MATHEMATICS

Student Textbook
Grade 8
 

Authors, Editors and Reviewers:

Gebreyes Hailegeorgis (B.SC.)
Basavaraju V.Mohan (Prof./Dr)

Evaluators:
Menberu Kebede
Abdella Mohe
 

FEDERAL DEMOCRATIC REPUBLIC OF ETHIOPIA

MINISTRY OF EDUCATION  

 
 

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ISBN: 978‐99944‐2‐056‐8
 

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Table of Contents
UNIT 1: SQUARES, SQUARE ROOTS, CUBES AND
CUBE ROOTS
1.1 The Square of a Number ...................................................... 2
1.2 The Square Root of a Rational Number ............................ 12
1.3 Cubes and Cube Roots ...................................................... 22
Summary ............................................................................. 33
Miscellaneous Exercise ..................................................... 34

UNIT 2: FURTHER ON WORKING WITH VARIABLES

2.1 Further on Algebraic Terms and Expressions ................. 38
2.2 Multiplication of Binomials ................................................ 49
2.3 Highest Common Factors ................................................... 56
Summary ............................................................................. 63
Miscellaneous Exercise ..................................................... 64

UNIT 3: LINEAR EQUATION AND INEQULITIES

3.1 Further on Solutions of Linear Equations ...................... 68
3.2 Further on Linear Inequalities ........................................... 81
3.3 Cartesian Coordinate System ............................................ 84
Summary ............................................................................. 96
Miscellaneous Exercise ..................................................... 98

UNIT 4: SIMILAR FIGURES

4.1 Similar Plane Figures ........................................................ 102
4.2 Similar Triangles................................................................ 111
Summary ........................................................................... 131
Miscellaneous Exercise ................................................... 133

UNIT 5: CIRCLES
5.1 Further On Circle .............................................................. 136
5.2 Angles in the Circle ........................................................... 144
Summary ........................................................................... 161
Miscellaneous Exercise ................................................... 162
 

UNIT 6: INTRODUCTION TO PROBABILITY

6.1 The Concept of Probability ............................................... 167
6.2 Probability of Simple Events ............................................ 171
Summary ........................................................................... 178
Miscellaneous Exercise ................................................... 178

UNIT 7: GEOMETRY AND MEASUREMENTS

7.1 Theorems on the Right Angled Triangle ......................... 180
7.2 Introduction to Trigonometry ........................................... 196
7.3 Solids Figures .................................................................... 208
Summary ........................................................................... 213
Miscellaneous Exercise ................................................... 215
UNIT
1 SQUARES, SQUARE
ROOTS, CUBES AND
CUBE ROOTS

Unit outcomes
After completing this unit, you should be able to:
 understand the notion square and square roots and cubes
and cube roots.
 determine the square roots of the perfect square
numbers.
 extract the approximate square roots of numbers by
using the numerical table.
 determine cubes of numbers.
 extract the cube roots of perfect cubes.

Introduction
What you had learnt in the previous grade about multiplication will be used in
this unit to describe special products known as squares and cubes of a given
numbers. You will also learn what is meant by square roots and cube roots and
how to compute them. What you will learn in this unit are basic and very
important concepts in mathematics. So get ready and be attentive!
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 Grade 8 Mathematics [SQUARES, SQUARE ROOTS, CUBES AND CUBE ROOTS ]

1.1 The Square of a Number

1.1.1 Square of a Rational Number
Addition and subtraction are operations of the first kind while multiplication and
division are operation of the second kind. Operations of the third kind are
raising to a power and extracting roots. In this unit, you will learn about
raising a given number to the power of “2” and power of “3” and extracting
square roots and cube roots of some perfect squares and cubes.

Group Work 1.1

Discuss with your friends
1. Complete this Table 1.1. Number of small squares
Standard Factor Power
1
Form Form Form
a) 1
1 1×1 12

b) 2

2
4 2×2 22

3
c)

3
. . .

d) 4

4 . . .

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 Grade 8 Mathematics [SQUARES, SQUARE ROOTS, CUBES AND CUBE ROOTS ]

e)
. . .

5
Figure 1.1
2. Put three different numbers in the circles so that when you add the numbers at
the end of each – line you always get a square number.

Figure 1.2
3. Put four different numbers in the circles so that when you add the numbers at
the end of each line you always get a square number.

Figure 1.3

Definition 1.1: The process of multiplying a rational number by itself is

called squaring the number.

For example some few square numbers are:

a) 1 × 1 = 1 is the 1st square number. c) 3× 3 = 9 is the 3rd square number.
b) 2× 2 = 4 is the 2nd square number. d) 4×4 = 16 is the 4th square number.

a) b) c) d)

Figure 1.4 A square number can be shown as a pattern of squares

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 Grade 8 Mathematics [SQUARES, SQUARE ROOTS, CUBES AND CUBE ROOTS ]
If the number to be multiplied by itself is ‘a’, then the product (or the result
a × a) is usually written as a2 and is read as:
 a squared or
 the square of a or
 a to the power of 2
In geometry, for example you have studied that the area of a square of side
length ‘a’ is a × a or briefly a2.
When the same number is used as a factor for several times, you can use an
exponent to show how many times this numbers is taken as a factor or base.

