In ABD and CBD :
ˆ B
ˆ  90
(1) B
proved
(2) AD  CD
(3) BD  BD
ABD  CBD
given
common
RHS
2
adjacent 's on a line
1
Alternatively:
In ABD and CBD :
(1) AD  CD
ˆ B
ˆ  90
(2) B
2
(c)
1
ˆ C
ˆ
(3) A
ABD  CBD
ˆ D
ˆ
D
1
2
given
proved
 's opp equal sides
SAA
ABD  CBD
 BD bisects D̂
EXERCISE 4
(a)
(1)
(2)
Prove that ABC  DEF
Why is AC  DF ?
35
(b)
(1)
(2)
Prove that ABC  EDC
ˆ D
ˆ?
Why is B
(c)
(1)
(2)
Prove that ABC  EDC
Why is Cˆ 1  Cˆ 2 ?
35
159
70
35
35
70
(2)
Prove that QPR  EDF
ˆ?
Why is Pˆ  D
(e)
(1)
(2)
(3)
Prove that ADB  CDB
Why is AB  CB ?
Show that ADC  BD
(f)
ABCD is a kite.
(1) Prove that ADC  ABC
(use two different methods)
ˆ .
(2) Show that AC bisects DCB
(g)
AB||DE and DB bisects AE.
(1) Prove that ABC  EDC
(2) Show that AE bisects DB.
(h)
PQRS is a parallelogram and TQUS is
a rectangle.
(1) Prove that PQS  RSQ
(2) Prove that PTQ  RUS
(i)
PQRS is a rhombus.
Prove that PQT  PST by using
all four conditions of congruency.
(j)
In an isosceles trapezium, the base
angles are equal and the pair of
non-parallel sides are equal.
In the diagram, ABCF is an isosceles
trapezium and ABDE is a rectangle.
(1) Prove that AEF  BDC
(2) Show that FC  2AB
if AB  10 cm and FE  5 cm .
(d)
(1)
160
Example 12
In the diagram, ECB bisects ACD at C. ED||AB.
(a)
Prove that ACD bisects EB at C.
(b)
Prove that ED  BA .
In order to prove that ACD bisects EB you
will need to prove that CE  CB .
Therefore first prove that CDE  CAB .
(a)
Statement
In CDE and CAB :
(1) CD  CA
ˆ
(2) D̂  A
Reason
given
alt  's equal; ED||BA
alt  's equal; ED||BA
SAA
CDE  CAB
CDE  CAB
ˆ
(3) Eˆ  B
CDE  CAB
 CE  CB
(b)
ED  BA
EXERCISE 5
(a)
In the diagram, P̂  Sˆ and QT  RT .
Prove that PQ  SR
(b)
In the diagram, AD  CD and BD bisects
ˆ . Prove that:
ADC
ˆ C
ˆ
(1) A
2
(2)
2
ˆ C
ˆ
A
1
1
(c)
In the diagram, CB  DB , AB  AD
and AE||BC.
Prove that AE bisects BD at E.
(d)
PQRS is a quadrilateral with PS||QR and
PS  QR .
(1) Prove that PQ  RS
(2) Prove that PQ||SR
(3) Why is PQRS a parallelogram?
161
Ĉ1
Ĉ 2
D̂
Â
Ê
B̂
(e)
ABCD is an isosceles trapezium.
Prove that:
(1) AC  BD
(2) BE  CE
(3) AE  DE
(f)
PQRS is a square. G is the midpoint of SR
and F is the midpoint of QR.
Prove that PG  SF .
(g)
ˆ C
ˆ and Fˆ  Fˆ
In the diagram, A
1
1
2
3
Prove that:
(1) Fˆ1  Fˆ2  Fˆ3  Fˆ4
(2) ABC is isosceles
In the next questions, you will be required to work with triangles in circles. Remember that
the lengths of radii in a circle are equal in length.
(h)
In the diagram, O is the centre of the circle
with AD  BC . Prove that D̂  Cˆ .
(i)
In the diagram, O is the centre of the circle.
ˆ .
BO bisects AOC
Prove that:
(1) OB bisects AC at B.
(2) OB  AC
(j)
In the diagram, O is the centre of the circle
passing through A, B and C. AB  AC .
Prove that:
ˆ O
ˆ
(1) O
1
2
(2) AO  BC
(3) ABC is a right-angled triangle.
162
SIMILAR SHAPES
Two shapes are similar if they have the same shape but not necessarily the same size. The one
shape is an enlargement (or reduction) of the other.
Revision of the similarity of polygons (Grade 8)
Consider the following examples:
(a)
Polygon EFGH is an enlargement of polygon ABCD. The sides of ABCD have been
multiplied by a factor of 2. The two polygons are similar.
ˆ  Eˆ  125
A
ˆ  Fˆ  85
B
ˆ G
ˆ  60
C
125
125 90
85
90
85
60
ˆ H
ˆ  90
D
60
EF 40 mm

2
AB 20 mm
FG 94 mm

2
BC 47 mm
GH 78 mm

2
CD 39 mm
EH 60 mm

2
AD 30 mm
The conclusion is as follows:
The corresponding angles are equal and the corresponding sides are in the same
proportion.
(b)
Rectangle ABCD is a reduction of rectangle EFGH. The sides of EFGH have been
multiplied by a factor of 12 .
ˆ  90
Ê  A
90
ˆ  90
Fˆ  B
90
ˆ C
ˆ  90
G
90 90
ˆ D
ˆ  90
H
AB 30 mm 1


EF 60 mm 2
90 90
BC 20 mm 1


FG 40 mm 2
CD 30 mm 1


GH 60 mm 2
90
90
AD 20 mm 1


EH 40 mm 2
The conclusion is as follows:
The corresponding angles are equal and the corresponding sides are in the same
proportion.
163
(c)
Consider the two polygons below:
ˆ  Eˆ  90
A
ˆ  Fˆ  90
B
90
90
90
90
90 90
ˆ G
ˆ  90
C
90 90
ˆ H
ˆ  90
D
EF 60mm

3
AB 20mm
FG 40mm

2
BC 20mm
GH 60mm

3
CD 20mm
HE 40mm

2
DA 20mm
The conclusion is as follows:
Polygon EFGH is not an enlargement of polygon ABCD.
The two polygons have different shapes. ABCD is a square and EFGH is a rectangle.
Even though the corresponding angles are equal, the ratios of the corresponding
sides are not the same.
(d)
Consider the two polygons below:
90
65
115
230
130
115
65
F̂  65
Ĝ  230
Ĥ  65
Î  90
Ĵ  90
  115
B̂  130
Ĉ  115
D̂  90
Ê  90
AB 20 mm 1


FG 40 mm 2
DE 35 mm 1


IJ 70 mm 2
90
90
90
ˆ
 F̂  A
ˆ B
ˆ
G
ˆ
 Ĥ  C
ˆ
 ˆI  D
 Jˆ  Eˆ
BC 20 mm 1


GH 40 mm 2
AE 30 mm 1


FJ 60 mm 2
CD 30 mm 1


HI 60 mm 2
The conclusion is as follows:
Polygon ABCDE is not a reduction of polygon FGHIJ.
The two polygons have different shapes. Even though the ratios of the
corresponding sides are equal, the corresponding angles are not equal.
164
Let’s now summarise what we learnt in Grade 8:
In order for two polygons to be similar, the following two conditions must BOTH be
satisfied:
(a)
The corresponding angles must be equal.
(b)
The ratios of the corresponding sides must be in the same proportion.
It is important to note that two polygons may well have their corresponding angles equal,
but the corresponding sides may not necessarily be in the same proportion (consider (c)). It
is also important to note that two polygons may well have their corresponding sides in the
same proportion but not all of their corresponding angles may be equal (consider (d)). This
is why it is necessary for both conditions to be satisfied.
EXERCISE 6
Show with reasons, why the following polygons are similar.
(a)
(b)
100
105
(c)
50
80
100
80
50
105
120
120
60
60
60
60
120
120
110
(d)
ABCD and EFGH are kites
110
55
85
55
85
EXERCISE 7
Explain why the following polygons are not similar:
(a)
ABCD and EFGH are rhombuses.
150
135
45
30
165
(b)
ABCD and EFGH are parallelograms.
(c)
PQRS and TUVW are kites.
130
28
151
74
30
82
Similarity of triangles
Consider the following example:
DEF is an enlargement of ABC . The two triangles have the same shape and are therefore
similar.
78
78
69
33
69
ˆ D
ˆ  78
A
DE 50 mm

2
AB 25 mm
ˆ  Eˆ  69
B
EF 88 mm

2
BC 44 mm
33
ˆ  Fˆ  33
C
DF 84 mm

2
AC 42 mm
The conclusion is as follows:
ABC and DEF are similar since the following conditions hold true:
(a)
ˆ D
ˆ  78
A
ˆ  Eˆ  69
B
ˆ  Fˆ  33
C
(b)
DE 50 mm

2
AB 25 mm
EF 88 mm

2
BC 44 mm
DF 84 mm

2
AC 42 mm
BC 44 mm 1


EF 88 mm 2
AC 42 mm 1


DF 84 mm 2
Alternatively:
AB 25 mm 1


DE 50 mm 2
Note:
There are no two triangles which can have corresponding angles equal with the ratios of their
corresponding sides being unequal. Also there are no two triangles where the ratios of the
corresponding sides are equal and their corresponding angles are unequal.
166
Some similarity notation with regard to triangles
If ABC is similar to DEF then we write this as follows: ABC|||DEF
If ABC|||DEF then the following conclusions can be made:
ABC ||| DEF
(a)
The triangles are equiangular which means that:
ˆ D
ˆ  Fˆ
ˆ
ˆ  Eˆ
A
C
B
(b)
The corresponding sides are in the same proportion which means that:
AB BC AC
DE EF DF


or


ABC ||| DEF
DE EF DF
AB BC AC
Summary of the conditions for similarity
Two polygons are similar if their corresponding angles are equal AND their corresponding
sides are in the same proportion. It is important to note that two polygons may well have
their corresponding angles equal, but the corresponding sides may not necessarily be in the
same proportion. It is also important to note that two polygons may well have their
corresponding sides in the same proportion but not all of their corresponding angles may be
equal.
With triangles, only one of the above two conditions needs to be true in order for the two
triangles to be similar. This is unique to triangles only. In other words, two triangles are
similar if their corresponding angles are equal OR if their corresponding sides are in the
same proportion. If the corresponding angles of two triangles are equal (triangles are
equiangular) then the ratios of their corresponding sides will always be in the same
proportion and the two triangles will be similar. Conversely, if the ratios of the
corresponding sides are in the same proportion, then the corresponding angles of the two
triangles will be equal and the triangles will be similar.
Example 13
Show that the following triangles are similar.
(a)
Â
Ĉ
Ê1
Ê 2
B̂
D̂
Statement
In AEB and CED:
ˆ C
ˆ
A
Reason
given
Eˆ 1  Eˆ 2
both equal 90
ˆ D
ˆ
B
AEB|||CED
sum of the  's of 
corr  's of the two triangles are equal
167
(b)
Redraw the triangles so that the corresponding
equal sides align.
AEB ||| CED
Statement
In AEB and CED:
AE 4cm 1


CE 8cm 2
EB 3cm 1


ED 6cm 2
AB 5cm 1


CD 10cm 2
AE EB AB 1




CE ED CD 2
AEB|||CED
Reason
given
given
given
corr sides in prop
EXERCISE 8
Show that the following triangles are similar.
(a)
(b)
45
(c)
45
55
44
55
ABC and ADE
(d)
P
44
A
D
D
R
Q E
1
1 E
F
B
168
C
(e)
(g)
(f)
PDE and PQR
(h)
ACE and BCD
(j)
ABD and BCD
P
3 cm
D
2,6 cm
2 cm
E
9 cm
=
(i)
10,4 cm
=
Q
6 cm
R
In the diagram, EC  2AE,
1
AB 1
BE  DE and
 .
2
DC 2
A
2
D
1
1
2
C
B
Example 14
If ABC|||EDC , calculate the value of x.
Statement
ABC|||EDC
AB BC AC


ED DC EC
x
2 cm 3 cm 1




12 cm 6 cm 9 cm 3
x
1


12 cm 3
x
1

 12 cm   12 cm
12 cm
3
 x  4 cm
169
Reason
given
corr sides in prop
given
LCD  12 cm
EXERCISE 9
The following pairs of triangles are similar. Calculate the value of x and/or y in each case.
P
(a)
(b)
D
9cm
6 cm
y
3 cm
E
Q
(c)
PQR ||| TSR
(d)
F
3 cm
R
9 cm
DBZ ||| XYZ
X
5
D
12
y 10
x
Y
B
9
(g)
ADE ||| ECB
DEFG is a square.
(f)
AEF ||| CDB
ABCD is a parallelogram.
170
=
PQR ||| PDE
=
(e)
Z
REVISION EXERCISE
(a)
ABCD is a parallelogram.
ˆ , Ĉ  60 and AE  EB.
AE bisects BAD
B̂2  2 x  20 and D̂ 2  3 x  30 .
(1)
(2)
(3)
3x  30
60
Calculate the size of Ê 3
Calculate the value of x.
Prove that D̂ 2  90
2 x  20
A
(b)
In quadrilateral ABCE, Ĉ1  75 ,
2
B̂1  20 , Ê 2  x , F̂1  y ,
AF  BF and FE  FC .
(1)
Calculate the size of x and y.
(2)
Prove that ABCE is a trapezium
by using three different methods.
y1 F
2
1
B
(c)
1
20
Calculate the length of PS.
Prove that PT||QR.
Prove that PQRT is a parallelogram.
2
3
1
2
3 cm
R
(e)
O is the centre of the circle. ABCD is
ˆ A
ˆ  30 .
a kite and A
1
2
Prove that:
(1)
AOB  AOD .
(2)
BOC is equilateral.
DOC is equilateral.
(3)
(4)
BCDO is a parallelogram.
(5)
BCDO is a rhombus.
ˆ  2BAD
ˆ .
(6)
BOD
(7)
AB  BC.
30 30
Consider the two triangles below.
(1)
Prove that ABC|||DEF
(2)
Prove that the triangles are
equiangular.
171
2
1 75
C
120 T
6 cm
Q
(d)
E
x2
1 2
PQ  6 cm , QR  3 cm and T̂  120 and
ˆ .
SP bisects QPT
(1)
(2)
(3)
3
P
In quadrilateral PQRT, PTS is an isosceles
triangle and QRS is an equilateral triangle.
2
1
1
S
D
(f)
In the diagram, AB||DE .
(1)
Prove that ABC|||DEC
(2)
If DE  4 cm , EC  2 cm and BC  4 cm ,
calculate the length of AB.
(g)
In ABC , BE  AC and BE bisects AD.
(1)
Prove that AEB  DEB
(2)
Prove that ABC|||AEB
(3)
Calculate the value of x if AE  4cm
and DC  1 cm
(4)
Calculate the length of BD.
(5)
Calculate the length of BC
(rounded off to one decimal place).
(h)
ABCD is a rectangle and
DECF is a rhombus.
Prove that:
(1)
ABF  DCE
(2)
FE||BC
ACD|||FDG
(3)
SOME CHALLENGES
(a)
(b)
ABCD is a parallelogram with
AB  AE  EC and Ê1  60 .
Calculate the size of:
(1)
x
Ê 2
(2)
60
P
In PQR , Q̂  90 , QS  PR and
R̂  θ .
(1)
Show that Q̂1  θ
(2)
(3)
(c)
2x
1
Show that Q̂3  90  2θ
There is only one value of θ
ˆ Q
ˆ . Calculate
for which Q
1
3
this value for θ .
S
2
1
1 2
3
Q
In PQR , TP||QRS.
(1)
Express y in terms of θ .
(2)
Calculate the size of θ .
2
T
θ
R
2θ  40
θ  10
172
2θ  40
(d)
D
In the diagram, DE  GF and DF  GE .
Prove that:
(1)
DEF  GFE
Eˆ  Fˆ
(2)
1
G
1
H
1
2
1
1
2
(e)
2
F
E
In ABC , DH  HE and BH  HC .
Prove that:
(1)
DB  EC
ABE  ACD
(2)
(3)
AB  AC by using two different
methods.
A
D 1
2
1
B
2
1
H
1
2
2
2
E
1
C
(f)
Calculate the height, h metres, at which the tennis ball must be hit so that it will just
pass over the net and land 6 metres away from the base of the net.
(g)
Tumelo wants to mount a TV antenna,
AB, on the roof of his house. The top of
the antenna has to be 3 metres higher
than the tree in order for him to have a
clear reception. The height of the
tree (FG) is 12 metres, CD  11 metres,
AH  7 metres and AB||FG.
(1)
(2)
(h)
Determine the length of HF.
If EF  13 metres, calculate the length of HE, correct to two decimal places.
A fish pond is in the form of an ellipse.
Using the information on the diagram,
(1)
Calculate the length (BC) and width (DF)
of the pond.
(2)
Show that ABC|||ADE and then use these
similar triangles to calculate the length of AB.
173
(i)
The diagram below shows how similar triangles relate to human sight. An image
similar to a viewed object appears on the retina. The actual height of the object (h) is
proportional to the size of the image as it appears on the retina (r). The distances
from the object to the lens of the eye (d) and from the lens to the retina (25 mm) are
also proportional.
(1)
An object that is 10 metres away appears on the retina as 1 mm tall. Calculate
the actual height of the object.
(2)
An object that is 1 metre tall appears on the retina as 1 mm tall. How far away
is the object?
25 mm
d
h
r
retina
lens
(j)
D
2
Suppose that in the diagram, m  n .
3
Calculate the size of m.
B m
1
30
A
(k)
1
n
C
E
In the quadrilateral ABCD, DC  BD , CD||AB, Cˆ 1  Cˆ 2 , D̂1  20 and Â1  30 .
ˆ B
ˆ .
The question required that learners calculate the size of B
1
The attempts of two learners, James and
C
Sarah, are correct but they obtain different
answers. Why is this so? Investigate any
possible false statements that might have
affected the answers. The attempt of each
learner is provided below.
2
2
2 1
1
B
30
20
1 2
D
James
ˆ C
ˆ  180
ˆ B
ˆ A
B
1
2
1
2
ˆ
ˆ
B̂  C  C  80
2
1
2
ˆ C
ˆ  40 and   30
C
1
2
1
ˆ B
ˆ  30  40  180
B
1
Angles opp equal sides; sum of angles of
triangle
Given
2
ˆ B
ˆ  110
B
1
2
Sarah
ˆ C
ˆ  180
ˆ B
ˆ C
B
1
2
1
2
ˆC  C
ˆ  80
1
Sum of the angles of a triangle
2
Co-interior angles supplementary; CD||AB
Angles opp equal sides; sum of angles of
triangle
ˆ B
ˆ  80  180
B
1
2
ˆ B
ˆ  100
B
1
2
174
2
1
A
CHAPTER 12: SHAPE AND SPACE (GEOMETRY)
TOPIC: GEOMETRY OF STRAIGHT LINES
REVISION OF THE GEOMETRY OF LINES AND ANGLES
Two angles are called adjacent angles if they have a common
vertex and a common arm between them. For example, the
two angles in the diagram are adjacent angles since they
share a common vertex B and a common arm BD.
ˆ or DBA
ˆ or B̂ .
The top angle can be named as ABD
1
ˆ or CBD
ˆ or B̂ .
The bottom angle can be named as DBC
2
ˆ B
ˆ B
ˆ .
Notice that ABC
1
2
Complementary angles add up to 90 . If the two angles
are adjacent angles, then we say that these angles are
adjacent complementary angles.
For example, B̂1 and B̂2 are complementary angles
ˆ B
ˆ  90 . They are also adjacent
because B
1
60
2
30
complementary angles since they share a common
vertex B and common arm BD.
ˆ is a right angle and this is indicated on the diagram using .
ABC
We say that the arms AB and BC are perpendicular and write this as AB  BC . AB and BC
form a right angle where they meet.
Supplementary angles add up to 180 . If the two angles
are adjacent angles, then we say that these angles are
adjacent supplementary angles.
For example, B̂1 and B̂2 are supplementary angles
120
60
ˆ B
ˆ  180 . They are also adjacent
because B
1
2
supplementary angles since they share a common
vertex B and common arm BD. Notice that the adjacent angles on the straight line ABC
add up to 180 .
Property 1
Adjacent angles on a straight line are supplementary.
ˆ B
ˆ  180 or
 If ABC is a straight line then B
1
2
ˆ
ˆ
 If B  B  180 then ABC is a straight line
1
2
Property 2
If two lines AB and CD cut each other (intersect) at E,
then the vertically opposite angles are equal.
Eˆ 1  Eˆ 3 and Eˆ 2  Eˆ 4 .
Property 3
The angles around a point add up to 360 .
ˆ B
ˆ B
ˆ B
ˆ  360
B
1
2
3
4
175
Parallel lines
Straight lines that are the same distance from each other along
the whole of their lengths are called parallel lines. Arrows on
the two lines indicate that they are parallel. A transversal is
a line that intersects the two parallel lines. In the diagram,
AB is parallel to CD and we write this as AB||CD.
The transversal is EF.
Corresponding angles
Corresponding angles lie either both above or both below the parallel lines and on the same
side as the transversal. They are the angles in matching corners and are equal. Always look
out for the F shape.
Alternate angles
Alternate angles lie on opposite sides of the transversal and between the parallel lines. They
are equal in size. Always look out for the Z or N shape.
Co-interior angles
Co-interior angles lie on the same side of the transversal between the parallel lines.
These angles are supplementary. Always look out for the U shape.
ˆ H
ˆ  180
G
1
1
ˆ H
ˆ  180
G
1
1
ˆ H
ˆ  180
G
1
1
ˆ H
ˆ  180
G
1
1
Example
(a)
Calculate the value of the angles indicated by small letters.
Statement
a  45
b  55
c  55
d  55
e  125
f  125
g  55
125
45
176
Reason
alt  's equal ; AB||CD
co-int  's suppl ; AB||CD
corr  's equal ; ST||VW
vert opp  's equal
adjacent 's on a line
corr  's equal ; AB||CD
adjacent 's on a line
(b)
AB and CD are straight lines
cut by transversal ST.
Show that AB||CD.
Statement
5 x  70  2 x  5  180
 7 x  75  180
 7 x  105
 x  15
Ê1  2(15)  5  35
5 x  70
2 x  5
Reason
adjacent 's on a line
F̂1  3(15)  10  35
 Eˆ  Fˆ
3x  10
1
1
 AB||CD
alt 's equal
REVISION EXERCISE
(a)
Calculate the value of the angles indicated by small letters.
(1)
(2)
(3)
32
86
44
(4)
(5)
(6)
40
x
100
144
(7)
(8)
(9)
160
146
85
(10)
(11)
(12)
113
81
136
(13)
(14)
(15)
116
(16)
120
124
(17)
63
94
86
177
(18)
(19)
46
2y
32
45
2 y  80
(20)
(21)
210
a  20
3a  30
6a  60
(22)
2a  31
59
(b)
Show that ABC is a straight line in each of the following diagrams:
(1)
(2)
20
49
(c)
160  2x
41
ˆ  60 E
ˆ  30 , CAB
In the diagram, CD||EF, EFC
ˆ  150 . Prove that AB||CD.
and ACF
30
C
A
(d)
ˆ  82 ,
In the diagram, QR||ST, Ŝ  42 , VTU
ˆ  x and TRS
ˆ  y , SRQ
ˆ  x  40 .
RTS
(1)
(2)
1
x
D
150
60
B
82
Prove that RS||TU.
Calculate the size of y.
42
x  40
(e)
F
ˆ  44 ,
In the diagram, LM||NP, QCM
ˆ  68 and ABV
ˆ  112 .
NDC
Prove that EF||TS.
112
44
68
178
The next questions involve the properties of triangles and quadrilaterals.
(f)
(g)
In ABC , EF||BC. BA is produced to D.
(1)
Calculate a and hence show that
AE  AF .
(2)
Calculate, with reasons, the value of b, c,
d and e.
5a  40
2a
80
2 x  40
In ABC , ED||BC. Calculate:
(1)
the value of x.
(2)
Calculate the size of D̂1 .
x  10
70
(h)
ABCD is a parallelogram with AE  BC .
Calculate, with reasons, the value of
a, b , c and d.
A
b d
F
B
(i)
a
c
ABCD is a parallelogram with B̂1  90 and
70
20
A
ABCD is a trapezium with AD||BC.
AD  BD , BC  CD and   78 .
Determine the unknown angles.
P
PQRS is a parallelogram.
Calculate the value of x and y.
b
78
B
(k)
C
E
  70 .
(1)
Calculate, with reasons, the size of
ˆ ,D
ˆ ,B
ˆ and B̂ .
D
1
2
2
3
(2)
Prove that BECD is a parallelogram.
(3)
Prove that BCED is a rectangle.
(j)
D
75
2
a c
1
y
ABCD is a rectangle and DECF is a
ˆ  90 and AF  DF .
parallelogram. AFD
(1)
Calculate the size of Ê .
(2)
Show that ED  DC .
(3)
Show that Fˆ1  Fˆ3 .
179
d
C
Q
x
2 x  50
T 110 S
(l)
e
D
1
2
R
(m)
In the diagram, BC||GK, Ê 3  60 ,
ˆ and BE  BD .
Ê  30 , GE bisects HEF
5
(1)
(2)
(3)
(4)
(5)
(n)
Show that BE  GK .
Show that BED is equilateral.
Show that EF||DB.
Calculate the size of Ĉ .
Show that BD  DC .
30
60
In DEF , GH||EF and DGKH is a kite.
Ĝ 2  40 , Ê  50 , K̂ 3  x and D̂  y .
(1)
Calculate, with reasons, the value of x
and y.
(2)
Show that GK  DE .
40
50
SOME CHALLENGES
(a)
ˆ  20 ,
In the diagram, AB||CD, GEC
ˆ  40 . Calculate the size of angle  .
AFG
Hint: Draw a third line that is parallel
to AB and CD.
40
20

(b)
In the diagram, AB  BD and
Ĉ 2  170  2 x .
Express the following angles in terms
of x:
B̂3
(1)
(2)
(c)
B̂2
ˆ θ,
In the diagram, BA||DC, NGA
ˆ  θ  20 and BHF
ˆ  1θ.
HNG
6
θ  20
θ
Calculate the size of θ .
(d)
1
6
ˆ  y,
In the diagram, PU||QT, PNR
ˆ  4 x  65 .
ˆ  4 x  5 and UST
QRN
(1)
(2)
(e)
170  2x
Express y in terms of x.
If x  30 , then show that
NR||US.
4x  5
In the diagram, AB||CD.
ˆ  250  4 x , GWB
ˆ  2x
TVC
ˆ  5 x  30 .
and WZV
(1)
Calculate the value of x.
(2)
Show that EF cannot be parallel to
GH.
2x
5 x  30
250  4x
180
4 x  65
CHAPTER 13: MEASUREMENT
TOPIC: THE THEOREM OF PYTHAGORAS
In this chapter, we will revise the famous Theorem of Pythagoras that you
studied in Grade 8. This theorem was first proved by the ancient Greek
mathematician Pythagoras of Samos, who lived from about 569 BC to
about 475 BC. The theorem was known to the ancient Babylonians 1000
years prior to Pythagoras. Ancient Indians (800 BC) and Chinese
mathematicians (500 BC) were also aware of this theorem. This theorem is
used extensively in so many areas of Mathematics such as Algebra,
Trigonometry, Measurement, Analytical and Euclidean Geometry.
Let’s briefly revise some important concepts from Grade 8 before exploring this theorem in
more detail.
Naming the sides of triangles
The sides of ABC can be named in terms of a, b and c .
In the diagram, the side opposite  is BC and we let BC  a .
The side opposite B̂ is AC and we let AC  b .
The side opposite Ĉ is AB and we let AB  c .
How the Theorem of Pythagoras works.
Suppose that in ABC , B̂  90 , AB  3 cm
and BC  4 cm .
On side AB create a square made up of 9 little squares
each with an area of 1 cm 2 .
A
On side BC create a square made up of 16 little
squares.
5c
It is now possible to create a square on the hypotenuse
m
made up 25 little squares.
4 cm
The area of the square on side AB
C
B
 (3 cm)(3 cm)  (3 cm) 2  9 cm 2
The area of the square on side BC
 (4 cm)(4 cm)  (4 cm)2  16 cm 2
The area of the square on side AC
 (5 cm)(5 cm)  (5 cm)2  25 cm2
From the above it should be clear that the area of the square on the hypotenuse is equal to the
sum of the areas of the squares on the other two sides.
If we now consider the sides of the triangle:
AB2  BC2  (3 cm)2  (4 cm)2  9 cm 2  16 cm 2  25 cm 2 and AC2  (5 cm) 2  25 cm 2
 AC 2  AB2  BC 2
This relationship between the hypotenuse and the other two sides is referred to as the
Theorem of Pythagoras. Let’s state a rule that applies to all right-angled triangles.
The Theorem of Pythagoras
In any right-angled triangle, the square on the
hypotenuse is equal to the sum of the squares
on the other two sides.
AB2  BC 2  AC 2 or c 2  a 2  b 2
BC2  AB2  AC 2 or a 2  c 2  b 2
AC2  AB2  BC 2 or b 2  c 2  a 2
181
Example 1
Example 2
Calculate the length of AB.
A
Calculate the length of DE.
Solution
ˆ  90
In ABC, C
9 cm
Solution
In DEF, Eˆ  90
C
 DE 2  DF2  EF2
 AB2  BC2  AC2 B
 AB2  (12) 2  (9) 2
12 cm
 DE 2  (17) 2  (15) 2
 AB2  144  81
 DE 2  289  225
 AB2  225
 DE 2  64
 AB  225
 AB  15 cm
Note: AB  15 cm
since AB is a length
which is always positive.
Example 3
Calculate the length of PR.
Solution
In PQR, Rˆ  90
 PR 2  PQ 2  RQ 2
 PR 2  (51) 2  (45) 2
 PR 2  2 601  2 025
2
 PR  576
 PR  576
 PR  24 units
 DE  64
 DE  8 m
Example 4
ˆ  90, c  18 units and
In ABC, B
a  24 units. Calculate the length of b.
Solution
First draw a rough sketch of the triangle
using the method of naming the sides as
discussed. The unknown side is b.
b2  a 2  c 2
 b 2  (24) 2  (18) 2
 b 2  900
 b  900
 b  30 units
Example 5
In ABC , Ĉ  90, AC  3 and BC  5.
(a)
Calculate the length of AB without using a calculator.
Leave your answer in surd form.
(b)
Now use your calculator to calculate the length of AB.
Round off to two decimal places.
Solution
(a)
ˆ  90
In ABC, C
 AB2  BC2  AC2
 AB2  (5) 2  (3) 2
 AB2  25  9
 AB2  34
 AB  34 units
(b)
Using a calculator:
AB  5,83 units
182
A
3
B
5
C
EXERCISE 1
(a)
Calculate the length of the unknown side in each of the following triangles.
(The triangles are not drawn to scale).
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
24
x
40
70
y
9
(10)
(11)
(12)
26
24
z
y
(13)
(14)
96
100
(15)
x
P
R
29
5
13
Q
20
(16)
119
169
(17)
(18)
A
155
d
B
183
93
C
(19)
(20)
A 3cm D
x
4cm
B
(21)
13cm
C
ˆ  90, AC  8 units and BC  6 units. Calculate the length of AB.
(b) (1) In ABC, C
ˆ  90, a  10 cm and c  24 cm. Calculate the length of b.
(2) In ABC, B
(3) In DEF, Eˆ  90, DE  15 m and DF  17 m. Calculate the length of EF.
(4) In DEF, Fˆ  90, d  42 mm and f  82 mm. Calculate the length of e.
(5) In PQR, Rˆ  90, PQ  116 and p  84. Calculate the length of PR.
(6) In PQR, Pˆ  90, q  36 and r  15. Calculate the length of RQ.
(7) In ABC, Cˆ  90, AC  16 x units and BC  30 x units. Express the length of AB in
terms of x.
ˆ  90, BC  50 x units and AB  14 x units. Express the length of AC in
(8) In ABC, A
terms of x.
(c) Calculate the unknown sides in each of the following triangles. Write your answers in
surd form. Then use a calculator and round off your answers to two decimal places.
(1)
(2)
A
(3)
4 cm
B
(4)
(5)
(6)
11
x
10
(7)
(8)
(9)
The converse (opposite) of the Theorem of Pythagoras
The converse or opposite of the Theorem of Pythagoras
states that if the square on the hypotenuse is equal to the
sum of the squares on the other two sides, then the triangle
is right-angled.
If AB2  BC2  AC 2 or c 2  a 2  b 2
then Ĉ  90
184
6 cm
C
Example 6
A
In ABC, AB  10 units, BC  6 units and AC  8 units.
Show that ABC is right-angled and state which angle
is the right-angle.
10
8
Solution
B
Square all of the sides:
BC2  (6) 2  36
C
6
AC2  (8) 2  64
AB2  (10) 2  100
Now find out which two squares add up to the third square. Clearly 100 is the sum of 36
and 64.
 AB2  BC2  AC 2
 ABC is a right-angled triangle and the right-angle is at C.
EXERCISE 2
(a) In the triangles below, show that the triangle is right-angled and state which angle is the
right-angle.
(1)
(2)
(3)
(4)
In
In
In
In
ABC, AB  24, BC  7 and AC  25.
ABC, c  14, b  48 and a  50.
PQR, PQ  24 m, QR  70 m and PR  74 m.
PQR, p  80 cm, q  82 cm and r  18 cm.
(b) In ABC, AB  15, BC  9 and AC  12
and in ACD, AC  12 and AD  13.
(1) Show that ABC is right-angled at Ĉ1 .
(2) Why is Ĉ2  90 ?
(3) Calculate the length of CD.
(c) In ABC, AB  74, BC  24, AC  70 and DE  25 .
185
=
(d) ABCD and EBFD are parallelograms. AE  ED.
(1) Show that ABCD is a rectangle.
(2) Why is Ĉ  90 ?
(3) Calculate the length of FC.
(4) Calculate the perimeter of ABCD.
(5) Calculate the area of ABCD.
=
(1) Show that ABC is right-angled at Ĉ1 .
(2) If ACDF is a square, why is FDE
a right-angled triangle?
(3) Calculate the length of EF rounded off
to two decimal places.
(4) Calculate the perimeter of trapezium ABEF.
The relationship between the sides and angles of triangles that are not right-angled
The Theorem of Pythagoras works in a right-angled triangle. But what about other triangles
that are acute-angled or obtuse-angled? Let’s investigate this by considering the following
triangles.
Triangle
Square on
longest side
Sum of the
squares of the
other two sides
Conclusion
ABC is acute-angled
AB is the
longest side
opposite the
largest angle.
c 2  (5) 2  25
a 2  b2
AB is the
longest side
opposite the
largest angle.
c 2  (6) 2  36
a 2  b2
25  33, 21
The square on
the longest side
is less than the
sum of the
squares on the
other two sides.
c2  a2  b2
36  25
The square on
the longest side
is greater than
the sum of the
squares on the
other two sides.
c 2  a 2  b2
C
4,5
80
c
5
45
A
a
b
3,6
55
B
ABC is obtuse-angled
25
120
 (3, 6) 2  (4,5) 2
 12,96  20, 25
 33, 21
 (3) 2  (4) 2
 9  16
 25
35
Summary of the relationship between the sides and angles of triangles
 In a right-angled triangle, the square on the hypotenuse is equal to the sum of the
c 2  a 2  b2
squares on the other two sides:
 In an acute-angled triangle, the square on the longest side is less than the sum of the
c2  a2  b2
squares on the other two sides:
 In an obtuse-angled triangle, the square on the longest side is greater than the sum of
c 2  a 2  b2
the squares on the other two sides:
Example 7
(a) In ABC, AB  10 units, BC  7 units and AC  9 units.
Determine whether ABC is acute-angled or obtuse-angled.
Solution
C
Draw a rough sketch of the triangle.
Square all of the sides:
AB2  (10) 2  100
9
AC2  (9) 2  81
2
A
2
BC  (7)  49
7
10
B
The longest side is AB. Find out if AB2 is greater than or less than AC2  BC2 .
AB2  100 and AC2  BC2  81  49  130
Since AB2  AC2  BC2 , the triangle is acute-angled.
186
(b) In ABC, c  12 units, a  6 units and b  8 units.
Determine whether ABC is acute-angled or obtuse-angled.
Solution
Draw a rough sketch of the triangle.
Square all of the sides:
c 2  (12) 2  144
2
A
2
a  (6)  36
b 2  (8)2  64
12
The longest side is c. Find out if c 2 is
c
b 8
2
2
a

