MI ND ACTI ON SERI ES MATHEMATI CS 9 Text book cont i nuousl yupdat edt ocompl y wi t ht heNat i onalCur r i cul um and AssessmentPol i cySt at ement( NCAPS) wr i t t en/ compi l edby t opEducat or s cr eat i ve,i nt er act i ve, conci seappr oach M. D.Phi l l i ps,J.Basson& J.Odendaal 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 requested to bring the matter to the attention of the publisher so that the due 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 52 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 49 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 74 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 PS 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 67 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) 45 (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) 459 (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: 28 and 23 5 20 the division line and the root line work like 28 (2 8) (2 3) 10 5 2 and 23 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 95 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 77 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 ab d d d Subtraction a b a b d d d Multiplication a c ac b d bd Division a c a d ad b d b c bc 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 x2 3 6 2 (b) x 1 x 3 1 5 2 Solutions (a) (b) 2 1 1 x x2 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 x4 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 x3 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 42 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 75 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 mn 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 nm 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 mn 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 mn 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 mn here, we get 5 x35 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 an 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 mn Law 3: (a ) a Law 5: an a bn b Law 2: Law 4: n 61 am n a mn 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 b8 c (1) (2) (3) (4) x 7 x 12 a 8b6 c 3 (c) Simplify the following, leaving your answers with positive exponents: ( x3 )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 52 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 52 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 (22 ) 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 x1 36 (18) 23 x1 64 (21) 32( x1) 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 8x2 1 5.5 x5 5 9.3x 92 (4) 6x 1 4 1 3x 27 4x x1 343 (2) x (11) (15) 1 16 4 (19) 3 (3) 1 . 32 x 3 (6) (10) 9 27 x x1 3 9 x 1 (7) 81 (11) 13 (3 x1 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 x1 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 x1 (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, 49597870691108 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: 1,23×1014 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 1019 (C is the symbol for coulomb, the unit for charge.) Write this as a normal number. Solutions (a) 9,111028 g (b) 7,1511108 (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 10mn 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 1010 (3, 2 1010 ) (2,5 1011 ) (f) (e) 4 105 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) (105 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) (1010 105 ) 0,25 103 0,9 1010( 5) 2,5 101 103 0,9 105 2,5 104 9 101 105 (3,2 2,5) (1010 1011 ) 9 106 8 1021 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 1011 3 1013 (d) Solutions (a) (b) 3, 2 108 2 108 (d) (3, 2 1010 ) (2,5 1011 ) 2,11011 0, 03 102 1013 5, 2 108 2,11011 0, 03 1011 2,5 1014 0, 2 1014 2, 07 1011 2,3 1014 (c) (e) (e) (3, 2 1010 ) (2,5 1011 ) 3 105 2 108 3,2 1010 0, 25 101 1011 0,003 103 105 2 108 3,2 1010 0, 25 1010 0,003 108 2 108 3, 45 1010 2,003 108 66 EXERCISE 4 (a) (b) (c) (d) (e) Write the following numbers out in full: 1, 218 104 (1) (2) 3, 254 104 (3) 1, 2345 107 (4) 3,6 108 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 ) (2 108 ) (4) (5 108 )(4 109 ) (5) (7) (f) This exercise is to be done without the use of a calculator. 2 103 5 1015 4, 2 1012 3 1013 12 5,6 10 (8) (1,5 107 )(4 1020 ) 3 106 2,05 1014 3 1012 (10) 1,3 1019 2,3 1020 (6) 13 (9) 4,3 10 (11) 3,57632 1013 2,32 1014 (12) 3 108 5 106 (13) 3 109 2 1010 7,9 1020 1 1021 (14) 2,03 1017 3 1015 4 108 2,5 1014 5 1015 (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 1011 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 3x1 3 33n1 81 2 3x1 18 (3) (6) (9) (10) 2 72 n 98 0 (11) 125n 25n1 (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: 3x1 1 (2) (1) 2n 3 81 (5) (4) 1n 5 3 45 (8) (7) 0 (b) (5) x 4n 64 2k 12 16k 1 5n 0 1 3x2 9 x 1 8 2 11 1 (17) (18) 2 11x1 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 1012 m . Write this number out in full. Calculate the following and leave your answer in scientific notation: 3 107 2 108 3,1 108 2,3 106 (2) (1) (16) (c) (d) (e) (f) (3) (2,5 109 )(4 1011 ) (4) (5) 5,1 107 2,3 106 1,066 1012 (6) 2 104 8 107 2,8 1017 3 1018 2 106 3 106 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 x2 (1) 27 x1 9 x1 315 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 3n1 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 2n1 (b) Tn 3 2n1 Tn 3 2n1 (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 n2 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 x2 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 x0 7(0) 4 y4 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 x5 2 (5) 1 2 x7 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 x2 y 14 (14 1) 5 x3 y 19 (19 1) 5 x4 y 24 (24 1) 5 x5 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 x2 5 4 3 2 x4 6 12 18 24 4x 3 7 1 9 17 (4) 1 3 (5) x3 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) 6n4 2 (6) Tn n 2 (7) Tn 2n (8) Tn 2n1 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 20C to F (1) Convert 35C to F (3) Convert 104F to C The circumference of a circle is given by the formula C 2r . (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 TC 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 40C , what is the height of the aircraft? The surface area of a sphere is given by the formula S 4r 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 2r (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 