Exponent
49 = 72 Base
Standard
numeral form
Power form

Note: 72 is read as
 7 squared or
 the square of 7 or
 7 to the power of 2

Example 1: Find the square of each of the following.

a) 8 b) 10 c) 14 d) 19
Solution
a) 82 = 8 × 8 = 64
b) 102 = 10 × 10 = 100
c) 142 = 14 × 14 = 196
d) 192 = 19 × 19 = 361

Example2: Identify the base, exponent, power form and standard form of
the following expression.
a) 102 b) 182
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 Grade 8 Mathematics [SQUARES, SQUARE ROOTS, CUBES AND CUBE ROOTS ]

Solution exponent
a)
100 = base
Standard
numeral form Power form

b) exponent
324 = base
Standard
numeral form Power form

Note: There is a difference between a2 and 2a. To see this distinction

consider the following examples of comparison.

Example3: a) 302 = 30 × 30 = 900 while 2 × 30 = 60

b) 402 = 40 × 40 = 1600 while 2 × 40 = 80
c) 522 = 52 × 52 = 2704 while 2 × 52 = 104
Hence from the above example; you can generalize that a2 = a × a and
2a = a + a, are quite different expressions.

Definition 1.2: A rational number x is called a perfect square, if and only

if x = n2 for some n ∈ Q.

Example 4: 1 = 12, 4 = 22, 9 = 32, 16 = 42, 25 = 52 . Thus 1, 4, 9, 16 and 25

are perfect squares.

Note: A perfect square is a number that is a product of a rational

number times itself and its square root is a rational number.

Example5: In Table 1.2 below some natural numbers are given as values
of x. Find x2 and complete table 1.2.

x 1 2 3 4 5 10 15 20 25 35
x2
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 Grade 8 Mathematics [SQUARES, SQUARE ROOTS, CUBES AND CUBE ROOTS ]
Solution When x = 1, x = 12 = 1 × 1 = 1
2

When x = 2, x2 = 22 = 2 × 2 = 4
When x = 3, x2 = 32 = 3 × 3 = 9
When x = 4, x2 = 42 = 4 × 4 = 16
When x = 5, x2 = 52 = 5 × 5 = 25
When x = 10, x2 = 102 = 10 × 10 = 100
When x = 15, x2 = 152 = 15 × 15 = 225
When x = 20, x2 = 202 = 20 × 20 = 400
When x = 25, x2 = 252 = 25 × 25 = 625
When x = 35, x2 = 352 = 35 × 35 = 1225

x 1 2 3 4 5 10 15 20 25 35
2
x 1 4 9 16 25 100 225 400 625 1225

You have so far been able to recognize the squares of natural numbers, you also
know that multiplication is closed in the set of rational numbers. Hence it is
possible to multiply any rational number by itself.

Example 6: Find x2 in each of the following where x is rational number

given as:
a) x = b) x = c) x = d) x = 0.26

Solution a) x2 =

b) x2 =

c) x2 =

d) x2 =

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Grade 8 Mathematics [SQUARES, SQUARE ROOTS, CUBES AND CUBE ROOTS ]

Note:
i. The squares of natural numbers are also natural numbers.
ii. 0 × 0 = 0 therefore 02 = 0
iii. We give no meaning to the symbol 00
iv. If a ∈ and a ≠ 0, then a0 = 1
v. For any rational number ‘a’, a × a is denoted by a2 and read as
“a squared” or “a to the power of 2” or “the square of a”.

Exercise 1A
1. Determine whether each of the following statements is true or false.
a) 15 2 = 15 × 15 d) 812 = 2× 81 g) x2 = 2x
b) 202 = 20 × 20 e) 41 × 41 = 412 h) x2 = 22x
c) 192 = 19 × 19 f) - (50)2 = 2500 i) (-60) 2 = 3600
2. Complete the following.
a) 12 × = 144 d) (3a)2 = ___ × ___
b) 51 × _____ = 2601 e) 8a = .+ .
c) 60 = ____ × ____
2
f) 28 × 28 = .
3. Find the square of each of the following.
a) 8 b) 12 c) 19 d) 51 e) 63 f) 100
4. Find x2 in each of the following.
− 50
a) x = 6 c) x = - 0.3 e) x= g) x = 0.07
3
1
b) x = d) x = -20 f) x = 56
6
5. a. write down a table of square numbers from the first to the tenth.
b. Find two square numbers which add to give a square number.
6. Explain whether:
a. 441 is a square number. c. 1007 is a square number.
b. 2001 is a square number.

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 Grade 8 Mathematics [SQUARES, SQUARE ROOTS, CUBES AND CUBE ROOTS ]

Challenge Problems
7. Find
a) The 8th square number. c) The first 12 square numbers.
b) The 12th square number.
8. From the list given below indicate all numbers that are perfect squares.
a) 50 20 64 30 1 80 8 49 9
b) 10 21 57 4 60 125 7 27 48 16 25 90
c) 137 150 75 110 50 625 64 81 144
d) 90 180 216 100 81 75 140 169 125
9. Show that the difference between any two consecutive square numbers is
an odd number.
10. Show that the difference between the 7th square number and the 4th square
number is a multiple of 3.

Theorem1.1: Existence theorem

For each rational number x, there is a rational
number y (y ≥ 0) such that x2 = y.

Example 7: By the existence theorem, if

a) x = 9, then y = 92 = 81 c) x = -17, then y = (-17)2 = 289

b) x = 0.5, then y = (0.5)2 = 0.25 d) x = , then y =

Rough calculation could be carried out for approximating and checking the
results in squaring rational numbers. Such an approximation depends on
rounding off decimal numbers as it will be seen from the following examples.