b
greater than or less than
c 2  144 and a 2  b 2  36  64  100
a
C
B
Since c 2  a 2  b 2 , the triangle is
6
obtuse-angled at C.
You might like to redraw the triangle to show all of this information.
EXERCISE 3
(a)
For each of the following triangles, determine whether the triangle is right-angled,
acute-angled or obtuse-angled. If the triangle is right-angled, state which angle is the
right angle. If the triangle is obtuse-angled, state which angle is the obtuse angle.
(1)
(2)
(3)
(4)
(5)
(b)
In
In
In
In
ABC, AB  12, BC  5 and AC  13.
ABC, a  9, b  10 and c  11.
DEF, DE  12, EF  5 and e  15 .
PQR, PQ  8 mm, QR  6 mm and q  11 mm .
(6)
(7)
ABCD is a kite.
(1)
Show that ABD is an obtuse-angled triangle.
(2)
Show that BCD is an acute-angled triangle.
(3)
Calculate the length of EC.
A
3
6
E
B
5
C
187
D
REVISION EXERCISE
In this exercise you will need to know the properties of quadrilaterals in order to answer some
of the questions. Refer to the chapter on the Geometry of 2D shapes.
(a)
Calculate the length of the unknown side in each of the following triangles. Round
off your answers to two decimal places when necessary.
(1)
(2)
(3)
2 cm
x
1 cm
(4)
(b)
(1)
(2)
(c)
(d)
(e)
(5)
(6)
ˆ  90, AB  28 m and BC  21 m. Calculate the length of AC.
In ABC, B
ˆ  90, g  82 cm and h  80 cm. Calculate the length of f.
In FGH, G
ˆ  90, DC  14 mm and d  15 mm. Calculate the length of c.
In BCD, C
(3)
Determine whether the following triangles are right-angled, acute-angled or obtuseangled.
(1)
In ABC, AB  36, BC  60 and AC  48 .
(2)
In ABC, AB  36, BC  60 and AC  50 .
(3)
In ABC, AB  26, BC  59 and AC  48 .
(1)
Calculate the length of QS.
(2)
Calculate the length of CD.
In the diagram below, AC  CE  48cm .
(1)
Show that ABC is right-angled.
(2)
Why is Ĉ2  90 ?
(3)
Calculate DC if DE  52cm
A
(f)
ABCD is a square. Calculate BD.
Leave your answer in surd form.
D
7 cm
B
188
C
(g)
ABDE is a trapezium.
(1)
Why is Ĉ1  90 ?
(2)
Calculate the length of ED.
E
A
18
26
B
(h)
PQRS is a rhombus.
(1)
Calculate the length of PN.
(2)
Calculate the length of SRT.
(i)
ABCD is a rectangle. AD  84 and
AB  80 . Calculate the length of AE.
(j)
DEFG is a rhombus. EG  16 cm and
DF  12 cm . Calculate the perimeter
of DEFG.
(k)
ABCD is a kite. BD  48, AE  18
and EC  32 . Calculate the perimeter
of ABCD.
10
1
2
C
3
D
50
A
B
D
E
C
The centre of the circle is O. AO  15 cm, AM  9 cm
and OM  12 cm.
(1)
Show that OMA is a right-angled triangle.
(2)
Prove that OM bisects AB at M.
189
A
M
9m
15 m
12 m
(l)
O
B
SOME CHALLENGES
(a)
(b)
The length of diagonal BD in square
ABCD is equal to the length of diagonal
FH in rectangle EFGH. Calculate the value
of x and hence the length of one side of the
square.
2x
In ABD, AC  BD, AB  15, BC  9, EC  x
and AE is three times EC.
(1)
Calculate the length of AE.
(2)
Calculate the length of BE rounded off
to two decimal places.
x
D
(c)
10 cm
B
A
50
The diagram shows the cross section of a
wooden log of radius 50 cm floating in water.
Calculate the length of AB.
cm
C
(d)
(e)
(1)
(2)
2
D
18 cm
C
E
A
2
d  a b c
A cuboid measures 4 cm by 5 cm by 8 cm.
Calculate the length of the diagonal FC.
Round off your answer to two decimal
places.
d
c
G
D
b
B
The Pythagorean spiral is a spiral composed of right
triangles. Using the information provided on the spiral,
calculate the value of x.
F
F
Prove that the length of the longest diagonal
of the cuboid is given by:
2
(f)
E
ABCDEF is a triangular prism. AB  9 cm, BC  12 cm
and CD  18 cm. Calculate the lengths of:
A
(1)
AC
(2)
CF (in surd form)
9 cm
(3)
AD (rounded off to two decimal places)
B
12 cm
C
a
1
1
1
1
1
1
1
x
1
1
1
1
1
190
1
1
1
1
1
CHAPTER 14: MEASUREMENT
TOPIC: AREA AND PERIMETER OF 2D SHAPES
PERIMETER AND AREA OF 2D SHAPES
In this module, we will revise the perimeter and area of squares, rectangles, triangles and
circles and then discuss the perimeter and area of other polygons including quadrilaterals.
SQUARES, RECTANGLES, TRIANGLES AND CIRCLES (GRADE 8 REVISION)
2D Shape
Square
Perimeter
A square has four equal sides.
The perimeter is the sum of
these equal sides. If one side is s
units in length, then the
perimeter is:
P  s  s  s  s  4s
Area
A square has four equal sides. If
one side is s units in length, then
the area is:
Area(A)  s  s  s 2
Rectangle
A rectangle has two pairs of
opposite sides equal
(two lengths and two breadths).
The perimeter is the sum of
these sides. If the length is l and
the breadth (width) is b, then the
perimeter is:
P  l  b  l  b  2l  2b
The perimeter of a triangle is the
sum of the three sides.
A rectangle has two pairs of
opposite sides equal
(two lengths and two breadths).
If the length is l and the breadth
(width) is b, then the area is:
A  length  breath  l  b
Triangles
Scalene
If the triangle is scalene and the
sides are a, b and c, then the
perimeter is:
P  abc
Isosceles
If the triangle is isosceles, then
the perimeter is:
P  a  a  b  2a  b
Equilateral
If the triangle is equilateral,
then the perimeter is:
P  a  a  a  3a
The area of a triangle is:
1
A  (base)(height)
2
1
 A  (b)(h)
2
The height can be drawn from
any vertex of the triangle.
b
h
191
2D Shape
Circle
The perimeter of a circle is called
its circumference. A diameter is
a line passing through the centre
and joining two points on the
circle. It divides the circle into
two semi-circles.
A radius is a line from the centre
to a point on the circle. An arc of
a circle is a portion of the
circumference of a circle.
A sector of a circle is the area
bounded by two radii and an arc.
It resembles a slice of pizza.
A semi-circle is a half of a full
circle.
d
A quarter-circle is a quarter of a
full circle.
Perimeter
The length of the
circumference is
calculated using the
formula C  2r where r
represents the radius of
the circle or C  d
where d represents the
diameter of the circle.
Remember that for any
C
circle:  
d
Area
The formula for the area
of a circle is A  r 2
where r represents the
radius of the circle.
The perimeter of the semi- The area of the semi-circle
circle is a half of the
is a half the area of the full
circumference of a full
circle:
circle:
1
A  r 2
1
2
C   2r   r
2
If the diameter is a solid
line, it forms part of the
perimeter. The formula is
therefore:
C  r  d
The perimeter of the
quarter-circle is a quarter
of the circumference of a
full circle:
1
C   2r 
4
If the radii are solid lines,
they form part of the
perimeter. The formula is
therefore:
1
C   2r   r  r
4
192
The area of the quartercircle is a quarter the area
of the full circle:
1
A  r 2
4
Converting between different units using the International System of Units (SI)
It is so important for you to understand how to do conversions in Mathematics. The following
method will really help you to get to grips with doing conversions with km, m, cm and mm.
Whenever you are required to perform perimeter or area calculations, it is important to make
sure that all units are the same.
Converting lengths
Each semi-circular arc represents one ten.
If you are converting from km to m to cm
10
to mm, you are moving right and so you
km
multiply by the number of arcs. If you are
converting from mm to cm to m to km, you
are moving left, so you divide by the number of arcs.
Let’s look at some examples.

10
10

m
10
10
10
cm mm
(a)
Convert 5 m to cm.
You are moving from m to cm which is right. Therefore multiply 5 cm by 2 arcs
(10 10  100) .
 5 m 100  500 cm
(b)
Convert 3 000 mm to km
You are moving from mm to km which is left. Therefore divide 3 000 mm by 6 arcs
(10 10 10 10 10 10  1000 000) .
3 000 mm

 0, 003 km
1 000 000
Converting areas
Each semi-circular arc represents one hundred.
If you are converting from km 2 to m 2 to cm 2 to
mm 2 you are moving right and so you multiply
by the number of arcs. If you are converting from
mm 2 to cm 2 to m 2 to km 2 , you are moving left, so
you divide by the number of arcs.

100
km2
100

100
100
m2
100
100
cm2 mm2
Let’s look at some examples.
(a)
Convert 5 m 2 to cm 2 .
You are moving from m to cm which is right. Therefore multiply 5 m 2 by 2 arcs
(100 100  10 000) .
 5 m 2  10 000  50 000 cm 2
(b)
Convert 3 000 mm 2 to m 2 .
You are moving from mm to m which is left. Therefore divide 3 000 mm 2 by 3
arcs (100 100 100  1000 000) .

3 000 mm 2
 0, 003 m 2
1 000 000
193
Example 1
Determine the perimeter and area of the given shape
if each square block has a side equal to 1 cm .
Round off your answers to two decimal places.
Solution
Fill in the dimensions on the diagram.
The shape is made up of part of a rectangle
(one length and two widths) and a semi-circle
(diameter not included).
Perimeter of part of rectangle
 242
 8 cm
Circumference of semi-circle
1
  2r 
2
1
  2(2) 
2
1
  4
2
 (2)cm
 Perimeter of shape
 8  (2)
 14, 28 cm
Area of rectangular part
 4 2
 8 cm 2
Area of semi-circle
1
  r 2 
2
1
  (2)2 
2
1
  (4) 
2
 (2) cm 2
Area of shape
 Area of rectangle  Area of semi-circle
 8  (2)
 14, 28 cm 2
194
Example 2
Determine the perimeter and area of the given shape if each
square block has a side equal to 1 cm .
Round off your answers to two decimal places.
Solution
Fill in the dimensions on the diagram. The shape is made up
of a quarter-circle and a right-angled triangle.
Perimeter of shape:
Perimeter of quarter-circle  circumference of quarter-circle  length of two radii
1
  2(2)   2  2 (radii are solid lines and are included in the perimeter)
4
1
1
  (4)   4
 2(2)
4
4
 ( )  4
 7,141592654 cm
To calculate the perimeter of the triangle, first use Pythagoras to calculate the hypotenuse.
Let x be the hypotenuse.
x 2  (2) 2  (2) 2
 x2  8
Area of shape:
 x  8 cm
 Area of quarter circle  Area of triangle
1
1
 [(2) 2 ]  (2)(2)
4
2
1
 (4)  2
4
 (  2) cm 2
Perimeter of triangle
 8 22
 6,828427125 cm
Perimeter of the whole shape
 7,141592654  6,828427125
 13,97 cm
 5,14 cm 2
Example 3
Determine the shaded area (in cm 2 ) if a circle diameter 6 m
is inside a square. Round off your answers to two decimal places.
Solution
The shaded area is the area of the square minus the area of the circle.
Area of square
 (6  6)  36 m 2
Area of circle

 (3) 2  (9) m 2
100 100 100
100
100 100
2
2
2
mm2
cm
m
km
Shaded area

 36  (9)
7, 725666118 m 2 100 100
 7, 725666118 m 2
 77 256, 66 cm 2
 77 256, 66 cm 2
195
EXERCISE 1
(a)
Calculate the perimeter and area of each of the following shapes. Make sure all units
of measurement are the same and round off your answers to two decimal places where
appropriate.
(1) ABCD is a square
(2) ABCD is a rectangle
(3)
(4) ABC is given
(5) ABC is given
(6) DEF is given
(7)
ABC is given
(8) DEF is given
(9) ABC is isosceles
(10)
ABC is given
(11) ABC is isosceles
(12) PQR is given
60
(13)
AEB is isosceles
EBCD is a square
(14)
(15)
(16)
60
60
196
(17)
(18)
(19)
(20)
(21)
(22)
(23)
(24)
(26)
(27)
14 m
(25)
(28)
60
(b)
(29)
(30)
(31)
(32)
(33)
(34)
Calculate the shaded areas rounded off to two decimal places.
(1)
(2)
(3)
(4)
(5)
197
Example 4
Example 5
2
The area of a triangle is 45 cm and its A circle has a circumference of 15 m. Find
the radius of the circle (in mm) rounded off
height is 50 mm. Calculate the base.
to two decimal places.
Solution
Solution
A  45 cm 2 and h  50 mm
C  2πr
Convert 50 mm into cm:
50 mm  10  5 cm
15  2r
 h  5 cm
15
r 
1
A  (b)(h)
2
2
 r  2,387324146 m
1
 45  (b)(5)
Now convert 2,387324146 m into mm:
2
2,387324146 m 10 10 10
 90  5b
 2 387,324146 mm
 b  18 cm
 r  2 387,32 mm
Example 6
A circle has an area of 144 cm 2 . Find the
of the diameter rounded off to two decimal
places.
Example 7
A circle has an area of (169) m 2 . Find the
length of the circumference rounded off to
two decimal places.
Solution
Solution
A  r 2
A  r
2
144  r 2
144
r2 

144
r 
(r  0)

 r  6, 770275003 cm
 d  13,54 cm
169  r 2
169  r 2
 r 2  169
 r  13 m (r  0)
C  2(13)
 C  81, 68 m
EXERCISE 2
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
The area of a triangle is 48 m 2 and the base is 16 m. Calculate the height.
The area of a square is 36 m 2 . Calculate the length of one side.
The area of a rectangle is 80 m 2 . The length is 10 cm. Calculate the width.
The perimeter of a rectangle is 32 cm and its length is 90 mm. Calculate the width.
The area of a square is 225 m 2 . Calculate the perimeter.
A circle has a circumference of 18 cm. Find the length of the radius of the circle
rounded off to two decimal places.
A circle has an area of 33 m 2 . Find the length of the radius rounded off to two
decimal places.
A circle has an area of 86 cm 2 . Find the length of the diameter rounded off to two
decimal places.
A circle has an area of 76 cm 2 . Find the length of the circumference rounded off to
two decimal places.
A circle has an area of (121) cm 2 . Find the length of the radius rounded off to two
decimal places.
198
(k)
(l)
(m)
A circle has an area of (196) cm 2 . Find the length of the diameter rounded off to
two decimal places.
A circle has an area of (81) cm 2 . Find the length of the circumference rounded off to
two decimal places.
A semi-circle has a perimeter of 100 m. Calculate the length of the diameter in cm.
We already know how to calculate the perimeter and area of a circle, triangle, square and a
rectangle. Let’s now focus on the other quadrilaterals: parallelograms, rhombuses, trapeziums
and kites.
PARALLELOGRAMS AND RHOMBUSES
The perimeter of a parallelogram is
the sum of the length of all four sides.
In the given diagram, the perimeter is
P  2a  2b
The area of a parallelogram is derived using a rectangle and shifting a triangular part inside
the rectangle. In the rectangle given below, a shaded triangle is drawn. If the shaded triangle
is shifted left as shown below, then a parallelogram is formed. The length (l) for both figures
is the same. For the parallelogram, the length now becomes the base (b) and the width (w)
becomes the height (h).
Area of rectangle  l  w
 Area of parallelogram  b  h
Note:
In a parallelogram, the height is always perpendicular to the chosen base.
The perimeter of a rhombus is the sum of the length of all
four equal sides. In the given diagram, the perimeter is
P  4s
Since a rhombus is a parallelogram, the area of a rhombus
is determined using the same formula as a parallelogram:
Area of rhombus  b  h  s  h
In fact, the area is the length of any side multiplied by the
height since all four sides are equal.
199
Example 8
Example 9
Calculate the perimeter and area of
the following quadrilateral:
Calculate the perimeter and area of
the following quadrilateral:
Solution
Solution
ABCD is a parallelogram
DEFG is a parallelogram
 Opposite sides are equal.
 DE  8 m
Use Pythagoras to calculate DG:
DG 2  (3) 2  (4) 2
Perimeter of ABCD
 2(24)  2(6)
 48  12
 60 cm
Area ABCD
 base  height
 DG 2  25
 DG  5 m
 FE  5 m
Perimeter of DEFG
 2(5)  2(8)
 10  16
 26 m
 24  10
 240 cm 2
Area DEFG
 base  height
 8 4
 32 m 2
Example 10
Calculate the perimeter and area of the following quadrilateral:
Solution
PQRS is a rhombus
The area of a parallelogram (and a rhombus)
is equal to the chosen base multiplied by the
height.
Perimeter of PQRS
 4  side
 4  13
 52 units
Area ABCD
 side  height
 13  12
 156 units 2
200
EXERCISE 3
Calculate the perimeter and area of each of the following parallelograms and rhombuses.
Make sure all units of measurement are the same and round off your answers to two decimal
places where appropriate.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l)
(m)
(n)
ABFG is a parm
BCDE is a rhombus
BEHJ is a square
3
2
TRAPEZIUMS
In any trapezium, the perimeter is
the sum of the length of all four sides.
In the given diagram, the perimeter is
P  abcd
The area of trapezium ABCD is obtained
by drawing diagonal BD and creating two
triangles. The two triangles have the
same height and the sum of the areas
of these triangles is the area of the
trapezium.
Area of trapezium ABCD
 Area ABD  Area BCD
1
1
  AD  h   BC  h
2
2
1
 h(AD  BC)
2
1
 (AD  BC)  h
2
201
Example 11
Calculate the perimeter and area of the following trapezium:
Solution
EF  6 cm (opp sides of rectangle)
 BC  3  6  2, 2  11, 2 cm
Convert 35 mm to cm:
35 mm  10  3,5 cm
Perimeter of ABCD
 3,5  11,2  3  6
 23,7 cm
The parallel sides are AD and BC.
AD  6 cm
BC  11, 2 cm
 Area of ABCD
 12 (AD  BC)  AE
 12 (6  11, 2)  2
 17, 2 cm 2
EXERCISE 4
Calculate the perimeter and area of each of the following trapeziums rounded off to two
decimal places where appropriate:
(a)
(b)
(c)
(d)
(e)
(f)
202
(g)
(h)
(i)
(j)
(k)
(l)
KITES
In any kite, the perimeter is the sum of the length of all
four sides. The adjacent sides are equal. In the given diagram,
the perimeter of kite ABCD is P  2a  2b
The area of kite ABCD is obtained by drawing the diagonals
AC and BD and then determining the area of ABC and ADC
Area of kite ABCD
 Area ABC  Area ADC
1
1
  AC  BE   AC  ED
2
2
1
 AC(BE  ED)
2
The area is equal to half the
1
product of the diagonals.
 AC  BD
2
Example 12
Calculate the perimeter and area of the following kite:
125
Solution
AE  EC  5 units (BD bisects AC at E)
BE  AE  5 units (given)
AB2  (5) 2  (5) 2
 AB2  50
50
 AB  50 units
125
 BC  50 units
50
Since AD  CD , we know that
CD  125
203
125
Perimeter of ABCD
 2 50  2 125
 36,50 units
125
50
Area ABCD
1
 (product of diagonals)
2
1
1
 (BD  AC)  (15 10)  75 units 2
2
2
50
125
EXERCISE 5
Calculate the perimeter and area of each of the following kites rounded off to two decimal
places where appropriate:
(a)
(b)
(c)
(d)
(e)
(f)
60
Example 13
The area of kite ABCD is 120 000 cm 2 . Diagonal BD is
equal to 12 m. Calculate the length of diagonal AC in mm.
Solution
First convert 120 000 cm 2 to m 2 :
120 000 cm 2  10 000  12 m 2
1
 (AC)(12)  12
2
 (AC)(6)  12
 AC  2 m
 AC  2000 mm

