CURRICULUM GRADE 10 -12
DIRECTORATE
NCS (CAPS)
TEACHER SUPPORT DOCUMENT
GRADE 10
MATHEMATICS
STEP AHEAD PROGRAMME
2022
This document has been compiled by the KZN FET Mathematics Subject Advisors.
1
PREFACE
This support document serves to assist Mathematics teachers on how to deal with curriculum
gaps and learning losses as a result of the impact of COVID-19 since 2020. It also captures
the challenging topics in the Grade 10 – 12 work. The lesson plans should be used in
conjunction with the 2022 Recovery Annual Teaching Plans. Activities should serve as a
guide on how to assess topics dealt with in this document. It will cover the following:
TABLE OF CONTENTS
TOPICS
PAGE NUMBERS
1.
ALGEBRA
1 – 58
2.
EUCLIDEAN GEOMETRY
59 – 85
3.
TRIGONOMETRY
86 – 111
4.
ANSWERS
112 -
2
Term
Sub-topics
1
TOPIC: ALGEBRA
Weighting: (30/100 marks from Paper 1)
LESSON 1: NUMBER SYSTEM
1 hour
10
Duration
Grade
Date
Number system.
Converting recurring decimals to the form
a
.
b
RELATED CONCEPTS/ TERMS/VOCABULARY
Real and non-real numbers
Rational and irrational numbers
Integers, counting numbers and natural numbers
Finite, recurring and non-recurring decimals
PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE
Knowledge of the Number System from earlier grades.
Knowledge of fractions and decimal fractions from earlier grades.
RESOURCES
Gr. 10 textbooks :
Siyavula; Platinum, Survival Series, Classroom Maths and Mind Action Series
The Answer Series 3 in 1 Study Guide for Gr. 10
ERRORS/MISCONCEPTIONS/PROBLEM AREAS
22
That  
, and therefore a rational number.
7
METHODOLOGY
 Explain to learners that all numbers are either Real or Non-real.
Non-real numbers are e.g. square roots of negative numbers. E.g. Consider  16 .
2
4 2  16 and  4  16 as well.
No number, when squared, equals  16 .
So:  16 is not a real number. It is Non-real!
They are Non-real, because a number  a number should have the number itself as answer.
And multiplying any number by itself can never have as answer a negative number.

As shown in the sketch below Real numbers are made up of Rational numbers ( Q ) and Irrational
numbers ( Q) .

Rational numbers are either terminating or recurring decimals, and can always be written in the form
a
, with a and b integers and b  0 .
b
3
Terminating (or finite) decimals: E.g.
3
 0,75 (the decimals stop at some point, in this case at the
4
second decimal).
1
 0,3333333....  0,3
3
5
 0,1515151515....  0,1 5
33
275
 0,275275275275275....  0,2 75
999
2
 0,13333333.......  0,13
15
(the decimals continue, but there is a recurring pattern in the decimal
expansion)
Included under Rational numbers are: Integers (Z) = .....  3;  2  1; 0 ; 1; 2 3 .....
Included under Integers are:
Counting numbers ( N 0 ) =  0 ; 1; 2 3 .....
Recurring decimals: E.g.
Included under Counting numbers are:


Natural numbers (N) = 1; 2 3 .....
Irrational numbers are non-terminating and non-recurring decimals, and cannot be written in the form
a
, with a and b integers and b  0 . Examples are surds and  .
b
E.g. 6  2,449489743.....
  3,141592654.....
(no recurring pattern in the decimal expansion)
Irrational numbers do exist (i.e. they are real), but one can only get an approximate value for them.
The decimals just go on and on.
a
How to convert an infinite recurring decimal to the form :
b

Example 1: 0,5
0,5  0,555555555555555........
Let x  0,555555555555555........
Line 1
Then:
Line 2
10x  5,555555555555555........
9x  5
Line 2 – Line 1:
5
x
9
The question could also have been: Show that 0,5 is a rational number.
Example 2: 0,1 8
0,1 8  0,181818181818.......
Let x  0,18181818181818.......
Then:
100x  18,18181818181818.......
Line 2 – Line 1: 99 x  18
18 2
x

99 11
Line 1
Line 2
4
ACTIVITIES/ASSESSMENTS
1.1.1
Classify the following numbers by placing a tick in the appropriate column(s):
N0
Non-real
Real
Irrational Rational
Integer
N
5
–2
4,1
1
2
3
0,7
9
3 8
10
 1
9
16
8
3
8
3
9
0
5
5
0

1.1.2
Write each recurring decimal in the form
(a)
(b)
(c)
0,2222222.......
0,27272727.......
0,7
a
:
b
(d)
(e)
(f)
0,2 13
0,1 5
0,54
5
TOPIC: ALGEBRA
Weighting: (30/100 marks from Paper 1)
LESSON 2: SURDS AND PRODUCTS
1 hour
10
Duration
Grade
1
Term
Date
To establish between which two integers a given simple surd lies.
Products of Algebraic expressions: Revision of Gr. 9 work.
RELATED CONCEPTS/ TERMS/VOCABULARY
Surd
Perfect square
Monomial, binomial and trinomial
Like terms
PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE
Multiplication of monomials and binomials.
Adding like terms.
RESOURCES
Gr. 10 textbooks :
Siyavula; Platinum, Survival Series, Classroom Maths and Mind Action Series
The Answer Series 3 in 1 Study Guide for Gr. 10
ERRORS/MISCONCEPTIONS/PROBLEM AREAS
Leaving out the middle term when determining the product of two binomials, e.g. to say that
a  b2  a 2  b 2
Sub-topics
METHODOLOGY
 To establish between which two integers a given simple surd lies:
o Example: the surd 40 .
o Search for the highest perfect square below 40.
Also: search for the first perfect square above 40.
To do that learners can list the perfect squares:
12  1
2 2  4 32  9 42  16 52  25
o So: 40 lies between 6 2 36 and 7 2 49 .

62  36
7 2  49
o Therefore: 62  40  7 2
And: 6  40  7 (taking square roots throughout)
Revision of Gr. 9 work on products of algebraic expressions:
First explain/revise the methods for monomial  binomial and binomial  binomial:
o a b  c   abc
[Just one term as answer]
o a b  c   ab  bc [The distributive law (can check the correctness of the answer by
substituting random values of a, b and c in the LHS and the RHS).]
o a  b c  d   a  b c  a  b d [Still the Distributive law]
 ac  bc  ad  bd
[Four terms in the answer]
o Product of two binomials can also be done using the FOIL method:
F: firsts; O: outers; I: inners; L: lasts.
O
F
I
L
This method has the advantage that the like terms, if any, will be next to each other and it will
be easy to add them.
6
E.g.: 4 x  2 y 3x  y   12 x 2  6 xy  4 xy  2 y 2  12 x 2  10 xy  2 y 2
Even when squaring binomials this same method may be used – just write the same bracket
twice, to make it easier to see what to do.
Learners are now given an exercise to consolidate gr. 9 skills before progressing to gr. 10
content.
ACTIVITIES/ASSESSMENTS
1.2.1
(a)
(b)
(c)
1.2.2
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
Between which two integers do each of the following surds lie?
7
70
 10
Expand and simplify the following:
4 x2 x  3
 x5 x  1
2 x6 x 2  x  8
x  5x  4
x  5x  4
x  5x  4
7 x  12 x  3
x  3 y 4 x  6 y 
(j)
(k)
(l)
(m)
(n)
(o)
(p)
(q)
5 x  25 x  2
4a  3b 4a  3b 
x  42
a  3b2
 2a  5b 2
2
 32 x  y 
2m  4n2
2 x  3 y 3x  2 y 
4
2
4
2
1 
1

 x   x  
2 
3

7
1
Term
TOPIC: ALGEBRA
Weighting: (30/100 marks from Paper 1)
LESSON 3: PRODUCTS CONTINUED
1 hour
10
Duration
Grade
Date
Products of Algebraic expressions: Binomial multiplied by trinomial.
Mixed exercise on products.
RELATED CONCEPTS/ TERMS/VOCABULARY
Monomial, binomial and trinomial
Like terms
PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE
Multiplication of monomials and binomials.
Adding like terms.
RESOURCES
Gr. 10 textbooks :
Siyavula; Platinum, Survival Series, Classroom Maths and Mind Action Series
The Answer Series 3 in 1 Study Guide for Gr. 10
ERRORS/MISCONCEPTIONS/PROBLEM AREAS
To leave out brackets where they are needed, e.g. to write x 2  x  3x  1  x 2  x 2  4 x  3  4 x  3 ,
instead of x 2  x  3x  1  x 2  x 2  4 x  3  4 x  3 .
Sub-topics
METHODOLOGY
 The product of a binomial and a trinomial:
a  b c  d  e   ac  d  e   bc  d  e   ac  ad  ae  bc  bd  be
Six terms in the answer, but where there are like terms, they can be combined.
 Example:
like terms indicated
x  3x 2  x  2  x 3  x 2  2 x  3x 2  3x  6
 x3  2x 2  x  6
 It is very important that learners get enough practice in determining products, and to ensure that they
are exposed to different types, also to those where there are more than one product in an expression (as
in 1.3.2 below).
ACTIVITIES/ASSESSMENTS
1.3.1
(a)
Expand and simplify:
x  1x 2  x  2
(g)
a  b3
x  y x 2  xy  y 2 
x  3 y x 2  3xy  9 y 2 
x 4  x 2  5 y x 2  5 y 
(e)
(d)
3c  22c 2  3c  1
x  2 y 3x 2  xy  2 y 2 
a  ba  b2
1.3.2
(a)
Expand and simplify the following:
5 y  12  3 y  42  3 y 
(d)
(b)
2 x  y 2  3x  2 y 2  x  4 y x  4 y 
(e)
(c)
8m  3n 4m  n   n  3m n  3m 
(f)
(b)
(c)
(f)
2
1

x 
x

2 
3

 x   x  
x 
x

8
Term
TOPIC: ALGEBRA
Weighting: (30/100 marks from Paper 1)
LESSON 4: FACTORISATION: REVISION OF GR. 9 FACTORISATION
1
1 hour
10
Duration
Date
Grade
Factorisation: Revision of Gr. 9 work: Common factor and Difference
between two squares.
RELATED CONCEPTS/ TERMS/VOCABULARY
Factorisation and determining a product as opposite/reverse processes.
PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE
Multiplication of monomials and binomials.
RESOURCES
Gr. 10 textbooks :
Siyavula; Platinum, Survival Series, Classroom Maths and Mind Action Series
The Answer Series 3 in 1 Study Guide for Gr. 10
ERRORS/MISCONCEPTIONS/PROBLEM AREAS
To attempt to factorise the sum of two squares.
METHODOLOGY
 Factorisation is the opposite/reverse process of determining a product.
4 x  3  4 x  12
E.g.:
Sub-topics
x  3x  2  x 2  5x  6


FACTORS → TERMS (multiplying)
FACTORS ← TERMS (factorisation)
Factorisation Method 1: Taking out a common factor:
E.g.: 3x 2  6 x
Determine the highest common factor between the terms, namely 3 x , and we can take this out as a
.
common factor and create a bracket: 3x 2  6 x  3x
To determine what has to be written in the bracket, divide (reverse of multiply) each term by the
common factor: 3x 2  6 x  3xx  2 .
After factorising: Take your final answer, multiply out and check to see if you get what you have
started with.
Factorisation Method 2: Difference between two squares:
When learners were calculating products, they had some examples of the type:
a  ba  b   a 2  ab  ab  b 2  a 2  b 2 .
The outers and inners (from FOIL) cancel each other and the middle term therefore falls away.
In the reverse (i.e. factorisation): The factors of a 2  b 2  a  b a  b  .
In other words, the factors of the difference between two perfect squares are
1st term  2nd term 1st term  2nd term .