Examples 8: Find the approximate values of x2 in each of the following:

a) x = 3.4
b) x = 9.7
c) x = 0.026

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 Grade 8 Mathematics [SQUARES, SQUARE ROOTS, CUBES AND CUBE ROOTS ]

Solution
a) 3.4 ≈ 3 thus (3.4)2 ≈ 32 = 9
b) 9.7 ≈ 10 thus (9.7)2 ≈ 102 = 100

c) 0.026 ≈ 0.03 thus (0.026)2 ≈ = 0.0009

Exercise 1B
1. Determine whether each of the following statements is true or false.
a) ( 4.2)2 is between 16 and 25 d) (9.9)2 = 100
b) 02 = 2 e) ( -13)2 = -169
c) 112 > (11.012)2 > 122 f) 81 × 27 = 92 × 9×3
2. Find the approximate values of x2 in each of the following.
a) x = 3.2 c) x = -12.1 e) x = 0.086
b) x = 9.8 d) x = 2.95 f) x = 8.80
3. Find the square of the following numbers and check your answers by
rough calculation.
a) 0.87 c) 12.12 e) 25. 14 g) 38.9
b) 16.45 d) 42. 05 f) 28. 23 h) 54. 88

1.1.2 Use of Table of Values of Squares

Activity 1.1
Discuss with your friends / partners/
Use table of square to find x2 in each of the following.
a) x = 1.08 b) x = 2.26 c) x = 9.99
d) x = 1.56 e) x = 5. 48 f) x = 7. 56
 To find the square of a rational number when it is written in the form of
a decimal is tedious and time consuming work. To avoid this tedious
and time consuming work a table of squares is prepared and presented
in the “Numerical tables” at the end of this book.
 In this table the first column headed by x lists numbers starting from
1.0. The remaining columns are headed respectively by the digits 0 to 9.
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 Grade 8 Mathematics [SQUARES, SQUARE ROOTS, CUBES AND CUBE ROOTS ]
Now if you want to determine the square of a number for example 2.54 proceed
as follows.
Step i. Under the column headed by x, find the row with 2.5.
Step ii. Move to the right along the row until you get the column under 4, (or
find the column headed by 4).
Step iii. Then read the number at the intersection of the row in (i) and the column
( ii), (see the illustration below).
Hence (2.54)2 = 6.452
X 0 1 2 3 4 5 6 7 8 9
1.0

2.0
2.5 6.452
3.0

4.0

5.0

6.0

7.0

8.0

9.0
Figure 1.5 Tables of squares

Note that the steps (i) to (iii) are often shortened by saying “2.5 under 4”.
 Mostly the values obtained from the table of squares are only
approximate values which of course serves almost for all practical
purposes.

Group work 1.2

Discuss with your group.
Find the square of the number 8.95
a) use rough calculation method.
b) use the numerical table.

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 Grade 8 Mathematics [SQUARES, SQUARE ROOTS, CUBES AND CUBE ROOTS ]

c) by calculating the exact value of the number.

d) compare your answer from “a” to “c”.
e) write your generalization.

Example9: Find the square of the number 4.95.

Solution: Do rough calculation and compare your answer with the value
obtained from the table.
i. Rough calculation
4.95 ≈ 5 and 52 = 25
(4. 95)2 ≈ 25
ii. Value obtained from the table
i) Find the row which starts with 4.9.
ii) Find the column headed by 5.
iii) Read the number, that is (4. 95)2 at the intersection of the row in (i) and
the column in ( ii) ; ( 4. 95)2 = 24. 50.
iii. Exact Value
Multiply 4.95 by 4.95
4.95 × 4.95 = 24. 5025
Therefore (4.95)2 = 24. 5025.
This example shows that the result obtained from the “Numerical table” is
an approximation and more closer to the exact value.

Exercise 1C
1. Determine whether each of the following statements is true or false.
a) (2.3)2 = 5.429 c) (3.56)2 = 30. 91 e) (5. 67)2 = 32
b) (9.1) 2 = 973. 2 d) (9.90)2 = 98. 01 f) (4. 36)2 = 16.2
2. Find the squares of the following numbers from the table.
a) 4. 85 c) 88.2 e) 2. 60 g) 498 i) 165
3
b) 6.46 d) 29. 0 f) h) 246
2
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 Grade 8 Mathematics [SQUARES, SQUARE ROOTS, CUBES AND CUBE ROOTS ]

1.2 The Square Root of a Rational Number

Group Work 1.3
Discuss with your Friends
Find the square root of each of the following numbers.
1 64 4 25
a) 81 b) 324 c) d) e) f)
4 49 49 625
 So far you have studied the meaning of x2 when x is a rational number. It
is now logical to ask, whether you can go in the reverse (or in the
opposite direction) or not. In this sub unit you will answer this and related
questions in a more systematized manner.

Definition 1.3: Square roots

For any two rational numbers a and b if a2 = b, then a is called the
square root of b.

Example 10:
a) 4 is the square root of 16, since 42 = 16.
b) 5 is the square root of 25, since 52 = 25.
c) 6 is the square root of 36, since 62 = 36.

Example 11: The area of a square is 49m2. What is the length of each side?
Solution:
S
×w=A

s × s = 49 m2
s2 = 49 m2 A = 49m2
S S
s = 7m

S
Figure 1.6
The length of each side is 7 meters. This is one way to express the mathematical
relationship “7 is the square root of 49” because 72 = 49.
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 Grade 8 Mathematics [SQUARES, SQUARE ROOTS, CUBES AND CUBE ROOTS ]

Note:
i. The notion “square root” is the inverse of the notion “square of
a number”.
ii. The operation “extracting square root” is the inverse of the operation
“squaring”.
iii. In extracting square roots of rational numbers, first decompose the
number into a product consisting of two equal factors and take one of
the equal factors as the square root of the given number.
iv. The symbol or notation for square root is “ ” it is called radical sign.
v. For b ≥ 0, the expression is called radical b and the number b is
called a radicand.
vi. The relation of squaring and square root can be expressed as follows:

a Squaring a2 = b, for b ≥ 0

a Square root a2

vii. a is the square root of b and written as a = .