100
km2
100
100

100
m2
100
100
cm2 mm2

10
km
10
10

204
m
10
10
10
cm mm
EXERCISE 6
(a)
(b)
(c)
(d)
(e)
(f)
The area of kite ABCD is 140 m 2 . Diagonal BD is equal to 14 m. Calculate the length
of diagonal AC in cm.
The area of kite ABCD is 900 000 cm 2 . Diagonal AC is equal to 18 m. Calculate the
length of diagonal BD in mm.
The area of a parallelogram is 75 cm 2 . Its height is 500 mm. Calculate the length of
the base in m.
The area of a parallelogram is 125 m 2 . Its base is 50 cm. Calculate the length of the
height in mm.
The area of a rhombus is 64 cm 2 and its height is 16 cm. Calculate the base of the
rhombus in mm.
The area of a rhombus is 100 m 2 and its height is 2000 cm. Calculate the perimeter of
the rhombus in mm.
DOUBLING THE DIMENSIONS OF TWO-DIMENSIONAL SHAPES
Example 14
If the length and breadth of rectangle ABCD are doubled, a new larger rectangle EFGH is
formed.
(a)
Calculate the perimeter of ABCD and EFGH. What do you notice?
Solution
Perimeter of ABCD  2(2)  2(3)  10 cm
Perimeter of EFGH  2(4)  2(6)  20 cm
It is clear that:
Perimeter of EFGH  2  Perimeter of ABCD.
(b)
Calculate the area of ABCD and EFGH. What do you notice?
Solution
Area of ABCD  (3)(2)  6 cm 2
Area of EFGH  (6)(4)  24 cm 2
It is clear that:
Area of EFGH  4  Area of ABCD.
Notice that rectangle ABCD can fit four
times into rectangle EFGH.
205
Example 15
Suppose that only the length of rectangle ABCD is doubled.
(a)
Calculate the perimeter of ABCD and EFGH. What do you notice?
Solution
Perimeter of ABCD  2(2)  2(3)  10 cm
Perimeter of EFGH  2(2)  2(6)  16 cm
It is clear that:
Perimeter of EFGH  Perimeter of ABCD  6 cm
(b)
Calculate the area of ABCD and EFGH. What do you notice?
Solution
Area of ABCD  (3)(2)  6 cm 2
Area of EFGH  (6)(2)  12 cm 2
It is clear that:
Area of EFGH  2  Area of ABCD.
Notice that rectangle ABCD can fit twice into rectangle EFGH.
EXERCISE 7
(a)
If the sides and height of rhombus ABCD
are doubled, a new larger rhombus EFGH is
formed.
(1)
Calculate the perimeter of ABCD
and EFGH. What do you notice?
(2)
Calculate the area of ABCD
and EFGH. What do you notice?
(b)
If the sides and height of square ABCD
are doubled, a new larger square EFGH is
formed.
(1)
Calculate the perimeter of ABCD
and EFGH. What do you notice?
(2)
Calculate the area of ABCD
and EFGH. What do you notice?
206
(c)
If the sides of trapezium ABCD are doubled
a new larger trapezium EFGH is formed.
Show that EFGH is four times the
size of ABCD.
(d)
If the diagonals of kite ABCD are doubled
a new larger kite EFGH is formed.
Show that EFGH is four times the
size of ABCD.
AC  6 cm
BD  4 cm
(e)
If the sides and height of parallelogram PQRS
are doubled, a new larger parallelogram TUVW
is formed.
(1)
Calculate the perimeter of
PQRS and TUVW.
What do you notice?
(2)
Calculate the area of PQRS
and TUVW. What do you notice?
(f)
The radius of circle centre A is half the
length of the radius of circle centre C.
The radii of the circles meet at B such
that ABC  9 units.
Show that the large circle is four times
the size of the small circle.
(g)
LMN is halved in size to form PQR .
Area of PQR
Calculate the ratio
.
Area of LMN
Interpret your answer.
(h)
The perimeter of a rectangle is 14 m and the area is 12 m 2 . If the sides of the
rectangle are doubled in length,
(1)
what will the perimeter of the enlarged rectangle be?
(2)
what will the area of the enlarged rectangle be?
(i)
A square has an area of 9 x 2 .
(1)
Write down the length of a side in terms of x.
(2)
Write down the perimeter in terms of x.
(3)
If the sides are doubled, write down the new perimeter in terms of x.
(4)
If the sides are doubled, write down the new area in terms of x.
(5)
If the area is doubled, write down the new perimeter in terms of x.
207
REVISION EXERCISE
(a)
Calculate the area and perimeter of the following shapes. Assume that the shapes are
made up of squares, rectangles, triangles or circles. Round off your answers to two
decimal places where appropriate.
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
(14)
208
(15)
(17)
(b)
(16)
ABCDEF is a regular hexagon
BCDE is a square
(18)
ABCDE is a regular pentagon
Calculate the unknown length(s) in each of the following shapes:
(1)
PQRS is a parallelogram with an area of 243 cm 2 and PS  27 cm .
(2)
ABCD is a parallelogram with an area of 340 cm 2 and height of 17 cm.
(3)
ABCD is a kite with an area of 525 m 2 , BD  35 m and AC  x .
(4)
ABCD is a rectangle with an area of 32 cm 2 and a length of 8 cm.
(5)
ABCD is a square with an area of 49 cm 2 .
209
(c)
(6)
ABC is a right-angled triangle with an area of 30 cm 2 and BC  12 cm .
(7)
A semi-circle with a circumference of (8) cm .
(8)
ABCD is a trapezium with an area of 48 m 2 has a height of 6 m. One of its
parallel sides is three times the length of the other parallel side.
In the given diagram, AD||BC.
(1)
Prove that Ê1  90
(2)
(3)
(4)
Calculate the size of Ĉ
Calculate the perimeter of ABCD.
Calculate the area of ABCD.
(d)
ABCD is a rhombus and EFGH is a square.
(1)
Calculate the area of ABCD using
the formula for the area of a rhombus.
(2)
Calculate the area of ABCD by
using the formula for the area of
a kite.
(3)
Calculate the area of EFGH using the formula for the area of a square.
(4)
Calculate the area of EFGH by using the formula for the area of a kite.
(5)
What can you conclude about ABCD and EFGH?
(e)
ABC is an equilateral triangle with a perimeter of 6 cm. The length of each side of
the triangle is increased by 2 cm to form DEF .
(1)
Show that the perimeter of DEF is double that of ABC .
(2)
Show that the area of DEF is four times that of ABC .
(f)
The perimeter of a square is equal to 48 cm. The length of each side is doubled.
(1)
What is the length of one of the sides of the enlarged square?
(2)
What is the area of the original square?
(3)
What is the area of the enlarged square?
(4)
How many times is the enlarged square bigger than the original square?
(g)
One of the congruent triangles in a regular hexagon has an area of 20 m 2 . If the sides
of the hexagon are doubled, what will the area of the enlarged hexagon be?
210
SOME CHALLENGES
A square is drawn inside a circle so that its vertices
lie on the circle.
One side of the square is equal to 20 cm.
(1)
Calculate the diameter of the circle.
(2)
Calculate the shaded area rounded off
to one decimal place.
(b)
A cross section of a toy is shown in the diagram.
(1)
Show that the external perimeter is equal to
12(  2)
(2)
Calculate the internal area rounded off to
one decimal place.
8
4 7
(a)
(c)
A small circle is drawn inside a large circle. Determine
the ratio of the area of the large circle to the smaller circle.
(d)
Calculate the shaded area inside the square.
(e)
Identical circles are cut out of a rectangular cardboard.
The length of the rectangle is 90 cm and the width
is 30 cm. The radius of each circle is 5 cm. Calculate:
(1)
the number of circles that can be cut out.
(2)
the total area of all the circles.
(3)
the area remaining once the circles have been
cut out of the rectangle.
(f)
The area of the small circle is 4 and the area of the
larger circle is 94 times the area of the small circle.
Calculate the length of the rectangle.
211
CHAPTER 15: PATTERNS, FUNCTIONS AND ALGEBRA
TOPIC: GRAPHS
INTERPRETING GRAPHS
In this chapter, we will revise different types of real-world graphs called global graphs that
you studied in Grade 8. We will then sketch the graphs of linear functions (straight lines).
REVISION OF DISCRETE AND CONTINUOUS GLOBAL GRAPHS
Discrete data is data that is counted. Counted data can only be whole numbers. For example,
the number of goals scored by Portugal in each match played in the recent world cup is
discrete data. You count the number of goals per match. This can be represented in a graph.
Dots are used to indicate the number of goals scored per match.
Continuous data is data that is measured. This data can involve real numbers rather than
just whole numbers. For example, suppose that the weight, in kilograms, of people going to
gym is related to the amount of nutrients taken by those people per day (in grams). The
weight and nutrients taken is measured rather than counted. It is possible for a person to
weigh 61,85kg, which is not a whole number. The weight of the person is not restricted to
whole numbers only. The amount of nutrients taken by that person can be 4,56g and not just
restricted to whole number grams. Other examples include the cost of material per metre or
the distance travelled against time. The global graph will not only consist of a few dots, but
millions of dots forming a solid straight line or curve.
Example 1
When bacteria are grown in a closed system like a test tube, the
population of cells almost always grow according to the pattern
shown in the graph below. The graph represents the number of
cells in the test tube (in millions) against time in hours. The
number of cells is recorded at hourly intervals.
No of cells (in millions)
[http://textbookofbacteriology net/growth_3 html]
18
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
1
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
2
3
4
5
6
7
8
9
How many cells are there in the test tube initially?
How many cells are there in the test tube after 1 hour?
Explain what is happening between A and B.
After how many hours are there 7 million cells in the test tube?
Explain what is happening between B and C.
Explain what is happening between C and D.
After how many hours are there 4 million cells in the test tube?
Explain what is happening between D and E.
Is the data discrete or continuous? Explain.
Can you read off from the graph the number of cells after one and a half hours?
212
10
(k)
What is the maximum number of cells in the test tube during the 10-hour time
period?
Solutions
(a)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
2 million
(b)
2 million
The initial cell population remains the same since the cells are adapting to their
new environment. The trend is linear (straight line).
After 3 hours.
The cells are dividing regularly and are growing in a non-linear way.
Growth cannot continue forever in the test tube. There is a limit to the available
nutrients for the cells to survive and there is also a limited amount of space in the
test tube. Therefore the cells stop dividing between C and D. The trend is linear.
After 2 hours and 9 hours.
The cell population starts decreasing (dying off) between D and E. At E, there are
no viable cells in the test tube.
The data is discrete since the number of bacteria is restricted to whole numbers.
No. The graph shows the number of cells at hourly intervals only. It is not possible
to know exactly how many cells are in the test tube at any time between these time
intervals based on the graph.
15 million.
Example 2
A leading expert in Cape Town believes that eating fewer
carbohydrates can help a person lose weight. Carbohydrates are foods
like pasta, cereal or bread. A researcher investigated this claim by
cutting out his carb-intake over a period of 12 weeks. The graph
below shows his body weight over time.
95
94
93
92
91
90
89
88
87
86
85
84
83
82
81
80
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
0
1
2
3
4
5
6
7
8
9
10
11
What was his weight at the start of the research?
What was his maximum and minimum weight during the 12 week period?
What was his weight after one week?
What was his weight approximately after 4 weeks?
During which week was his weight 84,5kg?
After how many weeks did his weight drop to 81kg?
Explain the change in his weight from O to A.
Explain the change in his weight from A to B.
Can you predict what his weight was during the 12th week?
Is the data discrete or continuous? Explain.
213
12
Solutions
(a)
(c)
(g)
(h)
(i)
(j)
94kg
(b)
Max weight: 94kg Min weight: 81kg
90kg
(d)
83,2kg
(e)
Third week
(f)
12 weeks
In just one week there appears to be a rapid decrease is his weight.
There is a steady decrease which is less rapid than during the first week.
His weight probably was 81kg.
The data is continuous since the weights are real numbers and it is possible to
determine an approximate measure of weight at any time throughout the 12 weeks.
Note: Not everyone agrees with the expert that few or no carbs is the way to go for losing
weight. Some people believe that carbs are essential in a diet. As an interesting
exercise, go on the internet and Google search what other people are saying about
this. Form your own opinion and maybe your class can have a discussion about this.
The trends in a discrete or continuous graph can be linear or non-linear. Graphs with a linear
trend follow a straight-line pattern. Graphs with a non-linear pattern follow a curved pattern.
Example 3
For each of the following graphs:
(1)
State whether the graph is discrete or continuous.
(2)
State whether the graph has a linear or non-linear trend.
(3)
State whether the graph is increasing, decreasing or constant.
(4)
Write down the maximum and minimum values (as read from the vertical axis).
(a)
(1)
(2)
(3)
(4)
Continuous
Non-linear
Decreasing
Min: 20
Max: 70
(b)
(1)
(2)
(3)
(4)
Continuous
Linear
Decreasing
Min: 0
Max: 60
(c)
(1)
(2)
(3)
(4)
Continuous
Linear
Increasing
Min: 10
Max: 70
(d)
(1)
(2)
(3)
(4)
Discrete
Non-linear
Decreasing
Min: 0
Max: 80
(e)
(1)
(2)
(3)
(4)
Continuous
Non-linear
Increasing
Min: 10
Max: 60
(f)
(1)
(2)
(3)
(4)
Discrete
Non-linear
Increasing
Min: 20
Max: 80
(g)
(1)
(2)
(3)
(4)
Continuous
Linear
Constant
No min or max
(h) )
(1)
(2)
(3)
(4)
Discrete
Linear
Increasing
Min: 10
Max: 70
214
EXERCISE 1
(a)
South Africa is said to hold over two-thirds of the world
population of rhinos. The Black rhino is a critically
endangered species and is at grave risk of extinction due to
poaching. Over the past few years, rhino poaching in South
Africa has increased significantly. This increase in poaching
is driven by a belief that rhino horn has medicinal powers. The increased value of
rhino horn has enticed well-organised, well-financed and highly-mobile criminal
groups to become involved in rhino poaching. According to the World Wildlife Fund
(WWF), the Kruger National Park, South Africa’s world-renowned game reserve lost
146 rhinos to poaching in 2010. The graph below represents the number of rhinos
poached from 2008 to 2014. The dots represent the number of rhinos poached for that
year(s).
[www.environment.gov.za/mediarelease/rhinopoaching_statistics]
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(b)
How many rhinos were poached in 2010?
What is the minimum number of rhinos poached per year starting from 2009?
What is the maximum number of rhinos poached per year starting from 2009
to the end of 2013?
In which year did poachers slaughter 668 rhinos?
It seems as if there was a drop in the number of rhinos poached from 2008 to
2009. Is this true? Explain.
Is the trend of poaching linear or non-linear?
Is the data discrete or continuous? Explain.
From the graph, it seems as if the number of rhinos poached dropped from
2013 to 2014. Is this true? Explain.
The South African government has put various anti-poaching units in place to
arrest and charge poachers. In your opinion, do you feel that the government
is justified in charging poachers, bearing in mind that Rhino horn is a valuable
product in many countries of the world? Some people feel that rhinos could be
killed in a more humane way.
Global warming is a gradual increase in the overall temperature of
the earth's atmosphere caused by increased levels of carbon
dioxide and other pollutants in the atmosphere. World-leading
scientists have concluded that humans have caused all or most of
the current warming. The effects of global warming threaten life on
planet Earth. Climate change has an effect on the ecosystem, water
supplies and food production. Millions of animal species could become extinct.
Weather patterns will change giving rise to stronger and more frequent storms. A
graph of overall global averages temperatures for given year periods is given below.
215
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(c)
What was the average global temperature for 1921-1930?
During which years was the temperature 14 C ?
During which years was the temperature a minimum?
During which years was the temperature a maximum?
During which years did the temperature remain constant?
During which years was there an increase in temperature?
During which years was there a decrease in temperature?
Is the data discrete or continuous? Explain.
What do you predict the temperature to be from 2015-2020?
There some global warming sceptics that disagree with the scientists. They
argue that the worldwide rise in surface temperature is due to natural
cycles and is not caused by humans. Google search global warming on the
internet to find out what people are saying. What is your opinion of global
warming? Is it caused by nature or by humans?
The following graph represents the distance of an object from O. The vertical
axis represents the distance in metres of the object from O. The horizontal axis
represents the time.
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
How far away from O is the object at 07h00?
How far away from O is the object at 08h00?
How far has the object moved in the first hour?
How far away from O is the object at 10h30?
Explain what is happening between A and B.
During which times was the object’s distance away from O increasing?
During which times was the object’s distance away from O decreasing?
What is the maximum distance of the object away from O?
At what time will the object’s distance away from O be zero?
What is the total distance moved by the object from 07h00 to 12h00?
216
(d)
When a diver dives down into the ocean, the water above
exerts a pressure (P) on the diver. For every 10,6 metres
under water, the pressure on the diver increases by 6,6kg. In
the deepest ocean, the pressure can be equivalent to having
the diver support 50 commercial airplanes. The equation
representing the pressure against depth of the diver under
water is given by P  0, 6D  0, 24 where D is the depth under the water in metres and
P is the pressure of the water above the diver as the depth increases or decreases. The
graph of a diver’s movement under water is shown in the graph below.
50
45
40
35
30
25
20
15
10
5
(1)
(2)
(3)
(4)
(5)
(6)
(7)
Show that the pressure is 6,6 kg if the depth is 10,6m.
Calculate the pressure at a depth of 21,5m. Show this on the graph.
Calculate the depth of the diver when the pressure is 15,5. Show this on the
graph.
What is the maximum depth that the diver dives? Explain why the diver
doesn’t continue increasing the depth.
Explain what is happening from O to A.
Explain what is happening from A to B.
Is the graph discrete or continuous? Explain.
REVISION OF PLOTTING POINTS IN THE CARTESIAN PLANE
A Cartesian plane consists of a
horizontal number line called the
x-axis and a vertical number line
called the y-axis. The two axes
intersect at right angles at 0,
which is called the origin.
The axes divide the plane into
four different quadrants. It extends
infinitely in all directions. To show
7
this we put arrows at the ends
of the axes.
A point on the Cartesian plane is
represented in terms of:
 its position relative to the origin, or
 its distance from the two axes.
6
5
4
3
2
1
6
5
4 3 2
1
0
1
2
3
4
5
6
217
1
2
3
4
5
6
7
Example 4
Consider the following points:
A(3 ; 4) , B(  2 ; 4) , C(  4 ;  5) ,
D(4 ;  4) , E(0 ; 5) , F(7 ; 0) ,
G(0 ;  6) and H(  7 ; 0)
(a)
(b)
(c)
(d)
(e)
6
B(  2 ; 4)
E(0 ; 5)
5
A(3 ; 4)
4
4
3
2
5
4
If we consider how the
 7 F(7 ; 0)
2 1
3
7
H(  7 ; 0)
first quadrant point A(3 ; 4)
3 4
2
5
1
6
7 6 5 4 3 2 1 0
7
lies relative to the origin:
4
4
1
The number 3 tells us that
2
the point at the origin first moves
5
4
3
3 units to the right. The number
4 tells us that the point then moves
4
D(4 ;  4)
4 units up.
5
C(  4 ;  5)
If we consider the position of the
6 G(0 ;  6)
point P(3 ; 4) relative to the axes:
The number 3 tells us that the point is 3 units away from the vertical axis. This
value can be read on the x-axis and the number 3 is the x-coordinate. The number
4 tells us that the point is 4 units away from the horizontal axis. This value can be
read on the y-axis and the number 4 is the y-coordinate.
If we consider how the second quadrant point B(  2 ; 4) lies relative to the origin:
The number 2 tells us that the point at the origin first moves 2 units to the left. The
number 4 tells us that the point then moves 4 units up.
If we consider the position of the point B(  2 ; 4) relative to the axes:
The number 2 tells us that the point is 2 units away from the vertical axis. This
value can be read on the x-axis and the number 2 is the x-coordinate. The number
4 tells us that the point is 4 units away from the horizontal axis. This value can be
read on the y-axis and the number 4 is the y-coordinate.
If we consider how the third quadrant point C(  4 ;  5) lies relative to the origin:
The number 4 tells us that the point at the origin first moves 4 units to the left. The
number 5 tells us that the point then moves 5 units down.
If we consider the position of the point C(  4 ;  5) relative to the axes:
The number 4 tells us that the point is 4 units away from the vertical axis. This
value can be read on the x-axis and the number 4 is the x-coordinate. The number
5 tells us that the point is 5 units away from the horizontal axis. This value can be
read on the y-axis and the number 5 is the y-coordinate.
If we consider how the fourth quadrant point D(4 ;  4) lies relative to the origin:
The number 4 tells us that the point at the origin first moves 4 units to the right. The
number 4 tells us that the point then moves 4 units down.
If we consider the position of the point D(4 ;  4) relative to the axes:
The number 4 tells us that the point is 4 units away from the vertical axis. This
value can be read on the x-axis and the number 4 is the x-coordinate. The number
4 tells us that the point is 4 units away from the horizontal axis. This value can be
read on the y-axis and the number 4 is the y-coordinate.
If we consider point E(0 ; 5) relative to the origin:
The number 0 tells us that the point doesn’t move horizontally. The number 5 tells
us that the point simply moves up 5 units. The x-coordinate is 0 and the y-coordinate
is 5.
218
(f)
If we consider point F(7 ; 0) relative to the origin:
The number 7 tells us that the point moves 7 units right. The number 0 tells doesn’t
move vertically. The x-coordinate is 7 and the y-coordinate is 0.
If we consider point G(0 ;  6) relative to the origin:
The number 0 tells us that the point doesn’t move horizontally. The number 6 tells
us that the point simply moves down 6 units. The x-coordinate is 0 and the ycoordinate is 6 .
If we consider point H(  7 ; 0) relative to the origin:
The number 7 tells us that the point moves 7 units left. The number 0 tells doesn’t
move vertically. The x-coordinate is 7 and the y-coordinate is 0.
(g)
(h)
THE LINEAR FUNCTION
The graph of a straight line is called a linear function. The equation of a linear function has
the form y  mx  c . Here are some examples of equations of straight lines.
For y  2 x
For y  x  5
For y  3x  1 :
The coefficient of x is 2
The coefficient of x is 1
The coefficient of x is 3
m  3
 m  2
m 1
The constant term is 1
The constant term is 1
The constant term is 5
c  1
c  0
c  5
Sketching lines using the table method and dual-intercept method
Sketching lines using a table is something you are already familiar with.
y
Example 5
Sketch the graph of y  2 x  4 .
y  2x  4
Solution
(3 ; 2)
(2 ; 0)
.
Let’s select x  3 ;  2 ;  1; 0 ;1; 2 ; 3 and then
substitute these x-values into the equation
to get the corresponding y-values.
y  2(3)  4  10
For x  3
y  2(2)  4  8
For x  2
y  2(1)  4  6
For x  1
y  2(0)  4  4 (y-intercept)
For x  0
y  2(1)  4  2
For x  1
y  2(2)  4  0 (x-intercept)
For x  2
y  2(3)  4  2
For x  3
Fill these values in a table, plot the points on
a Cartesian plane and then draw the line.
x
y
3
10
2
8
1
6
0
4
1
2
2
0
(1 ;  2)
(0 ;  4)
( 1 ;  6)
( 2 ;  8)
.
x
x-intercept
y0
y-intercept
x0
( 3 ;  10)
3
2
There are many more points that
can be found. All these points
form the graph of the straight line.
Notice that where the line cuts the y-axis, the x-value is 0 and where the line cuts the x-axis,
the y-value is 0. It is possible to sketch a line if we know the intercepts with the axes.
To get the y-intercept, we let x  0 in the equation and solve for y.
To get the x-intercept, we let y  0 in the equation and solve for x.
This method is called the dual-intercept method and it is discussed in the next example.
219
Example 6
Sketch the graph of y  2 x  4 using the dual-intercept method.
Solution
From the previous example, we know the following:
To get the y-intercept let x  0
To get the x-intercept let y  0
Using this dual-intercept method, let’s draw the graph of the given line.
y
To get the y-intercept let x  0
y  2x  4
 y  2(0)  4
 y  4
The coordinates of the y-intercept are (0 ;  4)
To get the x-intercept let y  0
y  2x  4
0  2x  4
 4  2x
x  2
The coordinates of the x-intercept are (2 ; 0)
Using the two intercept points, we can draw
the graph of the line.
y  2x  4
. (2 ; 0)
x
.(0 ;  4)
Example 7
Sketch the graph of y  3x  7 by using:
(a)
the table method
(b)
the dual-intercept method
Solutions
(a)
It is only necessary to select three
x-values when using the table method.
Select say x  1; 0 ;1
For x  1 y  3(1)  7  10
For x  0 y  3(0)  7  7
For x  1 y  3(1)  7  4
x
y
1
10
0
7
1
4
y
.(1 ; 10)
.(0 ; 7)
.(1 ; 4)
x
Note: It is usually required that both the
x-intercept and the y-intercept be shown
on the graph. The dual-intercept method
will give both intercepts.
(b)
y  3 x  7
Using the dual-intercept method:
To get the y-intercept let x  0
y  3 x  7
 y  3(0)  7
y 7
The coordinates of the y-intercept are (0 ; 7)
220
To get the x-intercept let y  0
y  3 x  7
 3x  7
x 
7
3
y  3 x  7
 2 13
The coordinates of the x-intercept are (2 13 ; 0)
y
.(1 ; 10)
.(0 ; 7)
.(1 ; 4)
.(2
1
3
; 0)
x
EXERCISE 2
(It is extremely important that you do this exercise thoroughly as you will need to refer the
graphs in this exercise later on in Exercise 5)
(a)
Sketch the following lines on the same set of axes using the table method:
yx
(2)
y  2x
(3)
y  3x
(1)
(4)
y  4x
(5)
y  x
(6)
y  2 x
(7)
y  3 x
(8)
y  4 x
(9)
y  5 x
y  12 x
(11)
y   12 x
(10)
Use x  2 ; 0 ; 2 in (10) and (11)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
Sketch the following lines on the same set of axes using the dual-intercept method:
y  x 1
y  x 1
y  x2
(2)
(3)
(1)
y  x2
y  x3
y  x 3
(4)
(5)
(6)
y  x4
y  x4
y  x 5
(7)
(8)
(9)
Sketch the following lines on the same set of axes using the dual-intercept method:
y  x 1
y   x 1
y  x  2
(2)
(3)
(1)
y  x  2
y  x  3
y  x  3
(4)
(5)
(6)
y  x  4
y  x  4
y  x  5
(7)
(8)
(9)
Sketch the following lines on the same set of axes using the dual-intercept method:
y  2x  2
y  2x  2
y  2x  4
(2)
(3)
(1)
(4)
y  2x  4
(5)
y  2 x  4
(6)
y  2 x  4
(7)
y  2x  6
(8)
y  2x  6
(9)
y  2 x  6
Consider y  5 x
(1)
Determine the intercepts with the axes using the dual-intercept method.
(2)
Why is it not possible to draw the graph?
(3)
Now draw the graph using the table method.
Consider y   13 x
(1)
Determine the intercepts with the axes using the dual-intercept method.
(2)
Why is it not possible to draw the graph?
(3)
Now draw the graph using the table method.
Using the graphs drawn in (a), (b), (c), (d) and (e) and their equations, what can you
conclude about the constant term (c) in each equation?
Sketch the following lines on the same set of axes using the dual-intercept method:
y  3x  1
(2)
y  3x  1
(3)
y  2x  3
(1)
(4)
y  2x  3
(5)
y  4 x  5
(6)
y  4 x  5
(7)
y  4x  8
(8)
y  4x  8
(9)
y  4 x  6
y  x7
(11)
y  x7
(12)
(10)
y  12 x  2
221
From the previous exercise, the following conclusions can be made:
 The constant term (c) in the equation y  mx  c represents the y-intercept.
 The y-intercept can be found by letting x  0 in the equation and solving for y.
 The x-intercept can be found by letting y  0 in the equation and solving for x.
 The dual-intercept method is not helpful when the y-intercept is 0.
Example 8
Without actually sketching the following lines, determine the coordinates of the intercepts
with the axes using the dual-intercept method:
y  5 x  9
(b)
4 x  3 y  12
(a)
Solutions
(a)
(b)
c  9 (y-intercept)
The coordinates of the y-intercept are (0 ; 9)
For the x-intercept let y  0
0  5 x  9
5x  9
9
4
x  1
5
5
4 x  3 y  12
To get the x-intercept let y  0
To get the y-intercept let x  0
4 x  3 y  12
4 x  3(0)  12
 4(0)  3 y  12
 4 x  12
 3 y  12
x  3
y 4
The coordinates of the y-intercept
The coordinates of the x-intercept
are (3 ; 0)
are (0 ; 4)
EXERCISE 3
(a)
Without actually sketching the following lines, determine the coordinates of the
intercepts with the axes using the dual-intercept method:
(1)
y  3x  4
(2)
y  3x  4
(3)
y  6x  2
y  6x  2
(5)
y  7 x  4
(6)
y  7 x  4
(4)
(7)
y  x9
(8)
y  x 9
(9)
y   x  11
y   x  11
(10)
(11)
y  13 x  3
(12)
y  13 x  3
(13)
(16)
(19)
(b)
y   15 x  1
2x  3y  6
4 x  3 y  24
(14)
(17)
(20)
y   15 x  1
x  4y  8
3x  3 y  6
(15)
(18)
(21)
y  72 x  4
3x  y  3
1 x  1 y 1
2
4
Sketch the graph of each of the following using the dual-intercept method:
y  6 x  3
(2)
y  6 x  3
(3)
y  5x  8
(1)
1
x  3y  6
5 x  10 y  20
(4)
y  3 x2
(5)
(6)
You must have noticed from (e) and (f) in the previous Exercise 2 that the dual-intercept
method is not always helpful particularly if the line cuts the axes at the origin where the xintercept and y-intercept are the same. This leads us to the third method of sketching lines. It
is called the point-intercept method.
222
Sketching lines using the point-intercept method
The point-intercept method involves using the y-intercept and then finding another point by
selecting any x-value, substituting it into the equation to get the corresponding y-value.
Example 9
Sketch the graphs of the following lines:
(a)
y  2 x
(b)
y  13 x
y
Solutions
(a)
y  2 x  0 (Add in 0)
y  2 x
The y-intercept is c  0
The x-intercept is therefore also 0.
The dual-intercept method is not helpful
here at all since you need two different
(0 ; 0)
x
intercepts to be able to draw the graph.
With the point-intercept method, simply
select any value for x and substitute it
into the equation to get the y-value.
(1 ;  2)
Choose x  1 say.
 y  2(1)  0  2
Another point on the line is (1;  2)
Plot the two points on a Cartesian
y
plane and then draw the line.
1
y  x0
(Add in 0)
3
The y-intercept is c  0
1
y x
Now select a value for x and substitute
3
into the equation to get the y-value.
(3 ; 1)
Choose x  3
x
(0
;
0)
1
 y  (3)  0  1
3
Another point on the line is (3 ;1)
Plot the two points and then draw the line.
Notice that we chose x  3 to avoid getting a fraction for y.
By the way, you could have also used the table method to draw the graph. However,
the point-intercept method saves a lot of time.
.
(b)
EXERCISE 4
(a)
Sketch the following lines using the point-intercept method:
yx
(2)
y  2x
(3)
(1)
(4)
y  4x
(5)
y  x
(6)
(7)
y  3 x
(8)
y  4 x
(9)
(10)
(b)
y   12 x
(11)
y  34 x
(12)
y  3x
y  2 x
y  12 x
y   23 x
Consider x  2 y
(1)
Determine the intercepts with the axes using the dual-intercept method.
(2)
What do you notice?
(3)
Now draw the graph of this line using either the table method or pointintercept method.
223
Increasing and decreasing linear functions
A line is said to be increasing if the y-values increase as the x-values increase.
A line is said to be decreasing if the y-values decrease as the x-values increase.
Example 10
The graphs of y  2 x  4 and y  3x  6 are drawn below.
y y  2x  4
y  3 x  6
y
( 1 ; 9)
(3 ; 2)
(2 ; 0)
(0 ; 6)
x
.(1; 3)
.(2 ; 0)
(1 ; 2)
(0 ;  4)
( 1 ;  6)
.
.
( 2 ;  8)
x
(3 ; 3)
( 3 ;  10)
For the graph of y  2 x  4
x  3 ;  2 ;  1; 0 ;1; 2 ; 3
For the graph of y  3 x  6
x  1; 0 ;1; 2 ; 3
As the values of x increase,
the y-values increase.
The line is an increasing function.
It slopes up from left to right.
As the values of x increase,
the y-values decrease.
The line is a decreasing function.
It slopes down from left to right.
y  10 ;  8 ;  6 ;  4 ;  2 ; 0 ; 2
y  9 ; 6 ; 3 ;  4 ; 0 ;  3
EXERCISE 5
Refer to the graphs of the lines sketched in Exercise 2 (a)-(d).
(a)
For each line, state whether the line is increasing or decreasing.
(b)
How does the coefficient of x in the given equation relate to whether the line is
increasing or decreasing?
(c)
What do you notice about lines which have equations where the coefficient of x is
the same?
From the previous exercise, the following should be clear to you:
If the coefficient of x in the equation y  mx  c is positive ( m  0 ), then the line is

increasing (slopes up) as read from left to right.
If the coefficient of x in the equation y  mx  c is negative ( m  0 ), then the line is

decreasing (slopes down) as read from left to right.
Lines which have equations with the same coefficient of x are parallel.

224
Gradient of a line
The gradient (or slope) of a line is defined to be the rate at which the y-values change as the
change in y -values (vertical change)
.
x-values change. It is expressed as the ratio
change in x-values (horizontal change)
Example 11
Consider the graph of the line y  3 x .
Consider the points (1;  3) and (3 ; 9) :
Moving from left to right, the y-values
increase by 12 (from 3 to 9)
and the x-values increase by 4 (from 1 to 3)
change in y -values 12


3
change in x-values 4
Consider the points (0 ; 0) and (2 ; 6) :
Moving from left to right, the y-values
increase by 6 (from 0 to 6)
and the x-values increase by 2 (from 0 to 2)
change in y -values 6


3
change in x-values 2
y
y  3x
4
2
12
( 1 ;  3)
.
(3 ; 9)
.(2 ; 6)
6
.(0 ; 0)
x
Notice that the value of the gradient is the same (constant) no matter which two points on
the line are used. The coefficient of x in the equation of the line y  3 x represents the
gradient. Since the y-values increase as the x-values increase, the line is increasing. The
gradient is also positive.
Let’s summarise these findings:
 Between any two points on the line the gradient (slope) is the same or constant.
 The coefficient of x in the equation of the line represents the gradient.
 The line is increasing (slopes up) as read from left to right since the y-values increase as
the x-values increase. The gradient is positive.
Example 12
Consider the graph of the line y   12 x .
Consider the points (2 ;1) and (2 ;  1) :
y
( 6 ; 3)
Moving from left to right, the y-values
1
y x
decrease by 2 (from 1 to 1 )
2
and the x-values increase by 4
( 2 ;1)
(from 2 to 2)
5
change in y -values 2
1



x
2
(0 ; 0)
change in x-values 4
2
(2 ;  1)
Consider the points (6 ; 3) and (4 ;  2) :
4
Moving from left to right, the y-values
10
(4 ;  2)
decrease by 5 (from 3 to 2 ) and the
x-values increase by 10 (from 6 to 4)
change in y -values 5
1



change in x-values 10
2
Notice that the value of the gradient is the same (constant) no matter which two points on
the line are used. The coefficient of x in the equation of the line y   12 x represents the
.
.
.
.
.
gradient. Since the y-values decrease as the x-values increase, the line is decreasing. The
gradient is also negative.
225
Let’s summarise these findings:
 Between any two points on the line the gradient is the same or constant.
 The coefficient of x in the equation of the line represents the gradient.
 The line is decreasing (slopes down) as read from left to right since the y-values
decrease as the x-values increase. The gradient is negative.
EXERCISE 6
For each of the following lines:
(a)
Determine the gradient by using any two points on the line.
(b)
State whether the gradients are positive or negative.
(c)
Write down the equation in the form y  mx .
(d)
State whether the lines are increasing or decreasing. Explain.
y
y
y
B
.
(7 ; 7)
.(4 ; 2)
.
(0 ; 0)
x
( 2 ;  4)
( 6 ;  3)
y
G
(7 ; 7)
E
(3 ; 3)
( 4 ; 2)
.
(0 ; 0)
.
.(5 ;10)
.(2 ; 6)
y
( 5 ;10)
.
(0 ; 0)
H
.(3 ; 6)
.(0 ; 0)
(0 ; 0)
x
.
x
(0 ; 0)
x
D
.(3 ; 9)
.(3 ; 3)
A
x
(0 ; 0)
y
C
y
.(3 ; 9)
.(2 ; 6)
.(0 ; 0)
(6 ;  3)
The steepness of a line (optional enrichment if time permits)
Before discussing the steepness of a line, let’s briefly summarise what we know about the
gradient of a line thus far. The graphs of the lines in the previous exercise are drawn on the
same set of axes below. Graphs A – D will be drawn on one set of axes and graphs E – G will
be drawn on another set of axes.
y
m3
D
m2
y
C
m 1
B
m
A
1
2
x
x
H
m  3
The gradients of the lines are positive and
the lines are increasing from left to right.
E
1
m
F
2
m  1
G
m  2
The gradients of the lines are negative and
the lines are decreasing from left to right.
226
x
x
The gradient of a line is the rate at which the y-values change with respect to the x-values.
If the y-values increase as the x-values increase, the gradient is positive and the line is
increasing. If the y-values decrease as the x-values increase, the gradient is negative and the
line is decreasing. In other words, gradient tells us the direction of the line. If the gradient is
positive, the line slopes upwards. If the gradient is negative, the line slopes downwards.
However, the lines slope upwards or downwards in different ways.
For example, the lines A – D all slope upwards and have a positive gradient but the way they
slant upwards is different for each line. Likewise, the lines E – H all slope downwards and
have a negative gradient but the way they slant downwards is different for each line.
The steepness of a line is the way that the line slants upwards or downwards from left to
right. The steepness of a line depends on the following factors:
 The size of the angle between the line and the horizontal (x-axis)
 The values of the vertical and horizontal changes
Comparing the steepness of lines using angles between the lines and the horizontal:
D
C
B
Lines slope downwards
E
F
A
27
63
45
H
G
45
27
63
72
72
Lines slope upwards
As the angle between the lines and the horizontal (x-axis) increase in size, the steeper the
lines become. Line D is steeper than line C which, in turn is steeper than line B. Line B is
steeper than line A. The athlete would find it easy to run up line A, less easy to run up line B.
However, due to lines C and D being too steep upwards, it would be impossible for him to
run up these lines. Likewise, line H is steeper than line G which, in turn, is steeper than line
F. Line F is steeper than line E. The athlete would find it easy to run down line E, less easy to
run down line F. However, due to lines G and H being too steep downwards, it would be
dangerous for him to run down these lines!
Comparing the steepness of lines using the values of the vertical and horizontal changes:
(a)
You can compare the steepness of different lines by comparing the positive value of
the vertical changes. The horizontal changes must be kept constant.
The gradients of the previous lines will be changed into fractions that can be used to
compare the positive vertical changes. Negative signs are ignored for lines E – H.
y
For line A:
m
1
2
( 12 )
For line B:
m 1
( 22 )
( 42 )
( 62 )
For line C:
m2
For line D:
m3
D
6
m   12
( 12 )
For line F:
m  1
( 22 )
For line G:
m  2
( 42 )
For line H:
m  3
( 62 )
D
C
2
4
.
.
2
1
G
F
E
4
2
A
B
2
2
1
x
A
y
.H
6
.
4
.
.
H
G
6
2
F
4
2
1
2
227
C
6
B
2
For line E:
2
x
2
2
E
1
2
2
For line A:
vertical change
horizontal change
 12
For line E:
vertical change
horizontal change
 12
For line B:
vertical change
horizontal change