Very important: Always first check if an expression contains a common factor before attempting any
other method of factorisation.
ACTIVITIES/ASSESSMENTS
1.4.1 Factorise the following:
 3x  15
(a)
(f)
9a 2  25b 2
(b) 16a 4b8  8ab7  36a 2b3
(g)
12a 2  3b 2
k 2  p2
(c)
(h)
x4 1
9
1 2 1 2
(d)
(i)
x2  2
x  y
x
4
81
2 2
2
x y x y
(e)
9
Term
Sub-topics
1
TOPIC: ALGEBRA
Weighting: (30/100 marks from Paper 1)
LESSON 5: FACTORISING TRINOMIALS
2 hours
10
Duration
Grade
Date
Factorisation: Trinomials.
RELATED CONCEPTS/ TERMS/VOCABULARY
Factorisation
Trinomials
PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE
Determining all the different pairs of factors of a specific number.
Factorisation of trinomials of the format x 2  bx  c .
RESOURCES
Gr. 10 textbooks :
Siyavula; Platinum, Survival Series, Classroom Maths and Mind Action Series
The Answer Series 3 in 1 Study Guide for Gr. 10
ERRORS/MISCONCEPTIONS/PROBLEM AREAS
Not being able to determine the relevant pairs of factors of a and c in ax 2  bx  c .
Not multiplying out after factorisation to ensure that the factors are correct.
METHODOLOGY
 Factorisation Method 3: Trinomials.
Factorisation of trinomials of the format x 2  bx  c is taught in Gr. 9, but should be revised with
learners in Gr. 10.
In Gr. 10 factorisation of trinomials is taken further to also include those of the format ax 2  bx  c ,
where a  1 .
A worksheet to guide learners in the factorisation of both these types of trinomials is included with this
lesson plan (Activity 1.5.1).
 After completion of the worksheet, teachers should go through the examples in the worksheet with
their learners, re-enforcing the concepts and skills therein and then give them a comprehensive
exercise to do on this topic. (Could start with Activity 1.5.2)
10
ACTIVITIES/ASSESSMENTS
Activity 1.5.1:
GRADE 10 MATHEMATICS
WORKSHEET – FACTORISING TRINOMIALS
PART A: Factorising 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄, with 𝒂 = 𝟏:
1. The formula 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 represents a general quadratic expression.
Given: the quadratic expression 𝑥 2 + 3𝑥 + 2.
In this case: 𝑎 = 1.
1.1
Write down the values of 𝑏 and 𝑐.
……………………………………………………………………………………
1.2
Determine the integers m en n for which 𝑚 × 𝑛 = 𝑐 and 𝑚 + 𝑛 = 𝑏.
……………………………………………………………………………………
1.3
Use the values of 𝑚 and 𝑛 (from 1.2) to form the algebraic expression  x  m  x  n  .
……………………………………………………………………………………
1.4
Determine the product of your answer to 1.3.
……………………………………………………………………………………
1.5
Compare the given quadratic expression to your answer to 1.4. What do you notice?
……………………………………………………………………………………
 The steps in question 1 describe a method of factorising an expression of the
format 𝑥 2 + 𝑏𝑥 + 𝑐.
 We determine integers m and n such that x 2  bx  c  x  m x  n  .
 x  m  and x  n  are factors of the expression x 2  bx  c .
2. Factorise the following expressions where b and c are positive.
2.1
𝑥 2 + 5𝑥 + 6 = ……………………………………………………………………
2.2
𝑥 2 + 8𝑥 + 12 = ………………………………………………………………….
2.3
𝑥 2 + 7𝑥 + 12 = ………………………………………………………………….
2.4
Comment on the signs of 𝑚 and 𝑛. …………………………………………….
3. Factorise the following expressions where b is negative and c is positive.
3.1
𝑥 2 − 5𝑥 + 6 = …………………………………………………………………..
3.2
𝑥 2 − 3𝑥 + 2 = …………………………………………………………………..
3.3
𝑥 2 − 8𝑥 + 15 = …………………………………………………………………
3.4
Comment on the signs of 𝑚 and 𝑛. …………………………………………….
3.5
Compare the signs of 𝑚 and 𝑛 with that of 𝑏. ………………………………..
11
4. Factorise the following expressions where b is positive and c is negative.
5.
4.1
x 2  x  12  …………………………………………………………………….
4.2
x 2  5 x  6  ……………………………………………………………………
4.3
x 2  2 x  15 …………………………………………………………………….
4.4
Comment on the signs of 𝑚 and 𝑛. …………………………………………….
Factorise the following expressions where b and c are negative.
5.1
x 2  x  2  ……………………………………………………………………..
5.2
x 2  x  12  …………………………………………………………………….
5.3
x 2  3 x  10  ……………………………………………………………………
5.4
Comment on the signs of 𝑚 and 𝑛. ……………………………………………..
6. Using your answers to 2.4 and 3.4: what do you observe about the signs of m and n
when c is positive? …………………………………………………………………….
7. Using your answers to 4.4 and 5.4: what do you observe about the signs of m
and n when c is negative? …………………………………………………………….
8. Using your answers to questions 4 and 5: which one of m and n has the same
sign as b? The bigger one, or the smaller one? …………………………………………
9. Complete the summary below by choosing the correct word:
While factorising the quadratic expression 𝑎𝑥 2 + 𝑏𝑥 + 𝑐, with 𝑎 = 1:
9.1
If c is positive, the signs in the two brackets are (the same / different).
9.2
If the signs in the brackets are different, then the bigger one of m and n
will have the sign of (the middle term b / the last term c).
12
PART B: Factorising 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄, with 𝒂 ≠ 𝟏:
We now need to pay attention to the possible factors of both a and c, and not only those of c, as was the
case in Part A .
TYPE 1: Factorising 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄, where both b and c are positive.
Trinomial no. 1: 𝟐𝒙𝟐 + 𝟏𝟑𝒙 + 𝟐𝟏:
Possible factors of 2𝑥 2 are 2𝑥 and 1𝑥.
Possible factors of 21 are 7 and 3; or 21 and 1.
Which combination of factors of 2𝑥 2 and 21 will lead to the correct factorisation of 2𝑥 2 + 13𝑥 + 21?
The different possible combinations are:
(2𝑥 + 7)(1𝑥 + 3); (2𝑥 + 3)(1𝑥 + 7); (2𝑥 + 21)(1𝑥 + 1) and (2𝑥 + 1)(1𝑥 + 21).
Now, multiply out each of these products:
(2𝑥 + 1)(1𝑥 + 21) = ………………………………………………………………………………….
(2𝑥 + 3)(1𝑥 + 7) = ……………….………………………………………………………………….
(2𝑥 + 21)(1𝑥 + 1) = …………………………………………………………..……………..……….
(2𝑥 + 7)(1𝑥 + 3) = ……………………….…………………………………………………..……….
All four these products give the correct first and third terms, namely 2𝑥 2 and 21, but only one gives the
correct middle term of 13𝑥, and is therefore the correct factorisation.
In this case there were only four possibilities to choose from. In many other instances there are many
more possibilities and it is not practical to write them all down and multiply out again to check. We need
a shorter method.
Trinomial no. 2: 15𝑥 2 + 19𝑥 + 6:
𝟑𝒙
𝟓𝒙
𝟓𝒙
𝟑𝒙
different possible factors of:
𝟏𝟓𝒙
𝟏𝒙
𝟏𝒙
𝟏𝟓𝒙
15𝑥 2
6
1
3
2
6
Now try different combinations, in search of the correct middle term of 19𝑥:
E.g.:
𝟑𝒙
6
i.e. factorising as (3𝑥 + 6)(5𝑥 + 1)
𝟓𝒙
1
However, this gives a middle term of 3𝑥 + 30𝑥 = 33𝑥, which is not correct.
More combinations can be tried. Another combination will be:
𝟓𝒙
3
i.e. factorising as (5𝑥 + 3)(3𝑥 + 2)
𝟑𝒙
2
This combination gives a middle term of 10𝑥 + 9𝑥 = 19𝑥, which is correct; and therefore this
factorisation is correct.
13
Trinomial no. 3: 𝟓𝒙𝟐 + 𝟐𝟖𝒙 + 𝟏𝟓:
𝟓𝒙
𝟏𝒙
different possible factors of:
𝟏𝒙
𝟓𝒙
5𝑥 2
15
1
5
3
15
Now try different combinations, in search of the correct middle term of 28𝑥:
E.g.:
𝟓𝒙
15
i.e. factorising as (5𝑥 + 15)(1𝑥 + 1)
𝟏𝒙
1
However, this gives a middle term of 5𝑥 + 15𝑥 = 20𝑥, which is not correct.
More combinations can be tried. Another combination will be:
𝟏𝒙
5
i.e. factorising as (𝑥 + 5)(5𝑥 + 3)
𝟓𝒙
3
This combination gives a middle term of 25𝑥 + 3𝑥 = 28𝑥, which is correct; and therefore this
factorisation is correct.
Trinomial no. 4: 𝟕𝒙𝟐 + 𝟐𝟕𝒙 + 𝟏𝟖:
Now factorise this trinomial in a similar way:
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
………………………………………………………………………………………………………...
14
TYPE 2: Factorising 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄, where b is negative and c is positive.
Trinomial no. 1: 𝟐𝒙𝟐 − 𝟏𝟑𝒙 + 𝟐𝟏:
𝟐𝒙
𝟏𝒙
7
𝟏𝒙
3
21
1
2𝑥 2
different possible factors of:
𝟐𝒙
𝟏𝒙
𝟐𝒙
7
3
21
From above: 6𝑥 + 7𝑥 = 13𝑥.
But to get a middle term of −13𝑥, the factors will be: (2𝑥 − 7)(𝑥 − 3).
Trinomial no. 2: 𝟓𝒙𝟐 − 𝟐𝟖𝒙 + 𝟏𝟓:
Now factorise this trinomial in a similar way:
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
………………………………………………………………………………………………………...
TYPE 3: Factorising 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄, where b is positive and c is negative.
Trinomial no. 1: 𝟐𝒙𝟐 + 𝒙 − 𝟐𝟏:
To obtain the −21 (a negative number) for the last term, the signs in the two brackets have to be different,
the one positive and the other negative.
𝟐𝒙
𝟏𝒙
different possible factors of:
𝟐𝒙
7
𝟏𝒙
3
𝟏𝒙
𝟐𝒙
2𝑥 2
21
1
7
3
21
7𝑥 − 6𝑥 = 1𝑥
Attempting to factorise: 2𝑥 2 + 𝑥 − 21 = (2𝑥 7)(1𝑥
3)
To get a middle term of +𝑥, it will have to be: (2𝑥 + 7)(1𝑥 − 3).
Trinomial no. 2: 𝟕𝒙𝟐 + 𝟏𝟔𝒙 − 𝟏𝟓:
Now factorise this trinomial in a similar way:
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
………………………………………………………………………………………………………...
15
TYPE 4: Factorising 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄, where both b and c are negative.
Trinomial no. 1: 𝟐𝒙𝟐 − 𝟏𝟏𝒙 − 𝟐𝟏:
To obtain the −21 (a negative number) for the last term, the signs in the two brackets have to be different,
the one positive and the other negative.
𝟐𝒙
𝟏𝒙
2𝑥 2
different possible factors of:
𝟏𝒙
7
𝟐𝒙
3
𝟏𝒙
𝟐𝒙
21
1
7
3
21
3𝑥 − 14𝑥 = −11𝑥
Attempting to factorise: 2𝑥 2 + 𝑥 − 21 = (𝑥 7)(2𝑥
3)
To get a middle term of −11𝑥, it will have to be: (𝑥 − 7)(2𝑥 + 3).
Trinomial no. 2: 𝟒𝒙𝟐 − 𝟗𝒙 − 𝟗:
Now factorise this trinomial in a similar way:
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
Activity 1.5.2:
Factorise the following:
p 2  8 p  15
(a)
(l)
2a 2  7a  3
(b)
p 2  8 p  15
(m)
2a 2  13a  15
(c)
p 2  2 p  15
(n)
4 y 2  12 y  5
(d)
p 2  2 p  15
(o)
6a 2  13ab  6b 2
(e)
x 2  7 x  10
(p)
12x 2  x  6
(f)
2a 2  14a  20
(q)
8 x 2  14 xy  15 y 2
(g)
b 2  3b  10
(r)
22 y 2  9 y  1
(h)
 y 2  3 y  10
(s)
3  2x  x2
(i)
a 2  5a  84
(t)
a 2  10a  25
(j)
x 2  26a  48
(u)
4 p2  4 p  1
(k)
2 x 2  5x  3
(v)
9a 2  30ab  25b 2
16
1
Term
TOPIC: ALGEBRA
Weighting: (30/100 marks from Paper 1)
LESSON 6: FACTORISATION BY GROUPING IN PAIRS
1 hour
10
Duration
Date
Grade
Sub-topics
Factorisation: Grouping in pairs.
RELATED CONCEPTS/ TERMS/VOCABULARY
Switch around
Ratio
PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE
Factorisation by means of taking out a common factor
RESOURCES
Gr. 10 textbooks :
Siyavula; Platinum, Survival Series, Classroom Maths and Mind Action Series
The Answer Series 3 in 1 Study Guide for Gr. 10
ERRORS/MISCONCEPTIONS/PROBLEM AREAS
To incorrectly leave out the + or – sign in the second line:
ax  bx  ay  by  xa  b   y a  b 
 xa  b  y a  b 
instead of
  x  y a  b 
METHODOLOGY
 Factorisation Method 4: Grouping in Pairs:
Sometimes when we have 4 terms, we need to group them before we can factorise:
ax  bx  ay  by  xa  b   y a  b 
[Take out x as common factor from the 1st two terms and
y from the last two terms.]
 a  b [ x  y ]
[Now take a  b  out as common factor.]
 Some “switch arounds”:
o From b  a  to a  b  :
a  b  b  a , therefore xa  b   y b  a   xa  b   y a  b   a  b  x  y 
o From b  a  to a  b  :
b  a  a  b  , therefore xa  b   y b  a   xa  b   y a  b   a  b  x  y 
o From  ay  by to ay  by :
 ay  by   y a  b  , therefore ax  bx  ay  by  xa  b   y a  b   a  b  x  y 
 The first step is to decide on the pairs of terms to group. A useful hint is to observe the ratio of the
coefficients. The ratio of the coefficients in the two pairs has to be the same. E.g. 5 : 3 = 10 : 6.
 Examples:
o 4 x 2  8 x  2a  ax  4 x 2  8 x  ax  2a
[ratios: 4 : 8 = 1 : 2]
 4 x x  2   a x  2 
 4 x  a  x  2 
o 2ac  6bd  3ad  4bc  2ac  3ad  4bc  6bd
[ratios: 2 : 3 = 4 : 6]
 a 2c  3d   4bc  6bd 
 a 2c  3d   2b2c  3d 
 a  2b 2c  3d 
ACTIVITIES/ASSESSMENTS
1.6.1 Factorise the following:
mx  my  nx  ny
(a)
(d)
(b)
mp  mq  p  q
(e)
(c)
x  4 y  3ax  12ay
(f)
6bx  15cx  10cy  4by
6 x 2  12ac  9ax  8cx
ac  cd  ab2  b 2 d
17
TOPIC: ALGEBRA
Weighting: (30/100 marks from Paper 1)
LESSON 7: FACTORISATION: SUM AND DIFFERENCE OF CUBES
1
2 hours
10
Duration
Date
Grade
Term
Factorisation: Sum and difference of cubes.
Factorisation: A mixed exercise to consolidate factorisation skills.
RELATED CONCEPTS/ TERMS/VOCABULARY
Cubes
Factorisation
PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE
Multiplication of a binomial by a trinomial
RESOURCES
Gr. 10 textbooks :
Siyavula; Platinum, Survival Series, Classroom Maths and Mind Action Series
The Answer Series 3 in 1 Study Guide for Gr. 10
ERRORS/MISCONCEPTIONS/PROBLEM AREAS
Because the sum of cubes can be factorised some learners attempt to also factorise the sum of squares.
Sub-topics
METHODOLOGY
 Factorisation Method 5: Sum and Difference of Cubes:
As introduction give learners the following products to multiply out:
o x  y x 2  xy  y 2 
o

o
They should obtain the following answers:
o x  y x 2  xy  y 2   x 3  y 3
o

x  y x 2  xy  y 2 
2 x  3 y 4 x 2  6 xy  27 y 2 
x  y x 2  xy  y 2   x 3  y 3
2 x  3 y 4 x 2  6 xy  27 y 2   8 x 3  27 y 3
o
Remind learners that factorisation is the reverse process of multiplication; and give them the challenge
to use the products they have just calculated and to come up with a “method” to factorise the sum and
difference of cubes. Guide them a bit if necessary to obtain the following method:
o From x 3  y 3  x  y  x 2  xy  y 2
(first)3 +(last)3 =(first + last) [(first)2 – (first)(last) + (last)2 ] where first = x and last = y
o From x 3  y 3  x  y  x 2  xy  y 2 
(first)3 – (last)3 =(first – last) [(first)2 + (first)(last) + (last)2 ] again: first = x and last = y
Learners will now be given an exercise to practice factorising the sum and difference of cubes.
Thereafter: the teacher will remind learners of the different methods of factorisation, and then give
them a mixed factorisation exercise to do, requiring them in each case to decide for themselves which
method(s) to use.




18
ACTIVITIES/ASSESSMENTS
1.7.1
Factorise the following:
(a)
x3 1
(d)
64x 3  125
(b)
1 8y3
(e)
2 x 3  16 y 6
(c)
a 3  27b3
(f)
 5a 9  5b9
1.7.2
Factorise completely:
(a)
x 2  x  12
(g)
16 p 2  56 p  49
(b)
5a 2  80
(h)
ax 2  3by 2  3bxy  axy
(c)
12 x 2  19 xy  21y 2
(i)
x2  2
(d)
4b3  8b 2  ab  2a
(j)
x2  8 
(e)
x3
1000
 3
1000
y
(k)
x6  y6
(f)
y 3  y 2  y 1
(l)
36x 2  100
1
4
16
x2
19
Term
1
TOPIC: ALGEBRA
Weighting: (30/100 marks from Paper 1)
LESSON 8: SIMPLIFICATION OF ALGEBRAIC FRACTIONS
1 hour
10
Duration
Date
Grade
Simplification of Algebraic fractions.
Multiplication and Division of Algebraic fractions.
RELATED CONCEPTS/ TERMS/VOCABULARY
Factors and terms
PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE
Ability to simplify, multiply and divide fractions.
RESOURCES
Gr. 10 textbooks :
Siyavula; Platinum, Survival Series, Classroom Maths and Mind Action Series
The Answer Series 3 in 1 Study Guide for Gr. 10
ERRORS/MISCONCEPTIONS/PROBLEM AREAS
In a fraction identical factors may be cancelled out; but not identical terms.
Sub-topics
METHODOLOGY
 Simplifying fractions:
o Discuss with the learners how (whether) the following two fractions can be simplified:
ab
ab
and
.
b
b
ab
b
 a 1  a , because  1 .

b
b
Conclusion: identical factors in the numerator and denominator may be cancelled out.
ab
 In the case of
the two b’s may not be “cancelled out”, because there are two terms in
b
the numerator and both of them have to be divided by b.
Conclusion: identical terms in the numerator and denominator may not be cancelled out.
ab
a b a
[Note:
could also be rewritten as    1 .]
b
b b b
o If there is more than one term in the numerator of an algebraic fraction, the strategy most of the
time will be to factorise the numerator and then to cancel identical factors in the numerator and
denominator.
o Examples:
4 x 2  10 x 2 x2 x  5
4 x 2  10 x 4 x 2 10 x

 2 x  5 or


 2x  5

2x
2x
2x
2x
2x
4 x 2  1 2 x  12 x  1 2 x  1



2 x2 x  1
2x
4x 2  2x
20

Multiplying and dividing algebraic fractions:
a c ac
 
o Multiplication:
.
b d bd
a c a d ad
   
o Division:
b d b c bc
o Examples:
p2  p  2 6  3 p


2p 4
2 p
 p  2 p  1  32  p 

2 p  2
2 p
 p  2 p  1   3 p  2 [See “switch arounds” in Lesson 6.]

2 p  2 
p2
 3 p  1

2
3x  9 y
3


2 x  5 4 x  10
3x  3 y  4 x  10


2x  5
3
3x  3 y  22 x  5


2x  5
3
 2 x  3 y 
ACTIVITIES/ASSESSMENTS
1.8.1
(a)
(b)
(c)
(d)
Simplify:
4x2 y 2  2 y 2
2 y2
6x2  x  1
3x 2  5 x  2
8a 3  27
6a 2  27a  27
(e)
(f)
(g)
x 2  x  6 x 2  16
 2
2x  8
x  2x
2k 2  k  6 k 4  9

 2k  3
k 2
2k 2  6
3x 2
3
x 2 y  xy 2  y 3


2 y 2 2 xy  2 y 2
x2
1  x 2 10 x

5x
x 1
21
Term
Sub-topics
TOPIC: ALGEBRA
Weighting: (30/100 marks from Paper 1)
LESSON 9: ADDING AND SUBTRACTING ALGEBRAIC FRACTIONS
1
2 hours
10
Duration
Date
Grade
Adding and Subtracting Algebraic fractions.
RELATED CONCEPTS/ TERMS/VOCABULARY
Lowest common denominator (LCD)
PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE
All the different methods of factorising algebraic expressions.
Adding and subtracting fractions.
Determining the lowest common denominator of numbers and of algebraic expressions.
RESOURCES
Gr. 10 textbooks :
Siyavula; Platinum, Survival Series, Classroom Maths and Mind Action Series
The Answer Series 3 in 1 Study Guide for Gr. 10
ERRORS/MISCONCEPTIONS/PROBLEM AREAS
Not using brackets after a minus sign.
METHODOLOGY
 Revise the process of adding and subtracting fractions in which no variables occur:
1 1
Example:  :
2 3
First step is to write them as equivalent fractions, i.e. with the same denominator. For that purpose we
need the lowest common denominator, i.e. the smallest number that both 2 and 3 can divide into, in
this case 6.
1 3
1 2
 and 
2 6
3 6
1 1 3 2 5
3 2
3 2
Therefore:     .
[  can also be written as
]
6 6
2 3 6 6 6
6
1 1 3 2 1
Similarly:    
2 3 6 6 6
22

Adding and subtracting algebraic fractions:
The same steps and reasoning can now also be applied to adding and subtracting algebraic fractions:
5
3 2
 2 
o
LCM of 4x, x 2 and 5 = 20x 2
4x x
5
 25x
60
8x 2 
This step could be left out.