Example 12: Find the square root of x, if x is:

a) 100 b) 125 c) 169 d) 256 e) 625 f) 1600

Solution a) x = 100 = 10 × 10
x = 10 2, thus the square root of 100 is 10.
b) x = 225 = 15 × 15
x = 152, thus the square root of 225 is 15.
c) x = 169 = 13 × 13
x = 132, thus the square root of 169 is 13.
d) x = 256 = 16 × 16
x = 162, thus the square root of 256 is 16.
e) x = 625 = 25 × 25
x = 252, thus the square root of 625 is 25.
f) x = 1600 = 40 × 40
x = 402, thus the square root of 1600 is 40.

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 Grade 8 Mathematics [SQUARES, SQUARE ROOTS, CUBES AND CUBE ROOTS ]

Exercise 1D
1. Determine whether each of the following statements is true or false.
900 30
a) 0 =0 d) − 121 = −11 g) − =-
961 31
36 1
b) 25 = ± 5 e) − =
324 3
1 1 324 18
=± f) =
c) 4 2 625 25
2. Find the square root of each of the following numbers.
a) 121 c ) 289 e) 400 g) 484
b) 144 d) 361 f) 441 h) 529
3. Evaluate each of the following.
1
a) d) - 576 g) 729
25
1 529
b) e) h) − 784
81 625
36 16
c) - f) − 676 i)
144 25

Challenge Problems
x x2 y 2
4. If = −2. Find +
y y 2 x2
5. Simplify: (81)2 + (49)2
6. If x = 16 and y = 625. Find (2 x + y ) .
2

Definition 1.4 : If a number y ≥ 0 is the square of a positive number x

(x ≥ 0), then the number x is called the square root of y.
This can be written as x = y.

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 Grade 8 Mathematics [SQUARES, SQUARE ROOTS, CUBES AND CUBE ROOTS ]

Example13: Find
a) 0.01 c) 0.81 e) 0.7921 g) 48.8601
b) 0.25 d) 0.6889 f) 0.9025

Solution
a) 0.01 = 0.1× 0.1 = 0.1 e) 0.7921 = 0.89 × 0.89 = 0.89
b) 0.25 = 0.5 × 0.5 = 0.5 f) 0.9025 = 0.95 × 0.95 = 0.95
c) 0.81 = 0.9 × 0.9 = 0.9 g) 48.8601 = 6.99 × 6.99 = 6.99
d) 0.6889 = 0.83 × 0.83 = 0.83

Exercise 1E
Simplify the square roots.
a) 35.88 c) 89.87 e) 62.25
b) 36.46 d) 99.80 f) 97.81

1.2.1 Square Roots of Perfect Squares

Group work 1.4
Discuss with your group.
1. Find the prime factorization of the following numbers by using the
factor trees.
a) 64 c) 121 e) 324 g) 625 i) 700
b) 81 d) 289 f) 400 h) 676

Note: The following properties of squares are important:

2
a a2
(ab) = a ×
2 2 b2 and   = (where b ≠ 0).
b b2
2
3 32 9
Thus ( 2×3)2
= × = 36 and   = 2 = .
22 32
4 4 16
Remember a number is called a perfect square, if it is the square of a
rational number.
The following properties are useful to simplify square roots of numbers.
Properties of Square roots, for a ≥ 0, b ≥ 0.

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 Grade 8 Mathematics [SQUARES, SQUARE ROOTS, CUBES AND CUBE ROOTS ]

If a and b represent rational numbers, then

a a
ab = a b and = where b ≠ 0.
b b

Example: 14 Determine whether each of the following numbers is a perfect

square or not.
16
a) 36 c) 81 e) g) 11
625
49
b) 49 d) f) 7
25
Solution: a) 36 is a perfect square, because 36 = 62 .
b) 49 is a perfect square, because 49 = 72.
c) 81 is a perfect square, because 81 = 92.
2
49 49  7 
d) is a perfect square, becaus =   .
25 25  5 
2
16 16  4 
e) is a perfect square, because =  .
625 625  25 
f) 7 is not a perfect square since there is no rational number whose
square is equal to 7. In other words there is no rational number n
such that n2 = 7.
g) 11 is not a perfect square since there is no rational number
whose square is equal to 11. In short there is no rational number
n such that n2 = 11.
Example 15: Use prime factorization and find the square root of each of the
following numbers.
a) 324 b) 400 c) 484
Solution: a) 324 = 2 × 2 × 3 × 3 × 3 × 3
Now arrange the factors so that 324 is a product of two identical
sets of prime factors.
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 Grade 8 Mathematics [SQUARES, SQUARE ROOTS, CUBES AND CUBE ROOTS ]

i.e 324 = (2 × 2 × 3×3 × 3×3) 324

= (2 × 3 × 3) × (2 × 3 × 3)
2 162
= 18 × 18 = 182
So, 324 = 18 × 18 = 18 2
81

3 27

b) 400 = (2 × 2 × 5) × (2 × 2 × 5) 3 9
400
Now arrange the factors so that 400 is a 3 3
product of two identical sets of prime
2
factors. 200

i.e 400 = ( 2× 2 × 5)×(2 × 2×5) 100

= 20 × 20 2
= 202 2 50
So 400 = 20 × 20
2 25
= 20
c) 484 = 2×2× 11×11, now arrange the 484 5
factors so that 484 is a product of two 5
identical sets of prime factors. 2 242
i.e 484 = 2×2 × 11×11
= (2 × 11) × (2 × 11) 121
2
= 22 × 22 = 222
11
So 484 = 22 × 22 11