2
2
For line F:
vertical change
horizontal change

2
2
For line C:
vertical change
horizontal change

4
2
For line G:
vertical change
horizontal change

4
2
For line D:
vertical change
horizontal change

6
2
For line H:
vertical change
horizontal change

6
2
When comparing lines, if the positive vertical change gets increasingly greater, then
the lines get progressively steeper.
Line D is steeper than line C. Line C is steeper than line B and A is the least steep.
Notice that for line D, the vertical change is 6 and the vertical change is 4 for line C.
Line D is therefore steeper than line C.
Line H is steeper than line G. Line G is steeper than line F and E is the least steep.
Notice that for line G, the vertical change is 4 and the vertical change is 2 for line F.
Line G is therefore steeper than line F.
(b)
You can also compare the steepness of lines by relating the positive vertical
change to the horizontal change. If the positive vertical change is greater than the
horizontal change for one line and the other way round for another line, then the first
line is steeper than the second line. For example, consider line A and line D.
vertical change
For D: horizontal
 62 . The vertical change is greater than the horizontal change.
change
For A:
vertical change
horizontal change
 12 . The vertical change is less than the horizontal change.
Therefore, line D is steeper than line A.
(c)
You can also compare the steepness of lines by comparing the coefficients of x
(ignoring negative signs). If the coefficient of x in one line is greater than the
coefficient of x in another line, then the first line is steeper than the second line.
For example, ignoring negative signs, line G is steeper than line F since the
coefficient of x in the equation of line G is greater than the coefficient of x in the
equation of line F. This is the easiest of all three methods.
EXERCISE 7
(a)
By comparing the positive value of the vertical changes, determine which of the two
lines in each case is the steeper of the two?
y  15 x ; y  52 x
(2)
y   54 x ; y   52 x (3)
y  73 x ; y  53 x
(1)
(4)
(b)
(c)
(d)
y  52 x ; y  3 x
(5)
y   75 x ; y  2 x
y  3 x ; y  52 x
(6)
By relating the positive vertical change to the horizontal change, determine which of
the two lines in each case is the steeper of the two?
y   95 x ; y   98 x
y  23 x ; y  32 x
(2)
y  73 x ; y  52 x
(3)
(1)
(4)
y  3 x ; y  14 x
(5)
y   75 x ; y  2 x (6)
y  4 x ; y   76 x
By comparing the coefficients of x (ignoring negative signs), determine which of the
two lines in each case is the steeper of the two?
y  10 x ; y  8 x
y  53 x ; y  75 x
(2)
y   23 x ; y   13 x (3)
(1)
(4)
y  101 x ; y  18 x
(5)
y  5 x ; y  52 x
Match the following equations with the given lines A – H.
yx
(2)
y  4x
(3)
y  12 x
(1)
(4)
(7)
y  7x
y  4 x
(5)
(8)
y  x
y   12 x
228
(6)
y  7 x
C
B
A
D
y
E
F
G
H
x
Summary of the main features of a linear function
 The equation of a linear function is given by y  mx  c where the coefficient of x
represents the gradient (m) and the constant term represents the y-intercept (c).
 If the gradient is positive ( m  0 ) then the line is increasing (slopes up) as read from
left to right. As the gradient increases, so does the steepness upwards.
 If the gradient is negative ( m  0 ) then the line is decreasing (slopes down) as read
from left to right. As the gradient increases (ignore negative signs) so does the steepness
downwards.
 Between any two points on the line the gradient is constant.
 For the y-intercept let x  0 and solve for y.
 For the x-intercept let y  0 and solve for x.
Example 13
Given: y   13 x  1
(a)
Determine the gradient and then state whether the line is increasing or decreasing.
(b)
Write down the coordinates of the y-intercept and x-intercept.
(c)
Draw the graph of the line on a set of axes.
Solutions
(a)
(b)
The gradient is m   13 which is negative.
The line is decreasing.
The y-intercept is c  1
The coordinates are (0 ;1)
For the x-intercept let y  0
0   13 x  1
y
1
y   x 1
3
(0 ; 1)
(3 ; 0)
x
 13 x  1
(c)
x  3
The coordinates are (3 ; 0)
We have the intercepts with the axes so it is easy to draw the graph.
Example 14
Given: y  4 x
(a)
Determine the gradient and then state whether the line is increasing or decreasing.
(b)
Write down the coordinates of the y-intercept.
(c)
Determine the coordinates of the x-intercept.
(d)
Draw the graph of the line on a set of axes.
y y  4x
Solutions
(1 ; 4)
(a)
The gradient is m  4 which is positive.
The line is increasing.
(b)
The coordinates are (0 ; 0) since y  4 x  0
(c)
The x-intercept is also (0 ; 0) since the
intercepts at the origin are equal.
(d)
Use the point-intercept method to draw the
graph. Choose x  1
(0 ; 0)
 y  4(1)  4
A point on the graph is (1; 4)
229
x
Example 15 (Rewriting equations in the form y  mx  c )
Given: 2 x  3 y  6
(a)
Determine the gradient and then state whether the line is increasing or decreasing.
(b)
Determine the coordinates of the y-intercept.
(c)
Determine the coordinates of the x-intercept.
(d)
Draw the graph of the line on a set of axes.
Solutions
(a)
The given equation is not in the form y  mx  c so it is not possible to determine
the gradient. However, if we get y on its own it will be possible to find the gradient.
2x  3y  6
 3 y  2 x  6
3 y 2 x 6
 

3
3
3
2
y 3 x2
2
which is negative. The line is decreasing.
3
There are two equations now available to use when finding the y-intercept.
2
2 x  3 y  6 or y   x  2 . You can use either one of these equations.
3
 Using 2 x  3 y  6 :
For the y-intercept let x  0
2(0)  3 y  6
The gradient is m  
(b)
3 y  6
y 2
The coordinates are (0 ; 2)
For the x-intercept let y  0
2 x  3(0)  6
 2x  6
x  3
The coordinates are (3 ; 0)
 Using y   23 x  2 :
(d)
y
The y-intercept is c  2 .
2x  3y  6
The coordinates are (0 ; 2)
For the x-intercept let y  0
2
0 x2
(0 ; 2)
3
 0  2 x  6
 2x  6
x  3
The coordinates are (3 ; 0)
We have the intercepts with the axes so it is easy to draw the graph.
230
(3 ; 2)
x
EXERCISE 8
For each of the following linear functions:
(1)
Determine the gradient and then state whether the line is increasing or decreasing.
(2)
Write down the coordinates of the y-intercept and x-intercept.
(3)
Draw the graph of the line on a set of axes.
y  2x  8
y  4 x  8
y  3x  9
(b)
(c)
(a)
y  5x  1
y  x  3
y  2 x  3
(e)
(f)
(d)
1
y  6x
(h)
y6x
(i)
y   52 x
(g)
(j)
7 x  2 y  14
(k)
3 x  4 y  24
(l)
x  5y  5
(m)
x y 4
(n)
2x  2 y  4
(o)
1
4
(p)
x  3y
(q)
x  4 y
(r)
x  15 y
x  2y 1
Horizontal and vertical lines
Example 16
y
Consider the points shown on the given graph.
(a)
What do you notice about the y-values?
(b)
What do you notice about the x-values?
(c)
Write down the equation of the line
(3 ; 2) (2 ; 2) ( 1; 2)
joining these points.
(d)
What is the gradient of this horizontal line?
.
. . .
(0 ; 2) (1 2) (2 ; 2) (3 ; 2) (4 ; 2)
Solutions
(a)
(b)
(c)
(d)
y
Consider the points shown on the given graph.
(a)
What do you notice about the x-values?
(b)
What do you notice about the y-values?
(c)
Write down the equation of the line
joining these points.
(d)
What is the gradient of this vertical line?
Solutions
(a)
(b)
(c)
x
The y-values are the same value ( y  2 )
The x-values are different and x  R
The equation of the horizontal line joining
the points is written as y  2 .
We say that the equation of the line is y  2
Take any two points on the line.
(1; 2) and (3 ; 2)
change in y -values 0

 0
change in x-values 2
The gradient of the horizontal line y  2 is zero.
Example 17
y2
(1; 4)
(1; 3)
(1; 2)
(1;1)
(1; 0)
The x-values are the same value ( x  1 )
The y-values are different and y  R
The equation of the vertical line joining
the points is written as x  1 .
We say that the equation of the line is x  1
231
(1;  1)
x 1
x
(d)
Take any two points on the line.
(1; 4) and (1;1)
change in y -values 3


which is undefined.
change in x-values 0
The gradient of the vertical line x  1 is undefined.
Conclusion
The equation of a horizontal line is given by the equation: y  number
The gradient of a horizontal line is zero.
x  number
The equation of a vertical line is given by the equation:
The gradient of a vertical line is undefined.
Example 18
Sketch the graphs of the lines x  3  0 and 2 y  4  0 on the same set of axes.
State the value of the gradients of these two lines.
y
Solutions
x3 0
(0 ; 2)
 x  3
The graph is a vertical line cutting
the x-axis at the point (3 ; 0)
The gradient is undefined.
2y  4  0
y2
( 3 ; 0)
x  3
2y  4
y 2
The graph is a horizontal line cutting the y-axis at the point (0 ; 2)
The gradient is zero.
EXERCISE 9
Sketch the graphs of the following lines and indicate the coordinates of the x-intercept or yintercept. Also state the value of the gradient of each line.
x6
(b)
x  3
(c)
y5
(a)
2 x  14
(f)
2y  4
(d)
y  4
(e)
5x  2  8
(h)
6 y  1  19
(i)
4  6 x  10
(g)
Summary of the methods of sketching straight line graphs and finding their gradients
Lines of the form y  mx  c
 The y-intercept is c (or let x  0 and solve for y).
 The x-intercept is obtained by letting y  0 and solving for x.
 The gradient of the line is the coefficient of x (value of m).
 The gradient of a line is constant between any two points on the line.
 If the gradient is positive ( m  0 ) then the line is increasing (slopes up) as read from
left to right.
 If the gradient is negative ( m  0 ) then the line is decreasing (slopes down) as read
from left to right.
232
x
Lines of the form y  mx :
 The y-intercept is 0 and so is the x-intercept.
 Use the point-intercept method (or table method) to determine another point on the line.
 The gradient of the line is the value of m and the same information applies as before.
Lines of the form px  qy  rx :
 Use the dual-intercept method to obtain the intercepts with the axes:
 To get the y-intercept let x  0 and solve for y.
 To get the x-intercept let y  0 and solve for x.
 To get the gradient, rewrite the equation in the form y  mx  c .
Lines of the form x  number and y  number :
 A vertical line has the equation x  number and the gradient is undefined.
 A horizontal line has the form y  number and the gradient is zero.
Finding the equations of lines given the points on the graph
Example 19
(Given the y-intercept and a point)
The graph of a line is given. Determine the equation of the line.
y
Solution
Method 1
(0 ; 3)
The y-intercept is 3.
 y  mx  3 ( c  3 )
( 2 ;1)
The gradient is the ratio
change in y -values 2
 1
change in x-values 2
m  1
.
x
The equation of the line is y  1x  3
or simply y  x  3
Method 2
The y-intercept is 3.
 y  mx  3 ( c  3 )
To get the gradient (m) substitute the point (2 ;1) into the equation and solve for m.
 y  mx  3
1  m(2)  3 since x  2 and y  1
1  2m  3
 2m  2
m  1
The equation of the line is y  x  3
233
Example 20
(Given the y-intercept and the x-intercept)
y
The graph of a line is given.
Determine the equation of the line.
Method 1
The y-intercept is 2.
 y  mx  2 ( c  2 )
The gradient is the ratio
change in y -values 2
1


change in x-values 4
2
1
m  
2
The equation of the line is y   12 x  2
2
4
x
Method 2
The y-intercept is 2.
 y  mx  2 ( c  2 )
To get the gradient (m) substitute the x-intercept (4 ; 0) into the equation and solve for m.
 y  mx  2
 0  m(4)  2 since x  4 and y  0
 0  m(4)  2
 0  4m  2
4m  2
2
1
m    
4
2
The equation of the line is y   12 x  2
Example 21
(Given a table of values)
Determine the equation of the line passing through the following points:
x
y
1
7
0
5
1
3
2
1
3
1
4
3
Solution
Method 1
The y-intercept is (0 ;  5)
 y  mx  5
Select any other point. Let’s choose the point (1;  3)
Substitute this popint into the equation to get c
3  m(1)  5
3  m  5
m  2
The equation of the line is y  2 x  5
234
Method 2
Use the method discussed in Chapter 7
x-value
The constant difference
between x-values
mutiplied by the x-value
What to do to get y-value
y-value
1
0
1
2
3
4
x
2(1)
2(0)
2(1)
2(2)
2(3)
2(4)
2( x)
5
7
5
5
5
3
5
1
5
1
5
3
5
2x  5
The equation of the line is y  2 x  5
EXERCISE 10
Determine the equation of the following lines:
(a)
(b)
(1; 2)
(c)
(d)
(1; 6)
4
(3 ;1)
2
(e)
4
(2 ;  2)
(f)
(g)
(h)
4
4
6
3
8
4
2
(i)
x
y
1
4
0
2
1
0
2
2
3
4
(j)
x
y
1
1
0
3
1
5
2
7
3
9
(k)
x
y
1
9
0
3
1
3
2
9
3
15
(l)
x
y
1
11
0
6
1
1
2
4
3
9
(m)
x
y
1
14
0
10
1
6
2
2
3
2
(n)
x
y
1
3
0
4
1
5
2
6
3
7
2
REVISION EXERCISE
(a)
For each of the following:
(i)
Determine the coordinates of the intercepts with the axes.
(ii)
Determine the gradient and state whether the line is increasing, decreasing,
zero or undefined.
(iii) Sketch the graph of the line, indicating the intercepts with the axes.
(1)
y  4x  4
(2)
y  4x  2
(3)
y  4 x  2
(4)
y  14 x  1
(5)
y  4x
(6)
y   14 x
(7)
(10)
(13)
(16)
x y 3
x7 0
4 x  0
x  4y
(8)
(11)
(14)
(17)
4 y  2x  8
y 5  0
2y  6  x
2 y  3x  0
235
(9)
(12)
(15)
(18)
5 x  3 y  15
4 x  y
2y  6  2
2 y  3x  6
(b)
For each of the following:
(i)
Determine the gradient.
(ii)
Sketch the graph of the line, indicating the intercepts with the axes.
(1)
3 x  2 y  3
(2)
3  2 y  3
(3)
3x  2 y  0
x
(4)
3x  2  3
(5)
y  2 4
(6)
 y  4x  1
(c)
The lines given below each pass through two points. Determine
line. Which of three lines is the steepest? Explain.
Line A: (3 ; 2) and (6 ;  4)
Line B: (3 ; 4) and (3 ;  4)
Line C: (3 ;  5) and (3 ; 5)
The lines given below each pass through two points. Determine
line. Which of three lines is the steepest? Explain.
Line A: (1; 6) and (2 ;  6)
Line B: (2 ; 8) and (1;  1)
Line C: (3 ;  1) and (6 ; 4)
Match the following equations with the given lines (A – L).
(1) y  x  4
(2) y  x  2
(3) y  x
(5) y  1
(6) x  4
(7) y   x  1
(9) y  2 x  1
(10) y  5 x
(11) y  5 x
(d)
(e)
(f)
In the diagram alongside line A, B and D
intersect on the x-axis. Determine:
(1)
the equation of line A
(2)
the coordinates of the x-intercept of line A
(3)
the equation of line B
(4)
the equation of line C and D
(5)
the equation of line E if this line
(4 ;  3)
passes through the origin and the point
of intersection of line C and D
236
the gradient of each
the gradient of each
(4) y   x
(8) y  2 x
(12) y  12 x  1
2
4
SOME CHALLENGES
(a)
(b)
(c)
(d)
(e)
Show that the x-intercept of the line y  mx  c can be determined using the following
c
expression: x    
m
Now use the above expression to write down the x-intercept of the following lines:
(1)
y  3x  6
(2)
y  2 x  3
(3)
y  4 x  6
Match the information on the left to the lines on the right.
(1)
m0 ;c0
(2)
m0 ;c0
(3)
m0 ;c0
(4)
m0 ;c0
Match the equations on the left to the lines on the right. Assume that b  0 .
y  a x where a  b
(1)
b
y  a x where a  b
(2)
b
y  a x where a  b
(3)
b
y  a x  c where a  0
(4)
b
Determine the equation of the line passing through the points (2 ; 4) and (3 ;  1) .
[Hint: Plot the points on a Cartesian plane and work from there]
A straight line passes through the points (a ; b) and (c ; d )
(c ; d )
Determine an expression for the gradient of the line in
(a ; b)
terms of a, b, c and d.
Hence use this expression to determine the gradient of the
line passing through the points (2 ; 6) and (5 ;12)
The length of a line segment is always positive.
In the diagram on the right, it is clear that the length
of OA and CB both equal 3 units. However, the
length of OC and AB is equal to 2 units even
though the horizontal movement is negative.
B(  2 ; 3)
(h)
(i)
Write down the co-ordinates of each point.
Determine the following lengths:
(1) OA
(2) OC
(3) BC
(4) AB
(5) OD
(6) EF
(7) OF
(8) ED
(9) OG
(10) HI
(11) OI
(12) GH
(13) OK
(14) KJ
(15) IJ
(16) HJ
Calculate the area of rectangles:
(1)
OABC
(2)
EFOD
(3)
GHIO
(4)
OIJK
Calculate the area of
(1)
OAB
(2)
GHI
237
2
3
Now answer the following questions based on the
given diagram below:
(f)
(g)
A(0 ; 3)
3
1
2
C(  2 ; 0)
2
1
2
1
y
6
5
4
B
C
3
E
2
F
7
6
D
1
G
5
4 3 2
1
O
0
A
1
2
3
4
5
K
6
7
1
2
H
3
4
5
6
I
J
x
CHAPTER 16: MEASUREMENT
TOPIC: SURFACE AREA AND VOLUME OF 3D SHAPES
In Grade 8, you studied the surface area and volume of cubes, rectangular prisms (cuboids)
and triangular prisms. We will revise this work and then focus on the surface area and
volume of cylinders.
However, before proceeding with any further, let’s revise converting between units (lengths,
areas, volumes and capacities. The Grade 8 textbook explains how to use these diagrams.
Converting between lengths

10
10
km
10

m
10
10
Converting between areas

100
10
cm mm
km
Converting between volumes
km
3
1000

1000
1000
m
3
cm
3

1000
1000
1000
1000
100
100
m2
100
100
cm2 mm2
Converting between capacities


1000
100
2
mm
m
kl
3
3

l
cm
ml
3
1 ml  1 cm3
1 kl  1 m 3
EXERCISE 1
(a)
(b)
(c)
(d)
Convert:
(1)
6 km to m.
(2)
(4)
80 mm to cm
(5)
(7)
70 000 cm to km
(8)
Convert:
(1)
9 km 2 to m 2
(2)
2
2
(4)
2 500 mm to cm
(5)
2
2
(7)
500 000 cm to km (8)
Convert:
(1)
4 cm3 to mm3
(2)
3
3
(4)
0,000 009 m to mm (5)
(7)
9 000 000 cm3 to m3 (8)
Convert:
(2)
(1)
30 kl to l
(4)
70 000 l to kl
(5)
3
(7)
40 ml to cm
(8)
3
(10) 0,018 kl to cm
(11)
3
(13) 0,123 m to l
(14)
0,06 m to cm
0,08 cm to m
7 mm to m
(3)
(6)
(9)
60 cm to mm
80 m to km
0,07 mm to cm
0,009 m 2 to cm 2
45 cm 2 to m 2
65 000 mm 2 to m 2
(3)
(6)
(9)
30,4 cm 2 to mm 2
80 000 m 2 to km 2
1,25 km 2 to m 2
0,002 m3 to cm3
(3)
0,002 km3 to m3
7 600 000 m3 to km3 (6)
750 mm3 to cm3
0,000 000 000 000 005 km3 to mm3
3 kl to ml
70 ml to l
40 kl to m3
13 000 cm3 to l
0,123 m3 to ml
(3)
(6)
(9)
(12)
(15)
0,003 l to ml
8 000 ml to kl
6 l to cm3
13 000 cm3 to kl
0,123 kl to m3
Revision of the Grade 8 formulae for calculating surface area and volume
Prism
Cuboid (Rectangular
prism)
Surface area
Sum of the areas of the six
rectangles:
Surface area
 2ab  2ac  2bc
Volume
Area of a chosen base multiplied
by the distance moved by the
base (height).
Volume
 (ab)  c
 abc
238
Cube
Sum of the areas of the six
squares:
Surface area
 2(a )(a)  2(a)(a)  2(a)(a)
 6a 2
Triangular prism
Area of a chosen base multiplied
by the distance moved by the
base (height).
Volume
 (a  a)  a
 a3
Sum of the areas of two triangles
and three rectangles.
Surface area
1

 2  (b  h)   ad  bd  cd
2

 bh  ad  bd  cd
 bh  d (a  b  c)
Area of a chosen base multiplied
by the distance moved by the
base (height).
Volume
1
 (b  h)  d
2
1
 bhd
2
Note: The surface area and volume of a prism can be calculated without these formulae. It is
not necessary to learn them off by heart. It is highly recommended that you
understand how to calculate surface area and volume using common sense rather that
learning formulae off by heart without understanding how they work.
The surface area of a cube, cuboid and triangular prism (Revision)
A prism is a three-dimensional shape with two congruent parallel polygonal faces at opposite
ends. These faces are referred to as the bases (or ends) of the prism. In a cube, the faces are
squares and in a rectangular prism (cuboid), the faces are rectangles. A triangular prism is
made up of two congruent triangles and three rectangles. One of the triangles is the base.
The surface area of a three-dimensional shape is the total exterior area of the shape. Shapes
such as prisms are made up of flat polygonal surfaces. It is easy to calculate the area of each
of these polygons and to then add up the areas to get the surface area of the prism. The units
of measurement for surface area are km 2 , m 2 , cm 2 and mm 2 .
Example 1
A cube with one side equal to 8 cm is given.
Calculate the surface area of the cube if the cube is
closed on all sides.
Solutions
If the cube is opened up and the faces are folded down so that the prism is made into a flat
surface, then we call this flat surface a net.
239
The area of one of the squares is (8)(8)  (8) 2
There are six congruent squares.
The surface area is the sum of the area of these six squares.
Surface area
 (8) 2  (8) 2  (8) 2  (8) 2  (8) 2  (8) 2
 6(8) 2
 6  64
 384 cm 2
Example 2
A cuboid (rectangular prism) is given.
Calculate the surface area of this cuboid if:
(a)
the cuboid is closed on all sides
(b)
the cuboid is open at the top and bottom.
Solutions
(a)
First convert 200 mm to cm to ensure
that all units are the same.
200 mm 10  20 cm
Now open up the cuboid and create a net.
The surface area is the sum of the areas of all six rectangles.
Area of rectangle A  (20)(8)  160
Area of rectangle B  (20)(8)  160
Area of rectangle C  (12)(8)  96
Area of rectangle D  (12)(8)  96
Area of rectangle E  (20)(12)  240
Area of rectangle F  (20)(12)  240
 Surface area
 2(20)(8)  2(12)(8)  2(20)(12)
 2(160)  2(96)  2(240)
 992 cm 2
240
(b)
If the cube is open on the top and bottom, then there are only four rectangles since
rectangle E (bottom) and F (top) will be missing. The surface area will then be the
sum of the areas of rectangle A, B, C and D.
 S  2(160)  2(96)  512 cm 2
Example 3
A triangular prism is given.
Calculate the surface area of this prism.
Solution
The prism is made up of two triangular bases
and three rectangles BCFE, ABED and ACFD.
1
Area of ABC  (8)(6)  24 cm 2
2
1
Area of DEF  (8)(6)  24 cm 2
2
Area BCFE  (14)(8)  112 cm 2
Area ABED  (14)(6)  84 cm 2
Use Pythagoras to get the length of AC:
AC2  (6) 2  (8) 2
Using the formula:
Surface area
 bh  d (a  b  c)
 AC2  36  64
 AC2  100
 AC  10 cm
 (8)(6)  (14)(6  8  10)
 48  (14)(24)
 48  336
Area ACFD  (14)(10)  140 cm 2
Surface area of triangular prism
 24  24  112  84  140
 384 cm 2
 384 cm 2
EXERCISE 2
(a)
Calculate the surface areas of the following closed cubes, cuboids and triangular
prisms.
(1)
(2)
(3)
(4)
(5)
(6)
241
(9)
11
cm
6 cm
(8)
9 cm
(7)
15 cm
10 cm
(11)
(12)
(13)
(14)
(15)
c
00
16
(10)
m
12 cm
0,02 m
(b)
A cube with one side equal to 14 cm is given.
Calculate the surface area of the cube if:
(1)
the cube is closed on all sides.
(2)
the cube is open on top.
(c)
The inside of an olympic-sized pool is to be
painted on the inside. Calculate the surface
area to be painted.
(d)
A triangular prism is open on top. Calculate the
surface area of the prism.
(e)
A tent in the form of a triangular prism has an
equilateral triangle as one of its faces.
Calculate the amount of material used to manufacture
the tent if no material is on the ground.
Round off your answer to two decimal places.
(Hint: Draw the height of ABC ).
242
36 m
x
The volume and capacity of a cube, cuboid and triangular prism (Revision)
The volume of a prism is the amount of space that the shape occupies. It is obtained by
multiplying one of the bases by the height of the prism (or the distance moved by the base).
The capacity of a three-dimensional shape is the amount of substance (liquid) that the shape
can hold.
Whenever you are required to calculate the volume of a prism, isolate a side (called the base)
that will be able to move through the prism as it moves parallel to itself. Multiply the area of
this base by the distance moved by the base (called the height).
Example 4
A closed rectangular prism has a length of 0,12 m, breadth of 6 cm and height of 8 cm.
Calculate the volume of the prism in cm3 .
How much water can the prism hold in millilitres?
How much water can the prism hold in litres?
8 cm
Solutions
It is first necessary to ensure that all units
are the same. Convert 0,12 m to cm.
0,12 m 100  12 cm
base
The volume is calculated as follows:
Volume
 area of base  distance moved by base
 area of base  height
 (8)(6) 12
0,12 m
8 cm
6c
m
(a)
6c
m
(a)
(b)
(c)
 576 cm3
12 cm
Note:
The base could have been chosen differently.
Volume
 area of base  distance moved by base
 area of base  height
 (12)(6)  8
8 cm
base
6 cm
12 cm
 576 cm3
(b)
The amount of water that can be contained in the prism is the capacity of the prism.
[ 1 cm3  1 ml ]
The capacity of the prism  576 ml
(c)
Convert 576 ml to litres
576 ml  1000  0,576 l
The capacity of the prism is 0,576 litres.
243

1000
1000
m3
kl

l
cm3
ml
1 ml  1 cm3
1 kl  1 m 3
Example 5
A triangular prism is given.
(a)
(b)
(c)
(d)
(e)
Calculate the volume of this prism
in cubic centimetres.
Calculate the capacity of this prism
in millilitres.
Calculate the capacity of this prism
in litres.
Calculate the volume of this prism
in cubic metres.
Calculate the capacity of this prism
in kilolitres.
Solutions
First convert all units to cm.
0,14 m 100  14 cm
80 mm 10  8 cm
(a)
The base in this prism is ABC .
Volume
 area of base  distance moved by base
 area of base  height
1
 (8)(6)  14
2
 336 cm3

10
10
km
10

m

1000
1000
m
kl
3

l
cm3
ml
(b)
Volume  336 cm3
 Capacity  336 ml
(c)
Capacity
 336 ml  1000
 0,336 l
(d)
Volume in m3
(e)
Volume  0, 000 336 m3
 336 cm3  1000 000
10
10
10
cm mm
1 ml  1 cm3
1 kl  1 m 3
 Capacity  0, 000 336 kl
 0, 000 336 m3
EXERCISE 3
(a)
Calculate the volume of the following closed cubes, cuboids and triangular prisms.
(1)
(2)
(3)
(4)
(5)
(6)
244
(7)
(8)
(9)
11
cm
15 cm
10 cm
(11)
(12)
(13)
(14)
(15)
c
00
16
(10)
(b)
A rectangular sand box has a length of 900 mm, breadth
of 60 cm and height of 0,12 m.
(1)
Calculate the volume of the prism in m3 .
(2)
How much sand can the prism hold in kl?
(3)
Calculate the volume in cm3 .
(4)
How much sand can the prism hold in ml?
(5)
What is the capacity in l?
(c)
A triangular prism is given.
(1)
Calculate the volume of this prism
in cubic centimetres.
(2)
Calculate the capacity of this prism
in millilitres.
(3)
Calculate the capacity of this prism
in litres.
(4)
Calculate the volume of this prism
in cubic metres.
(5)
Calculate the capacity of this prism
in kilolitres.
(d)
A triangular prism has an isosceles triangle as
one of its faces. Calculate the volume of the prism.
Round off your answer to two decimal places.
(Hint: Draw the height of ABC ).
245
36 m
m
0,02 m
Surface area and volume of a cylinder
A cylinder is a solid with two circles as bases and a curved surface which is not a flat surface.
Therefore, although the cylinder is a prism (has two congruent parallel bases), it is not a
polyhedron (not all faces are flat polygons). If the cylinder is opened up and flattened, the net
will be made up of two identical circles and a rectangle. The rectangle has the same length as
the circumference of the circles which is 2r . The width of the rectangle will be equal to the
height of the cylinder (h).
2πr
The surface area of the cylinder
 area of two circles  area of the rectangle
 r 2  r 2  2r  h
 2r 2  2rh
The volume of the cylinder
 area of chosen base  height (distance moved by the base)
 r 2  h
 r 2 h
Example 6
Calculate the surface area and volume of the cylinder if:
(a)
the cylinder is closed on all sides
(b)
the cylinder is open on top.
Round off your answers to two decimal places.
Solution
(a)
Surface area of the cylinder
 area of two circles  area of the rectangle
The radius is 7 cm which is half of the diameter.
 Area of circle
 (7) 2
The area of the curved surface is calculated using the
formula 2rh .
The length of the curved surface is 2(7) and the
width is 15 cm (same as the height of the cylinder).
Area of curved surface
 length  width
 (2r )  h
 2(7)(15)
246
2πr
The surface area of the cylinder
 2r 2  2rh
The volume of the cylinder
 area of chosen base height
 2(7) 2  2(7)(15)
 (98)  (210)
 r 2  h
 (7) 2 (15)
 (49)(15)
 (735)
 (308)
 967, 61 cm 2
 2309, 07 cm3
(b)
The top circle is missing.
Surface area
 (7) 2  2(7)(15)
 (49)  (210)
 (259)
2πr
 813, 67 cm 2
The volume will remain the same.
EXERCISE 4
(a)
(b)
Calculate the surface area and volume of the following cylinders. Round your answers
off to two decimal places.
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
A cylindrical drinking glass is made up of a solid base
and a top curved part which is hollow and open on top.
(1)
Calculate the total volume of the drinking glass.
(2)
What is the capacity of the drinking glass in ml?
(3)
What is the capacity of the drinking glass in l?
(4)
Calculate the surface area of the top curved part.
247
(c)
A cylindrical plastic straw has a length of 21 cm and a diameter of 7 mm.
Calculate the amount of plastic required to make a packet of 100 straws.
7 mm
21 cm
(d)
52 rectangular sheets of soft paper called roller
towels are joined together and wrapped around
a hollow cardboard cylinder. The dimensions of
each roller towel are 275 mm  220 mm. The roller
towels and cardboard cylinder form a larger cylinder
with a diameter of 10 cm and height 28 cm.
The distance between the outer and inner
circumferences is 3 cm.
(1)
Calculate the volume of space taken up by the
roller towels.
(2)
Calculate the total amount of soft paper used.
(3)
Calculate the amount of cardboard used to
manufacture the cardboard cylinder.
The relationship between surface area and volume
Example 7
The dimensions of a cylinder A are doubled to form a larger prism B.
(a)
Calculate the surface area of A.
(b)
Calculate the surface area of B.
Surface area of B
(c)
Determine the ratio
.
Surface area of A
(d)
Compare the surface area of A and B.
What do you notice?
(e)
Calculate the volume of A.
(f)
Calculate the volume of B.
Volume of B
.
(g)
Determine the ratio
Volume of A
(h)
Compare the volume of A and B.
What do you notice?
(i)
What is the relationship between the surface area and volume if the dimensions are
doubled?
Solutions
(a)
Surface area of A
 2(2) 2  2(2)(5)
 (8)  (20)
(b)
Surface area of B
 2(4) 2  2(4)(10)
 (32)  (80)
 (28) cm 2
 (112) cm 2
(c)
Surface area of B (112) cm2