2
20 x 2 20 x 2 
 20x
Shortcut way from 1st line: divide e.g. 4x into 20x 2 ;
2
25x  60  8 x
answer is 5x, multiply by the numerator: 5 x  5 , and

20 x 2
get the new numerator: 25x
o
 8 x 2  25x  60

20 x
x2
3
 2
2x  2 x 1
x2
3


2 x  1  x  1 x  1
x  2x  1  6

2x  1x  1

x 2  3x  2  6
2x  1x  1

x 2  3x  8
2x  1x  1
LCM of 2 x  1 and  x  1 x  1 is 2 x  1 x  1
For numerator: divide e.g. 2 x  1 into 2 x  1 x  1 ;
answer is  x  1 , multiply by  x  2  , and get the new
numerator:  x  1 x  1 .
ACTIVITIES/ASSESSMENTS
1.9.1 Write as a single fraction and simplify as far as possible:
3
2
x  3 x  2 x 1
 2
(a)
(e)


2
4a  9 4a  4a  3
3
2
6
3 a
a3
a
x y
(b)
(f)



2
3
4  2a  a
8a
a2
y x
1
2
1
2 3
4
 2
 2
(c) 1   2  3
(g)
2
2
a  ab b  a
a  ab
x x
x
(d)
2a 2a 2  3b 2 3b


3b
6ab
4a
(h)
x y
1
 2
3
3
x y
x  y2
23
Term
1
Sub-topics
TOPIC: ALGEBRA
Weighting: (30/100 marks from Paper 1)
LESSON 10: EXPONENTIAL LAWS
1 hour
10
Duration
Grade
Date
Revision of the exponential laws and definitions
RELATED CONCEPTS/ TERMS/VOCABULARY
Grade 8 and 9 exponents
PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE
 Definition of the exponent.
 Laws of exponents
RESOURCES
Textbooks: Mind Action Series; Answer Series and Study and Master study guide
Previous question papers.
ERRORS/MISCONCEPTIONS/PROBLEM AREAS
Incorrect use of exponential laws:
e.g. 23.32  632  65
METHODOLOGY
1. Revision of laws of exponents done in grade 9 and other lower grades.
2. Examples will be done for each law to ensure that learners understand and know how that law can
be applied:
Laws of Exponents:
 When multiplying powers with equal bases, keep the base the same and add the exponents:
Multiplying powers with equal bases: 𝑥 𝑚 × 𝑥 𝑛 = 𝑥 𝑚+𝑛 𝑥, 𝑦 > 0; 𝑚, 𝑛 ∈ 𝑍
e.g. 23 × 22 = 23+2
= 25

When dividing powers with equal bases, keep the base the same and subtract the denominator
exponent from the numerator exponent:
Dividing powers with equal bases:
𝑥 𝑚 ÷ 𝑥 𝑛 = 𝑥 𝑚−𝑛 𝑥, 𝑦 > 0; 𝑚, 𝑛 ∈ 𝑍
e.g. 23 ÷ 22 = 23−2
= 21

When raising a base with one power to another power, keep the base the same and multiply the
exponents.
Power of a power:
(𝑥 𝑚 )𝑛 = 𝑥 𝑚𝑛 𝑥, 𝑦 > 0; 𝑚, 𝑛 ∈ 𝑍
e.g. (23 )2 = 23×2
= 26

When multiplying different powers that have the same exponent:
Multiplying powers with the same exponents: 𝑥 𝑚 × 𝑦 𝑚 = (𝑥𝑦)𝑚 𝑥, 𝑦 > 0; 𝑚, 𝑛 ∈ 𝑍
e.g. 23 × 33 = (2.3)3 = 63 = 216
24
Definitions:

Negative exponents:
1
𝑥 −𝑛 = 𝑥 𝑛 , 𝑥 ≠ 0
1
1
e.g. 2−3 = 23 = 8

Power with exponent zero:
𝑥 0 = 1, 𝑥 ≠ 0
e.g. 20 = 1
Other Rules:

1𝑛 = 1 × 1 × 1 × … … . (𝑛𝑡𝑖𝑚𝑒𝑠) = 1
e.g. 1365 = 1

(−𝑎)𝑛 = 𝑎𝑛 if 𝑛 is even
e.g. (−2)4 = 24 = 16

(−𝑎)𝑛 = −𝑎𝑛 if 𝑛 is odd
e.g. (−2)3 = −23 = −8
1.10 ACTIVITIES/ASSESSMENTS
Classwork
Activity 1.10.1
Complete:
23  24 
a)
b)
22 x 3  32 x 3 
c)
53  52 
d)
23 x

22 x
e)
 4x y  
f)
 5x 
 4x 
 
g)
5 x0   2 x  
2
3 2
2

0
Homework
Activity 1.10.2
Simplify:
a)
b)
34
5.35
817.84
  .4
125  5 .3 .3
3. 312
4
c)
d)
e)
2
2
7
2
154
9 x 1
32 x.81
18 x.8 x 1
9 x 1.42 x 1
25
TOPIC: ALGEBRA
Weighting: (30/100 marks from Paper 1)
LESSON 11: EXPONENTIAL LAWS (CONTINUED)
1 hour
10
Duration
Date
Grade
1
Term
Multiplication and division
Sub-topics
RELATED CONCEPTS/ TERMS/VOCABULARY
Laws of exponents
PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE
Addition and subtraction of exponents
RESOURCES
Textbooks : Mind Action Series; Answer Series and Study and Master study guide
Previous question papers
METHODOLOGY
 Few examples will be done on the board with the learners.
Examples :
1. Multiplication bases variables
 a b  .a b
 ab  a
4 2
3


2
4 3
4 3
6
a8b 4 .a 4 b3
a 4 b12 a 2
a12 b7
a 2 b12
 a12  2 b7 12  a10 b 5

a10
b5
2. Multiplication bases numerical
182 n 1.9 2 .16n 1
1.
81n 3.64n  2
3 .2  .3  . 2 

3  . 2 
2
2 n 1
4


2
n 3
1.
2
6
4
n 1
n2
34 n  2 22 n 1.3.24 n  4
34 n 12.26 n 12
26 n 3.34 n  3
26 n 12.34 n 12
2
6 n 3 6 n 12 
4 n  3 4 n 12 
.3
 215.315

15
3
 
215  2 
315
26
3. Multiplication bases combined

1 1 
27 m 1 3  m 3 




81m 
2

1
2
1 1 
3 m 1 3  m 3 

3



4


1

4 2 2
3 m



4

1
4
3m 3 m 3
32 m 1
1 4
  1
 31 2.m 3 3
 33 m 2
1.11 ACTIVITIES/ASSESSMENTS
Classwork
Activity 1.11.1:
Simplify the following expressions:
a)
    2 xy 
3 x y 
x y 
2 xy 2
2
b)
c)
2
2
2
1 2
3
9 x 1.12 x 1
27 x.4 x 1
10 x.25 x 1.2
50 x 1.5 x
Homework
Activity 1.11.2:
Simplify the following expressions:
a)
9n.12n 1
4.6n
b)
2n  2.4n 3
8n  2
c)
4 x 1.8 x 1
16 x  2
27
TOPIC: ALGEBRA
Weighting: (30/100 marks from Paper 1)
LESSON 12: EXPONENTIAL LAWS (CONTINUED)
1 hour
10
Duration
Date
Grade
1
Term
Exponents that are fractions
Sub-topics
RELATED CONCEPTS/ TERMS/VOCABULARY
Laws of exponents
RESOURCES
Textbooks : Mind Action Series; Answer Series and Study and Master study guide
Previous question papers
METHODOLOGY
 Examples will be done on the board for learners:
3
3
1
1  1 
1.     3  
2
8  2 
1
4
1
2. 81   34  4  3
1
3 2
 4x  x

1
2
x 3
1
3 2
2 x  x
3. 
2

1
2
x 3
3 1

2x 2 2 2x
 3  3
x
x
4
 2x
28
1.12 ACTIVITIES/ASSESSMENTS
Classwork:
Activity 1.12.1:
Simplify:
1
a)
 27  3
b)
125 3
1
c) 64
d)
1
6
1
2
2
1
4 4
2
9x  . x
4
16a  . x
e)
x
6
3
1
f)  27 y 6  3 y 1
Homework
Activity 1.12.2:
Simplify:
1
2 2
a)
36 x 
3x 1
1
b)
 64 x6 3
3x 1
1
c)
125m  3
  13 
25.  m 


2
1
d)
 49 x 2  4
1
3 2
 7  .x

3
2
 12 
9
x
.
  x 
 
e)
2
 12 
 3x 


1
3 2
 13 
x
  x 
 
2
 16 
x 
 
1
3 9
f)
1
2
29
TOPIC: ALGEBRA
Weighting: (30/100 marks from Paper 1)
LESSON 13: SIMPLIFYING EXPONENTIAL EXPRESSIONS USING FACTORISATION
1
1 hour
10
Term
Duration
Date
Grade
Simplification of exponents, also using factorisation.
Sub-topics
RELATED CONCEPTS/ TERMS/VOCABULARY

Factorisation by taking out the Common Factor.

Simplification using exponential laws.
PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE
Laws of exponents
RESOURCES
Textbooks : Mind Action Series; Answer Series and Study and Master study guide
Previous question papers.
ERRORS/MISCONCEPTIONS/PROBLEM AREAS
Incorrect cancellation:
e.g.
32𝑥 −3𝑥 .2
3𝑥
( Learners just cancel 3𝑥 𝑎𝑛𝑑 3𝑥 ) ending up with 32𝑥 − 2 as the answer
METHODOLOGY

Revision of laws of exponents done in previous grades.

The following examples will be done as examples:
1.
3.7 x 1  10.7 x
11 x
.7
7
=
3.7 x.7  10.7 x
11 x
.7
7
[Using laws of exponents]
7 x  21  10 
11 x
.7
7
11
= 11  7
=
[Common factor]
[Dividing by 7𝑥 ]
7
2.
4x  4
2x  2
22x  4
= 2x
2 2
[Change 4𝑥 to prime base]
 2  2 2  2  2  2
=
x
x
x
2x  2
[Numerator is the difference of 2 squares: so factorise]
30
3.
21026 −21024
√22044
=
21024 (22 −1)
21022
= 21024−1022 (4 − 1) = 22 (3) = 12
[HCF law used]
[Application of the division law of exponents]

Learners will be doing the Classwork Activity individually, with the assistance of the teacher.

Homework given.
1.13 ACTIVITIES/ASSESSMENTS
Classwork activity:
1.13.1 Simplify
3x 1  3x  2
a)
3x  3x  2
b)
2 x  2  2 x 3
12.2 x
c)
3x 1  3x  2
8.3x 1
d)
8 x 2 x  2.16 x 1
16 x 1
Homework activity :
1.13.2 Simplify:
2n  2  22 n
a)
22 n
3n.5  3n 1
b)
3n.4
5n 1  5n 1
c)
5n.10  5n 1
31
1
Term
TOPIC: ALGEBRA
Weighting: (30/100 marks from Paper 1)
LESSON 14: EXPONENTIAL EQUATIONS
1 hour
10
Duration
Grade
Date
Exponential Equations
Sub-topics
PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE
 Laws of exponents
 Factorisation
 Solving quadratic equations
RESOURCES
Textbooks : Mind Action Series; Answer Series and Study and Master study guide
Previous question papers
METHODOLOGY
 Examples done on the board with learners actively involved:
Example 1
5x  25
5 x  52
x2
Example 2
1
x2  8
2
 12 
2
x  8
 
x  64
Example 3
1
2 x3 
4
2 x  3  2 2
x  3  2
x 1
32
Example 4
1
2
1
4
x  3x  2  0
2
1
 14 
4
 x   3x  2
 
 14
  14

 x  2   x  1  0



1
4
1
4
x  2 or x  1
4
4
 14 
 14 
4
4
 x   2 or  x   1
 
 
x  16 or x  1
 Learners will then be given activities for classwork, as well as homework.
1.14 ACTIVITIES/ASSESSMENTS
Classwork:
Activity 1.14.1:
1.14.1 Solve for 𝑥:
a)
5𝑥 = 125
b)
5𝑥 = 1
c)
5𝑥 = 0,04
d)
5𝑥 = 4𝑥
e)
2. (12)𝑥+3 = 32
f)
36𝑥 = 216
g)
5. 4𝑥−2 = 160
h)
4. 9𝑥+1 = 108
i)
𝑥 2 = 100
Homework:
Activity 1.14.2:
1.14.2 Solve for x:
a)
𝑥 2 = −100
b)
𝑥 3 = 64
c)
𝑥 3 = −64
d)
𝑥 8 = 256
1
e)
𝑥2 = 4
f)
𝑥3 = 5
g)
𝑥3 = 9
h)
1
2
1
2
1
4
x  5x  6  0
1.14.3 If 6𝑥 = 5, find the value of
18 x
.
2x
33
TOPIC: ALGEBRA
Weighting: (30/100 marks from Paper 1)
LESSON 15: LINEAR EQUATIONS
1 hour
10
Duration
Grade
1
Term
Date
1. Revise the solution of linear equations
Sub-topics
RELATED CONCEPTS/ TERMS/VOCABULARY
solving an equation/ linear equation
PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE
Like terms/ transposing/ additive and multiplicative inverses/ factorization/ removing brackets
RESOURCES
Siyavula, Mind Action Series, Pinetown Document on Lesson Plans
ERRORS/MISCONCEPTIONS/PROBLEM AREAS
Incorrect use of signs when removing brackets and after transposing to the other side
METHODOLOGY
Solving an equation means determining the value(s) of a variable in an equation that will make the left
hand side and the right side equal to each other.
Examples:
1.
2𝑠 + 3 = 11
Solution:
2𝑠 + 3 = 11
2𝑠 + 3 − 3 = 11 − 3
[applying the additive inverse]
2𝑠 = 8
1
1
2𝑠. 2 = 8 . 2
s=4
Checking: Substituting s = 4 in 2𝑠 + 4 = 11 makes both sides equal to 11.
Hence: s = 4 is the solution.
2.
𝑥+2
4
𝑥−6
1
+ 3 =2
𝑥+2
𝑥−6
1
Solution:
+ 3 =2
4
3(x + 2) + 4(x – 6) = 6
3𝑥 + 6 + 4𝑥 − 24 = 6
7𝑥 − 24 = 0
7𝑥 − 24 + 24 = 24
7𝑥 = 24
1
1
.
7𝑥
=
24.
7
7
[multiplying through by the LCD]
[removing the brackets]
[applying the additive inverse]
[applying the multiplicative inverse]
24
𝑥= 7
1.15 ACTIVITIES/ASSESSMENTS
Classwork
Activity 1.15.1
Solve the following equations:
(a)
3𝑘 − 8 = 7
(b)
(c)
(d)
4 + 2𝑡 = −2
2+𝑥
3
1
−2=3
𝑑 + 3(𝑑 + 1) = 2(𝑑 + 8)
(e)
𝑥+5
3
+𝑥 =1
(f)
3 x  2   4 x  1  24  x   7
(g)
2 x  12  2 x  32 x  3
(h)
5 2
x 8

 4
4 3x
12x
34
Homework:
Activity 1.15.2:
Solve the following equations:
(a)
(b)
(c)
5𝑚 − 3 = 2𝑚 + 12
4𝑥
5
𝑥
+ 16 = 32
𝑥+1
+ 3 = 17
2
(d)
5
1 3
4
 x     2x    4
4
5 2
3
(e)
51  3 p   210  p   10 p
(f)
1
3
2
 
m 1 m m  2
35
1
Term
Sub-topics
TOPIC: ALGEBRA
Weighting: (30/100 marks from Paper 1)
LESSON 16: QUADRATIC EQUATIONS
1 hour
10
Duration
Grade
Date
2. Solve Quadratic Equations (by factorisation)
RELATED CONCEPTS/ TERMS/VOCABULARY
Solving an equation/ writing equation in standard form/ zero –factor law
PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE
Factorisation of polynomials/ difference of two squares/
RESOURCES
Siyavula, Mind Action Series, Pinetown Document on Lesson Plans, Study & Master study guide
ERRORS/MISCONCEPTIONS/PROBLEM AREAS
Incorrect use of signs when
 removing brackets and
 after transposing to the other side
METHODOLOGY
 A quadratic equation is an equation where the highest exponent of the variable is 2.
 To solve a quadratic equation
▫ Write the equation in standard form i.e. 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0
▫ Factorise the left hand side i.e. 𝑎𝑥 2 + 𝑏𝑥 + 𝑐.
▫ Apply the zero–factor law
▫ Write down the values of the variable (solutions).