= 22
Exercise 1F
1. Determine whether each of the following statements is true or false.
0
a) 64 × 25 = 64 × 25 =0
d) 1296
64 1296
b) =4 e) =1
4 0
32 32 729 27
c) = f) =
64 64 1444 38

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 Grade 8 Mathematics [SQUARES, SQUARE ROOTS, CUBES AND CUBE ROOTS ]
2. Evaluate each of the following.
1296 81
a) 0.25 c) e)
1024 324
625 144
b) 0.0625 d) f)
1024 400

Challenge Problem

3. Simplify a) 625 - 0 − 172 − 3

b) 81 × 625
2
 1 
c)  
 64 
4. Does every number have two square roots? Explain.
5. Which of the following are perfect squares?
{0, 1, 4, 7, 12, 16, 25, 30, 36, 42, 49}
6. Which of the following are perfect squares?
{50, 64, 72, 81, 95, 100, 121, 140, 144, 169}
7. Copy and complete.
a) 32 + 42 + 122 = 132 c) 62 + 72 + =
b) 52 + 62 + = d) x2 + (x + 1)2 + =

Using the square root table

The same table which you can use to determine squares of numbers can be used
to find the approximate square roots, of numbers.

Example 16: Find 17.89 from the numerical table.

Solution:
Step i. Find the number 17.89 in the body of the table for the function
y = x2 .

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 Grade 8 Mathematics [SQUARES, SQUARE ROOTS, CUBES AND CUBE ROOTS ]

Step ii. On the row containing this number move to the left and read
4.2 under x.
These are the first two digits of the square root of 17. 89
Step iii. To get the third digit start from 17.89 move vertically up ward
and read 3.
Therefore 17.89 ≈ 4.23
X 0 1 2 3 4 5 6 7 8 9
1.0

2.0
2nd
3.0

4.0
4.2 1st 17.89
5.0

6.0

7.0

8.0

9.9

Figure 1.7 Table of square roots

If the radicand is not found in the body of the table, you can consider the number
which is closer to it.
Example 17: Find 10.59

Solution: i) It is not possible to find the number 10. 59 directly in the table of
squares. But in this case find two numbers in the table which are
closer to it, one from left (i.e. 10.56) and one form right (10.63)
that means 10.56 < 10. 59 < 10.63.
ii) Find the nearest number to (10.59) form those two numbers. So
the nearest number is 10.56 thus 10. 59 ≈ 10.56 = 3.25.

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 Grade 8 Mathematics [SQUARES, SQUARE ROOTS, CUBES AND CUBE ROOTS ]

Example 18: Find 83.60

Solution: i. It is not possible to find the number 83. 60 directly in the table of
squares. But find two numbers which are closer to it, one from
left (i.e. 83.54) and one from right (i.e 83.72) that means
83.54 < 83. 60 < 83.72.
ii. Find the nearest number from these two numbers. Therefore the
nearest number is 83. 54, so 83. 60 ≈ 83.54 = 9.14.

Note: To find the square root of a number greater than 100 you can use
the method illustrated by the following example.

Example 19: Find the square root of each of the following.

a) 6496 b) 9801 c) 9880 d) 9506

Solution:
a) 6496 = 64.96 ×100 c) 9880 = 98.80 × 100
= 64.96 × 100 = 98.80 × 100
= 8.06×10 = 9.94×10
= 80.6 = 99.4

b) 9801 = 98.01 × 100 d) 9506 = 95.06 × 100

= 98.01 × 100 = 95.06 × 100

= 9.90×10 = 9.75×10
= 99 = 97.5
Example 20: Find the square root of each of the following numbers by using
the table
a) 98.41 c) 984100 e) 0.009841

b) 9841 d) 0.9841 f) 0.00009841

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Grade 8 Mathematics [SQUARES, SQUARE ROOTS, CUBES AND CUBE ROOTS ]

Solution:
a) 98.41 = 9.92 1
e) 0.009841 = 98.41×
10,000
b) 9841 = 98.41 × 100 1
= 98.41 ×
= 98.41× 100 10,000
= 9.92 × 10 1
= 99.2 = 9.92× 100
= 0.0992
c) 984100 = 98.41 × 10000 f) 0.00009841 = 98.41 ×
1
1,000,000
= 98.41 × 10000
1
= 9.92 ×100 = 9.92× 1, 000
= 992
= 0.00992
1
d) 0.9841 = 98.41 ×
100
1
= 98.41 ×
100
1
= 9.92 × 10
= 0.992
Exercise 1G
1. Find the square root of each of the following numbers from the table.
a) 15.37 d) 153.1 g) 997 j) 5494
b) 40.70 e) 162.8 h) 6034 k) 5295
c) 121.3 f) 163.7 i) 6076 l) 3874
2. Use the table of squares to find approximate value of each of the
following.

a) 6.553 c) 24.56

b) 8.761 d) 29. 78

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 Grade 8 Mathematics [SQUARES, SQUARE ROOTS, CUBES AND CUBE ROOTS ]

1.3 Cubes and Cube Roots

1.3.1 Cube of a Number
If the number to be cubed is ‘a’, then the product a × a × a which is usually
written as a3 and is read as ‘a’ cubed. For example 3 cubed gives 27 because
3 × 3 × 3 = 27.
The product 3×3×3 can be written as 33 and which is read as 3 cubed.