4
Surface area of A (28) cm 2
(Notice that 4  22 )
(d)
Surface area of B  (112) cm 2  4  (28) cm 2  4  Surface area of A
(e)
Volume of A  (2)2 (5)  (20) cm3
(f)
Volume of B  (4) 2 (10)  (160) cm3
248
(g)
Volume of B (160) cm3

 8  23
3
Volume of A (20) cm
(h)
Volume of B  (160)cm3  8  (20)cm3  8  Volume of A.
(i)
If the dimensions are doubled (multiplied by 2), then the surface area of B is
22  the surface area of A and the volume of B is 23  the volume of A.
(Notice that 8  23 )
EXERCISE 5
(a)
The dimensions of a cylinder A are doubled to form a larger prism B.
(1)
Calculate the surface area of A.
(2)
Calculate the surface area of B.
Surface area of B
(3)
Determine the ratio
.
Surface area of A
(4)
Compare the surface area of A and B.
What do you notice?
(5)
Calculate the volume of A.
(6)
Calculate the volume of B.
Volume of B
(7)
Determine the ratio
.
Volume of A
(8)
Compare the volume of A and B.
What do you notice?
(9)
What is the relationship between the surface area and volume if the
dimensions are doubled?
(b)
The dimensions of a triangular prism A are multiplied by 3
to form a larger prism B.
(1)
Calculate the surface area of A.
(2)
Calculate the surface area of B.
Surface area of B
(3)
Determine the ratio
.
Surface area of A
(4)
Compare the surface area of A and B.
What do you notice?
(5)
Calculate the volume of A.
(6)
Calculate the volume of B.
Volume of B
(7)
Determine the ratio
.
Volume of A
(8)
Compare the volume of A and B.
What do you notice?
(9)
What is the relationship between the
surface area and volume if the dimensions
are multiplied by 3?
Conclusion
From the previous example and exercise, it should be clear that if the dimensions of a cube,
cuboid and triangular prism are multiplied by a number k (called the scale factor), then the
relationship between the surface area and volume is as follows:
Surface area of enlarged prism  k 2  surface area of original prism
Volume of enlarged prism  k 3  volume of original prism
249
(c)
If the surface area of a cuboid is 52 cm 2 and its volume is 24 cm3 , determine the
surface area and volume of the cuboid formed if the dimensions of the original cuboid
are:
(1)
doubled
(2)
multiplied by 3
(3)
multiplied by 5
Summary of formulae for calculating surface area and volume
Prism
Cuboid
(Rectangular prism)
Surface area
Sum of the areas of the six
rectangles:
Surface area
 2ab  2ac  2bc
Volume
Area of a chosen base multiplied
by the distance moved by the
base (height).
Volume
 (ab)  c
 abc
Cube
Sum of the areas of the six
squares:
Surface area
 2(a )(a)  2(a)(a)  2(a)(a)
 6a 2
Triangular prism
Sum of the areas of two triangles
and three rectangles.
Surface area
1

 2  (b  h)   ad  bd  cd
2

 bh  ad  bd  cd
 bh  d (a  b  c)
Cylinder
Sum of the areas of two circles
and a curved surface.
Surface area  2r 2  2rh
Area of a chosen base multiplied
by the distance moved by the
base (height).
Volume
 a3
Area of a chosen base multiplied
by the distance moved by the
base (height).
Volume
1
 (b  h)  d
2
1
 bhd
2
Area of a chosen base multiplied
by the distance moved by the
base (height).
Volume  r 2 h
REVISION EXERCISE
This exercise will also include real-world applications of surface area and volume.
Calculate the surface area and volume of the following prisms.
(1)
(2)
(3)
(4)
(5)
(6)
32 m
(a)
250
(8)
(9)
(11)
(12)
36 m
cm
00
16
(7)
(10)
(b)
A company manufacturing solid chocolate bars has two new packaging containers
that will have same amount of chocolate inside (100 g). The one container is a
triangular prism with an equilateral triangle as a base. The other is a rectangular
prism. The company wants to cut down on the cost of the cardboard used for making
a container. Determine which container will be the least expensive to wrap.
(c)
A cheese factory manufactures two types of cheese products.
A solid cylindrical cheese in a rectangular
cardboard box.
8 cheese wedges in a cylindrical cardboard container.
(1)
(2)
(3)
(4)
(5)
Calculate the volume and surface area of the rectangular cardboard box.
Calculate the volume and surface area of the solid cylindrical cheese in the
rectangular box.
Calculate the volume and surface area of the cylindrical container.
Calculate the volume and surface area of one cheese wedge.
Which cardboard container will be the least expensive to manufacture?
251
(d)
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
The volume of a cuboid is 70 cm3 . The length is 7 cm and the breadth is
5 cm. Calculate the height.
The volume of a cuboid is 50 160 mm3 . The length is 76 mm and the height is
3 cm. Calculate the breadth.
The volume of a cube is 13 824 cm3 . Calculate one side in metres.
The volume of a triangular prism is 160 cm3 . If the height of the prism is
10 cm, calculate the area of the base.
The volume of a triangular prism is 200 cm3 . If the area of the base is
10 cm 2 , calculate the height of the prism.
The surface area of a rectangular prism is 136 cm 2 . If the length is 8 cm and
the width is 4 cm, calculate the height of the prism.
The surface area of a rectangular prism is 700 m 2 . If the height is 5 m and the
width is 10 m, calculate the length of the prism.
The surface area of a cube is 150 cm 2 . Calculate the length of a side.
The volume of a cylinder is (45) cm3 . If the radius is 3 cm, calculate the
height.
The volume of a cylinder is (96) cm3 . If the height is 6 cm, calculate the
radius.
The surface area of a closed cylinder is (130) cm 2 . If the diameter is 10 cm,
calculate the height.
The surface area of a closed cylinder is (240) cm 2 . If the height is 14 cm,
calculate the radius.
SOME CHALLENGES
(a)
=
A wooden block in the form of a cube is shown below. A part of the block is removed
from the original cube (rectangular prism with base HGFE). The base of the newly
formed figure is the shaded area ABKCDEFGH and BC  DC  DJ . It is also given
that AH  ED  GK  4 cm and GF  x .
J
(1)
Write down the length of BC and DC in terms of x.
(2)
Write down the length of EF in terms of x.
D
H E
A
4 cm
4 cm
(3)
Determine the area of the shaded base
in terms of x and simplify your answer.
(4)
Express the volume of the figure in terms of x.
=
(5)
If the length of GF is 2 cm, calculate the volume
G x F
of the figure.
B
=
4 cm
K
C
(b)
A cube and cylinder is given. The height of the
cylinder is x.
(1)
If the volume of the cube and cylinder
is the same, calculate x.
(2)
If the surface area of the cube and cylinder
is the same, calculate x.
(c)
A two-litre carton of milk must fill small plastic cups.
(1)
If the capacity of a small cup is 100 ml, how many full cups can you get from
the two-litre carton?
(2)
Suppose that the capacity of a small cup is 0,12 l. How many full cups can you
then get from the two-litre carton?
252
(d)
The surface area of a triangular prism is 84 cm 2 and its volume is 36 cm3 .
Determine the surface area and volume of the prism formed if the dimensions
of the original prism are doubled.
(e)
The surface area of a cube is 1 500 000 cm 2 and its volume is 125 m3 .
(1)
Determine the surface area and volume of the cube formed if the dimensions
(in metres) of the original cube are doubled.
(2)
Calculate the length of a side of the larger cube.
(3)
What is the capacity of the original cube in l and kl?
(f)
The surface area of a cube is 54x 2 and its volume is 27x3 .
(1)
Determine the surface area and volume of the cuboid formed if the dimensions
of the original cuboid are doubled.
(2)
Calculate the length of a side of the cube.
(g)
A small rectangular block of fudge has dimensions: 1,5 cm 1,5 cm  3 cm .
Blocks of fudge must be put into a large rectangular box with a capacity
of 162 ml.
(1)
Calculate the volume of a block of fudge.
(2)
How many blocks of fudge can be put into
the large rectangular box?
(3)
If the length of a block of fudge is 3 cm and
the breadth (B) of the large box is 3 cm,
determine the length (L) and height (H) of the
box.
253
CHAPTER 17: SHAPE AND SPACE (GEOMETRY)
TOPIC: TRANSFORMATION GEOMETRY
In Grade 8, the following types of transformations were studied:
(a) Transformations in which the original triangle ABC and its image A|B|C| are
congruent (identical in size and shape). These included translations, reflections about
the y-axis, reflections about the x-axis and rotations of 90 clockwise and anticlockwise about the origin and rotations of 180 in either direction about the origin.
The following area and perimeter relationships were investigated and the results were as
follows:
 AB  A|B| ; BC  B|C| ; AC  A|C|
A|B|
B|C| A|C|



1
AB
BC
AC
 Area of A|B|C|  Area of ABC
Area of A|B|C|

1
Area of ABC
 Perimeter of A|B|C|  Perimeter of ABC
Perimeter of A|B|C|

1
Perimeter of ABC
(b) Transformations in which the original triangle ABC and its image A|B|C| are similar
(identical in shape but not size). These included enlargements and reductions through
the origin by a scale factor of k.
The following area and perimeter relationships were investigated and the results were as
follows:
 The lengths of the sides of A|B|C| are k times the lengths of the sides of ABC :
A|B|  k  AB
B|C|  k  BC
A|C|  k  AC
 Area of A|B|C|  k 2  Area of ABC where k is the scale factor.
The area of the image A|B|C| is k 2 times greater or smaller than the area of the
original ABC . If the scale factor is greater than 1 ( k  1 ), then A|B|C| is an
enlargement of ABC . If the scale factor is a fraction between 0 and 1 ( 0  k  1 ),
then A|B|C| is a reduction of ABC .
Area of A|B|C|

 k2
Area of ABC
 Perimeter of A|B|C|  k  Perimeter of ABC
Perimeter of A|B|C|

k
Perimeter of ABC
In Grade 9, we will now discuss general rules for these transformations in terms of x and y.
We will also focus on a further type of transformation referred to as a reflection about the line
y  x.
TRANSLATIONS
If ABC is translated to form A|B|C| , the transformation involves a horizontal movement
left or right followed by a vertical movement up or down.
254
Translating points in the Cartesian plane
Example 1
In each of the following diagrams, a point has been translated by a horizontal move followed
by a vertical move to form its image. Describe the translation and then write down a general
rule in terms of x and y for translating any point in the Cartesian plane using the given
y
translation.
6 A(1; 5)
(a)
Point A moved 3 units right and then 4 units down
5
|
3 units right
to form A , the image of A. The x-coordinate
4
of A| was obtained by adding 3 to the x-coordinate
3
|
2
of A. The y-coordinate of A is obtained by
1
A| (4 ;1)
subtracting 4 from the y-coordinate of A.
x
5
0 1 2 3 4

3

2

1
In other words, the image A| is obtained
1
2
as follows: A(1; 5)  A| (1  3 ; 5  4)  A| (4 ;1)
.
.
y
3 units right and 4 units down is A( x ; y )  A| ( x  3 ; y  4) .
Point P moved 4 units left and then 6 units up
4
|
|
|
to form P , the image of P. The x-coordinate of P
P (  2 ; 2) 3
2
is obtained by subtracting 4 from the x-coordinate of P.
1
The y-coordinate of P| is obtained by adding 6
3 2 1 0 1
to the y-coordinate of P. In other words, the image
1
|
2
P is obtained as follows:
3
|
|
P(2 ;  4)  P (2  4 ;  4  6)  P (2 ; 2)
6 units up
4
We say that P(2 ;  4) is mapped onto P (2 ; 2)
A general rule for translating any point P(x ; y )
We say that A(3 ;1) is mapped onto A| (3 ; 6)
A general rule for translating any point A(x ; y )
y
.
.
3
5
x
P(2 ;  4)
5
4
3
2
1
0
A(3 ; 1)
no horizontal movement
1
2
3
4
5
1
5 units up is A( x ; y )  A| ( x ; y  5) .
To summarise:
If ( x ; y ) is translated to the point ( x  h ; y  v) where h is a horizontal shift and v is a
vertical shift.
If h  0 , the horizontal translation is to the right.
If h  0 , the horizontal translation is to the left.
If v  0 , the vertical translation is up.
If v  0 , the vertical translation is down.
255
4
A| (3 ; 6)
6
5 units up
4 units left and 6 units up is P( x ; y )  P| ( x  4 ; y  6) .
Point A did not move horizontally at all.
It just moved 5 units up. The x-coordinate
of A| is the same as A because there is no
horizontal movement. The y-coordinate of
A| is obtained by adding 5 to the y-coordinate
of A. In other words, the image A| is
obtained as follows: A(3 ;1)  A| (3 ;1  5)  A| (3 ; 6)
.
2
4 units left
|
(c)
4 unit down
3
We say that A(1; 5) is mapped onto A| (4 ;1)
A general rule for translating any point A(x ; y )
(b)
.
x
EXERCISE 1
(a)
y
Determine the translation if points
A, B, C, D E and F are mapped
onto their respective images.
State a general rule in the form
( x ; y )  ....
6
B|
A|
5
B
4
3
A
2
E|1
F|
7
6
5
4 3 2
1
O
0
F
1
2
3
1
6
7
3
4
|
C
(b)
5
D
2
C
4
5
6
E
D|
(1)
Plot the point A(5 ; 2) on a Cartesian plane.
(2)
Now translate A using the rule A( x ; y )  A| ( x  3 ; y  2) .
Plot the image point on the same set of axes as A and state the coordinates of
the image point A| .
Plot the point B(3 ; 4) on a Cartesian plane.
(3)
Now translate B using the rule B( x ; y )  B| ( x  2 ; y  4) .
Plot the image point on the same set of axes as B and state the coordinates of
the image point B| .
Plot the point C(3 ; 5) on a Cartesian plane.
(4)
Now translate C using the rule C( x ; y )  C| ( x  3 ; y  1) .
Plot the image point on the same set of axes as C and state the coordinates of
the image point C| .
Plot the point D(4 ;  2) on a Cartesian plane.
(5)
Now translate D using the rule D( x ; y )  D| ( x  6 ; y  2) .
Plot the image point on the same set of axes as D and state the coordinates of
the image point D| .
Plot the point E(  5 ;  4) on a Cartesian plane.
(6)
Now translate E using the rule E( x ; y )  E| ( x  5 ; y  4) .
Plot the image point on the same set of axes as E and state the coordinates of
the image point E| .
Plot the point F(  1; 6) on a Cartesian plane.
(7)
Now translate F using the rule F( x ; y )  F| ( x  2 ; y  6) .
Plot the image point on the same set of axes as F and state the coordinates of
the image point F| .
Plot the point G(  4 ;  5) on a Cartesian plane.
(8)
Now translate G using the rule G( x ; y )  G| ( x  4 ; y  2) .
Plot the image point on the same set of axes as G and state the coordinates of
the image point G| .
Plot the point H(4 ;  5) on a Cartesian plane.
Now translate H using the rule H( x ; y )  H| ( x ; y  8) .
Plot the image point on the same set of axes as H and state the coordinates of
the image point H| .
256
x
(9)
Plot the point J(4 ;  5) on a Cartesian plane.
Now translate J using the rule J( x ; y )  J| ( x  8 ; y ) .
Plot the image point on the same set of axes as J and state the coordinates of
the image point J| .
Translating triangles in the Cartesian plane
Example 2
(a)
Draw ABC with vertex coordinates of A( 4 ; 5) , B( 6 ; 3) and C( 2 ; 3) .
(b)
Draw the image of ABC , i.e. A|B|C| , if ABC is translated using the rule
( x ; y )  ( x  3 ; y  7) . Indicate the coordinates of A| , B| and C| .
DEF has vertex coordinates of D(4 ; 2) , E(2 ; 0) and F(6 ; 0) . Determine the
translation in the form ( x ; y )  .... if:
(c)
(1)
DEF is the image of ABC
DEF is the image of A|B|C| .
(2)
y
Solutions
(a)
See diagram for ABC .
(b)
A(4 ; 5)  A| (4  3 ; 5  7)
 A| (1;  2)
B(6 ; 3)  B| (6  3 ; 3  7)
 B| (3 ;  4)
6
A(  4 ; 5)
5
4
C(  2 ; 3) 3
B(  6 ; 3)
D(4 ; 2)
2
1
7
6
5
C(2 ; 3)  C| (2  3 ; 3  7)
 C| (1;  4)
4 3 2
1
A| (  1 ;  2)
B| (  3 ;  4)
O
E(2 ; 0)
0
1
2
F(6 0)
3
4
5
6
7
x
1
2
4
C| (1  4)
5
(c)(1) A(4 ; 5)  D(  4  8 ; 5  3)  D(4 ; 2)
B(6 ; 3)  E(  6  8 ; 3  3)  E(2 ; 0)
C(2 ; 3)  F(2  8 ; 3  3)  F(6 ; 0)
The translation is 8 units right and
3 units down.
 ( x ; y )  ( x  8 ; y  3)
(2) A| (1;  2)  D(  1  5 ;  2  4)  D(4 ; 2)
6
Notice that the three triangles are
identical in size and shape. The
corresponding angles and sides are
equal.
We can write this as follows:
ABC  A|B|C|  DEF
B| (3 ;  4)  E(  3  5 ;  4  4)  E(2 ; 0)
C| (1;  4)  F(1  5 ;  4  4)  F(6 ; 0)
The translation is 5 units to the right and 4 units up.
 ( x ; y )  ( x  5 ; y  4)
EXERCISE 2
(a)
On a set of axes, draw ABC with vertex coordinates A( 3 ; 5), B( 4 ; 1) and
C( 1; 2) . Now draw the images of ABC formed using the following translations:
(1)
( x ; y )  ( x  6 ; y  1)
(call the image A|B|C| )
(2)
( x ; y )  ( x ; y  5)
(call the image A||B||C|| )
(3)
( x ; y )  ( x  10 ; y )
(call the image A|||B|||C||| )
257
y
(b)
(c)
Determine the translation rule ( x ; y )  .....
5
in each case:
4
A
3
(1)
A is translated to B.
2
(2)
B is translated to C.
B
1
(3)
C is mapped onto D.
5
3 4
5 4 3 2 1 0 1 2
(4)
A is mapped onto D.
1
(5)
A is the image of B.
D 2
C
(6)
B is the image of D.
3
(7)
B is the image of C.
4
(8)
C is the image of A.
5
(9)
A is the image of C.
ABC with vertex coordinates of A(5 ; 3) , B(6 ; 1) and C(8 ;1) undergoes the
x
following three translations to form its image A|B|C| :
( x ; y )  ( x ; y  4) followed by ( x ; y )  ( x  2 ; y ) followed by
( x ; y )  ( x  6 ; y  8) . Write down the coordinates of A|B|C| .
REFLECTIONS
In Grade 8, we learnt that a reflection is a mirror image of a shape about a line of reflection.
vertical line of reflection
If the page is folded on the vertical line,
the panda bear on the left will be exactly
the same as the panda bear on the right.
We say that the vertical line of reflection
is the vertical axis of symmetry.
If the page is folded on the horizontal line,
the top rhinoceros will be exactly the same
as the bottom rhinoceros. We say that the
horizontal line of reflection is the horizontal
axis of symmetry.
horizontal line of reflection
Reflections about the y-axis and x-axis were discussed in Grade 8. Let’s briefly revise these
reflections.
Reflections about the y-axis
y
If ABC is reflected about the y-axis to
form A|B|C| , then the first coordinates of
the triangles differ in sign.
A(6 ; 5)  A| (6 ; 5)
A(  6 5)
C(  2 ; 5)
6
C| (2 5)
A| (6 ; 5)
5
4
3
2
B(  6 ; 2)
B(6 ; 2)  B| (6 ; 2)
B| (6 ; 2)
1
7
|
C(2 ; 5)  C (2 ; 5)
6
5
4 3 2
1
O
0
1
2
3
4
5
258
6
1
2
3
4
5
6
7
x
Reflections about the x-axis
y
If ABC is reflected about the x-axis
to form A|B|C| , then second coordinates
of the triangles differ in sign.
A(6 ; 5)  A| (6 ;  5)
A(  6 ; 5)
C(  2 ; 5)
6
5
4
3
2
B(  6 ; 2)
B(6 ; 2)  B| (6 ;  2)
1
C(2 ; 5)  C| (2 ;  5)
7
6
5
4 3 2
O
1
0
1
2
3
4
5
6
7
x
1
B| (  6 ;  2)
2
3
4
|
A (  6 ;  5)
|
C (  2 ;  5)
5
6
We will now discuss general rules in terms of x and y for these reflections.
A general rule, in terms of x and y, for reflecting a point about the y-axis is ( x ; y )  ( x ; y )
where the first coordinates differ in sign.
A general rule, in terms of x and y, for reflecting a point about the x-axis is ( x ; y )  ( x ;  y )
where the second coordinates differ in sign.
Example 3
Write down the coordinates of the image of the given point if the point is reflected about
the y-axis and the x-axis. State the general rule of reflection in each case.
(a)
A(3 ; 4)
(b)
B(3 ; 4)
(c)
C(3 ;  4)
(d)
D(3 ;  4)
Solutions
(a)
Reflection about the y-axis: First coordinates differ in sign: A(3 ; 4)  A| (3 ; 4)
General rule: ( x ; y )  ( x ; y )
Reflection about the x-axis: Second coordinates differ in sign: A(3 ; 4)  A| (3 ;  4)
General rule: ( x ; y )  ( x ;  y )
(b)
Reflection about the y-axis: First coordinates differ in sign: B(3 ; 4)  B| (3 ; 4)
General rule: ( x ; y )  ( x ; y )
Note: The general rule ( x ; y )  ( x ; y ) works for reflecting the point B(3 ; 4) about
the y-axis since B(3 ; 4)  B| ((3) ; 4)  B| (3 ; 4) .
Clearly, the first coordinates differ in sign.
Reflection about the x-axis: Second coordinates differ in sign: B(3 ; 4)  B| (3 ;  4)
General rule: ( x ; y )  ( x ;  y )
(c)
Reflection about the y-axis: First coordinates differ in sign: C(3 ;  4)  C| (3 ;  4)
General rule: ( x ; y )  ( x ; y )
Reflection about the x-axis: Second coordinates differ in sign: C(3 ;  4)  C| (3 ; 4)
General rule: ( x ; y )  ( x ;  y )
(d)
Reflection about the y-axis: First coordinates differ in sign: D(3 ;  4)  D| (3 ;  4)
General rule: ( x ; y )  ( x ; y )
Reflection about the x-axis: Second coordinates differ in sign:
General rule: ( x ; y )  ( x ;  y )
259
D(3 ;  4)  D| (3 ; 4)
Example 4
A(6 ; 5) , B(6 ; 2) and C(2 ; 5) are
the coordinates of the vertices of ABC .
(a)
(b)
(c)
A(  6 ; 5)
C(  2 ; 5)
6
5
4
If the rule of transformation is
( x ; y )  ( x ; y ) , draw A|B|C| .
State the type of transformation and
indicate the coordinates of the vertices
of A|B|C| .
3
2
B(  6 ; 2)
1
5
6
7
4 3 2
1
2
1
0
4
3
5
7
6
1
2
If the rule of transformation is
( x ; y )  ( x ;  y ) , draw A||B||C|| .
State the type of transformation and
indicate the coordinates of the vertices
of A||B||C|| .
3
4
5
6
Write down the value of the following ratios:
Area of A|B|C|
Area of A||B||C||
(2)
(1)
Area of ABC
Area of ABC
| | |
Perimeter of A B C
Perimeter of A||B||C||
(4)
(3)
Perimeter of ABC
Perimeter of ABC
| |
| |
| |
A B BC
AC
A||B|| B||C||
A||C||
,
and
(6)
,
and
(5)
AB BC
AC
AB
BC
AC
Solutions
(a)
See diagram.
The transformation is a reflection
about the y-axis.
A(6 ; 5)  A| (6 ; 5)
A(  6 ; 5)
(b)
B(6 ; 2)  B|| (6 ;  2)
6
C| (2 5)
4
3
2
B(  6 ; 2)
B| (6 ; 2)
1
7
6
5
4 3 2
1
0
1
2
3
1
||
B (  6  2)
2
3
4
||
A (  6 ;  5)
C (  2 ;  5)5
||
6
||
C(2 ; 5)  C (2 ;  5)
(c)
(1)
(3)
(5)
(6)
Area of A|B|C|
1
Area of ABC
Perimeter of A|B|C|
1
Perimeter of ABC
A|B|
B|C|
1
1
AB
BC
A||B||
B||C||
1
1
AB
BC
A (6 ; 5)
5
B(6 ; 2)  B| (6 ; 2)
C(2 ; 5)  C| (2 ; 5)
See diagram
The transformation is a reflection
about the x-axis.
A(6 ; 5)  A|| (6 ;  5)
C(  2 ; 5)
(2)
(4)
Area of A||B||C||
1
Area of ABC
Perimeter of A||B||C||
1
Perimeter of ABC
A|C|
1
AC
A||C||
1
AC
260
4
5
6
7
EXERCISE 3
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
Write down the coordinates of the image of the point P(5 ; 7) if P is reflected about
the y-axis and x-axis. State the general rule of reflection in each case.
Write down the coordinates of the image of the point Q(5 ; 7) if Q is reflected about
the about the y-axis and x-axis. State the general rule of reflection in each case.
Write down the coordinates of the image of the point R(5 ;  7) if R is reflected
about the about the y-axis and x-axis. State the general rule of reflection in each case.
Write down the coordinates of the image of the point S(5 ;  7) if S is reflected
about about the y-axis and x-axis. State the general rule of reflection in each case.
Write down the coordinates of the image of the point T(0 ;  7) if S is reflected
about about the x-axis. State the general rule of reflection.
Write down the coordinates of the image of the point U(7 ; 0) if S is reflected
about about the y-axis. State the general rule of reflection.
Write down the line of reflection in each case:
A(4 ; 3)  A| (4 ; 3)
B(6 ; 9)  B| (6 ;  9)
(2)
(1)
(3)
C(9 ; 7)  C| (9 ; 7)
(4)
D(8 ; 5)  D| (8 ;  5)
(5)
E(4 ;  5)  E| (4 ;  5)
(6)
F(4 ;  5)  F| (4 ; 5)
(7)
(1)
G(10 ;  9)  G| (10 ; 9)
G(9 ;  9)  G| (9 ;  9)
(8)
On a set of axes, draw ABC if A(7 ; 6) , B(7 ; 2) and C(3 ; 2) are the
coordinates of the vertices.
If the rule of transformation is ( x ; y )  (  x ; y ) , draw A|B|C| .
State the type of transformation and indicate the coordinates of the vertices
of A|B|C| .
If the rule of transformation is ( x ; y )  ( x ;  y ) , draw A||B||C|| .
State the type of transformation and indicate the coordinates of the vertices
of A||B||C|| .
Write down the value of the following ratios:
Area of A|B|C|
Area of A||B||C||
(i)
(ii)
Area of ABC
Area of ABC
| | |
Perimeter of A B C
Perimeter of A||B||C||
(iv)
(iii)
Perimeter of ABC
Perimeter of ABC
| |
| |
| |
A B BC
AC
A||B|| B||C||
A||C||
,
and
(vi)
,
and
(v)
AB BC
AC
AB
BC
AC
(2)
(3)
(4)
Let’s now discuss another type of transformation referred to as the reflection of a point or
shape about the line y  x .
y
Reflections about the line y  x
yx
4
B(2 ; 3)
3
||
Consider the point B( 2 ; 3) . If B is
reflected about the slanted line y  x ,
its image will be the point B| (3 ;  2) .
5
|
3
2
1
1
M
0
1
1
2
seen that BM  MB| .
3
261
4
2
3
4
||
Clearly, the line BMB is perpendicular
to the line y  x . Also, it can be
4
2
B'(3;  2)
5
x
EXERCISE 4
(a)
(b)
y
For each point in the given diagram,
draw the reflection of the point about
the line y  x and indicate the
coordinates of the image.
Rewrite and complete the following:
A(  3; 4)  A| ( ; )
6
4
2
1
7
6
5
4 3 2
O
1
4
3
2
1
0
5
6
C(0;3)  C ( ; )
2
D(6 ; 2)
3
|
D(6;  2)  D ( ; )
4
What do you notice?
5
Write down, in words, a rule for
6
reflecting the point about the line y  x .
State a general rule in terms of x and y for reflecting a point about the line y  x .
Conclusion
A general rule, in terms of x and y, for reflecting a point about the line y  x is
( x ; y )  ( y ; x ) where the first and second coordinates have interchanged.
Example 5
(a)
Draw ABC with vertex coordinates of A(6 ; 5) , B(5 ; 2) and C(2 ; 2) .
(b)
Draw the image of ABC , i.e. A|B|C| , if ABC is translated using the rule
( x ; y )  ( y ; x) . Indicate the coordinates of A| , B| and C| .
(c)
(d)
State whether ABC and A|B|C| are congruent or similar.
Write down the value of the following ratios:
Area of A|B|C|
Perimeter of A|B|C|
(2)
(1)
Area of ABC
Perimeter of ABC
| |
| |
| |
A B BC
AC
(3)
,
and
AB BC
AC
Solutions
(a)
(b)
y
See diagram.
A(6 ; 5)  A| (5 ;  6)
6
A(  6 ;5)
5
|
B(5 ; 2)  B (2 ;  5)
(c)
(d)
C(2 ; 2)  C| (2 ;  2)
See diagram.
Both triangles are congruent.
7
Area of A|B|C|
(1)
1
Area of ABC
Perimeter of A|B|C|
(2)
1
Perimeter of ABC
A|B| B|C| A|C|
(3)