Examples :
Solve for x:
1.
𝑥 2 − 2𝑥 + 2 = 2
𝑥 2 − 2𝑥 + 2 − 2 = 0
𝑥 2 − 2𝑥 = 0
𝑥(𝑥 − 2) = 0
𝑥 = or 𝑥 − 2 = 0
𝑥 = 0 or 𝑥 = 2
2.
[writing the equation in standard form]
[factorising the left hand side]
[applying the zero – factor law]
[writing the solutions]
12𝑥 2 − 4𝑥 = 0
[already in standard form]
4𝑥(3𝑥 − 1) = 0
[factorising the left hand side]
4𝑥 = 0 or 3𝑥 − 1 = 0 [applying the zero – factor law]
1
𝑥 = 0 or 𝑥 = 3
[solutions]
NOTE:
You could also have started by first dividing both sides by the common factor and
then continued to solve the equation
e.g 12𝑥 2 − 4𝑥 = 0
3𝑥 2 − 𝑥 = 0
[dividing through by 4, the common factor]
x(3𝑥 − 1) = 0
[factorising the left hand side]
𝑥 = 0 or 3𝑥 − 1 = 0 [applying the zero – factor law]
1
𝑥 = 0 or 𝑥 = 3
BUT never divide through by a variable that you are solving for, because you will lose one of the
solutions.
E.g 12𝑥 2 − 4𝑥 = 0
12𝑥 − 4 = 0
1
𝑥 = 3 !!!!!! Only one solution; the other is LOST.
36
1.16 ACTIVITIES/ASSESSMENTS
Solve the following equations:
Classwork
Activity 1.16.1:
(a) 15𝑥 2 − 5𝑥 = 0
(b) 3𝑥 2 − 9𝑥 = – 6
(c) 𝑥 2 − 10𝑥 + 25 = 0
(d) (𝑥 − 2)2 = 0
(e) (𝑥 − 3)(𝑥 − 2) = 12
(𝑥 − 2)2 = 16
(f)
Homework
Activity 1.16.2:
(a) 𝑥 2 − 4 = 0
(b) 2𝑥 2 + 14𝑥 − 16 = 0
(c) 6(1 − 𝑥)2 = 𝑥
(d) −𝑥 2 + 4𝑥 = 4𝑥 2 − 14𝑥 + 9
37
TOPIC: ALGEBRA
Weighting: (30/100 marks from Paper 1)
LESSON 17: QUADRATIC EQUATIONS (CONTINUED)
1 hour
10
Duration
Date
Grade
1
Term
2. Solve Quadratic Equations
Sub-topics
RELATED CONCEPTS/ TERMS/VOCABULARY
solving an equation/ writing equation in standard form/ zero –factor law/ division by zero (restrictions)
PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE
Algebraic fractions/ finding LCD/ factorization of polynomials/ meaning of restrictions
RESOURCES
Siyavula, Mind Action Series, Pinetown Document on Lesson Plans, Study & Master study guide
ERRORS/MISCONCEPTIONS/PROBLEM AREAS
Incorrect use of signs when


removing brackets and
after transposing to the other side
Dividing through by a variable thus


losing one of the solutions or
making the expression not defined if the variable is equal to zero
Not checking validity of solutions
METHODOLOGY
 Restrictions arise when the variable is in the denominator, e.g.
2
, (here 𝑥 ≠ 0 ‼); or
𝑥
3
𝑥−2
, (now 𝑥 ≠ 2 !!)
You should always be on the lookout for such restrictions when working with algebraic fractions
otherwise your solutions will be incorrect.

Examples:
3𝑥
5
7
1. 𝑥+1 + 𝑥 = 𝑥 2 +𝑥
Solution :
3𝑥
𝑥+1
5
7
+ 𝑥 = 𝑥(𝑥+1)
3𝑥 2 + 5(𝑥 + 1) = 7
[multiplying through by LCD]
3𝑥 2 + 5𝑥 + 5 = 7
[removing brackets]
3𝑥 2 + 5𝑥 − 2 = 0
[writing the equation in standard form]
(3𝑥 − 1)(𝑥 + 2) = 0
[factorising the left-hand side]
(3𝑥 − 1) = 0 or (𝑥 + 2) = 0
[applying the zero-factor law]
1
𝑥 = 3 or 𝑥 = −2
[solutions]
63
4
7
And if x  2 the LHS = RHS = .
2
1
Checking solutions: if 𝑥 = , the LHS = RHS =
3
38
2. x 
x  3 2x

x 1 x 1
𝑥−3
2𝑥
Solution: 𝑥 − 𝑥−1 = 𝑥−1, restriction: 𝑥 ≠ 1
𝑥(𝑥 − 1) − (𝑥 − 3) = 2𝑥
[multipying through by LCD]
2
𝑥 − 𝑥 − 𝑥 + 3 = 2𝑥
[removing brackets]
2
𝑥 − 4𝑥 + 3 = 0
[adding like terms and writing equation in standard form]
(𝑥 − 1)(𝑥 − 3) = 0
𝑥 = 1 𝑜r 𝑥 = 3
But 𝑥 = 1 is not a solution as stated in the restriction above‼!
So 𝑥 = 3 is the ONLY solution.
1.17 ACTIVITIES/ASSESSMENTS
Classwork:
Activity 1.17.1:
Solve the following equations:
2 x 2  5 x  12  0
(a)
(b)
(c)
(d)
1
2
2𝑥 2 = 3 𝑥 2 + 3𝑥 + 14 3
2𝑥
𝑥−3
5
𝑥−2
5𝑥−3
𝑥
+ 9−𝑥 2 = 𝑥+3
4
3
− 𝑥 = 𝑥+6
Homework:
Activity 1.17.2:
Solve the following equations:
(a)
(b)
(c)
2𝑥 2 – 2𝑥 = 12
7
2
𝑥2
1
5
− 𝑥 = 𝑥2 + 2
4
3𝑥−2
3
1
− 2𝑥−3 = 2𝑥−1
39
Term
1
TOPIC: ALGEBRA
Weighting: (30/100 marks from Paper 1)
LESSON 18: LITERAL EQUATIONS
1 hour
10
Duration
Grade
Sub-topics
Date
Literal equations
RELATED CONCEPTS/ TERMS/VOCABULARY
Literal Equations
A literal equation is one that has several different letters or variables, e.g. scientific formulas.
Letters are used as the coefficients and variables in a literal equations, e.g: A   r 2 , the formula for the
area of a circle with radius r. In this example A is called the subject of the formula.
PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE
How to solve linear equations.
RESOURCES
Gr. 10 Mathematics Answer series
Gr. 10 Platinum Mathematics
Gr. 10 Classroom Mathematics
KZN 2021 Gr. 10 Step Ahead Document
ERRORS/MISCONCEPTIONS/PROBLEM AREAS
Mistakes made in changing of signs when transposing terms.
METHODOLOGY
LESSSON INTRODUCTION
Learners will be given the following introductory activity to do:
1. Solve for x: 2 x  3  7
2. Solve for y: 4  2 y  8
3. Solve for y: 3x  4 y  7
Solve for x: 2 x  a  b
Solve for y: x  2 y  t
Solve for y: ax  by  z
LESSON DEVELOPMENT
Examples:
1. Make x the subject of the formula:
qx  mx  p
qx  mx  p
x q  m   p
p
x
if q  m
qm
[take out x as common factor on the LHS]
40
2. Make r the subject of the formula:

A   R2  r 2
A


 R2  r 2
r 2  R2 
r  R2 
A

A

3. Consider the formula A  P 1  ni  .
3.1
3.2
Make i the subject of the formula.
Hence determine the value of i if A= R 13 000, P = R10 000 and n= 3.
Answer:
3.1
A  P 1  ni 
A
 1  ni
P
A
 ni    1
P
A
ni   1
P
A 1
i    1
P n
3.2
A 1
i    1
P n
 13000  1
i  
 1
 10 000  3
i  0,1
SUMMARY:
Steps for solving a literal equation:
 Remember that the aim is to isolate the subject of the formula.
 Begin by moving any terms containing the required subject of the formula to the left-hand side of
the equation.
 Move all terms that do not contain the subject of the formula to the right-hand side.
 If the required subject of the formula appears in more than one term, factorise to isolate it.
 Divide through by the coefficient of the required subject of the formula.
ACTIVITIES/ASSESSMENTS
Activity 1.18.1:
Make x the subject of the formula:
(a)
ax  b  c
(c)
(b)
3a  2b  ax
(d)
x
2
3 2 1
 
x y z
p  m2 
41
Activity 1.18.2:
(a)
Consider the formula v  u  at :
(i)
Make a the subject of the formula.
(ii)
Hence determine the value of a, if v = 75, u = 15 and t = 5.
(b)
Given the formula S  2rh  2r 2 :
(i)
Make h the subject of the formula.
(ii)
Hence determine the value of h if S = 1 221, and r= 10,5.
(iii) The total surface area of which type of solid can be calculated using this formula?
(c)
Given the formula A  P 1  i  :
(i)
Make i the subject of the formula.
(ii)
Hence determine the value of i, if P = R 12 000, n = 6 years and A = R 24 983,42,
correct to two decimal places.
n
Activity 1.18.3:
(a)
(b)
4
The volume of the a sphere is given by V   r 3 :
3
(i)
Make r the subject of the formula.
(ii)
Determine the value of r , if V = 4 188,79 cm3
(i)
(ii)
(c)
9
Given F  C  32 which is the formula that is used to convert to convert temperatures from
5
degrees Celsius to degrees Fahrenheit.
(i)
Derive the formula that will convert degrees Fahrenheit to degrees Celsius, i.e. make C the
subject of the formula.
(ii)
(d)
n
a  l  .
2
Write a as the subject of the formula.
Evaluate a if s =38,5 ; n = 11 and l = 11,5.
Given the formula: s 
The temperature in Montreal in Canada is 25  F . Convert this temperature to degrees
Celsius.
Make f the subject of the formula in:
1 1 1
  .
u v f
x 1  i   1
 , make x the subject of the formula.
Given F  
i
V
Given r 
:
h
(i)
Make h the subject of the formula.
(ii)
Solve for h , if V= 400 and r = 5 (correct to two decimal places).
n
(e)
(f)
(g)
Solve for x in
mx  n nx  m

 mn.
n
m
42
Term
TOPIC: ALGEBRA
Weighting: (30/100 marks from Paper 1)
LESSON 19: LINEAR INEQUALITIES
2 hours
10
Duration
Grade
1
Date
Linear Inequalities
Sub-topics
RELATED CONCEPTS/ TERMS/VOCABULARY
1.
Linear inequalities are like equations; except that the equal sign is replaced by an inequality sign.
Practical examples:
 The speed limit on a highway.
 Minimum payments of credit cards.
2.
Infinity    refers to the absence of a limit, the absence of an end.
METHODOLOGY
LESSON INTRODUCTION
ACTIVITY
Recap the solving of linear equations:
Solve for x:
1. 3x  4  8
2. 4  x  3  x  3  x  2 
3. 2 x  7  5  x  1
LESSON DEVELOPMENT
Linear equation:
Linear inequality:
If x  7  13 , then x  6 .
If x  7  13 , then x  6 or x  6 ;   .
This means that any value of x that is greater than 6 will make the inequality true.
Illustrated on a number line:
0
6
If x  7  13 , then x  6 or x    ; 6
This means that any value of x that is less than or equal to 6 will make the
inequality true.
Illustrated on a number line:
0
6
Take note
 Round brackets: endpoint is excluded
 Square brackets: endpoint is included
43
THE LOGIC OF SOLVING THE LINEAR INEQUALITIES
Solving linear inequalities is very similar to solving linear equations.
Consider the following:
Given the inequality 4  8 , on both sides…..
Action:
Result
Is the result a true statement, or not?
59
Add 1
True
3

7
Subtract 1
True
8  16
Multiply by 2
True
Divide by 2
True
24
 8  16
Multiply by – 2
False; it should be:  8  16
Divide by – 2
False; it should be:  2  4
 2  4
– 16
–8
8
16
0
It can clearly be seen from the number line that e.g. 8  16 (8 is to the left of 16), but  8  16 (  8 is to
the right of  16 ).
Conclusion: The inequality signs changes direction when an inequality is multiplied or divided through
by a negative number.
EXAMPLES
Solve for x, and write the solution in interval notation. Also illustrate your solution on a number line.
1. 6 x  3 x  2   3  6 x
6x – 3(x+2 ) >3+6x
6x – 3x – 6 >3+6x
– 3x > 9
x < –3 OR
[simplify]
[divide by –3]
[solution]
(– ∞ ; –3)
–3
2.
x
x
1  2 
5
10
2x – 10 ≤ 20 – x
3x ≤ 30
x ≤ 10 OR (– ∞ , 10 ]
[multiply by 10 throughout the inequality]
[divide by 3]
[solution]
0
3.
10
2  x  3  10
1  x  7
–1
4.
0
 5  1  3 x  10
 6  3x  9
2  x  3
 3  x  2 OR
–3
0
7
 3 ; 2
0
2
44
ACTIVITIES/ASSESSMENTS
Activity 1.19.1 (Day 1 )
Classwork:
Solve for x , where x  R , giving the solution in interval notation. Also illustrate your solution on a
number line.
1.
2.
3.
4.
x + 6 > 10
1 – 2x > x – 2
5( 2x – 1)  35
5x  3  x  1  2  3x
5. 2  3  2 x   2  2 x  7 
Homework:
Solve for x , where x  R , giving the solution in interval notation. Also illustrate your solution on a
number line.
6. 2  x  3  5 x  76
x3
5 8
2
3  x  3 7  x  3

3
8.
2
4
4  x 2x 1
9.

 x2
2
3
3x  1 3x  3 19
10.


4
8
8
7.
ACTIVITY 1.19.2 ( Day 2 )
Classwork:
Solve for x , where x  R , giving the solution in interval notation. Also illustrate your solution on a
number line.
2  2 x  4  2
2   x  3  10
5  2 x  3  11
12  3  3x  6
x
5. 0   1  3
3
1.
2.
3.
4.
Homework:
Solve for the unknown, giving the solution in interval notation. Also illustrate your solution on a number
line.
m
6. 1  2   4
3
3x  2
7. 4 
8
2
1 x 1 1
8.  