Activity 1.2
Discuss with your friends
1. Copy and complete this Table 1.3
Number of small cubes
Standard Factor form Power form
form
a)

1 1×1×1 13

b)
8 2× × . 23

c)
27 × × .

Figure 1.8

2. a) Which of these numbers are cubic numbers?

64 100 125 216 500 1000
1728 3150 4096 8000 8820 15625
b) Write the cubic numbers from part (a) in power form.
3. Find a3 in each of the following.
a) a = 2 c) a = 10 e) a = 0.5
b) a = -2 d) a = f) a = 0.25
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 Grade 8 Mathematics [SQUARES, SQUARE ROOTS, CUBES AND CUBE ROOTS ]

Definition 1.5: A cube number is the result of multiplying a rational

number by itself, then multiplying by the number
again.

For example, some few cube numbers are:

a) 1 × 1 × 1 = 1 is the 1st cube number.
b) 2 × 2 × 2 = 8 is the 2nd cube number.
c) 3 × 3 × 3 = 27 is the 3rd cube number.
a) b) c)
Figure 1.9 A cube number can be
shown as a pattern of cubes
Example 21: Find the numbers whose cube are the following.
a) 4,913 b) 6,859 c) 9,261 d) 29,791
Solution:
a) 4,913 = 17 × 17 × 17 = 173 c) 9,261 = 21 × 21 × 21 =213
b) 6,859 = 19 × 19 × 19 =193 d) 29,791 = 31 × 31 × 31 = 313

Example 22: Identify the base, exponent, power form and standard numeral
form:
a) 403 b) 433
Exponent
64,000 base
Solution: a) Standard
numeral form Power form
Exponent
79,507 base
b) Standard
numeral form Power form

Example 23: In Table1.4 below integers are as values of x, find x3 and

complete the table 1.4.
x -4 -3 -2 -1 0 1 2 3 4 5 6
x3

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 Grade 8 Mathematics [SQUARES, SQUARE ROOTS, CUBES AND CUBE ROOTS ]
Solution:
When x = -4, x3 = (-4)3 = -4 × -4 × -4 = -64
When x = -3, x3 = (-3)3 = -3 × -3 × -3 = -27
When x = -2, x3 = (-2)3 = -2 × -2 × -2 = -8
When x = -1, x3 = (-1)3 = -1 × -1 × -1 = -1
When x = 0, x3 = 03 = 0 × 0 × 0 = 0
When x = 1, x3 = 13 = 1 × 1 × 1 = 1
When x = 2, x3 = 23 = 2 × 2 × 2 = 8
When x = 3, x3 = 33 = 3 × 3 × 3 = 27
When x = 4, x3 = 43 = 4 × 4 × 4 = 64
When x = 5, x3 = 53 = 5 × 5 × 5 = 125
When x = 6, x3 = 63 = 6 × 6 × 6 = 216
Lastly you have:
x -4 -3 -2 -1 0 1 2 3 4 5 6
x3 -64 -27 -8 -1 0 1 8 27 64 125 216

The examples above illustrate the following theorem. This theorem is called
existence theorem.

Theorem 1.2: Existence theorem

For each rational number x, there is a rational
number y such that y = x3.

Rough calculations could be used for approximating and checking the results in
cubing rational numbers. The following examples illustrate the situation.

Example 24: Find the approximate values of x3 in each of the following.

a) x = 2.2 b) x = 0.065 c) x = 9.54
Solution:
a. 2.2 ≈ 2 thus (2.2)3 ≈ 23 = 8
b. 0.065 ≈ 0.07 thus (0.065)3 ≈
= 0.000343
c. 9.54 ≈ 10 thus (9.54) ≈ 10 = 1,000
3 3

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 Grade 8 Mathematics [SQUARES, SQUARE ROOTS, CUBES AND CUBE ROOTS ]

Exercise 1H
1. Determine whether each of the following statements is true or false.
3
e)   =
4 64
a) 43 = 16 × 4 c) (-3)3 = 27
3 125
3
d)   = 64 = 4
3 3 27 3
b) 4 = 64 f)
4 16

2. Find x3 in each of the following.

-1
a) x = 8 c) x = -4 e) x =
5
1
b) x = 0.4 d) x = - f) x = -0. 2
4
3. Find the approximate values of x3 in each of the following.
a) x = -2.49 c) x = 2.98
b) x = 2.29 d) x = 0.025

Challenge Problem
4. The dimensions of a cuboid are 10xcm
4xcm, 6xcm and 10xcm. Find
a) The total surface area 4xcm
b) The volume 6xcm
Figure 1.10 Cuboid

Table of Cubes
Activity 1.3
Discuss with your friends
Use the table of cubes to find the cubes of each of the following.
a) 2.26 c) 5.99 e) 8.86 g) 9.58 i) 9.99
b) 5.12 d) 8.48 f) 9.48 h) 9.89 j) 9.10

To find the cubes of a rational number when it is written in the form of a decimal
is tedious and time consuming work. To avoid this tedious and time consuming
work, a table of cubes is prepared and presented in the “Numerical Tables” at
the end of this textbook.