2r 2 2rh 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 dx 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 x7 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 x2 (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. x34 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 abc 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 x62 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) 56r 3 8r 4 16r (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 x4 x 2 6 x 18 x 60 x 2 8x 8 x 2 40 x 6 x 2 18 x 60 8 x x2 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. x6 (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 x4 x7 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 73 The constant 3 “moved across” the equal sign and became 3. There was a sign change from 3 to 3 . x 10 (b) x5 7 If we subtract 5 from both sides we get: x 75 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 x25 (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: x3 x2 x2 2 3 (b) Solve for x: x3 x2 x2 2 3 Solutions (a) (b) The lowest common denominator (LCD) is 6. Always put brackets around the expressions in the numerators. x3 x2 x2 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) x2 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) x2 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: x4 (1) 7 (2) 8 3x 1 4 x (4) (5) 2 3 x5 (7) 11 x (8) 3 Solve the following: y2 (1) 3 4 2 4x 1 3x 2 3x (3) 4 2 2 3k 2 2k 3 (5) 1 7 2 2( x 3) x7 3x 8 (7) 5 3 2 (1) Simplify: (2) Solve: Example 5 Solve for x: 3 x7 7 5 2(1 2m) 6 3 4x 7 3(2 x 10) 2 (2) (4) (6) (8) (3) (6) (9) x4 0 4 3 (a 8) 6 4 w w 4 2 4 6 2a 1 a 50 5 2x 1 3x 2 x 3 6 2 m2 m6 1 4 3 2 1 1 1 (3 x 2) (2 x 1) 1 3 4 2 3x 8 2 x 1 x2 x 2 10 5 2 3x 8 2 x 1 x2 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 x2 (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 x6 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. x2 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 10r 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 Â , 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 Â , 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 Â , 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 Â 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 Â 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 Â 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 Â , 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 Â 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 Â 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 Â 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: Â 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 Â 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 CD 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 SR 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 Â 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 Â 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 Â 78 . Determine the unknown angles. P PQRS is a parallelogram. Calculate the value of x and y. b 78 B (k) C E Â 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 Â 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 abc 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 2r 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 2r 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 2r 4 If the radii are solid lines, they form part of the perimeter. The formula is therefore: 1 C 2r 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 242 8 cm Circumference of semi-circle 1 2r 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 22 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 2r 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 abcd 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 y0 y-intercept x0 ( 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: yx (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 x2 (2) (3) (1) y x2 y x3 y x 3 (4) (5) (6) y x4 y x4 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 x7 (11) y x7 (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 x9 (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 x2 (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 x0 (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: yx (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 m3 D m2 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: m2 For line D: m3 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. yx (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 x2 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 x2 (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) y6x (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 y2 (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 x3 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 y2 ( 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. x6 (b) x 3 (c) y5 (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 x7 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) m0 ;c0 (2) m0 ;c0 (3) m0 ;c0 (4) m0 ;c0 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 2r . 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 2r h 2r 2 2rh 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 2rh . 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 (2r ) h 2(7)(15) 246 2πr The surface area of the cylinder 2r 2 2rh 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 2r 2 2rh 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 yx 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 2r . 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 2rh . 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 (2r ) 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 2r 2 2rh (a) The volumes of both solids below are equal. (1) (2) (3) (4) (5) (b) Surface area of a sphere 4r 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). 23 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. 22 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 42 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 42 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 22 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 22 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: 0,3114,33 R4,44 0,1114,33 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? 318 (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? 319