2
6
12
1 x 1
1
9.
  1
4 12 3
6
10. ax  x  1  a
45
1
Term
Sub-topics
TOPIC: ALGEBRA
Weighting: (30/100 marks from Paper 1)
LESSON 20: SIMULTANEOUS LINEAR EQUATIONS
2 hours
10
Duration
Date
Grade
Solving Simultaneous Linear Equations, using the Substitution method.
RELATED CONCEPTS/ TERMS/VOCABULARY
 Equation is a statement that the values of two mathematical expressions are equal.
 Root is a solution to an equation usually expressed as a number, a single term or an expression.
 Simultaneous means together or at the same time.
 Simultaneous equations are two or more equations with two or more unknowns to be solved.
PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE
 Algebraic expressions
 Linear equations
 Solving simultaneous equations using elimination method
RESOURCES
 Answer Series Grade 10
 Study & Master Study guide Grade 10
 Siyavula Grade 10
 Classroom Mathematics Grade 10
ERRORS/MISCONCEPTIONS/PROBLEM AREAS
 Use of the six operation signs when solving equations
 Transposing a term from one side of an equal sign to the other side
 Removal of brackets
 Simplification of algebraic terms
METHODOLOGY
SUBSTITUTION METHOD
 Use the simplest of the two given equations to express one of the variables in term of the other.
 Substitute the above in the remaining equation and the equation will remain with one unknown.
 Solve the new equation using the rules of solving equations.
 Substitute the solution from above in the simpler equation to determine the value of the remaining
variable.
 Check the validity of your solutions.
Example 1
Solve simultaneously using the substitution method: 𝑦 = 2𝑥 – 3 and 𝑥 + 𝑦 = 3:
Solution:
𝑦 = 2𝑥 – 3 … … . (1)
𝑥 + 𝑦 = 3 … … . … … . . (2)
Substitute (1) in (2):
Substitute x  2 in (2):
2 y 3
𝑥 + 2𝑥 – 3 = 3
3𝑥 = 6
y 1
𝑥 = 2
Checking the validity of the solution:
From (1): RHS  22   3  1  LHS
From (2): LHS  2  1  3  RHS
46
Example 2:
Solve for x and y using the substitution method: 𝑥 + 2𝑦 = 11 and 2𝑥 – 3𝑦 = −6.
Solution:
𝑥 + 2𝑦 = 11 … … . (1)
2𝑥 – 3𝑦 = −6 … … (2)
From (1): x  11 2 y ………..(3)
Substitute (3) in (2):
2(11 − 2𝑦) – 3𝑦 = −6
22 – 4𝑦 – 3𝑦 = −6
22 – 7𝑦 = − 6
7𝑦 = 28
𝑦 = 4
Substitute y = 4 in (3):
𝑥 = 11 – 2 × 4 = 3
Checking the validity of the solution:
From (1): LHS  3  24   11  RHS
From (2): LHS  23  34   6  RHS
Example 3:
𝑥
𝑦
Solve for x and y using the substitution method: 3 + 2 = 1 and 𝑥 + 2𝑦 = 1.
x y
  1 …………………..(1)
3 2
𝑥 + 2𝑦 = 1 … … … … … … … . . (2)
From (2):
x  1 2 y ……………..(3)
Substitute (3) in (1):
1 2 y y
  1 ………………(4)
3
2
21  2 y   3 y  6
(4)  6 :
2  4 y  3y  6
y  4
Subst. y  4 in (2):
x  2 4   1
x9
Checking the validity of the solution:
9 4
 3  2  1  RHS
From (1): LHS  
3 2
From (2): LHS  9  2 4   1  RHS
47
1.20 ACTIVITIES/ASSESSMENTS
Activity 1.20.1 Classwork
(a) Solve the following simultaneously using the substitution method:
(i)
𝑝 + 2𝑞 = 1 and 3𝑝 – 𝑞 = 10
(ii)
2𝑥 = 3𝑦 – 4 and 𝑦 = 3 – 𝑥
(b) Solve simultaneously using the elimination method
𝑦
𝑥
1
1
1
+
1
=
and
𝑥
+
=
𝑦
2
5
4
2
3
Activity 1.20.2 Homework
(a) Solve the following simultaneously using substitution method:
(i) 𝑦 = 2𝑥 + 1 and 𝑥 + 2𝑦 + 3 = 0
(ii) 5 = 𝑥 + 𝑦 and 𝑥 = 𝑦 – 2
(iii) 3𝑥 − 4𝑦 = −4 and − 2𝑥 – 𝑦 = 10
(b) Solve the following simultaneously using any method:
𝑥
𝑦
𝑦
𝑥
(i) 4 = 2 – 1 and 4 + 2 = 1
(ii)
(iii)
𝑥+1
4
𝑥+𝑦
2
= 2𝑦 – 8 and
= 7 −
2𝑥−𝑦
3
𝑥+𝑦
2
and
−
𝑥−𝑦
𝑥−𝑦
4
3
−
= 1
𝑥+𝑦
3
1
+ 42 = 0
48
Term
Sub-topics
1
TOPIC 1: ALGEBRA
Weighting: (30/100 marks from Paper 1)
LESSON 21: SIMULTANEOUS EQUATIONS (CONTINUED)
1 hour
10
Duration
Date
Grade
Solving Simultaneous Linear Equations, using the Elimination method.
RELATED CONCEPTS/ TERMS/VOCABULARY
 An equation is a statement that the values of two mathematical expressions are equal
 A root is a solution to an equation, usually expressed as a number, a single term or an expression
 Simultaneous means together or at the same time.
 Simultaneous equations are two or more equations with two or more unknowns to be solved.
PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE
 Algebraic expressions
 Linear equations
 Removal of brackets
RESOURCES
 Siyavula Grade 10
 Classroom Mathematics Grade 10
 Study & Master Mathematics Study Guide Grade 10
 Answer Series Grade 10
ERRORS/MISCONCEPTIONS/PROBLEM AREAS
 Use of the six operation signs when solving equations
 Transposing a term from one side of an equal sign to the other side without changing its sign.
METHODOLOGY
When solving for two unknown variables, two variables are required as well as two or more equations and
these equations are called simultaneous equations.
The solutions are the values of the unknown variables which satisfy both equations simultaneously and
they are called the roots of the equations
Simultaneous equations can be solve algebraically using elimination method or algebraic method
Simultaneous equations can be solved graphically
Simultaneous equations are used to determine the point of intersection of graphs.
Elimination method:
 Step 1: Multiply each equation by a suitable number so that the two equations have the same
leading coefficient.
 Step 2: Subtract or add the two equations in order to eliminate one of the unknowns
 Step 3: Solve for the unknown variable that is left using the rules of solving equations
 Step 4: Substitute the value of the unknown solved in step 3 in one of the simple equations given to
determine the value of the remaining variable.
 Step 5: Check the validity of your solutions.
49
Example 1:
Solve for x and y
𝑥 + 𝑦 = 10 … … (1)
x – y = 2 …… (2)
Solution using the Elimination method:
(1)+ (2):
2x = 12
Divide both sides by 2:
x=6
Substitute x = 6 in (1):
6 + y = 10
𝑦 = 10 − 6
y=4
Checking the validity of the solution:
From (1):
6 + 4 = 10
From (2):
6–4=2
Example 2:
Solve for a and b
2a +3b = 8 ……
(1)
3a + 4b = 11……
(2)
Solution using the Elimination method
(1) × 3:
6a + 9b = 24 ……. (3)
(2) × 2:
6a + 8b = 22 …….. (4)
(3) – (4):
b=2
Substitute b = 2 in (1): 2a  32   8
a 1
Checking the validity of the solution:
From (1):
2+6=8
From (2):
3 + 8 = 11
Example 3:
Solve for x and y:
𝑥
𝑦
= 3 ………….(1)
5
𝑥
4
𝑦
= 2 – 1……….(2)
Solution using the Elimination method:
(2) × 4:
x = 2y – 4 ………(3)
5𝑦
(1) × 5:
𝑥 =
(3) – (4):
0 = 2𝑦 − 3 – 4….(5)
3
… … … … (4)
5𝑦
0 = 6𝑦 – 5𝑦 – 12
y = 12
Subst. y = 12 in (3):
𝑥 = 2 × 12 − 4 = 20
Checking the validity of the solution:
20
12
 4 and
4
From (1):
5
3
20
12
 5 and
1  5
From (2):
4
2
(5) × 3:
50
1.21 ACTIVITIES/ASSESSMENTS
1.21.1 Classwork
Solve the following simultaneously, using the substitution method:
(a) 𝑥 + 7𝑦 = 49 and 𝑥 + 3𝑦 = 9
(b) 𝑥 + 𝑦 = 8 and 3𝑥 + 2𝑦 = 21
(c) 3𝑥 – 14𝑦 = 0 and 𝑥 – 4𝑦 + 1 = 0
1.21.2 Homework
(a) Solve the following simultaneously, using the substitution method:
(i)
𝑥 + 𝑦 = 1 and 𝑦 – 2𝑥 = 3
(ii)
2𝑦 = 𝑥 + 1 and 𝑥 + 𝑦 – 2 = 0
(iii) 𝑦 – 3𝑥 = 2 and 2𝑦 − 5𝑥 – 10 = 0
𝑥
𝑦
𝑦
𝑥
(iv)
= 2 – 1 and 4 + 2 = 1
2
(b) Solve the following simultaneously, using any method
𝑎
𝑎
𝑏
(i) 2 + b = 4 and 4 – 4 = 1
1
1
(ii) 𝑥 + 𝑦 = 3 and
1
1
– = 11
𝑥 𝑦
51
Term
Sub-topics
1
TOPIC 1: ALGEBRA
Weighting: (30/100 marks from Paper 1)
LESSON 22: WORD PROBLEMS
1 hour
10
Duration
Grade
Date
Word problems, leading to equations.
RELATED CONCEPTS/ TERMS/VOCABULARY
 More means add
 Less means subtract
 Double means multiply by two
PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE
 Linear equations
 Simultaneous equations
 Distance = speed × time
 Changing the subject of the formula
RESOURCES
 Siyavula Grade 10
 Classroom Mathematics Grade 10
 Study & Master Mathematics Study Guide Grade 10
 Answer Series Grade 10
ERRORS/MISCONCEPTIONS/PROBLEM AREAS
 The use of correct units when forming equations
 Substitution
 Understanding of the language used
 Answering the question at the end
METHODOLOGY
Steps for approaching word problems:
 Read very carefully and make sure that you understand the given information
 Drawing your own diagram or table can be very helpful
 Decide what you have to determine and use a variable (usually x and/or y) to represent it
Hint: if more than one unknown is present, let the smallest value be x
 Write the other unknown in terms of x
 You may also choose to name the other unknown to be y
 Use the remaining information to set-up the equation/s
 Answer the question that was asked and discard a solution that is that is nor realistic
 Check your answers
Useful information[S1]
 Area of rectangle = length × breadth
Perimeter of a rectangle = 2(length + breadth)
 Profit = selling price – cost price
 Distance = speed × time
 If I am x years now; then: five years from now, I will be (x + 5) years old.
52
EXAMPLE 1: SPEED/DISTANCE/TIME
A boy covered 31 km in 4 hours, partly by cycling at 14 km/h and partly walking at 4 km/h.
How far has he walked?
Solution:
Let the distance he walked be x km
The distance he cycled will then be (31 – x) km
Speed for cycling = 14 km/h
Speed for walking = 4 km/h
distance 31  x