25
 Grade 8 Mathematics [SQUARES, SQUARE ROOTS, CUBES AND CUBE ROOTS ]
In this table the first column headed by ‘x’ lists numbers starting from 1.0. The
remaining columns are headed respectively by the digit 0 to 9.0. Now if we want
to determine the cube of a number, for example 1.95 Proceed as follows.
Step i. Find the row which starts with 1.9 (or under the column headed by x).
Step ii. Move to the right until you get the number under column 5 (or find the
column headed by 5).
Step iii. Then read the number at the intersection of the row in step (i) and the
column step (ii) therefore we find that (1.95)3 = 7.415. See the illus-
tration below.
X 0 1 2 3 4 5 6 7 8 9
1.0
1.9 7.415
2.0

3.0

4.0

5.0

6.0

7.0

8.0

9.0
Figure 1.11 Tables of cubes

Note that the steps (i) to (iii) are often shortened saying “1.9 under 5”
Mostly the values obtained from the table of cubes are only a approximate values
which of course serves almost for all practical purposes.
Group work 1.5
Find the cube of the number 7.89.
a) use rough calculation method.
b) use the numerical table.
c) by calculating the exact value of the number.
d) compare your answer from “a” to “c”.
e) write your generalization.
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 Grade 8 Mathematics [SQUARES, SQUARE ROOTS, CUBES AND CUBE ROOTS ]

Example 25: Find the cube of the number 6.95.

Solution:
Do rough calculation and compare your answer with the value obtained
from the table.
i. Rough Calculation
6.95 ≈ 7 and 73 = 343
( 6.95)3 ≈ 343
ii. Value Obtained from the Table
Step i. Find the row which starts with 6.9
Step ii. Find the column head by 5
Step iii. Read the number, that is the intersection of the row in (i) and the
column (ii), therefore ( 6. 95)3 = 335.75
iii. Exact Value
Multiply 6.95 ×6.95×6.95 = 335.702375
so (6.95)3= 335. 702375
This examples shows that the result obtained from the numerical tables is an
approximation and more closer to the exact value.

Exercise 1I
1. Use the table of cubes to find the cube of each of the following.
a) 3.55 c) 6.58 e) 7.02 g) 9.86 i) 9.90 k) 9.97
b) 4.86 d) 6.95 f) 8.86 h) 9.88 j) 9.94 l) 9.99

1.3.2 Cube Root of a Number

Group work 1.6
Discuss with your group.
1. In Figure 1.12 to the right, the volume
s
of a cube is 64 m3. What is the length
of each edge? V = 64m3
2. Can you define a “cube root” of
s
a number precisely by your own word?
s
3. Find the cube root of 12,167 × 42,875.
Figure 1.12 Cube
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 Grade 8 Mathematics [SQUARES, SQUARE ROOTS, CUBES AND CUBE ROOTS ]

Definition 1.6: The cube root of a given number is one of the

three identical factors whose product is the
given number.

Example 26:
a) 0× 0×0 = 0, so 0 is the cube root of 0.
b) 5×5×5 = 125, so 5 is the cube root of 125.
1 1 1 1 1 1
× × = , so is the cube root of .
c) 5 5 5 125 5 125

Note: i) 43 = 64, 64 is the cube of 4 and 4 is the cube root of 64

i.e 3
64 = 4.
ii) is a radical sign.
Index

Or symbolically: Radicand

Radical sign
iii) The relation of cubing and extracting cube root can be
expressed as follows:
a Cubing a3 = b

a Cube root a3 = b

iv) a is the cube root of b and written as a = .

When no index is written, the radical sign indicates a square root.

For example 3 512
is read as " the cube root of 512" .
The number 3 is called the index and 512 is called the radicand.
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 Grade 8 Mathematics [SQUARES, SQUARE ROOTS, CUBES AND CUBE ROOTS ]

Cube Roots of Perfect Cubes

Group work 1.7
Discuss with your group
1. Find the cube root of the perfect cubes.
1
a) 3
27 b) 3 c) 3 125 d) 3
- 64
27
2. Which of the following are perfect cubes?
{42,60,64,90,111,125,133,150,216}
3. Which of the following are perfect cubes?
{3,6,8,9,12,27,y3, y8, y9, y12, y27}

Note: The following properties of cubes are important: (ab)3 = a3×b3

3
a a3
and   = 3 (where b ≠ 0).
b b
3
3 3 3 27
Thus ( 2 × 2)3= × = 8 × 8 = 64 and
23 23   = = .
4 4 3 64
A number is called a perfect cube, if it is the cube of a rational number.

Definition 1.7: A rational number x is called a perfect cube if and only if x = n3

for some n ∈ .

Example 27:
1 = 13, 8= 23, 27 = 33, 64 = 43 and 125 = 53.
Thus 1, 8, 27 , 64 and 125 are perfect cubes.

Note: A perfect cube is a number that is a product of three identical factors of

a rational number and its cube root is also a rational number.

Example 28: Find the cube root of each of the following.

1
a) 216 b) c) -64 d) -27
8
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 Grade 8 Mathematics [SQUARES, SQUARE ROOTS, CUBES AND CUBE ROOTS ]
Solution:
a) 216 = 6 × 6 × 6 = 6 − 64 = 3 − 4 × −4 × −4 = −4
3 3 3
c)
1 3 1 1 1 1
b) 3 = × × d) 3
− 27 = 3 − 3 × −3 × −3 = - 3 -3
8 2 2 2 = 2

Exercise 1J
1. Determine whether each of the following statements is true or false.
1 1 −1 1
a) 3 17281 = 26 b) 3 = c) 3
− 64 = ± 4 d) 3 =
729 90 625 20

2. Find the cube root of each of the following.

1 g) 3
x3 −1
a) 0 c) 1000 e) i)
729 1331
b) 343 d) 0. 001 −1 0
f) 9261 h) 27

3. Evaluate each of the following.

a) 3 − 27 1 f) 3
− 1000 h) − − 64
3
d) − 3
1 64 −1
b) 3 −8 − 27 3
8 g) 3 i) 125
e) 3
3
27 27 64
c)