1
AB
BC
AC
7
1
|
(e)
C(0 ;3)
3
B(  5 ;2)
B(  5; 2)  B| ( ; )
(c)
(d)
y x
5
A(  3 4)
4
3
B(  5 2) C(  2
2)2
1
6
5
4 3 2
1
O
0
1
2
1
3
2
4
5
6
7
C| (2 ; 2)
3
4
5
6
262
B| (2 ; 5)
A| (5 ; 6)
x
x
EXERCISE 5
(a)
(b)
(c)
Write down the coordinates of the images of the following points if they are reflected
about the line y  x
A(5 ; 7)
B(5 ; 7)
C(5 ;  7)
(2)
(3)
(1)
D(5 ;  7)
E(4 ; 0)
F(0 ;  4)
(5)
(6)
(4)
Write down the line of reflection in each case:
A(7 ; 3)  A| (3 ; 7)
B(2 ;  8)  B| (8 ; 2)
(2)
(1)
(3)
C(7 ; 3)  A| (3 ;  7)
(5)
(1)
E(4 ;  5)  E| (4 ;  5)
(6)
F(4 ;  5)  F| (4 ; 5)
On a set of axes, draw ABC if A(6 ;  5) , B(6 ;  2) and C(2 ;  5) are the
coordinates of the vertices.
If the rule of transformation is ( x ; y )  ( y ; x) , draw A|B|C| .
State the type of transformation and indicate the coordinates of the vertices
of A|B|C| .
Write down the value of the following ratios:
Area of A|B|C|
(i)
Area of ABC
Perimeter of A|B|C|
(ii)
Perimeter of ABC
A|B| B|C|
A|C|
(iii)
,
and
AB BC
AC
(2)
(3)
(4)
D(4 ; 5)  D| (4 ;  5)
Let’s now summarise all three rules of reflection.
Reflection about the y-axis:
( x ; y )  (  x ; y ) (The first coordinates differ in sign)
Reflection about the x-axis:
( x ; y )  ( x ;  y ) (The second coordinates differ in sign)
Reflection about the line y  x :
( x ; y )  ( y ; x)
(The first and second coordinates have interchanged)
ENLARGEMENTS AND REDUCTIONS OF SHAPES
Whenever a shape is enlarged or reduced through the origin, the coordinates of the shape
move further away from or closer to the origin thereby enlarging or reducing the original
shape. The original shape and its image are similar to each other. A scale factor (k) is the
number multiplied by the coordinates of each point to form the enlarged or reduced image of
the original shape. In Grade 8, various area and perimeter ratios were established and these
must be revised and known in Grade 9.
The following examples are useful for revising this work. The general rules in terms of x and
y will be included in the examples.
263
y
Example 6
In the given diagram, ABC is
enlarged through the origin by a
scale factor of 2 to form its
image A|B|C| .
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
9
8
7
6
C(2 ; 5)
Determine the coordinates
5
| | |
of the vertices of A B C .
4
What do you notice about
3
the scale factor?
2
Write down a general rule
B|
A|
in terms of x and y for this
1 A(2 ; 1)
B(5 ; 1)
transformation.
9 10
3 4
2
2 1
5
7 8
0
1
6
Determine the lengths of the
sides of ABC and A|B|C| .
What do you notice?
Notice that the scale
A|B| B|C|
A|C|
factor of 2 is a number
,
and
.
Calculate the ratios
greater than 1.
AB BC
AC
What do you notice?
Calculate the area of ABC and A|B|C| .
Compare the areas of both triangles. What do you notice?
Area of A|B|C|
Calculate the ratio
. What do you notice?
Area of ABC
Calculate the perimeter of both triangles.
Compare the perimeter of ABC and A|B|C| . What do you notice?
Perimeter of A|B|C|
Calculate the ratio
. What do you notice?
Perimeter of ABC
Solutions
(a)
A| (4 ; 2) , B| (10 ; 2) , C| (4 ;10)
(b)
The scale factor of 2 is a
number greater than 1.
(c)
The general rule for enlarging
ABC through the origin by
a scale factor of 2 is
( x ; y )  (2 x ; 2 y )
(d)
C|
10
y
 BC  25
 BC  5 units
C| (4 ; 10)
10
9
8
7
6
10
8
C(2 ; 5)
5
AB  3 units A|B|  6 units
AC  4 units A|C|  8 units
BC2  (3) 2  (4) 2
2
x
4
4
3
2
1
2 1
0
5
|
6
A (4 ; 2)
A(2 ; 1) 3
B(5 ; 1)
1
2
3
4
5
6
7
B| (10 ; 2)
8
9
10
(B|C| ) 2  (6)2  (8) 2
 (B|C| ) 2  100
 B|C|  10 units
Notice that the length of each side of A|B|C| is double the length of each side of
ABC :
A|B|  2AB
B|C|  2BC
A|C|  2AC
264
x
(e)
(f)
(g)
A|B| 6 units
B|C| 10 units
A|C| 8 units

2

2

2
AB 3 units
BC
5 units
AC 4 units
The corresponding sides of ABC and A|B|C| are in the same proportion and
therefore the triangles are similar. The ratios represent the scale factor.
The lengths of the sides of ABC have been increased by a scale factor of 2.
Area ABC
Area A|B|C|
1
1
 AB  AC
 A|B|  A|C|
2
2
1
1
 (3)  (4)
 (6)  (8)
2
2
2
 6 units
 24 units 2
Area A|B|C|  24 units 2  4  6 units 2  4  Area ABC
The area of the enlarged image triangle is 4 times larger than the area of the original
triangle. In other words, four original triangles can be fitted into the image triangle.
ABC
ABC
ABC
ABC
ABC
A|B|C|
Notice too that Area A|B|C|  22  Area ABC .
This is the same as stating that:
Area A|B|C|  (scale factor)2  Area ABC .
(h)
(i)
(j)
(k)
Area of A|B|C|
24 units 2

 4  22 .
2
Area of ABC
6 units
This ratio represents (scale factor)2 .
Consider ABC :
AB  3 units
AC  4 units
From Pythagoras:
BC 2  (4) 2  (3) 2
Consider A|B|C| :
A| B|  6 units
A|C|  8 units
From Pythagoras:
(B|C| ) 2  (6) 2  (8) 2
 BC2  25
 BC  5 units
Perimeter of ABC
 (3  4  5)  12 units
 (B|C| ) 2  100
 B|C|  10 units
Perimeter of A|B|C|
 (6  8  10)  24 units
Perimeter of A|B|C|  24 units  2  12 units  2  perimeter of ABC
This is the same as stating that:
Perimeter of A|B|C|  scale factor  perimeter of ABC .
Perimeter of A|B|C|
24 units

 2 (This ratio represents the scale factor)
Perimeter of ABC
12 units
265
Example 7
y
In the diagram below, ABC is reduced
through the origin by a scale factor of 13
10
A(  9 ; 9)
9
to form its image A|B|C| .
8
(a)
Write down the coordinates of
7
the vertices of ABC and A|B|C| .
6
(b)
What do you notice about the
5
scale factor?
4
A| (  3 ; 3)
(c)
Write down a general rule in
3
C(  6 ; 3)
B(  9 ; 3)
terms of x and y for this
2
transformation.
|
|
B (  3 ; 1) C ( 2 1)1
(d)
Without calculating any areas,
10 9 8 7 6 5 4 3 2 1 0
write down the ratio
| | |
Area of A B C
. Use the previous example to assist you.
Area of ABC
(e)
Without calculating any lengths, write down the
following ratios:
Perimeter of A|B|C|
A|B|
(i)
(ii)
Perimeter of ABC
AB
Use the previous example to assist you.
B|C|
A|C|
(f)
What can you deduce about the ratios
and
?
BC
AC
Solutions
1
3
2
x
Notice that the scale
factor of 13 is a
fraction between 0
and 1.
(a)
(b)
A(9 ; 9)
A| (3 ; 3)
B(9 ; 3)
B| (3 ;1)
The scale factor of 13 is a fraction between 0 and 1.
(c)
The general rule for reducing ABC through the origin by a scale factor of
C(6 ; 3)
C| (2 ;1)
1
3
is
1 1 
( x ; y)   x ; y  .
3 3 
2
(d)
(e)
(f)
1
Area of A|B|C|
1
 (scale factor)2    
9
Area of ABC
3
| | |
Perimeter of A B C
1
 scale factor 
(i)
Perimeter of ABC
3
A|B|
1
 scale factor 
AB
3
| |
| |
| |
A B BC A C 1
Since the triangles are similar, we can deduce that



AB
BC
AC 3
since these ratios represent the scale factor.
(ii)
Conclusion
If ABC is enlarged or reduced through the origin by a scale factor of k so as to form its
image A|B|C| , then the two triangles are similar to each other. The triangles are identical
in shape but not size. The following applies to the two triangles:
(a)
The general rule in terms of x and y for an enlargement or reduction of a triangle (or
some other shape) through the origin is given by ( x ; y )  (kx ; ky ) where k is the
scale factor.
(b)
The lengths of the sides of A|B|C| are k times the lengths of the sides of ABC :
A|B|  k  AB
B|C|  k  BC
A|C|  k  AC
266
(c)
(d)
(e)
(f)
(g)
The corresponding sides are in proportion. The ratio of one side of A|B|C| to
the corresponding side of ABC is equal to the scale factor (k).
A|B|
B|C|
A|C|
(k is the scale factor)
k
k
k
AB
BC
AC
Area of A|B|C|  k 2  Area of ABC where k is the scale factor.
The area of the image A|B|C| is k 2 times greater or smaller than the area of the
original ABC .
If the scale factor is greater than 1 ( k  1 ), then A|B|C| is an enlargement of
ABC . If the scale factor is a fraction between 0 and 1 ( 0  k  1 ), then A|B|C| is a
reduction of ABC .
Area of A|B|C|
 k2
Area of ABC
Perimeter of A|B|C|  k  Perimeter of ABC
Perimeter of A|B|C|
k
Perimeter of ABC
EXERCISE 6
(a)
y
In the diagram below, ABC and DEF
12
are given.
1
10
(1)
Write down the coordinates of the
9
vertices of ABC and DEF .
8
D
ABC is enlarged through
(2)
7
the origin by a scale factor of 4.
6
On the given Cartesian plane,
5
| | |
4
draw A B C , the image of
C
3
ABC .
2
(3)
Calculate the following ratios:
1 A
B
A|B| B|C|
A|C|
0
1
2
3 4
2 1
(i)
,
and
AB BC
AC
| | |
Area of A B C
Perimeter of A|B|C|
(ii)
(iii)
Area of ABC
Perimeter of ABC
(4)
(5)
(b)
DEF is reduced through the origin by a scale factor of
F
E
5
1
2
6
7
10
9
8
x
13
12
11
.
On the given Cartesian plane, draw D|E|F| , the image of DEF .
Calculate the following ratios:
D|E| E|F|
D|F|
Area of D|E|F|
Perimeter of D|E|F|
(i)
,
and
(ii)
(iii)
DE EF
DF
Area of DEF
Perimeter of DEF
In the diagram below, ABC is enlarged
through the origin by a scale factor of 5 to
form its image A|B|C| .
(1)
Write down the coordinates
of the vertices of ABC
and A|B|C| .
(2)
On the Cartesian plane, draw
A|B|C| .
y
10 9
8 7
6
5 4
3 2
A
1
B
0
1
C 2
3
4
5
6
7
8
9
267
10
1
2
3
x
(3)
Without calculating the areas, write down the ratio
Use the conclusion to assist you.
Area of A|B|C|
.
Area of ABC
(4)
(c)
(d)
(e)
Without actually calculating any lengths, determine the following ratios:
Perimeter of A|B|C|
A|B|
(i)
(ii)
Perimeter of ABC
AB
Use the conclusion to assist you.
B|C|
A|C|
and
?
(5)
What can you deduce about the ratios
BC
AC
A picture of Car A is enlarged through the origin by a scale factor of 2. Car B is the
image of Car A. Car B is then enlarged through the origin by a scale factor of 3. Car C
is the image of Car B. The coordinates of the left front wheel of Car A are (1; 2) .
(1)
Write down the coordinates of
the left front wheel of Car B.
(2)
Complete the general rule:
A(x ; y )  B( ; )
(3)
Write down the coordinates of
the left front wheel of Car C.
(4)
Complete the general rule:
A(x ; y )  C( ; )
(5)
How much bigger is Car B than
Car A?
(6)
How much bigger is Car C than
Car B?
(7)
How much bigger is Car C than
(1 ; 2)
Car A?
Two triangles ABC and A|B|C| are given. State whether the triangles are
congruent or similar in each of the following cases:
A|B| B|C| A|C|
A|B| B|C| A|C|
(1)


1
(2)


6
AB
BC
AC
AB
BC
AC
Perimeter of A|B|C|
Perimeter of A|B|C|
(3)
1
(4)
9
Perimeter of ABC
Perimeter of ABC
Area of A|B|C|
Area of A|B|C|
(5)
(6)
1
5
Area of ABC
Area of ABC
Two triangles PQR and P|Q|R | are similar. Determine the scale factor if:
(1)
P|Q|  3PQ
(2)
P|Q|  13 PQ
(3)
P|Q|
5
PQ
(4)
P|Q| 1

PQ 5
(5)
(6)
Area of A|B|C|  36 Area of ABC
Perimeter of A|B|C|  36 Perimeter of ABC
(7)
Area of P|Q|R |
 81
Area of PQR
(8)
Area of P|Q|R |
1

Area of PQR
81
(9)
Perimeter of P|Q|R |
 81
Perimeter of PQR
(10)
Perimeter of P|Q|R |
1

Perimeter of PQR
81
268
SUMMARY OF ALL TRANSFORMATIONS
Transformations in which the original triangle and its image are congruent
If ABC is translated, reflected about the axes or rotated through the origin so as to form its
image A|B|C| , then the two triangles are congruent. The triangles are identical in shape
and size.
Translations
If ABC is translated to form A|B|C| , the transformation involves a horizontal movement
left or right followed by a vertical movement up or down.
We translate the point ( x ; y ) to the point ( x  h ; y  v) where h is a horizontal shift and v is
a vertical shift.
If h  0 , the horizontal translation is to the right.
If h  0 , the horizontal translation is to the left.
If v  0 , the vertical translation is upward.
If v  0 , the vertical translation is downward.
Reflections
 If ABC is reflected about the y-axis to form A|B|C| , then first coordinates of the
triangles differ in sign: ( x ; y )  (  x ; y )

If ABC is reflected about the x-axis to form A|B|C| , then second coordinates of the
triangles differ in sign: ( x ; y )  ( x ;  y )
If ABC is reflected about the line y  x to form A|B|C| , then the first and second
coordinates interchange: ( x ; y )  ( y ; x)
Area and perimeter ratios for transformations involving congruent triangles
A|B|
B|C| A|C|
Area of A|B|C|
Perimeter of A|B|C|



1
1
1
AB
BC
AC
Area of ABC
Perimeter of ABC
Transformations in which the original triangle and its image are similar
If ABC is enlarged or reduced through the origin by a scale factor of k so as to form its
image A|B|C| , then the two triangles are similar. The triangles are identical in shape but
not size. To get the coordinates of the image points, multiply the original coordinates by the
scale factor.
 The lengths of the sides of A|B|C| are k times the lengths of the sides of ABC :
A|B|  k  AB
B|C|  k  BC
A|C|  k  AC
 The corresponding sides are in proportion. The ratio of one side of A|B|C| to the
corresponding side of ABC is equal to the scale factor (k).
A|B|
B|C|
A|C|
k
k
k
(k is the scale factor)
AB
BC
AC
 Area of A|B|C|  k 2  Area of ABC where k is the scale factor.
The area of the image A|B|C| is k 2 times greater or smaller than the area of the
original ABC . If the scale factor is greater than 1 ( k  1 ), then A|B|C| is an
enlargement of ABC . If the scale factor is a fraction between 0 and 1 ( 0  k  1 ), then
A|B|C| is a reduction of ABC .
Area of A|B|C| (image)
Area of ABC (original) 1

 k2
Note:

Area of ABC (original)
Area of A|B|C| (image) k 2



Perimeter of A|B|C|  k  Perimeter of ABC
Perimeter of A|B|C|
Perimeter of ABC
1
k
Note:

| | |
Perimeter of ABC
k
Perimeter of A B C
269

The scale factor (k) can be determined in the following ways:
Area of image shape
One side of image shape
k2 
k
Area of original shape
Corresponding side of original shape
k 
Area of image shape
Area of original shape
k
Perimeter of image shape
Perimeter of original shape
REVISION EXERCISE
(a)
(b)
For each of the following, determine whether the transformation is a translation, a
reflection about the x-axis, a reflection about the y-axis, a reflection about the line
y  x , an enlargement or reduction through the origin.
(1)
A(4 ; 2)  A| (4 ;  2)
(2)
A(4 ; 2)  A| (4 ; 2)
(3)
A(4 ; 2)  A| (8 ; 4)
(4)
A(4 ; 2)  A| (2 ;1)
(5)
A(4 ; 2)  A| (2 ; 5)
(6)
B(4 ; 2)  B| (4 ;  2)
(7)
B(4 ; 2)  B| (4 ; 2)
(8)
B(4 ; 2)  B| (7 ; 9)
(9)
B(4 ; 2)  B| (12 ; 6)
(10)
B(4 ; 2)  B| (2 ;1)
(11)
C(5 ;  6)  C| (5 ; 6)
(12)
C(5 ;  6)  C| (15 ;  18)
(13)
C(5 ;  6)  C| (1 14 ;  1 12 )
(14)
C(5 ;  6)  C| (5 ;  6)
(15)
D(3 ; 5)  D| (5 ;  3)
(16)
E(1;  9)  E| (9 ;1)
(17)
F(3 ;  7)  F| (7 ;  3)
(18)
G(1;  2)  G| (1; 2)
(19)
(1)
G(1;  2)  G| (1;  2)
(20) H(0 ;  2)  H| (2 ; 0)
Draw PQR where the coordinates of the vertices are P(3 ;  1) , Q(2 ;  1)
and R(1;  3) .
(2)
Draw P|Q|R | , the image of PQR , after it has gone through the following
transformations:
( x ; y )  (  x ; y ) followed by ( x ; y )  ( x ;  y ) followed by
( x ; y )  ( x  3 ; y  2) followed by ( x ; y )  ( y ; x) .
Complete:
P(3 ;  1)  P| ( ; )
Q(2 ;  1)  Q| ( ; )
R(1;  3)  R | ( ; )
(3)
(c)
y
ABC , DEF and square
PQRS are represented in the
given diagram.
(1)
On the given Cartesian
plane, draw A|B|C| , the
image of ABC if ABC
is reduced by a scale factor
of 12 .
(2)
(3)
3
2
1
4
3 2
1
0
2
1
D
1
P
2
3
E
S
3
4
5
F
Q
R
On the given Cartesian plane,
6
draw D|E|F| , the image of
7
DEF if DEF is enlarged
8
by a scale factor of 4.
9
On the given Cartesian plane,
10
draw P|Q|R |S| , the image of
PQRS if PQRS is a square that enlarged by a scale factor of 2.
270
7
8
9 10
11
x
12
B
A
C
4
5
6
(4)
(5)
(6)
(7)
(8)
(9)
(10)
On the given Cartesian plane, draw D||E||F|| , the image of DEF if DEF is
reflected about the y-axis, then reflected about the x-axis and then translated 8
units right and 1 unit down.
Which figures are congruent?
Which figures are similar?
How much larger is D|E|F| than DEF ?
How much smaller is A|B|C| than ABC ?
Determine the following ratios:
A|B|
D|E|
P|Q|
(i)
(ii)
(iii)
PQ
AB
DE
Determine the following ratios:
Perimeter of A|B|C|
Area of A|B|C|
(i)
(ii)
Perimeter of ABC
Area of ABC
| | |
Perimeter of D E F
Area of D|E|F|
(iv)
(iii)
Perimeter of DEF
Area of DEF
|| || ||
Perimeter of D E F
Area of D||E||F||
(vi)
(v)
Perimeter of DEF
Area of DEF
| | | |
Perimeter of P Q R S
Area of P|Q|R |S|
(viii)
(vii)
Perimeter of PQRS
Area of PQRS
SOME CHALLENGES
(a)
The point P(a ; b) is mapped onto its image P| . Write down the coordinates of P| if
the transformation is:
(1)
a reflection about the x-axis.
(2)
a reflection about the y-axis.
(3)
a reflection about the line y  x .
(4)
a translation of 4 units right and 3 units down.
(5)
a translation of 2 units left and 5 units up.
(6)
an enlargement through the origin where the scale factor is 6.
(7)
a reduction through the origin where the scale factor is 16 .
(b)
Two triangles PQR and P|Q|R | are similar. Determine the scale factor if:
(1)
P|Q|  3PQ
(2)
PQ  3P|Q|
(3)
P|R |  15 PR
(4)
PR  15 P|R |
(5)
Area of P|Q|R |
 16
Area of PQR
(6)
Area of P|Q|R |
1

Area of PQR
25
(7)
*(9)
*(11)
Perimeter of P|Q|R |
 16
Perimeter of PQR
Area of PQR
1

9
Area of P|Q|R |
Perimeter of PQR
1

| | |
3
Perimeter of P Q R
(8)
*(10)
*(12)
271
Perimeter of P|Q|R |
1

Perimeter of PQR
25
Area of PQR
9
Area of P|Q|R |
Perimeter of PQR
3
Perimeter of P|Q|R |
(c)
(d)
(e)
(f)
(g)
DEF is an equilateral triangle with a perimeter of 6 cm. The length of each side of
the triangle is increased by 4 cm. Determine the following:
(1)
the perimeter of DEF
(2)
the perimeter of D|E|F|
Area of D|E|F|
(3)
the scale factor
(4)
Area of DEF
| | |
Perimeter of D E F
D|E|
(6)
(5)
Perimeter of DEF
DE
The perimeter of a square is equal to 48 cm. The length of each side is doubled.
(1)
What is the length of one of the sides of the enlarged square?
(2)
What is the scale factor?
(3)
What is the area of the original square?
(4)
What is the area of the enlarged square?
(1)
The area of a square is 3 x 2 . If the area of the square is enlarged by a scale
factor of 3, determine the area of the enlarged square.
(2)
The perimeter of a square is 2 x . If the perimeter of the square is enlarged by a
scale factor of 3, determine the perimeter of the enlarged square.
Four identical circles fit into a square. Their centres are the vertices of the smaller
square. The smaller square has an area of 4 cm 2 .
Determine:
(1)
the scale factor if square EFGH is an
enlargement of square ABCD.
Area of EFGH
(2)
Area of ABCD
Perimeter of EFGH
(3)
Perimeter of ABCD
In the diagram below, A, B, C and D are congruent rhombuses.
(1)
Write down an algebraic rule to transform figure A to figure C.
(2)
Describe the two transformations that will transform figure A to figure B.
Hence write down a single algebraic rule to describe the transformation of A
to B.
(3)
Describe the two transformations that will transform figure B to figure D.
Hence write down a single algebraic rule to describe the transformation of B
to D.
y
8
D
7
6
5
4
B
C
3
2
1
8
7
6
5
4
3
2
1
0
1
2
3
4
5
6
7
8
272
1
2
3
A
4
5
6
7
8
9
x
CHAPTER 18: SHAPE AND SPACE (GEOMETRY)
TOPIC: GEOMETRY OF 3D SHAPES
REVISION OF PRISMS, PYRAMIDS AND PLATONIC SOLIDS
A polygon is a two-dimensional figure made up of three or more straight sides. A polygon
made up of three sides is a triangle. Polygons made up of four sides are called quadrilaterals.
Quadrilaterals include parallelograms, rhombuses, rectangles, squares, trapeziums and kites.
Polygons with more than four sides include pentagons (five sides), hexagons (six sides) and
octagons (eight sides).
A regular polygon is a polygon in which all the sides are equal in length and all the interior
angles are equal in size. Equilateral triangles and squares are regular polygons since their
sides and angles are equal. Pentagons, hexagons and octagons can be regular polygons if their
interior angles and sides are equal. It is important to note a rhombus is not a regular polygon
since its interior angles are not all equal. A rectangle is not a regular polygon since its sides
are not equal in length. Polygons that are not regular are called irregular polygons.
Note:

The formula for calculating the sum of the interior angles of a polygon of n sides is
given by the formula: 180(n  2) (See Constructions page 140)

The size of an interior angle of a regular polygon is given by the formula:
180(n  2)
n
Here are the regular polygons.
Polygon
3 sides
60
60
Regular polygon
Equilateral
Triangle
Interior angles
The sum of the interior angles:
180(3  2)  180
The size of an interior angle:
180(3  2)
 60
3
Square
The sum of the interior angles:
180(4  2)  360
The size of an interior angle:
180(4  2)
 90
4
60
4 sides
b
b
b
b
5 sides
Pentagon
108
108
108
108 108
54 54
54
54
72 72
54
54
72 72
72
54
54
54 54
273
The sum of the interior angles:
180(5  2)  540 .
The size of an interior angle:
180(5  2)
 108
5
The five triangles in the pentagon
are congruent isosceles triangles.
The five angles at the centre each
equal 72 since 72 5  360 .
The base angles of each triangle
all equal 54 .
6 sides
Hexagon
120 120
120
120
120 120
60 60
60
60
60
60 60 60 60
60 60 60 60
60
60
60
60 60
8 sides
Octagon
135 135
135
135
135
135
135 135
67,5 67,5
67,5
67,5
67,5
45
45
45
67,5
67,5
67,5
45
45
67,5
45
45
45
67,5
The sum of the interior angles:
180(6  2)  720 .
The size of an interior angle:
180(6  2)
 120
6
The six triangles in the hexagon
are congruent equilateral triangles.
The six angles at the centre each
equal 60 since 60 6  360 .
All angles in each triangle equal
60 .
The sum of the interior angles:
180(8  2)  1080 .
The size of an interior angle:
180(8  2)
 135
8
The eight triangles in the octagon
are congruent isosceles triangles.
The eight angles at the centre each
equal 45 since 45 8  360 .
The base angles of each triangle
all equal 67,5 .
67,5
67,5
67,5
67,5
67,5 67,5
Note:
Scalene and isosceles triangles, rectangles, kites and trapeziums are irregular polygons since
their sides are not equal in length. Rhombuses are irregular polygons since their interior
angles are not equal.
Let’s now focus on solid objects in three-dimensions. In particular, we will consider threedimensional solids called polyhedrons (or polyhedra).
A polyhedron is a three-dimensional solid bound by polygons. These polygons are called
faces (or sides). An edge is a side of the polygonal face. It is the line along which two faces
meet. A vertex is a corner of the polyhedron. This is the point where two or more edges
meet.
Notice:
The polyhedron on the right is made
up of the following:
8 faces

12 edges

6 vertices

There are three types of polyhedrons:
Prisms

Pyramids

Regular polyhedrons (Platonic solids)