hours
Time cycling 
speed
14
Time walking 
distance x
 hours
speed
4
Time cycling + Time walking = 4hrs
31  x x
 4
14
4
231  x   7 x  428 
62  2 x  7 x  112
5 x  50
x  10
He has walked 10 km.
[multiply through by LCD = 28]
EXAMPLE 2: COST
A hotel charges R200 less per day for a child than for an adult. A family of 2 adults and 3 children stay
there for 6 days. The bill at the end is R11 400. How much does the hotel charge per day for a child?
Solution:
Let the charge per adult be Rx.
The charge per child will then be R(x – 200)
12 x  18 x  200   11 400
Total cost:
12x  18x  3600  11400
30x  15 000
x  500
The charge per adult per day is R500 and the charge per child per day is R500 – R200 = R300.
1.22 ACTIVITIES/ASSESSMENTS
1.22.1 Classwork:
(a) Two cars are travelling in the same direction and the distance between them is 0,1 km. The car in
front is travelling at 120 km/h and the car at the back at 125 km/h. After how long will the car at
the back overtake the car in front? Give your answer in minutes.
(b) A shop sells bicycles and tricycles. In total there are 7 cycles and 19 wheels. How many bicycles
and how many tricycles are there?
1.22.2 Homework:
(a) A fruit shake costs R2 more than a chocolate milk shake. If 3 fruit shakes and 5 chocolate milk
shakes cost R78, determine the individual prices.
(b) A train travels 120 km at a constant speed. If it had travelled 6 km/h faster, it would have taken
one hour less to complete the journey. Calculate its actual speed.
(c) 600 tickets have been sold for the school concert. The tickets for adults cost R30 each and the
tickets for learners R15 each. The total amount of money from the ticket sales was R13 200. How
many learner tickets have been sold?
53
Term
Sub-topics
1
TOPIC 1: ALGEBRA
Weighting: (30/100 marks from Paper 1)
LESSON 23: WORD PROBLEMS (CONTINUED)
1 hour
10
Duration
Date
Grade
Word Problems
RELATED CONCEPTS/ TERMS/VOCABULARY
 More means add
 Less means subtract
 Double means multiply by two
 Constant or uniform speed
 Factorisation
PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE
 Linear equations
 Quadratic equations
 Simultaneous equations
 Distance = speed x time
 Changing the subject of the formula
RESOURCES
 Siyavula Grade 10 Mathematics
 Classroom Mathematics Grade 10
 Study & Master Mathematics Study Guide Grade 10
 Answer Series Mathematics Grade 10
ERRORS/MISCONCEPTIONS/PROBLEM AREAS
 The use of correct units when forming equations
 Substitution
 Understanding of the language used
 Factorisation
METHODOLOGY
 Continuation from the previous lesson on Word problems.
More examples will be discussed with learners, and they will be given more activities to do on this
topic.
 EXAMPLE 1:
The combined ages of two people is 72 years. In 10 years’ time the first one will be 3 times as old as the
second one was 6 years ago. How old are the two people now?
Solution:
The first person is now x years old.
The second person is now (72 –x) year old.
In 10 years’ time the first person will be( x + 10) years old.
6 years ago the second person was ( 66 –x) years old.
Ten years from now:
x + 10 = 3(66 – x)
x  10  198  3 x
4 x  188
x  47
At present the first person is 47 years old.
The second person is (72 – 47) = 25 years old.
54
EXAMPLE 2:
A two-digit number is such that if the digits are interchanged, the resulting number is one less than double
the original number. Three times the sum of the digits is 7 less than the original number. Determine the
original number.
Solution:
Let the number be (10y + x).
10 x  y  210 y  x   1
8x  19 y  1 ……………..…(1)
3 x  y   10 y  x   7
2 x  7 y  7 ………….…….(2)
(2)  4:
8x  28 y  28 ………………(3)
(1) – (3):
9 y  27
y 3
2 x  73  7
Substitute y  3 in (2):
x7
The number is 37.
EXAMPLE 3:
The breadth of a rectangle is 3 cm less than the length and the area is 10 cm². Determine the dimensions
of the rectangle.
Solution:
Let the length be x cm and the breadth will then be (x – 3) cm.
Area = length × breadth
x  x  3  10
x 2  3 x  10
x 2  3 x  10  0
x  5x  3  0
x  5 or x  3
Since length cannot have a negative value, x  5 .
The length is 5 cm and the breadth is 2 cm.
1.23 ACTIVITIES/ASSESSMENTS
1.23.1 Classwork:
(a) John is 21 years older than his son, Andile. The sum of their ages is 37. How old is Andile?
(b) The sum of two consecutive odd numbers is 20. Determine the values of the two odd numbers.
(c) The product of two consecutive negative integers is 1 122. Determine the values of the two
integers.
1.23.2 HOMEWORK:
(a) Tapelo is currently four times as old as his daughter, Lindy. Six years from now, Tapelo will be
three times as old as Lindy. Calculate Lindy’s current age.
(b) A grocer mixes 10 kg of fudge with 20 kg of nougat that costs R11,00/kg more. He sells the
mixture at R62,50/kg. If he makes a profit of R275,00, what is the price of 1kg of fudge?
(c) The length of a rectangle is twice the breadth and its area is 128 cm². Calculate the length and the
breadth of the rectangle.
(d) The tens digit of a number is double the units digit. If 27 is subtracted from the number, the digits
of the number interchange in the new number. What was the original number?
55
TOPIC 2: EUCLIDEAN GEOMETRY
Weighting: (30/100 marks from Paper 2)
LESSON 1: REVISION OF LINES AND ANGLES FROM GR. 9 EUCLIDEAN GEOMETRY
1
1 hour
10
Term
Duration
Date
Grade
Sub-topics
Revision of lines and angles
RELATED CONCEPTS/ TERMS/VOCABULARY
Parallel lines, intersect, vertex
PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE
Types of angles
RESOURCES
Grade 10 Mind Action Series
Grade 10 Siyavula Maths
ERRORS/MISCONCEPTIONS/PROBLEM AREAS
Assuming that lines are parallel
METHODOLOGY
An angle is formed where two straight lines meet at a point, also known as a vertex.
Angles are labelled with a caret on a letter, for example, Â.
Angles can also be labelled according to the line segments that make up the angle, for example
𝐶Â𝐵 𝑜𝑟 𝐵Â𝐶.
The "∠" symbol is a short method of writing angle in geometry and is often used in phrases such as
“sum of ∠𝑠 in ∆".
Angles are measured in degrees which is denoted by °, a small circle raised above the text, similar to an
exponent.
Types of Angles
1. Adjacent angles on a straight line are supplementary. Supplementary angles add up to 180°
ˆ B
ˆ  180 .
B̂1 and B̂2 share a vertex and a common side. Hence B
1
2
2. If two lines intersect, vertically opposite angles are equal.Two lines intersect if they cross each other at
a point.
Vertically opposite angles are angles opposite each other when two lines intersect.They share a vertex
and are equal.
Ê1  Ê 3 and Ê 2  Ê 4
56
3. The angles around a point add up to 360° (A revolution).
Bˆ1  Bˆ 2  Bˆ3  Bˆ 4  3600
PARALLEL LINES
Parallel lines are always the same distance apart (equidistant) and they are denoted by arrow
symbols as shown below.
𝐶𝐷 ∥ 𝐴𝐵. EF is a transversal line. A transversal line intersects two or more parallel lines. Below are the
properties of the angles formed by the above intersecting lines.
1. Corresponding Angles
Corresponding angles lie either both above or both below the lines and on the same sideof the
transversal. If the lines are parallel, the corresponding angles will be equal.
2. Alternate Angles
Alternate angles lie on opposite sides of the transversal and between the lines. If the lines are parallel, the
alternate angles will be equal.
3. Co-interior Angles
Co-interior angles lie on the same side of the transversal between the lines.If the lines are parallel, the
co-interior angles are supplementary
If two lines are intersected by a transversal such that corresponding angles are equal; or alternateangles
are equal; or co-interior angles are supplementary, then the two lines are parallel.
57
2.1 ACTIVITIES
Exercise 1: Mind Action Series (Pg. 165)
Activity 2.1.1:
1. In ∆ABC, EF∥BC. BA is produced to D.
a) Calculate, with reasons, the value of 𝑎 and
hence show that AE = AF.
b) Calculate, with reasons, the value of 𝑏, 𝑐, 𝑑,
𝑎𝑛𝑑 𝑒.
2. In the diagram below, CD∥EF, DÊF = 28°,
B̂ = 48° and BD̂E  160  . Prove that
AB∥CD.
3. In the diagram below, PU∥ 𝑄𝑇, T̂  42 ,
RQ̂S  82 , PQ̂T  y , UPT = x and
QPT = x + 40°.
a) Prove that PT∥QS
b) Calculate y
58
TOPIC: EUCLIDEAN GEOMETRY
Weighting: (30/100 marks from Paper 2)
LESSON 2: REVISION OF TRIANGLES FROM GR. 9 EUCLIDEAN GEOMETRY
1
1 hour
10
Term
Duration
Date
Grade
Sub-topics
Revise Triangles: Classification, Congruency and
Similarity
RELATED CONCEPTS/ TERMS/VOCABULARY
PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE
Types of angles
RESOURCES
Grade 10 Mind Action Series
Grade 10 Siyavula Maths
ERRORS/MISCONCEPTIONS/PROBLEM AREAS
Using congruency conditions without understanding
METHODOLOGY
PROPERTIES OF TRIANGLES
A triangle is a three-sided polygon. Triangles can be classified according to sides and also be classified
according to angles.
1. TYPES OF TRIANGLES according to sides
a) Scalene Triangle:
b) Isosceles Triangle:
 Two sides are equal.
 Angles opposite equal sides are equal.
c) Equilateral Triangle
 All three sides are equal.
 All three interior angles are equal
59
2. TYPES OF TRIANGLES according to angles
a) Acute-angled Triangle
All 3 interior angles are less than 90°
b) Obtuse-angled Triangle
One interior angle is greater than 90°
The other two are acute
c) Right-angled Triangle
One interior angle is equal to 90°
The other two are acute
From Pythagoras
Theorem
2
AB  AC2 +BC2
AC2 =AB2  BC2
In triangle ABC, BC is produced to D.
  B̂  Ĉ 2  180
(Sum of ∠𝑠 of a ∆)
Ĉ1  Â  B̂
(ext. ∠ of ∆)
3. CONGRUENT TRIANGLES – 4 Conditions
a) If three sides of a triangle are equal in length to the
corresponding sides of another triangle, then the two
triangles are congruent.
Side, Side, Side (S, S, S)
b) If two sides and the included angle of a triangle are
equal to the corresponding two sides and
included angle of another triangle, then the two
triangles are congruent.
Side, Angle, Side (S, A, S)
c) If one side and two angles of a triangle are
equal to the corresponding one side and two
angles of another triangle, then the two
triangles are congruent.
Angle, Angle, Side (A, A, S)
d) If the hypotenuse and one side of a rightangled triangle are equal to the hypotenuse and
the corresponding side of another right-angled
triangle, then the two triangles are congruent.
90°, Hypotenuse, Side (R, H, S).
We use ≡ to indicate that triangles are congruent.
NOTE: The order of letters when labelling congruent triangles is very important.
60
4. SIMILAR TRIANGLES
Two triangles are similar if one triangle is a scaled version of the other. This means that their
corresponding angles are equal in measure and the ratio of their corresponding sides are in
proportion. The two triangles have the same shape, but different scales.
Congruent triangles are similar triangles, but not all similar triangles are congruent.
We use /// to indicate that two triangles are similar.
a) If all three pairs of corresponding angles of two triangles are equal, then the triangles are similar.
Angle, Angle, Angle (A, A, A)
b) If all three pairs of corresponding sides of two triangles are in proportion, then the triangles are
similar.
NOTE: The order of letters for similar triangles is very important. Always label similar triangles in
corresponding order.
5. THEOREM OF PYTHAGORAS
AB 2  AC 2  BC 2 if Ĉ  90 
2.2 ACTIVITIES:
Activity 2.2.1:
Exercise 1: Mind Action Series (Pg. 166)
1. AB∥DE and DC = CB
a) Prove that AC = CE and AB = DE
2. Prove that C1  90 using congruency.
3. Show that the following triangles are similar.
61
4. If ∆𝐴𝐵𝐶///∆𝐷𝐸𝐶, calculate x and y.
Activity 2.2.2:
TEST 1: LINES, ANGLES AND TRIANGLES
MARKS: 25
DURATION: 30 Min
QUESTION 1
In the diagram below, AB and DC are two parallel lines cut by two transversal lines at
X, Y and Z respectively.
A
X
𝑥
Z
60
D
B
Y
C
𝑥 − 20°
1.1.1 Determine giving reasons, the value of 𝑥 in the diagram:
(6)
1.1.2 Name one pair of co-interior angles
(1)
1.1.3 Name one pair of alternate angles
(1)
1.1.4 Complete: If two parallel lines are cut by a transversal, then the co-interior angles are
……………..
(1)
1.1.5 Complete: The size of angle XYD = ……………….
(2)
Reason .....................
62
QUESTION 2
In the diagram below, AD = CD and PQ∥RS. AR and FC are straight lines. RS and FC
intersect at E. PQ also intersects FC at B.
2.1
2.2
Determine the sizes of the following angles, giving appropriate reasons:
2.1.1
D̂1
(2)
2.1.2
B̂1
(2)
2.1.3
 2
(2)
ˆ ˆ
Show that REF=B
3
(3)
[9]
63
Term
Sub-topics
1
TOPIC 2: EUCLIDEAN GEOMETRY
Weighting: (30/100 marks from Paper 2)
LESSON 3: PROPERTIES OF QUADRILATERALS (1)
1 hour
10
Duration
Date
Grade
Properties of Quadrilaterals (sides, angles and diagonals)
RELATED CONCEPTS/ TERMS/VOCABULARY
Polygon, Straight line, diagonals, bisect, intersect, adjacent
PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE
Parallel lines, interior angles
RESOURCES
Grade 10 Mind Action Series
Grade 10 Siyavula Maths
ERRORS/MISCONCEPTIONS/PROBLEM AREAS
Treating parallelogram as a rectangle or as a rhombus.
METHODOLOGY
A quadrilateral is a closed shape (polygon) consisting of four straight line segments.A polygon is a
two-dimensional figure with three or more straight sides. The interior angles of a quadrilateral add up
to 360°.
1. PARALLELOGRAM
 A parallelogram is a quadrilateral with both pairs of opposite sides parallel.
2. RECTANGLE
 A rectangle is a parallelogram that has all four angles equal to 90°.
3. RHOMBUS
 A rhombus is a parallelogram with all four sides of equal length.
4. SQUARE
 A square is a rhombus with all four interior angles equal to 90°.
A square has all the properties of a rhombus.
OR
 A square is a rectangle with all four sides equal in length.
5. TRAPEZIUM
 A trapezium is a quadrilateral with at least one pair of opposite sides parallel.
NOTE: A trapezium is sometimes called a trapezoid.
6. KITE
 A kite is a quadrilateral with two pairs of adjacent sides equal.
64
PROPERTIES OF QUADRILATERALS
QUATRILATERAL
PARALLELOGRAM
RECTANGLE
RHOMBUS
RECTANGLE
TRAPEZIUM
PROPERTIES
• Both pairs of opposite sides are parallel.
• Both pairs of opposite sides are equal in length.
• Both pairs of opposite angles are equal.
• Both diagonals bisect each other.
• Both pairs of opposite sides are parallel.
• Both pairs of opposite sides are of equal length.
• Both pairs of opposite angles are equal.
• Both diagonals bisect each other.
• Diagonals are equal in length.
• All interior angles are equal to 90°
• Both pairs of opposite sides are parallel.
• Both pairs of opposite sides are equal in length.
• Both pairs of opposite angles are equal.
• Both diagonals bisect each other.
• All sides are equal in length.
• The diagonals bisect each other at 90°
• The diagonals bisect both pairs of opposite
angles.
• Both pairs of opposite sides are parallel.
• Both pairs of opposite sides are equal in length.
• Both pairs of opposite angles are equal.
• Both diagonals bisect each other.
• All sides are equal in length.
• The diagonals bisect each other at 90°
• The diagonals bisect both pairs of opposite
angles.
• All interior angles equal 90°.
• Diagonals are equal in length.
• One pair of opposite side are parallel.
• The diagonals of a trapezium intersect but don’t
bisect each other.
• Diagonals lie between parallel lines and
therefore, the alternate angles are equal.
65
KITE
• Diagonal between equal sides bisects the other
diagonal.
• One pair of opposite angles are equal (the
angles between unequal sides).
• Diagonal between equal sides bisects the
interior angles and is an axis of symmetry.
• Diagonals intersect at 90°
ACTIVITIES
Activity 2.3.1:
1. The following are properties of some quadrilaterals.
a) Having a pair of parallel sides.
b) Having two pairs of parallel sides.
c) Having four right angles.
d) Having four equal sides.
e) Having equal diagonals.
In the table below, mark a “” in the box if the quadrilateral has the property referred to in a) to e)
above:
PROPERTIES
QUADRILATERALS
a)
b)
c)
d)
e)
Kite
Trapezium
Parallelogram
Rectangle
Square
Rhombus
2. Referring to the figure below, use the names of the quadrilaterals to complete the sentences.
66
Term
1
TOPIC 2: EUCLIDEAN GEOMETRY
Weighting: (30/100 marks from Paper 2)
LESSON 4: PROPERTIES OF QUADRILATERALS (2)
1 hour
10
Duration
Date
Grade
Sub-topics
Properties of Quadrilaterals (sides, angles and diagonals)
RELATED CONCEPTS/ TERMS/VOCABULARY
Parallel, interior angles, bisect
PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE
Properties of quadrilaterals, naming quadrilaterals, Naming triangles
RESOURCES
Grade 10 Mind Action Series
Grade 10 Siyavula Maths
ERRORS/MISCONCEPTIONS/PROBLEM AREAS
Difference between parallelogram and rhombus
METHODOLOGY
EXAMPLES:
1. DELM is a parallelogram.
a) Calculate the value of 𝑥 and hence the sizes of the
interior angles.
b) If DE = 2DM and ML = 10 cm, determine the
length of the other sides of DELM.
SOLUTION:
2 x  x  180
a)
3x  180
x  60
Ê  M̂  60 
D̂  120 
L̂  120 
b) DE = 10 cm
DM = 5 cm
EL = 5 cm
[co-int ∠𝑠; DM∥EL]
[opp ∠𝑠 of parm equal]
[opp ∠𝑠 of parm equal]
[opp sides of parm]
[DE = 2DM]
[opp sides of a parm]
2. ABCD is a parallelogram. BH bisects AB̂C and HC
bisects BĈD . AB̂C  60  . F̂  120  . BH∥GC and
BG∥HC.
AD is produced to E such that AB = DE = 30 cm.
BC is produced to F.
Prove that
a) BGCH is a rectangle.
b) DCFE is a rhombus.
67
SOLUTION:
a) BCGH is a parallelogram
AB̂C  60 
B̂1  B̂2  30
BĈD  120 
Ĉ1  Ĉ 2  60
Ĥ 2  90
 BGCH is a rectangle
b)
[both pairs opp. sides are parallel]
[given]
[BH bisects AB̂C ]
[co-int ∠s ; AB∥DC]
[HC bisects BĈD ]
[sum of ∠𝑠 𝑜𝑓 ∆]
[BGCH is a parm with an interior ∠ = 90°]
F =120°
Ĉ1  Ĉ 2  120
 F̂  Ĉ1  Ĉ 2
 DC∥EF
AD∥BC
ADE and BCE are straight lines
 DE∥CF
DCFE is a parallelogram
DC = AB = 30 cm
 DC = DE = 30 cm
 DCEF is a rhombus
ACTIVITIES
Activity 2.4.1:
1. PQRS is a rhombus with S2  35 .
[corresponding angels are equal]
[opp. sides of parallelogram ABCD]
[given]
[both pairs opp sides are ∥
[opp sides of a parm are equal]
[DCEF is a parm with adjacent sides equal]
Calculate the sizes of all the interior angles
2.
ABCD is a square. AÊB  55 
Calculate F1
3.
In rectangle ABCD, AB = 3x and BC = 4x
Determine the length of AC and BD in terms of
x.
4.
ABCD is a trapezium with AD||BC. AB = AD
and BD = BC. Ĉ  80 
Determine the unknown angles.
68
5.
ABCD is a kite. The diagonals intersect at E.
BD = 30 cm, AD =7 cm and DC = 25 cm.
Determine:
a) AE
b) AC
c) B̂1 if Â1  20 .
Term
Sub-topics
1
TOPIC 2: EUCLIDEAN GEOMETRY
Weighting: (30/100 marks from Paper 2)
LESSON 5: PROPERTIES OF QUADRILATERALS (3)
1 hour
10
Duration
Date
Grade
Properties of Quadrilaterals (sides, angles and diagonals)
RELATED CONCEPTS/ TERMS/VOCABULARY
Triangle, quadrilaterals, parallel
PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE
Types of quadrilaterals and Properties of quadrilaterals
RESOURCES
Grade 10 Mind Action Series
Grade 10 Siyavula Maths
ERRORS/MISCONCEPTIONS/PROBLEM AREAS
Properties of a kite, stating reasons
METHODOLOGY
Example:
In trapezium ABCD, AD||BC with   D̂  70  and
EC = DC.
Prove that ABCE is a parallelogram.
Solution:
Ê 2  70
AB || EC
ABCE is a parallelogram
[∠s opp = sides]
[corresponding s are equal]
[both pairs opposite sides are equal]
69
ACTIVITIES
Activity 2.5.1:
1.
Determine the sizes of the interior angles of
parallelogram ABCD.
2.
In ∆ABC, Â  80  and Ĉ  35 
Calculate the interior angles of parallelogram MENB.
3.
In parallelogram ABCD, AB = BE = DE.
D̂1  x and Â1  28 .
Calculate x.
4.
AD = BD = BC.
Ĉ  75  and AD̂B  30  .
Prove that ABCD is a parallelogram.
70
Term
TOPIC 2: EUCLIDEAN GEOMETRY
Weighting: (30/100 marks from Paper 2)
LESSON 6: PROPERTIES OF QUADRILATERALS (4)
1 hour
10
Duration
Date
Grade
1
Sub-topics
The opposite Sides and Angles of a parallelogram are equal
RELATED CONCEPTS/ TERMS/VOCABULARY
Properties of parallelogram, congruent triangles
PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE
Diagonals, Parallel
RESOURCES
Grade 10 Mind Action Series
Grade 10 Siyavula Maths
ERRORS/MISCONCEPTIONS/PROBLEM AREAS
Naming angles as sides
METHODOLOGY
The proofs of the following theorems are examinable in Gr. 10 Euclidean Geometry:
1. The opposite sides and angles of a parallelogram are equal.
2. The diagonals of a parallelogram bisect each other.
3. The diagonals of a rectangle are equal.
4. The diagonals of a rhombus bisect each other at right angles and bisect the angles of the rhombus.
As an example the proof of the first of these theorems is given below:
Prove that the opposite sides and angles of a
parallelogram are equal.
PROOF:
Given:
Parallelogram ABCD.
R.T.P.:
AB = CD and AD = BC
2
1
BÂD  BĈD and B̂  D̂
Construction:
Draw diagonal AC.
Proof:
In ABC and CDA :
1.
 2  Ĉ1
[alt. ' s; AB || CD]
2.
Â1  Ĉ 2
[alt. ' s; AD || BC]
2 1
3. AC  AC
[common]
 ABC  CDA [ ;; s ]
 AB  CD and BC  AD and B̂  D̂ [  ' s ]
Also: Â1  Â 2  Ĉ1  Ĉ 2 [ Â 2  Ĉ1 ; Â1  Ĉ 2 ]
 BÂD  BĈD
71
ACTIVITIES
Activity 2.6.1
1.
KLMN is a parallelogram
Calculate the size of the interior angles
2.
In parallelogram ABCD, AB = 50 cm and E is a point
on AD such that AB = AE and CD = DE.
Determine:
a) DE
b) the perimeter of ABCD.
3.
ABCD is a parallelogram. AM bisects  . AB = AM.
Ĉ  120  .
Calculate the size of the interior angles.
72
Term
Sub-topics
1
TOPIC 2: EUCLIDEAN GEOMETRY
Weighting: (30/100 marks from Paper 2)
LESSON 7: PROPERTIES OF QUADRILATERALS (5)
1 hour
10
Duration
Date
Grade
The diagonals of a parallelogram bisect each other
RELATED CONCEPTS/ TERMS/VOCABULARY
Diagonals
PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE
Properties of a parallelogram, congruent triangles
RESOURCES
Grade 10 Mind Action Series
Grade 10 Siyavula Maths
ERRORS/MISCONCEPTIONS/PROBLEM AREAS
Choosing triangles to prove congruency.
METHODOLOGY
The proofs of the following theorems are examinable in Gr. 10 Euclidean Geometry:
1. The opposite sides and angles of a parallelogram are equal.
2. The diagonals of a parallelogram bisect each other.
3. The diagonals of a rectangle are equal.
4. The diagonals of a rhombus bisect each other at right angles and bisect the angles of the rhombus.
As a further example the proof of the second of these theorems is given below:
Prove that the diagonals of a parallelogram bisect
each other:
Given:
Parallelogram ABCD with diagonals
AC and BC intersecting in E.
R.T.P.:
AE = EC and BE = ED
Proof:
In ABE and CDE :
1. Â 2  Ĉ1 [alt.  ’s; AB || CD]
2. B̂1  D̂ 2 [alt.  ’s; AB || CD]
3. AB = CD [opp. sides of parm]
[AAS]
ABE  CDE
 AE = EC and BE = ED [  s ]
73
EXAMPLE:
Diagonals AC and BD of parallelogram ABCD
intersect at M. AP = QC and AC =600 mm,AB = 500
mm and AP =150 mm.
Prove that PBQD is a parallelogram
SOLUTION:
AM = MC
[diagonals of a parm]
But AC = 600 mm
[given]
AM = MC = 300 mm
AP =QC =150 mm
[given]
PM = MQ =150 mm
Also, BM = MD
[diagonals of parm]
∴ PM = MQ and BM =MD
∴ PBQD is a parallelogram [diagonal of quad bisect]
ACTIVITIES
Activity 2.7.1:
1.
In the diagram, BCDF, EDCF and ABCF are
parallelograms. BC = 4 units and CD = 6 units.
Prove that ABDE is a parallelogram.
2.
Parallelograms ABCD and ABDE are given
with DF = DB.
Prove that BCFE is a parallelogram.
3.
ABCD is a parallelogram with AE = FC.
Prove that BEDF is a parallelogram.
74
1
Term
TOPIC 2: EUCLIDEAN GEOMETRY
Weighting: (30/100 marks from Paper 2)
LESSON 8: PROPERTIES OF QUADRILATERALS (6)
1 hour
10
Duration
Date
Grade
If one pair of opposite sides of a quadrilateral are equal and parallel, then
the quadrilateral is a parallelogram.
RELATED CONCEPTS/ TERMS/VOCABULARY
Diagonal, parallel
Sub-topics
PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE
Congruent triangles
RESOURCES
Grade 10 Mind Action Series
Grade 10 Siyavula Maths
ERRORS/MISCONCEPTIONS/PROBLEM AREAS
Prove congruency and reasons for congruency
METHODOLOGY
TAKE NOTE: Learners need to be able to apply the theorem below, but the proof of the theorem is not
examinable. It is given here as enrichment.
Theorem: If one pair of opposite sides of a quadrilateral are equal and parallel, then the quadrilateral is a
parallelogram
Required to prove: ABCD is a parallelogram:
Construction: Draw diagonal AC.
Proof:
In ABC and CDA :
(a)
Ĉ 2  Â1
[alt. s ; AD || BC ]
(b) AC = AC
[common]
(c) BC = AD
[given]
 ABC  CDA
[SAS]
 2  Ĉ1
[  s ]
 AB || CD
[alt. s are equal]
 ABCD is a parallelogram
[both pairs opp. sides are ||]
75
ACTIVITIES
Activity 2.8.1:
1. In parallelogram ABCD, AB =AD and Ĉ  100  .
Calculate the sizes of all the interior angles.
2. ∆ABD and ∆BCD are two isosceles triangles.
Ĉ  75  and AD̂B  30  .
Prove that ABCD is a parallelogram.
3.
In quadrilateral LMNP, Ê1  62 , P̂1  68 ,
FP = FN and LE = LM.
Prove that:
a) LP||MN
b) LMNP is a parallelogram
4. In quadrilateral ABCD, AB = 5 cm, BC = 10 cm,
FD = 3 cm, BE = FD and AE = FC. AE ⊥ BC and
CF ⊥ AD.
Prove that ABCD is a parallelogram.
76
Term
Sub-topics
TOPIC 2: EUCLIDEAN GEOMETRY
Weighting: (30/100 marks from Paper 2)
LESSON 9: MIDPOINT THEOREM
2 hours
Duration
Grade
Midpoint Theorem
1
10
Date
RELATED CONCEPTS/ TERMS/VOCABULARY
Midpoint
PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE
Parallel lines, types of angles, properties of parallelogram
RESOURCES
Grade 10 Mind Action Series
Grade 10 Siyavula Maths
ERRORS/MISCONCEPTIONS/PROBLEM AREAS
Assuming that any point on a line is its midpoint
METHODOLOGY
The midpoint is the centre of a line segment (it bisects the line segment).
MIDPOINT THEOREM:
CONVERSE OF MIDPOINT THEOREM:
A
D
B
If AD = DB and AE = EC,
1
then DE∥BC and DE= BC
2
A
E
D
C
B
E
C
If AD = BD and DE∥BC
1
then AE = EC and DE= BC
2
77
ACTIVITIES
Activity 2.9.1
INVESTIGATING THE MIDPOINT THEOREM
Use a ruler and determine midpoint S of side PQ and midpoint T of side PR of triangle PQR.
Indicate on the sketch that PS = SQ and PT = TR. Draw ST.
1.
Measure each of the following angles, using a protractor:
1.1
Q̂  .......... ... degrees
1.3
PŜT  .......... ... degrees
1.2
R̂  .......... ... degrees
1.4
PT̂S  .......... ... degrees
2.
What do you notice concerning your answers in question 1?
(4)
………………………………………………………………………………………………(2)
3.
Using your answer to question 2, what can you conclude concerning ST and QR?
………………………………………………………………………………………………(1)
4.
Give a reason for your answer in question 3.
………………………………………………………………………………………………(1)
5.
Measure the lengths of the following sides:
5.1
ST = ………………. mm
6.
What do you notice concerning the lengths of ST and QR?
5.2
QR = ………………. mm
(2)
………………………………………………………………………………………………(1)
5.
Hence, make a conjecture regarding the line joining the midpoints of two sides of any triangle:
………………………………………………………………………………………………………
….……………………………………………………………………………………………………
………………………………………………………………………………………………(2)
78
In this part all answers have to be justified, using accurate measurements of distances and/or
angles.
Given: GHJ with K on GH, and L on GJ.
1.
Are K and L the midpoints of GH and GJ respectively?
…………………………………………………………………………………..…………… (2)
2.
Is KL || HJ?
……………………………………………………………………………………………….. (3)
3.
Is HJ = 2KL?
…………………………………………………………………………………………………(1)
Activity 2.9.2:
Given: AD = 5 cm and MC = 6 cm.
1.
A
Calculate, with reasons:
1.1
The length of BM
1.2
The length of DP
1.3
The length of DE
D
P
M
B
C
E
79
2.
In ∆ACD, AB = BC, GE =15 cm,
AF = FE = ED.
Calculate the length of CE.
3.
M, N and T are the midpoints of AB,
BC and AC of ABC . Â=60 and
B̂=80 .
A
60
°
Calculate the interior angles of MNT .
T
M
80
B °
N
C
Activity 2.9.3
1.
In ∆ABC, AD = DB and AE = EC. DE is
produced to F. DB||FC and BC = 32 mm.
a) Prove that DBCF is a parallelogram.
b) Calculate the length of DE.
2.
In ∆ABC, AE = EB and EF||BC. In ∆ ACD,
FG||CD.
Prove that AG = GD.
3.
In ∆DEF, DS = SE, EU = EF and ST||EF.
Prove that SEUT is a parallelogram.
80
TOPIC 2: EUCLIDEAN GEOMETRY
Weighting: (30/100 marks from Paper 2)
LESSON 10: CONSOLIDATION OF GR. 10 EUCLIDEAN GEOMETRY
1
2 hours
10
Term
Duration
Grade
Date
Sub-topics
Consolidation
ACTIVITIES/ASSESSMENTS
Activity 2.10.1
1. Study the diagram below and
calculate the unknown angles w, x,
y and z.
A
F
z
x
Give reasons for your statements.
y
53
C
B
2.
DBE NOV. 2015 GRADE 10
In the diagram below, PQRS is a
parallelogram having diagonals PR and
QS intersecting in M. B is a point on PQ
such that SBA and RQA are straight lines
and SB = BA. SA cuts PR in C and PA is
drawn.
2.1
Prove that SP = QA.
2.2
Prove that SPAQ is a
parallelogram
Prove that AR = 4MB.
2.3
3.
M
C
P
Q
B
A
A
B
AÔD
O
Calculate the length of AO.
36,87
3.3
R
Write down the size of the
following angles:
3.1.1 CD̂O
3.1.2
3.2
E
S
In the diagram, ABCD is a rhombus
having diagonals AC and BD intersecting
ˆ
in O. ADO=36,87
and DO = 8 cm.
3.1
w
74
D
If E is a point on AB such that
OE | | AD, calculate the length of
OE.
D
8 cm
C
81
4.
ABC is right angled at B. F and G are midpoints
of AC and BC respectively. H is the midpoint of
AG. E lies on AB such that FHE is a straight line.
4.1
Prove that E is the midpoint of AB.
4.2
If EH = 3,5cm and the area of AEH=9,5cm2 ,
calculate the length of AB.
4.3
Hence, calculate the area of ABC .
82
TOPIC 3: TRIGONOMETRY
Weighting: (40/100 marks from Paper 2)
LESSON 1: DEFINITIONS OF TRIGONOMETRIC RATIOS
1
1 hour
10
Term
Duration
Grade
Date
1. Define the trigonometric ratios sin  , cos  and tan  using rightSub-topic(s)
angled triangles.
RELATED CONCEPTS/
Pythagoras theorem, Sum of the angles of a triangle, Hypotenuse, Adjacent
TERMS/VOCABULARY
Side/ Angles, Opposite side, Reciprocals, Period, Amplitude, maximum
value, minimum value; decreasing function and increasing
PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE
Different types of Triangles, Linear equations, types of different angles,
RESOURCES
Calculators, Text Books
NOTES :
BASELINE ASSESSMENT :
Ratios and Theorem Of Pythagoras
Example 1
Given
Calculate the following ratios and write them in simplest form.
1.1
1.2
1.3
𝐴𝐵
𝐷𝐸
𝐵𝐶
𝐸𝐹
𝐴𝐶
𝐷𝐹
=
=
=
=
=
=
83
Example 2
Example 3
What is the length of the hypotenuse in the
following triangles
State which side of the following triangles
Is the Hypotenuse
Example 4
The three side length of two right- angled triangles are listed below. For each triangle state the length of the
hypotenuse
Opposite to
A
B
C
Adjacent to
Hypotenuse - The side opposite the 90° angle (longest side)
Opposite
- The side opposite the angle C
Adjacent
- The remaining side next to C
84
Definition of trig ratios:
The ratio
opp
opp
is called the sine of the angle θ and can be written as sin  
.
hyp
hyp
The ratio
adj
adj
is called the cosine of the angle θ and can be written as cos  
.
hyp
hyp
The ratio
opp
opp
is called the tangent of the angle θ and can be written as tan  
.
adj
adj
cos  
b.
c.
d.
sin  
e.
cos  
f.
tan  
tan  
A
Opposite to
Example:
Consider the following diagram and answer the
questions that follow.
State the following:
a.
sin  
B
C
Adjacent to
3.1 ACTIVITIES/ASSESSMENT
Activity 3.1.1:
3.1.1
Determine the following trig ratios
85
3.1.2
State the following
3.1.3
State the following
86
TOPIC 3: TRIGONOMETRY
Weighting: (40/100 marks from Paper 2)
LESSON 2: THE RECIPROCAL TRIGONOMETRIC RATIOS
1
1 hour
10
Term
Duration
Grade
Date
Define the reciprocals trigonometric ratios cosec , sec  and cot  using
Sub-topic(s)
right-angled triangles.
(These three reciprocals should be examined in grade 10 only.)
Pythagoras theorem, Sum of the angles in a triangle, Hypotenuse, Adjacent Side/
RELATED CONCEPTS/
Angles, Opposite side, Reciprocals, Period, Amplitude, maximum value,
TERMS/VOCABULARY
minimum value; decreasing function and increasing.
PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE
Different types of Triangles, Linear equations, types of different angles.
RESOURCES
Calculators, Text Books
NOTES :
Example :
1
2
3
2
has a reciprocal of
; and
has a reciprocal of
2
1
2
3
cosec  =
1
sin 𝜃
=
𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒
Leaners to complete:
sin R =
cosec R =
cos R =
sec R =
tan R =
cot R =
3.2 ACTIVITIES/ASSESSMENT
Activity 3.2.1
Refer to the diagram alongside to answer the following questions:
State the value of each of the following
a)
b)
c)
d)
e)
f)
sec A
cot A
cosec A
cot C
cosec C
sec C
87
TOPIC 3: TRIGONOMETRY
Weighting: (40/100 marks from Paper 2)
LESSON 3: SPECIAL ANGLE VALUES
1
1 hour
10
Term
Duration
Grade
Date
Derive values of the trigonometric ratios for the special angles (without
Sub-topic(s)
using a calculator),   0  ; 30  ; 45 ; 60  ; 90  .
Pythagoras theorem, Sum of the angles in a triangle, Hypotenuse, Adjacent
RELATED CONCEPTS/
Side/ Angles, Opposite side, Reciprocals.
TERMS/VOCABULARY


PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE
 Different types of Triangles.
 Linear equations.
 Types of different angles.
RESOURCES
 Grade 10 textbooks
 Ruler
 Protractor
 Pencil
NOTES :
Special Angles
Begin by constructing an isosceles right – angled triangle: (angles of 90 ; 45; 45)
sin 30 =
sin 60 =
cos 30 =
cos 60 =
tan 30 =
tan 60 =
88
3.3 ACTIVITIES/ASSESSMENT
Activity 3.3.1
Determine the value of each of the following without using a calculator:
1.
sin 2 30  cos 2 30
2.
sin 2 30  cos 2 60
3.
sin30.tan 45.cos 45
4.
sin 45
cos 45
5.
cos30.tan 60  cos ec 45.sin 60
6.
sin 30.sec 45
1
2
sin 60
2
2
Activity 3.3.2
Evaluate the following trigonometric ratios with the use of calculators. Round off all answers to two
decimal places.
1.
cos10 
2. sin312 
3.
sin35  sin 75 
4. sin 43  cos 43 
5.
cos 24

24
2
2
89
TOPIC 3: TRIGONOMETRY
Weighting: (40/100 marks from Paper 2)
LESSON 4: TRIGONOMETRIC EQUATIONS
1
1 hour
10
Term
Duration
Grade
Date
Solve simple trigonometric equations for angles between 00 and 900.
Sub – topic(s)
RELATED CONCEPTS/ Minimum value, maximum value.
TERMS/VOCABULARY
PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE
 Linear equations.
 Types of different angles.
RESOURCES
 Calculators.
 Text Books.
NOTES:
Solving of simple trigonometric equations for angles between 0 0 and 900 .
Example 1:
Consider cos   0,5
Steps: Use a calculator, shift button and enter cos ( cos 1 ).
  cos 1 0,5
  600
Examples:
1
2 sin   1  0
1
sin  
2
1
  sin 1  
2
  300
3.
sin 3  0,157
3  sin 1 0,157
3  9,0328......
  3,010
2 sin   3  0
3
sin  
2
3
  sin 1  
2
A calculator will show Math Error. This means the
equation has no solution. This is simple because the
ratios of cos and sin have a maximum value of 1, hence
they cannot be solved for values greater than 1.
2.
4.


cos x  600  0,5
x  60   cos0,5
0
x  600  600
x  600  600
x  00
90
3.4 ACTIVITIES/ASSESSMENT
Activity 3.4.1
Solve the following equations and round off your answers to 2 decimal places:
tan   0,357
1.
2cos   3
2.
3.
2sin x  1  2
5.
7.
9.
4.
2 tan    10   3  5
1
cos 3x  0,12
3
6.
3cos  2  12   2  1
3
sin x  cos 33
2
8.
sec  x  10   5, 649
tan 
 1  2,32
0,3
10.
4 cos   2  1
91
TOPIC 3: TRIGONOMETRY
Weighting: (40/100 marks from Paper 2)
LESSON 5: EXTEND THE DEFINITIONS OF TRIGONOMETRIC RATIOS TO 0     360  (1)
1
1 hour
10
Term
Duration
Grade
Date
Sub-topic(s)
Extend definitions of sin 𝜃, cos 𝜃 and tan 𝜃, for 0° ≤ 𝜃 ≤ 360°
RELATED CONCEPTS/ TERMS/ VOCABULARY
Cartesian plane, anticlockwise, radius
PRIOR KNOWLEDGE/ BACKGROUND KNOWLEDGE
Trigonometric ratios, intervals
RESOURCES
Mind Action Series, Maths handbook and Study Guide, Study & Master, Siyavula
ERRORS/MISCONCEPTIONS/PROBLEM AREAS
Choosing the correct quadrant in the Cartesian plane
METHODOLOGY
Angles in a Cartesian plane
Positive angles are measured in an anti-clockwise direction from the positive x-axis
Example 1:
Determine in which quadrants the following angles lie:
92
Trigonometric Ratios in the Cartesian Plane
The CAST Diagram
The CAST diagram summarizes where trigonometric ratios are positive.
Example 2
Determine in which quadrant does the terminal arm of the angle lie if:
a.
sin   0 and cos >0
d.
tan   0 and
b.
sin   0 and cos  0
e.
sin   0 and   90; 270
c.
tan   0 and
f.
cos   0 and 0    180
cos  0
cos  0
3.5 ACTIVITIES/ASSESSMENT
Activity 3.5.1
Give the quadrant in which the radius will lie for each of the following:
5
and   0;180
13
(b)
tan  
sin   
12
and   0;270
13
(d)
sec  
cot   
4
and   0;180
3
(a)
cos   
(c)
(e)
4
and   180;360
3
5
and   0;180
3
93
TOPIC 3: TRIGONOMETRY
Weighting: (40/100 marks from Paper 2)
LESSON 6: EXTEND DEFINITIONS OF TRIGONOMETRIC RATIOS TO 0     360  (2)
1
1 hour
10
Term
Duration
Grade
Date
Using diagrams to determine the numerical values of ratios for angles from
Sub-topic(s)
0  to 360
Pythagoras theorem, Sum of the angles in a triangle, Hypotenuse, Adjacent
RELATED CONCEPTS/
Side/ Angles, Opposite side, Reciprocals, Cartesian Plane, etc.
TERMS/VOCABULARY
PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE
 Different types of Triangles.
 Linear equations.
 Types of different angles.
 Quadrants
RESOURCES
 Grade 10 textbooks
 Ruler
 Protractor
 Pencil
METHODOLOGY
94
Examples
1.
If 3sin  1 and   90; 270 calculate
without the use of a calculator, and with the aid
of a diagram, the value of cos   sin  :
sin 
2.
Consider the diagram below. Point T  2;3
is a point on the Cartesian plane such that 
is the angle of inclination of OT.
y
1

r
3
sin is negative in 3rd quadrant
Calculate the following, without the use of a
calculator:
a. tanβ
3
b.
c.
sin   cos 
x2  y2  r 2
x 2   1  32
2
x2  9  1
x 8
y x

r r
1
8
 
3 3
a.
1 8

3
b.

13sinβ.cosβ
r  13
tan  
3
2
sin 2   cos 2 
2
 3   2 

 

 13   13 
9 4
 
13 13
1
c.
2
 3  2 
13sin  cos   13 

  6
 13  13 
ACTIVITIES/ASSESSMENT
Activity 3.6.1
95
2.1
Calculate the length of OR
2.2
State the values of the six trigonometric ratios of  .
96
1
Term
Sub-topic(s)
TOPIC 3: TRIGONOMETRY
Weighting: (40/100 marks from Paper 2)
LESSON 7: TRIGONOMETRIC GRAPHS (1)
1 hour
10
Duration
Grade
Date
Point-by-point plotting of basic trigonometric graphs defined by 𝑦 = sin 𝜃,
𝑦 = cos 𝜃 and 𝑦 = tan 𝜃 for 𝜃 ∈ [0°; 360°].
Angles on a Cartesian Plane, Special Angles, Intercepts, Turning points,
Maximum/Minimum points, etc.
RELATED
CONCEPTS/ TERMS/
VOCABULARY
PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE
 Different types of Triangles.
 Linear equations.
 Types of different angles.
 Quadrants
 Coordinates
RESOURCES





Grade 10 textbooks
Ruler
Protractor
Pencil
Graph paper (if possible)
METHODOLOGY :
97
ACTIVITY/ASSESSMENT
Activity 3.7.1:
Using point-by-point plotting, sketch the graphs of the following trigonometric functions:
𝑓(𝑥) = sin 𝑥, 𝑓(𝑥) = cos 𝑥, 𝑓(𝑥) = tan 𝑥 on different sets of axes, for θ∈[0°;360°].
98
Term
Sub-topic(s)
1
TOPIC 3: TRIGONOMETRY
Weighting: (40/100 marks from Paper 2)
LESSON 8: TRIGONOMETRIC GRAPHS (2)
1 hour
10
Duration
Grade
Date
Study the effect of a and q on the graphs defined by y  a sin   q ;


y  a cos   q ; and y  a tan   q ; for   0 ; 360 ..
RELATED CONCEPTS/
Transformation, amplitude, turning point.
TERMS/VOCABULARY
PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE
 Sketching trigonometric functions, intuitive understanding of the shapes of the graphs of sin, cos, tan .
 Minimum value and maximum value
RESOURCES
 Calculator
 Textbooks
METHODOLOGY:
Examples:
Teacher to draw the graph of y  tan x . Due to the difference in characteristics between the graph of
y  tan x and those of y  sin x and y  cos x , the concept of asymptotes needs to be reintroduced.
Sketching of the graph by point by point method:
99
ACTIVITIES/ASSESSMENT
Activity 3.8.1
100
TOPIC 3: TRIGONOMETRY
Weighting: (40/100 marks from Paper 2)
LESSON 9: SOLVING 2D-PROBLEMS INVOLVING RIGHT-ANGLED TRIANGLES
1
1 hour
10
Term
Duration
Grade
Date
Sub-topic(s)
Solve two-dimensional problems involving right-angled triangles.
Pythagoras theorem, Sum of the angles in a triangle, Hypotenuse, Adjacent
Side/ Angles, Opposite side, Reciprocals, Period, Amplitude, maximum
value, minimum value; decreasing function and increasing
PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE
Different types of Triangles, Linear equations, types of different angles.
RESOURCES
Calculators, Text Books
METHODOLOGY
RELATED CONCEPTS/
TERMS/VOCABULARY
Solving problems using Trigonometric ratios:
Examples:
1. Finding the length of a side:
In △FUN, FN=10, ∠U=90˚ and ∠N=28˚
Calculate the length of FU, rounded off to one decimal place.
Solution:
Let N be the point of reference (given angle)
opp
 sin 28
hyp
FU

 sin 28
10
 FU  4, 7 units
2. Calculating an angle:
Calculate the size of θ correct to one decimal place.
101
Angles of elevation and depression
1. Angle of elevation
102
Examples:
1. The angle of depression of a boat on the ocean from the top of a cliff is 55˚. The boat is 70 metres from
the foot of the cliff.
a.
b.
What is the angle of the elevation of the top of the cliff from the boat?
Calculate the height of the cliff.
Solutions
a. The angle of elevation of the top of the cliff from the boat is 55˚. ie ∠B=55˚
b. We calculate the height of the cliff as follows:
h
 tan 55
7
h  70 tan 55
h  100m
2.
In a soccer World Cup, a player kicked the ball from a distance of 11 metres from the goalposts
(4 metres high) in order to score a goal for his team. The shortest distance travelled by the ball is in a
straight line. The angle formed by the pathway of the ball and the ground is represented by θ.
a. Calculate the largest angle θ for the player will possibly score a goal.
b. Will the player score a goal if the angle θ is 22˚? Explain.
Solutions:
4m
tan  
a.
11m
tan   0.3636.....
  tan 1 0.3636.....
b.
 20 
No. The ball will go above the goalpost.
103
ACTIVITIES/ASSESSMENT:
Activity 3.9.1:
3.
ˆ  30 , BCD
ˆ  45 and BDC
ˆ  90
In the diagram, AB  10cm , BAC
104
105
ANSWERS TO QUESTIONS FROM ACITIVITIES
TOPIC 1: ALGEBRA
LESSON 1: NUMBER SYSTEM
Activity 1.1.1
5
Real; Rational; Z; N0 ; N
–2
Real; Rational; Z
4,1
Real; Rational
1
2
Real; Rational
3
Real; Rational
0,7
9
Real; Rational; Z; N0 ; N
3 8
Real; Rational; Z
10
Real; Irrational
9
16
Real; Rational
8
Non-real
8
Real; Rational; Z
9
0
5
5
0
Real; Irrational
 1
3
3
Real; Rational; Z; N0
Undefined

Real; Rational
Activity 1.1.2
2
(a)
9
3
(b)
11
2
(c)
9
71
(d)
333
5
(e)
33
49
(f)
90
106
LESSON 2: SURDS AND PRODUCTS
Activity 1.2.1
(a)
between 2 and 3
(b)
between 8 and 9
(c)
between – 4 and – 3
Activity 1.2.2
(a)
8x 2  12x
(b)
 5x 2  x
(c)
12x3  2 x 2  16x
(d)
x 2  9 x  20
(e)
x 2  x  20
(f)
x 2  9 x  20
(g)
14x 2  19x  3
4 x 2  18xy  18 y 2
(h)
1
1
x2  x 
(i)
6
6
2
(j)
25x  4
(k)
16a 2  9b 2
(l)
x 2  8x  16
(m)
a 2  6ab  9b 2
(n)
4a 2  20ab  25b 2
(o)
 12 x 2  12 xy  3 y 2
(p)
4m2  32mn  64n 2
(q)
6 x8  5x 4 y 2  6 y 4
LESSON 3: PRODUCTS (CONTINUED)
Activity 1.3.1
(a)
x 3  2 x 2  3x  2
(b)
6c 3  13c 2  3c  2
(c)
3x 3  7 x 2 y  4 y 3
(d)
a 3  a 2b  ab2  b3
(e)
a 3  3a 2b  3ab2  b3
x3  y3
(f)
x 3  27y 3
(g)
Activity 1.3.2
34 y 2  16 y  7
(a)
(b)
(c)
(d)
(e)
(f)
 4 x 2  16xy  19 y 2
41m2  4mn  4n 2
2 x 4  25 y 2
1
x2  2  2
x
6
x2 1  2
x
Page 107 of 120
LESSON 4: FACTORISATION:
REVISION OF GR. 9 FACTORISATION
Activity 1.4.1
 3 x  5
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
4ab3 4a 3b 5  2b 4  9a 
k  p k  p 
3 
3

 x   x  
x 
x

2
x y  y  1
3a  53a  5
32a  b 2a  b 
x  1x  1x 2  1
1  1
1 
1
 x  y  x  y 
9  2
9 
2
LESSON 5: FACTORISING
TRINOMIALS
Activity 1.5.2
 p  5 p  3
(a)
 p  5 p  3
(b)
 p  5 p  3
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l)
(m)
(n)
(o)
(p)
(q)
(r)
(s)
(t)
(u)
(v)
 p  5 p  3
x  5x  2
2a  5a  2 
b  5b  2
  y  5 y  2 
a  12 a  7 
x  24 x  2
2 x  1x  3
2a  1a  3
2a  3a  5
2 y  52 y  1
3a  22a  3
4 x  33x  2
4 x  32 x  5
11 y  12 y  1
3  x 1  x 
a  52
2 p  12
3a  5b2
Page 108 of 120
LESSON 6: FACTORISATION BY
GROUPING IN PAIRS
Activity 1.6.1
m  n x  y 
(a)
 p  q m  1
x  4 y 1  3a 
3x  2 y 2b  5c 
3x  4c 2 x  3a 
a  d c  b 2 
(b)
(c)
(d)
(e)
(f)
LESSON 7: FACTORISATION: SUM
AND DIFFERENCE OF CUBES
Activity 1.7.1
(a) x  1 x 2  x  1

(b)
(c)
(d)

1  2 y 1  2 y  4 y 2 
a  3ba 2  3ab  9b 2 
4 x  516x 2  20x  25
(e)
2x  2 y 2 x 2  2 xy 2  4 y 4 
(d)
4b  a b  2
(f)  5a  b a 2  ab  b 2 a 6  a 3b 3  b 6 
Activity 1.7.2
(a)  x  4  x  3
(b) 5a  4 a  4 
(c) 4 x  3 y 3 x  7 y 
(e)
(f)
(g)
(h)
(i)
2
 x 10  x 2 x 100 
  
  2 
 10 y  100 y y 
 y  1y 2  1
4 p  72
x  y ax  3b 
3 
3

 x   x  
2 
2

2
(k)
4

x  
x

x  y  x 2  xy  y 2 x  y  x 2  xy  y 2
(l)
4 9 x  25
(j)


2




Page 109 of 120
LESSON 8: SIMPLIFICATION OF
ALGEBRAIC FRACTIONS
Activity 1.8.1
(a)
2x2  1
2x  1
(b)
x2
4a 2  6 a  9
(c)
3a  3
 2 x  1
(d)
x  3x  4
(e)
2x
2
k 3
(f)
2
3
(g)
x  y3
LESSON 9: ADDING AND
SUBTRACTING ALGEBRAIC
FRACTIONS
Activity 1.9.1
1
(a)
6
x2  y2
(b)
xy
(c)
(d)
(e)
(f)
(g)
(h)
x 3  2 x 2  3x  4
x3
4a 2  15b 2
12ab
1
2a  32a  1
6  3a  a 2
2  a  4  2a  a 2
1
ab
3xy
x  y x  y x 2  xy  y 2 


Page 110 of 120
LESSON 10: EXPONENTS
Activity 1.10.1: Classwork
(a)
27
(b)
1
(c)
5
(d)
1
16x 6 y 4
16
(f)
25
(g)
6
Activity 1.10.2: Homework
3
(a)
5
27
(b)
4
125
(c)
3
1
(d)
9
1
(e)
2
(e)
LESSON 11: EXPONENTS
Activity 1.11.1: Classwork
8
(a)
3x3
16
(b)
3
1
(c)
625
Activity 1.11.2: Homework
(a)
32n1.2n
1
(b)
4
(c)
2 x7
Page 111 of 120
LESSON 12: EXPONENTS
Activity 1.12.1: Classwork
(a)
3
(b)
5
(c)
2
(d)
3
(e)
2a
(f)
3y
Activity 1.12.2: Homework
(a)
2x 2
2
(b)
x5
m
(c)
5
x2
(d)
7
1
(e)
3
(f)
1
LESSON 13: EXPONENTS
Activity 1.13.1: Classwork
6

(a)
10
1
(b)
1
(c)
2
31

(d)
16
Activity 1.13.2: Homework
(a)
3
1
(b)
2
26
(c)
25
Page 112 of 120
LESSON 14: EXPONENTS
Activity 1.14.1: Classwork
(a)
3
(b)
0
(c)
2
(d)
0
(e)
7
3
(f)
2
9
(g)
2
1
(h)
2
10; 10
(i)
Activity 1.14.2: Homework
(a)
no solution
(b)
4
(c)
4
(d)
2
(e)
16
(f)
125
(g)
27
(h)
81; 16
Activity 1.14.3: Homework
25
Page 113 of 120
LESSON 15: LINEAR EQUATIONS
Activity 1.15.1
(a)
5
3
(b)
1
8
(c)
2
1
6
(d)
2
1

(e)
2
3
(f)
5
(g)
2
1
(h)
2
Activity 1.15.2
(a)
5
(b)
20
(c)
20
(d)
1
5
(e)
2
(f)
LESSON 16: QUADRATIC EQUATIONS
Activity 1.16.1
1
0;
(a)
3
(b)
1;2
(c)
5
(d)
2
 1; 6
(e)
 2;6
(f)
Activity 1.16.2
 2; 2
(a)
 8 ;1
(b)
2 3
;
(c)
3 2
3
;3
(d)
5
Page 114 of 120
LESSON 17: QUADRATIC EQUATIONS
(CONTINUED)
Activity 1.17.1
3
 ;4
(a)
2
11
 ;4
(b)
5
 3 ;  1 is rejected
(c)
 2 ;12
(d)
Activity 1.17.2
 2;3
(a)
6
 2;
(b)
5
1
0;
(c)
4
LESSON 18: LITERAL EQUATIONS
Activity 1.18.1
cb
(a)
x
a
2b
(b)
x
3
a
(c)
x  2m 2  2 p
(d)
3 yz
x
2z  y
Activity 1.18.2
vu
(a)
(i) a 
t
(ii) a  12
S  2r 2
(b)
(i) h 
2r
(ii) 8,01
(iii) a closed cylinder
A
(i) i  n  1
(c)
P
(ii) 0,13
Page 115 of 120
Activity 1.18.3
(a)
(b)
(c)
(d)
(e)
(f)
3V
4
(ii) 10,00
2s  nl
(i) a 
n
9
(ii) 
2
5
(i) C  F  17,78
9
(ii)  3,89 C
uv
f 
uv
Fi
x
1  i n  1
V
(i) h  2
r
(i) r  3
(ii) 5,09
(g)
x
mn
mn
LESSON 19: INEQUALITIES
Activity 1.19.1
1.
4 ;  
2.
  ; 1
3.
4 ;  
4.
5 ;  
5.
 1 ;  
6.
  ; 10 
7.
3 ;  
8.
  ;  9
9.
2 ;  
 24

 9 ;  
Activity 1.19.2
1.
1 ; 3
2.
 7 ; 5
3.
 1 ; 7
4.
 1 ; 5
5.
 3 ; 6
10.
Page 116 of 120
Activity 1.19.3
6.
 9 ; 6
7.
 2 ; 6
8.
3

 2 ; 
2

9.
10.
 1 ; 10
  ;  1
LESSON 20: SIMULTANEOUS
EQUATIONS
Activity 1.20.1
(a)(i)
p  3 ; q  1
(ii)
x  1; y  2
(b)
x  10; y  6
1.20.2 Homework
Activity 1.20.2
(a)(i)
x  1; y  1
(ii)
3
7
x ;y 
2
2
(iii)
x  4 ; y  2
(b)(i)
4
12
x ;y 
5
5
(ii)
x  9 ; y  3
(iii)
x  5; y  7
LESSON 21: SIMULTANEOUS
EQUATIONS
Activity 1.21.1
(a)
x  21; y  10
(b)
x  5; y  3
(c)
3
x  7; y  
2
Activity 1.21.2
2
5
x ;y 
(a)(i)
3
3
(ii
x  1; y  1
(iii)
x  6; y  20
2
8
x ;y 
(iv)
3
3
16
4
(b)(i) x  ; y 
3
3
1
1
x ;y 
(ii)
7
4
Page 117 of 120
LESSON 22: WORD PROBLEMS
Activity 1.22.1
(a)
Time taken is 0,02 hours = 1,2
minutes
(b)
Number of bicycles is 2 and the
number of tricycles is 5
Activity 1.22.2
(a)
Cost of individual chocolate milk
shake is R9,00 and for fruit shake is
R11,00
(b)
The actual speed is 24km/h
(c)
320 learner tickets were sold
LESSON 23: WORD PROBLEMS
(CONTINUED)
Activity 1.23.1
(a)
Andile’s age is 8 years
(b)
The consecutive odd numbers are 9
and 11
(c)
The negative consecutive integers
are – 34 and – 33
Activity 1.23.2
(a)
Lindy’s present age is 12 years
(b)
The price of 1 kg of fudge is R46,00
(c)
The breadth is 8 cm and the length is
16 cm
(d)
The original number is 63
Page 118 of 120
TOPIC 3: TRIGONOMETRY
LESSON 1
3.1.1 1
3.1.3
cos A 
sin C 
sin α = 𝑞
b)
cos α = 𝑞
c)
tan 𝛼 = 𝑟
d)
sin 𝜃 = 𝑞
e)
cos 𝜃 = 𝑞
f)
tan 𝜃 = 𝑝
LESSON 2
3.2.1 a)
sec A = 3
b)
cot A =
c)
cosec A = 4
d)
cot C =
e)
cosec C =
f)
sec C = 4
c
b
c
b
c
tan A 
a
2.
sin Z 
cos Z 
tan Z 
sin Y 
cos Y 
tan Y 
120
5 745
65
5 745
 0,88
 0, 48
120
 1,85
65
65
5 745
120
5 745
 0, 48
 0,88
65
 0,54
120
3.1.2
a)
b)
c)
d)
e)
f)
LESSON 4
3.4.1 1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
c
a
b
cos C 
a
b
tan C 
c
b
sin B 
a
c
cos B 
a
c
tan B 
b
sin C 
  19,65
  30
x  30
  35
x  22,97
  6
x  33,99
x  69,8
  44,89
  41, 41
𝑝
a)
𝑟
𝑝
𝑟
𝑝
𝑟
5
3
4
5
4
3
5
3
5
LESSON 3
3.3.1
1
1
2
1
2
3.
1
2 2
4.
.5.
1
6.
2 2
3
9
2
3.3.2
1.
2.
3.
4.
5.
LESSON 5
3.5.1
(a)
(b)
(c)
(d)
(e)
0,985
0,743
1,54
1
0,038
quadrant 1
quadrant 3
quadrant 3
quadrant 2
quadrant 2
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LESSON 6
3.6.1
1.1
1.2
29
5
1.3
29
2
29
1.4
5
2
2.1
3.1
3.2
13
3.3
3.4
4.1
4.2
4.3
4.4
y  6
LESSON 9
3.9.1
1.1
1.2
2.1
2.2
3.1
3.2
4.1
4.2
4.3
16,35 units
14,72 units
94,65 units
86,78 units
71,57
18, 43
5,76 units
37,51
18,91 units
3
5
4
cos   
5
3
tan  
4
24
sin   
25
sin   
25
24
24
tan   
7
cos ec  
1
Page 120 of 120