274
Revision of prisms
A prism is a polyhedron with two
congruent parallel polygonal faces
at opposite ends of the polyhedron.
These faces are referred to as the
bases (or ends) of the prism.
The other faces are called lateral faces
and these faces are parallelograms.
Note:
The front base is parallel to the back base.
As the front base moves horizontally
towards the back base (or vertically
depending on the prism), the volume
of the prism is obtained. This is the
space occupied by the three-dimensional
prism. The surface area of the prism is
the sum of the areas of the two triangular
bases and the three lateral faces.
Irregular prisms
Irregular prisms do not have regular polygons
as bases. The lateral faces are parallelograms.
In the prism below, notice that the bases of the
prism are parallel, congruent but irregular.
Regular prisms
Regular prisms have regular polygons as
bases. The lateral faces are parallelograms.
In the prism on the right, notice that the
bases of the prism are parallel, congruent
and regular.
Right prisms
Whenever a prism (regular or irregular) has lateral rectangular faces that are perpendicular to
the bases, then the prism is called a right prism. Here are examples of right prisms.
275
Oblique prisms
Base
Late
ral fa
ce
Whenever a prism (regular or irregular) has
lateral faces that are not perpendicular to the
bases, then the prism is called an oblique
prism. Here is an example of an oblique prism.
Base
The angle between
the base and a lateral
face is not a right angle.
Revision of Pyramids
A pyramid is a polyhedron with a polygon as base. Three or more triangles are based on the
sides of the polygon and meet in one point, the apex of the pyramid. Right pyramids are such
that the apex is perpendicularly above the centre of the regular base. The slanted triangles
will therefore be congruent.
The pyramid in Figure 1 has a rectangular base (irregular polygon). The opposite triangles are
congruent and meet at the apex of the pyramid. The pyramid in Figure 2 has a pentagonal
base (regular polygon). All triangles are congruent and meet at the apex of the pyramid.
Nets of prisms and pyramids
If a prism is opened up and the faces are folded down so that the prism is made into a flat
surface, then we call this flat surface a net. Nets are useful for investigating the properties of
polyhedrons. Consider the following cube. It is possible to open up the cube, fold the faces
down and draw the net of this cube.
Example of the net of a prism
Open up the prism and flatten the squares. Then draw the net.
It is now possible to see that the cube has six square faces.
276
Example of the net of a pyramid
Open up the pyramid and flatten the four triangles. Then draw the net.
Summary of the different types of right prisms and pyramids
Right prism
Triangular prism
The bases are triangles (regular or irregular).
The lateral faces are rectangles.
The lateral faces are perpendicular to the
bases.
Rectangular prism
(Cuboid)
The bases are rectangles (irregular).
The lateral faces are rectangles.
The lateral faces are perpendicular to the
bases.
Cube
The bases are squares (regular).
The lateral faces are squares.
The lateral faces are perpendicular to the
bases.
Pentagonal prism
The bases are pentagons (regular or
irregular).
The lateral faces are rectangles.
The lateral faces are perpendicular to the
bases.
Hexagonal prism
The bases are hexagons (regular or
irregular).
The lateral faces are rectangles.
The lateral faces are perpendicular to the
bases.
Octagonal prism
The bases are octagons (regular or
irregular).
The lateral faces are rectangles.
The lateral faces are perpendicular to the
bases.
277
Right prism with trapezium bases
The bases are trapeziums (irregular)
The lateral faces are rectangles.
The lateral faces are perpendicular to the
bases.
Right prism with kite bases
The bases are kites (irregular)
The lateral faces are rectangles.
The lateral faces are perpendicular to the
bases.
Right Pyramids
Right triangular pyramid
Base is an equilateral triangle.
Apex is perpendicularly above the centre of
the base. The slanted faces are congruent
triangles.
Rectangular pyramid
(not a right pyramid)
Base is a rectangle.
Apex is perpendicularly above the centre of
the base. The slanted faces are triangles. The
opposite triangles are congruent.
Right square pyramid
Base is a square.
Apex is perpendicularly above the centre of
the base. The slanted faces are congruent
triangles.
Right pentagonal pyramid
Base is a regular pentagon.
Apex is perpendicularly above the centre of
the base. The slanted faces are congruent
triangles.
278
Right hexagonal pyramid
Base is a regular hexagon.
Apex is perpendicularly above the centre of
the base. The slanted faces are congruent
triangles.
Right octagonal pyramid
Base is a regular octagon.
Apex is perpendicularly above the centre of
the base. The slanted faces are congruent
triangles.
In Grade 8 you learnt about Euler’s law. This is the
equation connecting the number of faces (F), edges (E)
and vertices (V). The equation is V  E  F  2 .
Leonard Euler (1707-1783) was the first person to notice
that for all polyhedrons, this formula applies where
The value of this law is that it tells us whether or not a solid
is a polyhedron. For example, there is no polyhedron
with 10 faces, 17 vertices and 24 edges.
The reason for this is that:
V  E  F  17  24  10  3  2
[http://math2033.uark.edu/wiki/index.php/Leonhard_Euler]
The Platonic Solids and their nets
The five Platonic Solids have been known to us for thousands of years. These five special
polyhedrons are the tetrahedron, the hexahedron (cube), the octahedron, the icosahedron, and
the dodecahedron. The Platonic solids are regular polyhedrons in which all the faces are the
same regular polygon and where the same number of regular polygons meet at each vertex.
The sides are all equal in length and the angles are all equal.
These solids were discovered by the early Pythagoreans, perhaps by 450 BC. There is
evidence that the Egyptians knew about at least three of the solids; their work influenced the
Pythagoreans.
However it was Plato (427-347 BC) that studied these solids extensively. The five solids have
therefore been named after him. It was he who identified the solids with the elements
commonly believed to make up all matter in the universe. In Plato's times, people believed
that all things were made up of five different atoms. They were fire, air, water, earth, with the
fifth being the cosmos (the universe itself).
279
Tetrahedron
Icosahedron
Hexahedron (cube)
Dodecahedron
Octahedron
Plato
The nets of the five Platonic solids are revised below.
Hexahedron
Octahedron
Tetrahedron
Icosahedron
Dodecahedron
Notes on the faces, vertices and edges of the Platonic Solids
The faces of a hexahedron are squares. The faces of a tetrahedron, octahedron and icosahedron
are equilateral triangles. The faces of a dodecahedron are regular pentagons.
For all five Platonic solids, Euler’s law is true: V  E  F  2
The hexahedron and octahedron have the same number of edges (12). The number of faces and
vertices are interchanged. The hexahedron has 6 faces and 8 vertices whereas the octahedron
has 8 faces and 6 vertices. The dodecahedron and icosahedron have the same number of edges
(30). The number of faces and vertices are interchanged. The dodecahedron has 12 faces and
20 vertices whereas the icosahedron has 20 faces and 12 vertices.
280
EXERCISE 1 (Revision)
(a)
For each of the following polygons draw a net and state the number of faces.
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(b)
For each of the following nets, create and then draw the indicated polyhedron in your
workbook. Use the nets provided to you by your teacher to assist you in creating the
prisms. Cut out the nets from the cardboard sheets and fold them into the required
prism.
(1)
(2)
(3)
281
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(c)
(d)
Determine, using Euler’s law, whether or not a polyhedron can be formed if there are:
(1)
12 vertices, 14 edges and 4 faces.
(2)
16 vertices, 20 edges and 6 faces.
(3)
18 vertices, 17 edges and 3 faces.
(4)
15 vertices, 19 edges and 6 faces.
Verify Euler’s law for the following pyramids:
(1)
Triangular pyramid
(2)
Square pyramid
(3)
Rectangular pyramid
(4)
Pentagonal pyramid
(5)
Hexagonal pyramid
(6)
Octagonal pyramid
Note to the educator:
Excellent interactive lessons on solid geometry can be found at the following websites:
www.mathsisfun.com/geometry or www.yenka.com.
282
CYLINDERS
A cylinder is a solid with two circles as bases and a curved surface which is not a flat surface.
Therefore, although the cylinder is a prism (has two congruent parallel bases), it is not a
polyhedron (not all faces are flat polygons). If the cylinder is opened up and flattened, the net
will be made up of two identical circles and a rectangle. The rectangle has the same length as
the circumference of the circles which is 2r . The width of the rectangle will be equal to the
height of the cylinder (h).
2πr
Example
Calculate the area of the curved surface for the following cylinder.
Round off your answer to two decimal places.
Solution
The area of the curved surface is calculated using the
formula 2rh .
The radius is 7cm which is half of the diameter.
The length of the curved surface is 2(7) and
the width is 15cm (same as the height of the cylinder).
Area of curved surface
 length  width
 (2r )  h
 2(7) 15
2πr
 (210) cm 2
 659, 73 cm 2
SPHERES
A sphere is a perfectly round object in three-dimensional space that
resembles the shape of a completely round ball. It is not a polyhedron. The
r
points on the surface of the sphere are the same distance from the centre. The
radius is the straight line from any point on the sphere to its centre. Of all the
shapes, a sphere has the smallest surface area for a fixed volume and the
greatest volume for a fixed surface area. For example, if you blow up a
balloon, it naturally forms a sphere because it is trying to hold as much air as possible with as
small a surface area as possible. Spheres appear all around us. Here are some examples.
.
Pearls are spherical in shape, come from oysters and are made into expensive necklaces.
283
It is interesting to note that the Earth is not a perfect sphere because it is slightly squashed at
the poles. The Earth is called a spheroid. The sun is considered to be the most perfect natural
sphere. According to research published in the latest issue of the magazine “Science”, the Sun
is actually the most perfectly round natural object in the known universe.
A soccer ball is made from what is called a truncated icosahedron. This is a solid whose faces
are made up of two or more types of regular polygons. When a truncated icosahedron made
up of 12 regular pentagons and 20 hexagons is inflated, the flat regular surfaces become
curved and the solid becomes spherical in shape. A soccer ball is created in this way. There
are also 60 vertices and 90 edges.
EXERCISE 2
You may assume the following formulae in this exercise:
Volume of cylinder  r 2 h
Volume of a sphere  43 r 3
Surface area of a cylinder  2r 2  2rh
(a)
The volumes of both solids below are equal.
(1)
(2)
(3)
(4)
(5)
(b)
Surface area of a sphere  4r 2
(1)
(2)
If the cylinder has a height of 10 cm and a radius of 3 cm and the sphere has a
radius equal to 3 67,5 , show that the solids have the same volume.
Show that the sphere has a smaller surface area than the cylinder.
Draw a net for the cylinder. Indicate dimensions (not to scale).
Calculate the length of the curved surface rounded off to two decimal places.
Calculate the area of the curved surface rounded off to two decimal places.
If the surface area of a sphere is 36 , calculate the length of the radius.
If the height of a cylinder is 10 cm and the volume is 160 , calculate the
length of the radius.
284
(c)
A hemisphere is half of a sphere. It is in the shape of a dome.
(1)
(2)
Calculate the amount of paint (in litres) required to paint the exterior of the
dome with its cylindrical base. The dimensions are given above.
Calculate the volume of the dome with its cylindrical base.
REVISION EXERCISE
(a)
Match the following nets to their respective polyhedron. Name the polyhedron.
Net 1
Solid 1
Net 2
Solid 2
Net 3
Solid 3
Net 4
Solid 4
285
Net 5
Solid 5
Net 6
Solid 6
Net 7
Solid 7
Net 8
Solid 8
Net 9
Solid 9
Net 10
Solid 10
286
Net 11
Solid 11
Net 12
Solid 12
Net 13
Solid 13
Net 14
Solid 14
Net 15
Solid 15
Net 16
Solid 16
287
Net 17
(b)
Solid 17
Draw nets for the following three-dimensional solids.
SOME CHALLENGES
(a)
There are altogether 11 possible nets for a cube as shown in the diagram below.
Draw a net made up of six squares which cannot be folded into a cube.
(b)
The five Platonic solids are regular polyhedrons in which all the faces are the same
regular polygon and where the same number of regular polygons meet at each vertex.
The sides are all equal in length and the angles are all equal. Plato discovered that
there are only five Platonic solids.
Take note of the following information pertaining to Platonic solids:

The faces of each Platonic solid are identical regular polygons (squares,
triangles and pentagons)

At each vertex at least three faces or more must meet. Two polygons cannot
make a Platonic solid.

The formula for calculating the sum of the interior angles of a polygon of n
sides is given by the formula: 180(n  2)

The size of an interior angle of a regular polygon is given by the formula:
180(n  2)
n
288
(1)
Consider an icosahedron where five equilateral triangles meet at a vertex.
The net of the icosahedron is shown below. The five circled triangles are
flattened to form their own net.
Using the enlarged net provided to you by your teacher, cut out this net and
proceed as follows:
Drag the centre of this five-triangled net upwards and let the edges of the two
shaded triangles meet so that the five triangles once again meet at the vertex.
Now cut out the net on the right which has a further
triangle included. Try to drag the centre of this net
upwards as you did with the five-triangled net. Is it
possible to create a vertex where the six triangles meet?
(2)
Now answer the following questions:
(i)
Using the net with five triangles, calculate the sum of the five interior
angles meeting at the vertex.
(ii)
If the number of triangles increased to six, what would the sum of the
interior angles now be? Use the net with six triangles.
(iii)
Which net, the five-triangle or six-triangle one, made it possible to
drag the centre upwards so as to form a vertex where the triangles met?
Explain.
(iv)
What can you conclude about the sum of the interior angles at a vertex
of a polyhedron?
(v)
By calculating the sum of the interior angles at a vertex of the regular
polygonal faces of the other Platonic solids, verify your conclusion in
(iv).
(vi)
Explain why it is not possible to form a Platonic solid with a regular
hexagon.
(vii) Explain why it is not possible to form a Platonic solid with a regular
octagon.
(viii) Using your findings, explain why there can only be five Platonic
solids.
289
CHAPTER 19: DATA HANDLING
In this chapter, we will revise Data Handling concepts from Grade 8 and then discuss a new
type of graph called a scatter plot. There are three main topics in Data handling:
Topic 1:
Collect, organise and obtain summary statistics
Topic 2:
Represent data
Topic 3:
Analyse, interpret and report data.
Let’s briefly revise the main ideas in these three topics. Then we will revise the data cycle
used in research situations.
TOPIC 1: COLLECT, ORGANISE AND OBTAIN SUMMARY STATISTICS
Collecting data
Let’s briefly revise some important features of collecting data studied in Grade 8.
The starting point in any research project is to select a topic of research. A person might
consider starting a business and needs to research various aspects of the business in order to
assess whether it will be a worthwhile venture or not. Someone else might want to investigate
South African attitudes towards rhino poaching or e-tolls. Once the topic has been chosen,
methods of obtaining or collecting information relating to the research needs to be planned
and implemented. The following features are important to take into consideration when
collecting data:
Data:
The results of a statistical investigation are called data. For example,
the set of marks obtained by a Grade 9 class for a class test given is
called data. Raw data is data that is not organised in any meaningful
way. This data must be organised and presented in different ways so that
it can be interpreted and analysed.
Discrete data:
This is data that is counted. The data values are whole numbers. For
example, the number of spectators attending the different matches
during the 2014 FIFA World Cup is discrete data.
Continuous data: This is data that is measured. The data values are rational numbers. For
example, measuring the heights of the soccer players in the World Cup is
continuous data since the heights are not restricted to whole numbers.
The heights can be 1,85 m or 1,76 m.
Survey:
A survey is used to collect data from a selection of people in a
community. If data is to be collected from all people in that community,
then a census is taking place.
Questionnaire:
A questionnaire is used to collect data when doing a survey. The
following guidelines are important when designing a questionnaire:
Make sure the instructions are clear for the respondent (the person who
answers the questionnaire).
Use multiple choice responses to avoid ambiguous answers.
Avoid controversial questions that might offend the respondent.
Population:
A population is made up of the entire group of people studied. For
example, a population could be all women in South Africa. Another
population could be all the members of a particular religion.
Sample:
A sample of a population is a small selection from the population. It
must be representative, random and unbiased and must represent all
people in the larger population. People must be selected at random so as
to avoid bias leading to misleading results.
290
Organising data
Data can be organised using tallies, frequency tables, stem and leaf displays and class
intervals.
Obtaining summary statistics for the data
Summary statistics for data include the following:
(a)
Measures of central tendency:
(1) Mean:
This is the average of all data values.
(2) Median: This is the middlemost value.
(3) Mode:
This is the most frequently-occurring value.
(b)
Measures of dispersion:
(1) Range:
Difference between highest and lowest data values.
(2) Extremes: These are values that are significantly higher or lower than the other
data values. Extremes can affect the mean of the data and are
sometimes excluded from the data.
Example 1
(Ungrouped data, frequency tables and summary statistics)
A research survey was recently conducted by a university to investigate the profitability of
petrol stations in South Africa. The topic researched was the effect that Convenience Stores
at petrol stations have on petrol sales. The appearance of the Convenience Store, as well as
the service given by the staff are two factors that may well attract customers to the station
thereby increasing petrol sales. The researchers investigated this hypothesis by working with
two petrol stations and interviewing a sample of 10 000 customers per station over a period
of six months. Questionnaires were used in the research and the data was collected, recorded
and organised. The ratings of two selected loyal customers that regularly fill up at each
petrol station were recorded at different times. [www.actacommercii.co.za]
Rating scale scores
Extremely poor
(1)
Poor
(2)
Average
(3)
Convenience Store A
Appearance rating
Service rating
Score
Frequency
Score
Frequency
1
482
1
891
2
513
2
1 105
3
1 808
3
2 905
4
4 192
4
3 089
5
3 005
5
2 010
Customer at Store A
5
4
5
4
Appearance of store
5
2
5
4
Service given
Customer at Store B
1
2
1
2
Appearance of store
1
1
2
1
Service given
(a)
(b)
(c)
(d)
Outstanding
(5)
Convenience Store B
Appearance rating
Service rating
Score
Frequency
Score
Frequency
1
2 226
1
2 101
2
3 317
2
3 199
3
3 210
3
3 202
4
760
4
1 397
5
487
5
101
3
5
2
3
4
3
4
4
4
5
4
4
4
4
3
2
2
3
3
2
2
3
3
2
3
3
3
2
Calculate the mean, mode, median and range
given by the selected customer at Store A.
Calculate the mean, mode, median and range
given by the selected customer at Store B.
Calculate the mean, mode, median and range
of the sample of 10 000 for Store A.
Calculate the mean, mode, median and range
of the sample of 10 000 for Store B.
291
Good
(4)
4
4
for the appearance and service ratings
for the appearance and service ratings
for the appearance and service ratings
for the appearance and service ratings
Solutions
(a)
Customer at Store A
Appearance of the store:
2
3
4
4
4
4
4
4
4
5
5
43
Mean 
 3,91
Mode  4 (most frequently occurring score)
11
There is an odd number of values. The median is part of the data set.
Median  4 (bold number in the data set)
Range  Highest value  lowest value  5  2  3
Service given:
2
3
3
4
4
4
4
5
5
5
5
44
Mean 
4
There are two modes: 4 and 5 (data is bimodal)
11
There is an odd number of values. The median is part of the data set.
Median  4 (bold number in the data set)
Range  Highest value  lowest value  5  2  3
(b)
Customer at Store B
Appearance of the store:
1
1
2
2
2
2
3
3
3
3
3
4
29
Mean 
 2, 42
Mode  3 (most frequently occurring score)
12
There is an even number of values. The median is not part of the data set. It is the
average of the middle two values 2 and 3 (bold numbers in the data).
23 5
Median 
  2,5 (number inserted between 2 and 3)
2
2
1
1
2
2
2
2 2,5 3
3
3
3
3
4
Range  Highest value  lowest value  4  1  3
Service given:
1
1
1
2
2
2
2
2
3
3
3
4
26
 2,17
Mode  2 (most frequently occurring score)
Mean 
12
There is an even number of values. The median is not part of the data set. It is the
average of the middle two values 2 and 2.
22
Median 
 2 (number inserted between 2 and 2)
2
1
1
1
2
2
2 2 2
2
3
3
3
Range  Highest value  lowest value  4  1  3
(c)
Convenience Store A
Appearance rating:
(1 482)  (2  513)  (3 1 808)  (4  4 192)  (5  3 005) 38 725

 3,87
Mean 
10 000
10 000
Mode  4
292
4
Score
1
2
3
4
5
Appearance rating
Frequency
482
513
1 808
The first 2 803 scores are 3 or less
4 192
The first 6 995 scores are 4 or less
3 005
The median is the middlemost value. There are 10 000 data values. Therefore the
median is the average between the 5 000th and 5 001st value. Since the first 2 803
values have a rating of 3 or less ( 482  513  1 808  2 803 ) and the first 6 995
values have a rating of 4 or less ( 482  513  1 808  4 192  6 995 ), the median
must be 42 4  4 .
Range  Highest value  lowest value  5  1  4
Service rating:
(1 891)  (2 1 105)  (3  2 905)  (4  3 089)  (5  2 010) 34 222

 3, 42
Mean 
10 000
10 000
Mode  4
Service rating
Score
1
2
3
4
5
Frequency
891
1 105
2 905
3 089
2 010
The first 4 901 scores are 3 or less
The first 7 990 scores are 4 or less
The median is the middlemost value. There are 10 000 data values. Therefore the
median is the average between the 5 000th and 5 001st value. Since the first 4 901
values have a rating of 3 or less ( 891  1 105  2 905  4 901 ) and the first 7 990
values have a rating of 4 or less ( 891  1 105  2 905  3 089  7 990 ), the median
must be 42 4  4 .
Range  Highest value  lowest value  5  1  4
(d)
Convenience Store B
Appearance rating:
(1 2 226)  (2  3 317)  (3  3 210)  (4  760)  (5  487) 23 965

 2, 40
Mean 
10 000
10 000
Mode  2
Score
1
2
3
4
5
Appearance rating
Frequency
2 226
The first 2 226 scores are equal to 1
3 317
The first 5 543 scores are 2 or less
3 210
760
487
The median is the middlemost value. There are 10 000 data values. Therefore the
median is the average between the 5 000th and 5 001st value. Since the first 2 226
values have a rating of 1 and the first 5 543 values have a rating of 2 or less
( 2 226  3 317  5 543 ), the median must be 22 2  2 .
293
Range  Highest value  lowest value  5  1  4
Service rating:
(1 2 101)  (2  3 199)  (3  3 202)  (4 1 397)  (5 101) 24 198
Mean 

 2, 42
10 000
10 000
Mode  3
Score
1
2
3
4
5
Service rating
Frequency
2 101
The first 2 101 scores are equal to 1
3 199
The first 5 300 scores are 2 or less
3 202
1 397
101
The median is the middlemost value. There are 10 000 data values. Therefore the
median is the average between the 5 000th and 5 001st value. Since the first 2 101
values have a rating of 1 and the first 5 300 values have a rating of 2 or less
( 2 101  3 199  5 300 ), the median must be 22 2  2 .
Range  Highest value  lowest value  5  1  4
Example 2
(Stem-and-leaf displays and grouped data in class intervals)
The researchers in the previous example investigated the number of litres of petrol
purchased by 50 motorists at each of the two petrol stations on one Saturday morning in
April. The data is presented below (litres are rounded off to the nearest whole number).
Petrol Station A:
(Only raw data was made available)
82
71
64
94
72
50
98
84
66
80
64
78
66
95
81
55
88
80
72
74
Petrol Station B:
Class interval
30  x  40
40  x  50
50  x  60
60  x  70
70  x  80
80  x  90
90  x  100
(a)
(b)
(c)
(d)
(e)
(f)
(g)
49
96
88
81
83
44
77
72
82
87
52
75
71
88
79
59
54
65
73
38
68
57
68
67
77
74
56
97
72
39
(Data was made available as grouped data in class intervals)
Frequency
1
4
6
9
18
7
5
Draw a stem-and-leaf display for Station A.
Organise the data into class intervals.
Calculate the actual mean for Station A.
Why is it not possible to calculate the actual mean for Station B?
Calculate an estimated mean for Station B.
In what interval is the mode for Station A and Station B?
In what interval is the median for Station A and Station B?
294
Solutions
(a)
3
4
5
6
7
8
9
(b)
(c)
(d)
(e)
8
4
0
4
1
0
4
9
9
2
4
1
0
5
4
5
2
1
6
5
6
2
1
7
6
6
2
2
8
7
7
2
2
9
8 8
3 4 4 5 7 7 8 9
3 4 7 8 8 8
Class interval (litres)
30  x  40
40  x  50
50  x  60
60  x  70
70  x  80
80  x  90
90  x  100
Frequency
2
2
7
8
14
12
5
3 602
 72 litres (add up all values and divide by 50)
50
The actual litres in each class interval is not known.
Actual mean 
Class interval
Midpoint
30  x  40
90  x  100
30  40  35
2
40  50  45
2
50  60  55
2
60  70  65
2
70 80  75
2
80  90  85
2
90 100  95
2
Totals
-
40  x  50
50  x  60
60  x  70
70  x  80
80  x  90
Frequency
Midpoint  Frequency
1
35
4
180
6
330
9
585
18
1 350
7
595
5
475
50
3 550
3 550
 71
50
Mode for Station A is in the interval 70  x  80 .
Mode for Station B is in the interval 70  x  80 .
The median for both stations is the average between the 25th and 26th value. The
median for both stations is in the interval 70  x  80 .
Estimated mean 
(f)
(g)
295
EXERCISE 1
(a)
In Example 1, the ratings of two selected customers were recorded. The researchers
selected four further customers and the ratings for appearance at Store B are provided
below.
Customer 3
Customer 4
Customer 5
Customer 6
1
2
3
2
3
3
5
3
2
2
3
3
2
3
4
4
4
3
3
3
2
3
4
4
5
4
5
4
5
5
5
3
5
4
5
3
5
4
5
3
5
4
Calculate the mean, mode, median and range for the appearance ratings given by the
four additional customers.
(b)
The researchers decided to include a third petrol station (C) located on a national
highway in Gauteng in their research (see Example 1). The sample size increased to
100 000 customers. The data for the ratings are presented below.
Convenience Store C
Appearance rating
Service rating
Score
Frequency
Score
Frequency
1
2 820
1
3 910
2
3 130
2
10 050
3
18 080
3
19 050
4
45 922
4
45 890
5
30 048
5
21 100
Calculate the mean, mode, median and range for the appearance and service ratings
of the sample of 100 000 for Store C.
(c)
The researchers then decided to investigate the number of litres of petrol purchased
by 5000 customers at Station C in May. The data is presented as grouped data in class
intervals.
Class interval
Frequency
30  x  40
124
40  x  50
326
50  x  60
402
60  x  70
678
70  x  80
2 402
80  x  90
857
90  x  100
211
(1)
(2)
(3)
(4)
(d)
Why is it not possible to calculate the actual mean for Station C?
Calculate an estimated mean for Station C.
In what interval is the mode for Station C?
In what interval is the median for Station C?
The researchers investigated the number of litres of diesel purchased by 30 truck
drivers at Station C. The raw data is presented below (litres are rounded off to the
nearest whole number).
82
64
55
50
49
44
52
59
68
74
71
78
88
98
96
77
75
54
57
56
64
66
80
84
88
72
71
65
68
97
(1)
Draw a stem-and-leaf display for this data
(2)
Organise the data into class intervals.
(3)
Calculate the actual mean for this data.
(4)
Calculate an estimated mean for this data.
296
TOPIC 2: REPRESENT DATA
Once data has been collected, organised and summary statistics provided, the next step is to
represent the data graphically by means of bar graphs, histograms, pie charts and broken-line
graphs. This will give a visual picture of the data.
Example 3
In Example 1, the researchers obtained data involving the rating scores of customers at two
different petrol stations, A and B. We can represent the ratings of the sample groups in the
form of bar graphs. The data is presented below. Draw a double bar graph and broken-line
graphs for this data.
Convenience Store A
Appearance rating
Service rating
Score
Frequency
Score
Frequency
1
482
1
891
2
513
2
1 105
3
1 808
3
2 905
4
4 192
4
3 089
5
3 005
5
2 010
Convenience Store B
Appearance rating
Service rating
Score
Frequency
Score
Frequency
1
2 226
1
2 101
2
3 317
2
3 199
3
3 210
3
3 202
4
760
4
1 397
5
487
5
101
Solution
Bar graph
4500
Appearance
Service
4192
4000
Frequency
3500
3317
3199
2905
3000
2500
2226 2101
3210 3202
3089
3005
2010
1808
2000
1500
1397
1105
1000
500
760
891
513
482
487
101
B
A
Rating (1)
A
B
Rating (2)
A
B
Rating (3)
B
A
Rating (4)
B
A
Rating (5)
Broken-line graphs
4500
Appearance
Store B
Store A
4192
4000
Frequency
3500
3317
3210
3005
3000
2500
2226
2000
1808
1500
1000
500
482
760
513
487
Rating (1)
Rating (2)
Rating (3)
297
Rating (4)
Rating (5)
Store A
4500
Service
Store B
4000
Frequency
3500
3202
3199
3000
3089
2905
2500
2101
2000
2010
1500
1105
1000
1397
891
500
101
Rating (1)
Rating (2)
Rating (3)
Rating (4)
Rating (5)
Note:
The data represented in the bar graph and broken-line graphs is discrete. The data values are
whole numbers. The broken-line graphs show the trends in the data more clearly than the
bar graph. It is clear to see that the ratings are much better for Store A than for Store B.
Example 4
In Example 2, the researchers investigated the number of litres of petrol purchased by 50
motorists. The data for Station A is presented below. Draw a histogram for this data.
Class interval (litres)
30  x  40
40  x  50
50  x  60
60  x  70
70  x  80
80  x  90
90  x  100
Frequency
2
2
7
8
14
12
5
Solution
20
18
Frequency
16
14
12
10
8
6
4
2
0
Number of litres
298
Note:
In a histogram, the data is continuous since the number of litres is measured. The values can
be rational numbers. There can be 50,2 litres. In this example, the values measured have
been rounded off to the nearest whole number. The rectangular bars are next to each other.
Example 5
The researchers investigated what percentages of the petrol price are allocated to the
different stake-holders. This information is presented below. Represent the data in a piechart. The petrol price is 1433 cents per litre.
Item
Government
Oil refinery
Oil company
Petrol station owner
Total
Percentage per litre
31%
50%
11%
8%
100%
[www.investorcontacts.co.za]
Solution
In order to represent this information on a pie graph, you need to do the following:

Express the data values as fractions and percentages of the total.
Multiply each fraction by 360 to obtain angles.

Round off the angles to the nearest degree (all angles must add up to 360 )

Draw the angles in the pie graph (you may use estimation)

Use different colours to shade in each sector of the circle.

Name each sector as per the given data.

The price of a litre in rands is R14,33.
Government:
Oil company:
0,3114,33  R4,44
0,1114,33  R1,57 (true rounded off value is 1,58)
4,44
1,58

 360  112

 360  40
14,33
14,33
Refinery:
Petrol station owner:
0,50 14,33  R7,17
0, 08 14,33  R1,15
7,17
1,15

 360  180

 360  28
14,33
14,33
Note: R4,44  R1,57  R7,17  R1,15  R14,33 and 180  112  40  28  360
180
112
40
28
299
EXERCISE 2
(a)
(b)
(c)
(d)
Refer to Exercise 1(b) on page 296. Draw a bar chart and broken-line graph to
represent the data.
Refer to Exercise 1(c). Draw a histogram for the data.
Refer to Exercise 1(d). Draw a histogram for the data.
Motorists not only have to deal with increases in petrol prices, but also with having to
pay toll fees. Here is a breakdown of the toll fees for the four different classes of
vehicle for toll roads between Johannesburg/Pretoria and Cape Town. Draw a bar
graph to represent this data.
Toll plaza
Grasmere
Vaal
Verkeerdevlei
Huguenot
Total cost
[www.aa.co.za]
(e)
Class 2
(2 axle)
R46.00
R95.00
R87.00
R84.00
R312.00
Class 3
(3-4 axle)
R53.00
R115.00
R131.00
R131.00
R430.00
Class 4
(5  axle)
R70.00
R153.00
R184.00
R212.00
R619.00
In Example 1 and Exercise 1 (b), the researchers wanted to investigate the
profitability of owning a petrol station. Not only did they research the effect of factors
such as the appearance of the convenience stores and the quality of service provided,
they also investigated the total monthly revenue generated by petrol stations A, B and
C over the six-month period (Jan – Jun). The data is presented below.
Store
A
B
C
(1)
(2)
(f)
Class 1
(light vehicles)
R15.00
R50.50
R43.00
R30.00
138.50
Jan
5 302 100
2 866 000
5 732 100
Feb
5 373 750
3 009 300
6 305 200
Mar
5 732 000
2 579 400
6 591 800
Apr
6 448 500
2 006 200
7 165 000
May
6 735 100
1 433 000
7 594 900
Jun
5 732 000
1 146 400
7 021 700
On the same broken-line graph, represent the total monthly revenue generated
by the three petrol stations.
Discuss the trends for the three petrol stations.
The start-up costs for owning a petrol station business are as follows:
Petrol station construction
Equipment for petrol station
Office equipment
Office furniture
Working capital (first purchase)
Total cost
[www.investorcontacts.co.za]
R12 000 000
R2 500 000
R150 000
R 200 000
R1 911 654
R16 761 654
Draw a pie-chart to represent this data.
(g)
Petrol and diesel is moved from refineries by pipelines, rail, sea and road to
approximately 4600 petrol stations nationally. The average distribution of litres of
petrol per province per month is presented below. Draw a pie chart for this data.
Gauteng
336 251 830
Mpumalanga
71 575 310
North West
50 514 308
Western Cape
145 322 807
Eastern Cape
67 212 922
Limpopo
41 181 998
300
KwaZulu-Natal
142 435 035
Free State
51 995 253
Northern Cape
15 999 364
Representing data in scatter plots
Scatter plots are graphs that represent the relationship between two sets of data. The scatter
plot reveals if there is a relationship or correlation between the two sets of data. For
example, suppose that the two sets of data are the alcohol level of drivers and the number of
road accidents. The alcohol level can be represented on the x-axis and the number of
accidents on the y-axis. The correlation between alcohol level and number of accidents can be
investigated by using the scatter plot graph. Sometimes, data values can fall outside the
general trend of the other values. We refer to these values as outliers.
Example 6
A researcher interested in the relationship between blood alcohol levels and road accidents
examined the number of accidents for various blood alcohol levels in certain towns in South
Africa during 2014. The results are recorded in the following table:
Alcohol level
(g/100ml)
No. of accidents
Alcohol level
(g/100ml)
No. of accidents
(a)
(b)
(c)
(d)
0.005
0.010
0.015
0.020
0.025
0.030
0.035
0.040
0.045
100
170
260
300
320
380
420
450
500
0.050
0.055
0.060
0.065
0.070
0.075
0.080
0.085
0.090
550
570
590
900
600
630
650
625
700
Draw a scatter plot to represent the two sets of data.
Describe the trends in the scatter plot.
Which point is an outlier in the data? Explain possible reasons for this outlier.
Predict the possible number of accidents if the blood alcohol level is 0.1
Solutions
(a)
.
850
800
.
.
.
..
750
700
650
600
550
500
400
350
300
250
200
150
100
.
.
..
.
0.020
450
0.025
Number of accidents
900
..
.
..
.
.
0.085
0.090
0.080
0.075
0.070
0.060
0.065
0.050
0.055
0.045
0.040
0.030
0.035
0.010
0.015
0.005
50
Blood Alcohol levels
(b)
(c)
(d)
There seems to be a correlation between high alcohol levels and increased number
of accidents. The trend is linear (the pattern follows a straight line).
There is an outlier at a blood alcohol level of 0.065. The number of accidents are
extremely high and don’t follow the general linear trend of the other data values.
A possible explanation for the high number of accidents is that there might have
been an unusual amount of traffic on the roads with a whole lot of accidents
happening. A blood alcohol level of 0.065 was probably the most common for the
drivers.
This is difficult to know for sure but based on the scatter plot, it seems as if it might
increase to 750 accidents.
301
Note:
Data on a scatter plot could follow a linear positive or negative trend.
Sometimes there may be no correlation between the data.
Positive linear correlation
y
.
.
.
.
.. .
.
.. .
Negative linear correlation
y
No correlation
y
..
.
. .. .
. . ..
.
....
... ..
..
x
x
x
EXERCISE 3
(a)
The researchers (see previous examples) investigated the relationship between the
location of the petrol station and the average monthly sales of petrol. The location was
given a rating from 1-10 on the following criteria:
Relation to a major route:
3
Access to the station:
3
Geographical location rural or urban:
3
Relation to a shopping centre:
1
24 petrol stations were given a location rating and the average volume of petrol sold
per month was recorded. The raw data obtained is presented below.
Location
rating
Volume
of sales
Location
rating
Volume
of sales
Average monthly sales volume
Location
rating
Volume
of sales
3
4
4
4
5
6
6
6
50 000
100 000
120 000
180 000
150 000
200 000
250 000
600 000
7
7
7
8
8
8
8
9
140 000
190 000
300 000
310 000
350 000
370 000
400 000
100 000
9
9
9
9
9
10
10
10
395 000
440 000
500 000
540 000
580 000
560 000
600 000
640 000
700 000
600 000
500 000
400 000
300 000
200 000
.
100 000
1
2
3
...
4
5
6
Location rating of petrol station
302
7
8
9
10
(1)
(2)
(3)
(b)
The first four data values are represented on the scatter plot graph. Plot the
other data values on the graph.
Describe the trend of the data.
What petrol stations are outliers? What has possibly caused these stations to be
outliers?
The researchers also investigated the relationship between the number of pumping
bays and the average monthly sales. The raw data is presented below. The number of
pumping bays at 24 different stations was recorded as well as the sales figures for one
month.
Number
of bays
Volume
of sales
Number
of bays
Volume
of sales
Average monthly sales volume
Number
of bays
Volume
of sales
5
6
6
7
8
8
8
8
150 000
100 000
240 000
120 000
200 000
240 000
300 000
310 000
8
8
9
10
10
11
12
12
390 000
500 000
220 000
400 000
600 000
160 000
150 000
300 000
12
12
12
14
14
18
18
20
340 000
400 000
490 000
300 000
370 000
310 000
400 000
600 000
16
18
700 000
600 000
500 000
400 000
.
. . .
300 000
200 000
100 000
2
(1)
(2)
4
6
8
10
Number of pumping bays
12
14
The first four data values are represented on the scatter plot graph. Plot the
other data values on the graph.
Is there a correlation between the number of pumping bays and volume of
petrol sold?
303
20
TOPIC 3: ANALYSE, INTERPRET AND REPORT DATA
Whenever a statistical investigation is conducted, the researcher must follow the steps of
what is called the data cycle. The data cycle involves the following nine steps:
STEP 1
STEP 2
STEP 3
STEP 4
STEP 5
STEP 6
STEP 7
STEP 8
STEP 9
Select a topic of research
A person might be considering starting a business and needs to research
various aspects of the business to assess whether it will be a worthwhile
venture or not. Someone else might want to investigate South African attitudes
towards rhino poaching or e-tolls.
Collect the data
The researcher now needs to collect data for her investigation. She might want
to use questionnaires. Samples of a population must be random, representative
and unbiased.
Record the data
Once the researcher has collected the data, the next step will be to record the
data. This involves capturing the information from completed questionnaires.
Organise the data
The next step is to organise the data into frequency tables or class intervals.
Summary statistics can be obtained for the data.
Represent the data
The researcher now needs to decide which graphs are most suitable for
representing the data. Bar graphs, broken-line graphs, pie charts, histograms
and scatter plots are commonly used.
Analyse the data
The researcher will now study the graphs and look for trends in the data.
Measures of central tendency (mean, median and mode) and dispersion (range,
extremes and outliers) can be used to investigate different aspects of the data
(age, gender, quantity of products).
Summarise the data
The findings from the analysis of the data must now be summarised in a
sensible manner so that conclusions can be made in response to the question
posed in Step 1.
Interpret the data
In this step, conclusions are now made based on the summarised data. Whether
or not the business is going to work is decided on in this step.
Report the data
In this final step, a written report based on the findings is drawn up. This
report brings everything together so that final decisions can be made. If the
investigation was about a topic of interest like rhino poaching, then the report
will give a written conclusion about current attitudes and the way forward. The
report can also contain recommendations to interested parties. For example,
a medical aid group might have researched the effectiveness of a new pill for
treating headaches. Recommendations are made in the final report regarding
whether or not it should be made available in the market.
In the investigation that follows, you will be required to work through a data cycle as an
example. Once you have completed this investigation, you will have a better understanding of
what researchers do. You will then be given a project to do that will require you to select
your own research topic and to work through an entire data cycle. This project will take some
time and it will count for a lot of marks. Your teacher will provide you with the details of this
research project.
304
INVESTIGATION
Note to educator:
This investigation and its rubric are available in the Teacher’s Guide
on page 306.
In this investigation, you are required to research the profitability of owning a petrol station
in South Africa. The findings of the researchers in this chapter may be used in your research.
The first five steps of the data cycle have been done for you through the examples and
exercises. The remaining steps are left as an exercise for you to do. You will be required to
analyse, summarise and interpret the findings of the researchers and you will need to draw up
a final report in which you make conclusions about owning a petrol station based on the
research findings. The investigation is broken up into different tasks.
TASK 1
In this task, you will focus on a petrol station owner’s monthly income, expenses and profit.
This information is essential for any potential petrol station owner.
There are two types of petrol prices charged per litre:
The wholesale price is the price charged to the petrol station owner and the retail price is
the price charged to the consumer. The retail price is a mark-up on the wholesale price. The
dealer margin is the difference between these two prices. The dealer margin is the extra
profit per litre made by the owner of the petrol station.
The petrol station owner charges the customer the retail price. He keeps the dealer margin
and must then pay the following stake-holders a percentage of the wholesale price:
The Government: 31%
The Oil Company: 11%
The Oil Refinery: 50%
The owner gets to keep 8% of the wholesale price plus the dealer margin. This is called his
gross profit per litre. When multiplied by the number of litres sold, this is then his gross
income.
The owner then has other expenses which include:
Staff salaries:
Rental:
Water, electricity, advertising, interest, telephone:
50% of gross profit
15% of gross profit
26% of gross profit
Assume that the number of litres of petrol sold per month by the petrol station owner is
370 000 litres, the wholesale price is R12,93 and the retail price is R14,56.
(a)
What is the total revenue received by the owner using the retail price?
(1)
(b)
Calculate the dealer margin for the petrol station owner.
(2)
(c)
Calculate the amount paid per litre to the Government, the Oil Company and
the Refinery. How much profit per litre does the owner then make?
(4)
(d)
Calculate the owner’s total gross income.
(1)
(e)
How much money will the owner have to pay towards the other expenses?
(4)
(f)
What is the owner’s net profit after the other expenses?
(2)
(g)
Why is the dealer margin so important to the petrol station owner?
(1)
(h)
What is the effect of a strike in the petrol industry or a failure of the refineries to
deliver petrol to the petrol stations on time?
(2)
(i)
Draw a bar graph to represent the percentage allocation based on the wholesale
price.
(4)
(j)
Draw a pie chart to represent the other expenses and the owner’s net profit.
(4)
[25]
305
TASK 2
In this task, you will analyse the effect of petrol price changes on the amount of petrol sold
every year.
(a)
The graph below represents the average percentage change in the petrol price as well
as the average percentage change in the volume of petrol sold over the period 2002 –
2012. [www mbendi.co.za]
% change in volume
% change in price
35%
% change
30%
25%
20%
15%
10%
5%
0%
5%
6%
4%
2002
1%
2003
2004
2005
2006
2007
2008
2009
2010
2011
2012
4%
10%
Now answer the following questions based on the broken-line graph:
(1)
What was the lowest average percentage change in the petrol price over the
given period?
(1)
(2)
What was the highest average percentage change in the amount of petrol sold
over the given period?
(1)
(3)
By referring to trends in the graph, does the change in the average percentage
of the petrol price seriously affect the amount of petrol sold?
(4)
(b)
The average monthly fuel prices in cents per litre for unleaded petrol for 2013 and
2014 (up to July) are presented below.
2013
Jan
Feb
1186 1227
2014 Jan
Feb
1357 1396
[www.aa.co.za]
(1)
(2)
(3)
(4)
Mar
1308
Mar
1432
Apr
1320
Apr
1439
May
1247
May
1424
Jun
1239
Jun
1361
Jul
1323
Jul
1392
Aug
1355
Sep
1350
Oct
1330
Nov
1302
Draw a broken-line graph to represent the data for this time period.
Discuss the trends in the petrol price over this period.
Calculate the average percentage change in price from January 2013 to
July 2014. Will the percentage changes from month to month affect the
amount of petrol sold? Explain.
Draw a misleading graph to show that the fuel price is out of control.
[Hint: Do something weird with the scale and have fun]
306
Dec
1319
(3)
(2)
(5)
(4)
[20]
TASK 3
In this task, you are required to analyse, summarise, interpret and report the findings of the
researchers presented in the chapter.
The following are the key factors that might possibly affect the owning of a profitable petrol
station:
(1)
(2)
(3)
(4)
(5)
(6)
(7)
The appearance of and the service at the convenience stores.
(Refer to Example 1, Exercise 1(a), (b), Example 3)
The number of litres of petrol or diesel sold per month.
(Refer to Example 2, Exercise 1 (c), (d), Example 4, Exercise 2(g))
The petrol price.
(Refer to Task 2)
The total revenue generated per month.
(Refer to Exercise 2(e) and Task 1)
The expenses (stake-holders and other expenses).
(Refer to Example 5 and Task 1)
The location of the petrol station.
(Refer to Exercise 3(a))
The number of pumping bays.
(Refer to Exercise 3(b))
Other factors might include:
(1)
(2)
(3)
(4)
The start-up costs of owning a station.
(Refer to Exercise 2(f))
The inclusion of car wash facilities at the station.
The effect of industry strikes
Shortages of oil from overseas
Work through all of the above information and then do the following:
Write a final report in which you state the advantages and disadvantages of owning a petrol
[20]
station. Why would you own or not own a petrol station?
TOTAL MARKS: 65
307
CHAPTER 20: DATA HANDLING
TOPIC: PROBABILITY
In Grade 8 you learnt that Probability is all about calculating, estimating or predicting what
might happen in the future. It is about how likely something can happen or the chance of it
happening. For example, finding out the chance of it raining tomorrow based on weather
information is an example of calculating probability.
Let’s start off by revising some important probability language.
We will use two coins in our discussion. A coin will be tossed
first followed by the tossing of the other coin.
Experiment:
The experiment here involves the tossing of the
one coin followed by the tossing of the other.
Trial:
Each tossing of the one coin followed by the other is called a trial.
Outcome:
An outcome is the result of a trial. When one coin is tossed followed by the
tossing of the other coin, there are four possible outcomes:
 First coin is a head (H) and the second coin is a head (H)
 First coin is a head (H) and the second coin is a tail (T)
 First coin is a tail (T) and the second coin is a head (H)
 First coin is a tail (T) and the second coin is a tail (T)
Sample space: The set of all possible outcomes is called the sample space.
We write this as follows: S  HH, HT, TH, TT
Event:
There are a total of four possible outcomes in the sample space.
An event is a collection of one or more outcomes of an experiment. It
consists of one or more outcomes of the sample space.
Some possible events for the experiment above could be:
Event A  getting a head on the first toss  HH, HT
HH and HT are called favourable outcomes.
The number of outcomes in event A is 2.
Event B  getting three tails in any order  HT, TH, TT
The number of outcomes in event B is 3.
Note that each of the four outcomes in the sample space have an equal
chance of happening. If there are thousands of trials, each outcome has an
equal chance of happening. In fact, for every trial that takes place, each
outcome (HH, HT, TH or TT) has a one in four chance of happening. We
say that the probability of any one of the four outcomes in the sample space
happening is 14 . Each outcome has a 25% chance of happening. This can
also be written as a decimal 0,25. We call these outcomes equally likely
outcomes.
Let’s now revise the two ways of determining the probability of an event happening.
Empirical Probability
Using empirical probability, the probability of an event happening is determined by doing a
large number of trials in an experiment and then calculating the relative frequency of the
event happening.
Relative frequency: The relative frequency of an event happening is the number of times an
event occurs in relation to the number of trials done. We can represent
this as a fraction:
Relative frequency 
number of times the event happens
the total number of trials
308
Theoretical Probability
Using theoretical probability, the probability of an event happening is calculated by
determining the number of outcomes in an event divided by the total number of outcomes in
the sample space. Calculating the probability of an event happening by doing a whole number
of trials is extremely time-consuming. Using a theoretical probability approach is quicker and
far more accurate particularly if the outcomes are equally likely as in the case of the two
coins. We calculate the theoretical probability of an event happening by using the following
definition:
Probability 
number of favourable outcomes in the event
total number of equally likely outcomes in the sample space
The Probability scale
(a)
In the first event HH , there is only one outcome out of a possible four. The event of
getting two heads in the previous experiment has a probability of
(b)
1
4
 0,25.
If you calculate the probability of getting at least one tail in each trial, then the answer
is: Probability  34  0, 75 since the outcomes in this event are HT, TH, TT .
This means that the probability decimal is low in comparison to the event HH
which contains only one out of the four possible outcomes and therefore is a smaller
decimal.
(c)
If the event is two heads or two tails HH , TT , then the probability is said to be an
even chance or fifty-fifty event since: Probability  24  12  0,5  50%
(d)
If the event is the probability of getting a head or a tail in any order, then the
favourable outcomes are HH, HT, TH, TT . The probability is therefore:
Probability  44  1  100% since all of the outcomes are in the sample space. The
probability of getting the event HH, HT, TH, TT is a certain event.
(e)
The probability of getting a smartie when you toss the two coins is called an
impossible event since the experiment involves coins not smarties. The probability is
therefore: Probability  04  0  0% since there are no outcomes involving smarties.
Note: Since the number of favourable outcomes are always less than or equal to the total
number of outcomes in the sample space, all probabilities must lie between 0 and 1.
We can represent probabilities on what is called the probability scale.
All probabilities are expressed as fractions, decimal or percentages.
 If there is a high chance of an event happening, the probabilities will lie between
0,5 and 1.
 If there is a low chance of an event happening, the probabilities will lie between
0 and 0,5.
 If there is an even or 50-50 chance of an event happening, the probability will be 0,5.
 If there is no chance of an event happening, the probability will be 0.
 If the event always happens, the probability will be 1.
309
Revision examples using Theoretical Probability
Example 1
A bag contains three red pens, four blue pens and seven black pens. One pen is taken out of
the bag at random. What is the probability of choosing:
(a)
a red pen?
(b)
a blue pen?
(c)
a black pen
Solutions
(a)
(b)
(c)
There are a total of 14 pens in the bag of which there are 3 red pens.
3
The probability of choosing a red pen is therefore .
14
There are a total of 14 pens in the bag of which there are 4 blue pens.
4 2
The probability of choosing a blue pen is therefore  .
14 7
There are a total of 14 pens in the bag of which there are 7 black pens.
7 1
The probability of choosing a black pen is therefore  .
14 2
Example 2
A six-sided die is thrown. Determine the probability of rolling:
(a)
the number 5
(b)
an even number
(c)
a natural number
(d)
a natural number greater than 6
Solutions
(a)
There are six numbers (outcomes) in the sample space S  1, 2, 3, 4, 5, 6
The event is choosing the outcome 5 .
1
6
The event is choosing the outcomes 2, 4, 6 .
The probability is therefore
(b)
3 1

6 2
The event is choosing the outcomes 1, 2, 3, 4, 5, 6 .
The probability is therefore
(c)
6
1
6
The event is choosing the outcomes 7, 8, 9, ........ . A six-sided die doesn’t have
The probability is therefore
(d)
any of these numbers.
The probability is therefore
0
0
6
310
Example 3
There are 52 cards in a pack excluding the joker.
Clubs (13)
Spades (13)
Hearts (13)
Diamonds (13)
Determine the probability of drawing, at random:
(a)
a queen from the pack of cards.
(b)
a queen of hearts from the pack of cards.
(c)
a spade or a club.
(d)
a diamond excluding the king, queen and jack of diamond.
Solutions
(a)
(b)
(c)
(d)
4
1

52 13
1
There is one queen of hearts. The probability is therefore
52
There are four queens. The probability is therefore
13 13 26 1



52 52 52 2
There are 10 diamonds excluding the king, queen and jack of diamonds.
10 5
The probability is therefore
.

52 26
There are 13 spades and 13 clubs. The probability is therefore
Example 4
A die is rolled 900 times. Predict how many times you would expect to roll a number greater
than 2.
Solution
The numbers greater than 2 are 3, 4,5, 6 .
4 2
 .
6 3
Predicted number of times  (probability)  (number of trials)
2
  900
3
 600
The probability of rolling a number greater than 2 is
311
EXERCISE 1
(a)
(Revision of Grade 8)
A die was thrown 50 times. The results are recorded in the table below.
Number on dice
Number of throws
1
5
2
7
3
8
4
10
5
13
6
7
(1)
(2)
(3)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
Calculate the relative frequency that the number 5 will occur.
What is the theoretical probability that the number 5 will occur?
Explain the difference in value between the relative frequency and the
theoretical probability.
A bag contains six red balls, five green balls and eight white balls. Calculate the
probability of selecting, at random:
(1)
a red ball
(2)
a green ball (3)
a white ball (4) a blue ball
(5)
any ball
(6)
a red or white ball
(7)
a ball that is not red
Ten numbered balls are placed in a bag. One is taken out of the bag at random.
Determine the probability of selecting a ball:
(1)
numbered 3
(2)
numbered 9
(3)
with an odd number
(4)
with a number divisible by 2
Suppose that the number 7 ball is removed from the bag. Determine the probability of
then selecting a ball with a number greater than 6.
A card is drawn at random from a pack of cards. Determine the probability of
drawing:
(1)
a diamond
(2)
an ace of spades
(3)
a king
(4)
a heart or diamond (5)
the ten of clubs
(6)
the queen of clubs
(7)
a jack, queen or king (8)
the joker if it is included in the pack
A letter is chosen at random from the word ENGINEERING. Determine the
probability of choosing:
(1)
the letter R
(2)
the letter G (3)
the letter E
(4)
the letter N
(5)
a vowel
(6)
the letter I
(7)
the letter E or N
A die is rolled 300 times. Predict how many times you would expect to roll the
number 5.
A die is rolled 4000 times. Predict how many times you would expect to roll an odd
number.
The spinner shown has equal numbered segments.
(1)
What is the probability of obtaining the number 6 on
the first spin?
(2)
What is the probability of obtaining a factor of 6 on the
second spin?
(3)
How many 2’s would you expect in 1800 spins?
(4)
What is the probability of obtaining the number 2 in 1200 spins?
About one in five people in a company get really sick every month. How many sick
people would you expect to find in a company employing 100 people?
In a pack of 52 cards, all the aces, jacks, kings and queens are taken out as well as all
the number 6 cards. These cards are place on a table.
(1)
How many cards are on the table?
(2)
Suppose that a card is selected from the table and then replaced. This is done
100 times. How many times would you expect to select an ace from the table?
What is the probability in this situation?
6
(j)
(k)
312
1 2
5 4
3
COMPOUND EVENTS
A compound event consists of two or more simple events. For example, tossing a die once is
a simple event. The outcomes are {1, 2, 3, 4, 5, 6}. However, tossing the coin a second time
and stating the outcomes of tossing the coin twice is called a compound event. There are
many more outcomes in this compound event. For example, one outcome is getting the
number 3 followed by the number 6. We will discuss an example of this experiment later on
in the chapter.
The outcomes of a compound event can be represented in one of two ways:
 Two-way tables
 Tree diagrams
The examples that follow will illustrate how to determine the probability of a compound
event using two-way tables and tree diagrams.
Example 1
Suppose that a coin is tossed twice.
(a)
Draw a two-way table to represent all possible outcomes in the sample space.
(b)
Draw a tree diagram to represent all possible outcomes in the sample space.
(c)
Using the two-way table and tree diagram, determine the probability of obtaining:
(1)
a head on the first toss
(2)
a tail on the first toss
(3)
a head followed by a head
(4)
a head and a tail in any order
(5)
no heads
(6)
at least one head
Solutions
(a)
The sample space for this situation is S  HH, HT, TH, TT
We can represent these outcomes in a two-way table as follows:
Second
toss
First
toss
H
(b)
T
H
HH HT
T
TH
TT
We can represent these outcomes in a tree diagram as follows:
1
2
1
2
(c)
There are four outcomes in the
sample space. Each of the four
outcomes has an equally likely
chance of happening. In fact, there
is a one-in-four chance of any of
these outcomes happening.
(1)
1
2
1
4
1
2
1
2
1
4
1
4
1
2
1
4
On the first toss, there is
a one-in-two chance of
getting a head or tail.
On the second toss,
there is also a one-intwo chance of getting a
head or tail. The chance
of getting one of the
four outcomes in the
sample space is one-infour.
Using the two-way table, the probability of getting a head on the first toss is
2
4

1
. Using the tree diagram, the probability fraction is shown as 1 .
2
2
313
(2)
Using the two-way table, the probability of getting a tail on the first toss is
2
4
(3)

1
. Using the tree diagram, the probability fraction is shown as 1 .
2
2
The probability of getting a head followed by a head can be seen in the twoway table as HH. The probability of this happening is 1 .
4
Using the tree diagram, there are two ways of approaching this. You can look
at the HH at the end of the arrow to obtain a probability of 1 . Alternatively,
4
you can multiply the two probabilities along the HH branches of the tree
diagram.
The probability is therefore 1  1
2
(4)
2

1
.
4
The probability of getting a head followed by a tail in any order can be seen
in the two-way table as HT or TH.
The probability is therefore 1
4
 14  24 
1
2
Using the tree diagram, multiply the probabilities along the HT and TH
branches and then add the two answers.
Probability  1  1
 12  12  14  14  24 
The outcome of getting no heads is TT .
2
(5)
2
1
2
Using the two-way table, the probability is 1 .
4
Using the tree diagram, multiply the probabilities along the TT branch.
The probability is therefore 1  1
2
(6)
2

1
.
4
The probability of getting at least one head means that in any outcome there
must be one or two heads.
The outcomes here are HH, HT, HT .
Using the two-way table, there are three outcomes giving a probability of 3 .
4
Using the tree diagram, multiply the probabilities along the HH, HT and TH
branches and then add the three answers.
Probability
 12  12  12  12  12  12
 14  14  14 
3
4
Note:
The probabilities of each of the four outcomes in the sample space
add up to 1.
1
4
 14  14  14  44  1
We can therefore state the probability of getting at least one head as
follows:
Probability of getting at least one head
 1  probability of getting no heads (only tails)
 1 1
4
3
4
314
Example 2
Suppose that the coin is tossed three times.
(a)
Draw a tree diagram to represent all possible outcomes in the sample space.
(b)
Using the tree diagram, determine the probability of obtaining:
(1)
three heads
(2)
two tails followed by a head
(3)
two heads and a tail in any order
(4)
at least one tail (no heads)
Solutions
(a)
1
2
1
2
1
2
1
2
1
2
1
2
1
8
1
8
1
2
1
2
1
2
1
2
1
8
1
8
1
8
1
8
1
2
1
2
1
2
1
8
1
8
1
2
(b)
(1)
(2)
(3)
The probability of obtaining 3 heads is 1  1  1
 81
The outcome of two tails followed by a head is TTH .
The probability is 1  1  1  1
2 2 2
8
2
2
2
The outcomes of two heads and a tail in any order are:
HHT , HTH and THH . Therefore are three possible equally likely
outcomes.
The probability is 1
8
(4)
 18  18  83
The probability of getting at least one tail is all outcomes excluding HHH .
This means that there are seven outcomes containing at least one tail.
The probability is therefore 7 .
8
Note:
You can also use the fact that since all probabilities in the sample
space add up to 1:
Probability of getting at least one tail
 1  probability of no tails
 1  probability of heads
1 1  7
8
8
315
EXERCISE 2
(a)
Kaiser Chiefs will soon be playing two soccer matches against Orlando Pirates. The
outcomes per match are Win (W), Lose (L) or Draw (D).
(1)
Complete the following two-way table and tree diagram if we only consider
winning or losing as the outcomes of the two matches for each team. Write the
probability fractions on the tree diagram.
Second
match
First
match
W
(2)
(3)
(4)
(5)
L
W
L
Determine the probability of a team winning both matches.
Determine the probability of a team losing both matches.
Determine the probability of a team losing the first match but winning the
second match.
Determine the probability of a team winning at least one match.
(b)
Suppose that Draw (D) is included as an outcome. Fill in the probability fractions on
the given tree diagram.
Now determine the probability of a team:
(1)
winning both matches.
(2)
losing both matches
(3)
winning the first match and a draw
in the second match.
(4)
losing the first match.
(5)
winning the second matches.
(6)
winning or losing a match but
never a draw.
(7)
losing at least one match.
(c)
A student at university is to write three tests during the first semester.
The outcomes of each test are Pass (P) or Fail (F).
Draw a tree diagram to represents all possible outcomes.
You may use the tree diagram provided.
Now determine the probability that the student will:
(1)
pass the first test.
(2)
fail the last test.
(3)
pass all three tests.
(4)
fail the first test but pass the other
two tests.
(5)
pass only two tests
(6)
pass at least one test.
(d)
A car dealer has the following cars available for sale:
Red car with manual transmission; Red car with automatic transmission
Blue car with manual transmission; Blue car with automatic transmission
(1)
Draw a two-way table to represent all possible outcomes.
(2)
Determine the probability that a red car with manual transmission is sold.
(3)
Determine the probability that a car with automatic transmission is sold.
316
Example 3
A
A six-sided die is rolled twice. The given tree
diagram represents the following events:
A  getting a 4 and not A  not getting a 4
(a)
(b)
A
not A
Fill in the probability fractions on the tree diagram.
Determine the probability of:
(1)
getting a 4 on the first roll.
(2)
not getting a 4 on the first roll.
(3)
getting a 4 on the first roll and then not getting
a 4 on the second roll.
(4)
getting a 3 on the first roll and a 4 on the second roll.
(5)
not getting a 4 on the first roll and second roll.
(6)
getting at least one 4.
A
not A
not A
Solutions
(a)
The probability of getting a 4 is 1 . Therefore the probability of not getting a 4 is
6
1 1
6

5
. Notice that the two probabilities 1 and 5 add up to 1.
6
6
6
Outcomes
Roll 2
1 1 1
1
 
A,A
A
Roll 1
6 6 36
6
A
1
1 5 5
5
 
6
A , not A
not A
6 6 36
6
5
6
(b)
not A
1
6
A
not A , A
1 5 5
 
6 6 36
5
6
not A
not A , not A
5 5 25
 
6 6 36
(1)
The probability of getting 4 on the first roll is 1 .
(2)
The probability of not getting 4 on the first roll is 5 .
(3)
The probability of getting a 4 on the first roll and then not getting
6
6
a 4 on the second roll is 1  5
6
(4)
5
 36
The probability of not getting a 4 on the first roll and second roll is
5 5

6 6
(6)
5
 36
The probability of getting a 3 on the first roll and a 4 on the second roll is
5 1

6 6
(5)
6

25
36
The probability of getting at least one 4 is the same as calculating
1  (probability of not getting a 4 on both rolls)
1 5  5
6
6
11
 36
Alternatively, simply add up the probabilities 1
36
317
5
5
 36
 36
 11
36
EXERCISE 3
(a)
A wooden cube has each of its six faces painted with one the following numbers:
2
4
4
4
6
6
The wooden cube is rolled twice.
Fill in the probability fractions on the given
1
tree diagram.
6
Determine the probability of:
(1)
getting a 2 followed by a 4.
(2)
getting a 4 followed by a 4.
(3)
getting a 6 and a 2 in any order.
(4)
getting a least one 2 in both rolls.
(b)
A bag contains five blue pills and seven red pills.
(1)
What is the probability of taking out, at random, a blue pill?
(2)
What is the probability of taking out, at random, a red pill?
Suppose that a pill is taken out of the bag and put back into the bag.
Then a second pill is take out of the bag and then put back into the bag.
Draw a tree diagram to represent the probabilities for this situation.
Now determine the probability of:
(3)
taking out two blue pills.
(4)
taking out two red pills.
(5)
taking out a blue pill followed by a red pill.
(6)
taking out a red pill and blue pill in any order.
(7)
taking out at least one red pill.
(c)
It is winter in Cape Town. If the probability that there will be rain tomorrow if it is
raining today is 0,7 and the probability that it will rain tomorrow if it is clear today is
0,4. It is raining on Sunday. What is the probability that it will rain on Tuesday?
REVISION EXERCISE
(a)
(b)
(c)
(d)
A coin is tossed and a die is thrown.
Draw a tree diagram to show all possible outcomes.
Determine the probability of:
(1)
getting a head and the number 4.
(2)
getting a tail and the number 6.
(3)
getting the number 3.
(4)
getting even numbers.
Sally plans to give birth to two children in the future. Assume that the children born
will not be twins. Determine the probability that both children will be born on a
Saturday if births are equally likely on each of the seven days of the week. Use a tree
diagram to help you.
A box contains three i-pads, two laptops and nine mobile phones. At item is taken out
of the box, at random, and then put back into the box. A second item is then taken out
of the box and once again put back into the box. Draw a tree diagram to show all of
the outcomes.
Determine the probability that:
(1)
an i-pad is taken out of the box followed by a laptop.
(2)
both items taken out of the box are mobile phones.
(3)
at least one of the items taken out is an i-pad.
(4)
no laptops are taken out of the box.
In a marathon, there are prizes for first, second and third positions. Three athletes,
Simon, Neeran and Thabo, are in the lead at the end of the marathon and there is no
one behind them for kilometres. Use a tree diagram to help you determine the possible
outcomes of the race. What is the probability of Thabo, Simon and Neeran finishing
first, second and third, respectively?
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(e)
Two dice are rolled. Complete the following table which shows the possible outcomes
as well as the sum of the two numbers. A few outcomes are done for you.
1
2
1
2
3
4
5
6
3
4
5
6
(4 ; 3) sum  7
(6 ; 6) sum  12
Calculate the probability of obtaining:
(1)
two identical numbers on the two dice.
(3)
a sum of 10.
(5)
a sum of less than 9.
(2)
(4)
(6)
two prime numbers.
a sum of 9 or more.
at least one number 4.
SOME CHALLENGES
(a)
Ten numbered balls are put into a bag.
(1)
(b)
(c)
(d)
A ball is taken out of the bag and put onto a table. Determine the probability
of taking out a number 7 ball.
(2)
Suppose that a second ball is now taken out of the bag and placed next to the
number 7 ball. Determine the probability of taking out a number 4 ball.
(3)
Suppose that a third ball is now taken out of the bag and placed next to the
second ball. Determine the probability of taking out a number 3 or 8 ball.
One ball is selected at random from a bag containing 12 balls. Suppose that there are x
red balls in the bag.
(1)
Determine the probability of selecting a red ball?
(2)
Determine the probability of selecting a ball that is not red?
(3)
Now add a further 6 balls to the bag. The probability of now selecting a
red ball is doubled. Determine the value of x.
(1)
Using the information provided on the given diagram,
calculate the shaded area.
(2)
Now calculate the probability of throwing a dart
into the shaded area.
Two identical spinners are spun together and the numbers are added.
(1)
Complete the following two-way table to illustrate the outcomes.
A few outcomes are done for you in the table.
1
1
2
3
4
5
6
2
3
3
4
5
8
6
6
6
(2)
(3)
(e)
1 2
5 4
3
6
1 2
5 4
3
What is the probability of getting a sum of 7?
Suppose that we exclude from the sample space any outcome where the two
numbers are identical. What is the probability of obtaining a sum of 6 in this
situation?
(4)
What is the probability of obtaining a sum of 5 on three successive spins?
Use a tree diagram to help you.
There are 15 CD’s and 25 DVD’s in a container. Two items are randomly taken out of
the container. What is the probability that both items taken out of the container are
DVD’s?
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