Challenge Problem
2 5 ×7 2
4. Simplify: a) 5 18 − 3 72 + 4 50 b)
14 × 45
5. Simplify the expressions. Assume all variables represent positive rational
number.
3
y5 c) 16a 3 3
x5 y 11
a) 3 e) g) 3
27y 3 x2 y2

b) 3 16z 3 3
b4 f)
3
15m 4 n 22 h) 3
20s 15 t 11
d)
27b

30
 Grade 8 Mathematics [SQUARES, SQUARE ROOTS, CUBES AND CUBE ROOTS ]

Table of Cube Roots

The same table which you can used to determine cubes of numbers can be used
to find the approximate cube roots, of numbers.
3
Example 29: Find 64.48 from the numerical table.
Solution: Find the value using rough calculations.
64.48 ≈ 64; 3
64.48 ≈ 3
64
≈ 4× 4× 4 = 4
3

Step i: Find the number 64. 48 in the body of the table for the relation
y = x 3.
Step ii: Move to the left on the row containing this number to get 4.0
under x. These are the first two digits of the required cube root
of 64. 48.
Step iii: To get the third digits start from 64.48 and move vertically
upward and read 1 at the top.
There fore 3
64.48 ≈ 4.01 .

X 0 1 2 3 4 5 6 7 8 9
1.0

2.0

3.0

4.0 64.48

5.0

6.0

7.0

8.0

9.0
Figure 1.13 Tables of cube roots

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 Grade 8 Mathematics [SQUARES, SQUARE ROOTS, CUBES AND CUBE ROOTS ]
Example 30:
In Figure 1.14 below, find the exact volume of the boxes.
(a) (b)

w=



Figure 1.14
Solution
a) V =  × w × h
But the box is a cube, all the side of a cube are equal.
i.e  = w = h = s
V = s × s × s = s3
V = 3 5cm × 3 5cm × 3 5cm
V= ( 5cm )
3 3

3
 1
V= 5 3 
 
 
V = 5 cm3
Therefore, the volume of the box is 5cm3.
b) V = ℓ × w × h
V = 14m × 2m × 7 m
V = 14 m × 14 m 2
V = ( 14×14 )m 3
V = 14m3
Therefore, the volume of the box is 14m3.

Exercise 1k
1. Use the table of cube to find the cube root of each of the following.
a) 32.77 c) 302.5 e) 3114
b) 42.6 d) 329.5 f) 3238
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Grade 8 Mathematics [SQUARES, SQUARE ROOTS, CUBES AND CUBE ROOTS ]

Summary for unit 1

1. The process of multiplying a number by itself is called squaring the
number.
2. For each rational number x there is a rational number y (y ≥ 0) such that
x2 = y.
3. A square root of a number is one of its two equal factors.
4. A rational number x is called a perfect square, if and only if x = n2 for
some n ∈ .
5. The process of multiplying a number by itself three times is called cubing
the number.
6. The cube root of a given number is one of the three identical factors whose
product is the given number.
7. A rational number x is called a perfect cube, if and only if x = n3 for some
n∈ .
8. index

radicand
radical sign
9. The relationship of squaring and square root can be expressed as follows:

a Squaring a2 = b where b ≥ 0

Square root
a a2
• a is the square root of b and written as a =
10. The relationship of cubing and cube root can be expressed as follows:
a Cubing a3 = b

a Cube root a3 = b

• a is the cube root of b and written as a =

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 Grade 8 Mathematics [SQUARES, SQUARE ROOTS, CUBES AND CUBE ROOTS ]

Miscellaneous Exercise 1
1. Determine whether each of the following statements is true or false.
3 8 −3 2 125 1 3
a) = c) = 2.5 e) =
2 32 4 5 8 3 3
−1
7
1
=
64 f) 0.25 =
b) 1 2
9 9 d) × 63 = ±3
7 g) 0.0036 = 0.06
2. Simplify each expression.
36 25  3 
a) c) 8 e) 2 2  + 2
324 4  2 
50 16
b) d)
2 4
3. Simplify each expression.
a) 600 d) 3 ( 3+ 6 ) g) 2 ( 2+ 6 )
b) 50 + 18 e) 19 2 h) 2 ( 3+ 8 )

c) 5 6( ) 2
f) 64 + 36
4. Simplifying radical expressions ( where x ≠ 0).
3
32 12x 4
a) c) e) 3 p17q18
3
−4 3x
3
162x 5
3
b) 3 2 d) 80n 5
3x

5. Study the pattern and find a and b

34 36 35 311 37 a 32 b
32

Figure 1.15

34
 Grade 8 Mathematics [SQUARES, SQUARE ROOTS, CUBES AND CUBE ROOTS ]

6. Study the pattern and find a, b, c and d.

2x2 a

6 6x3 -5x5
b
X3
-5x4 c
2x2
2x5

4x7 d

Figure 1.16

7. An amoeba is a single cell animal. When

the cell splits by a process called “fission”
there are then two animals. In a few hours
a single amoeba can become a large
colony of amoebas as shown to the right.
Figure 1.17
Number of splits Number of amoeba cells
0 1
1 2
2 4 = 2 × 2 = 22
3 8 = 2 × 2 × 2 = 23
How many amoebas would there be
a) After 4 splits? c) after 6 splits?
b) After 5 splits? d) after 10 splits?
8. Using only the numbers in the circular table, write down, all that are:
a) square numbers
9261 1225 27
b) cube numbers 625
324 441 4096

Figure 1.18

35
 Grade 8 Mathematics [SQUARES, SQUARE ROOTS, CUBES AND CUBE ROOTS ]

9. Find the exact perimeter of a square whose side length is 5 3 16 cm.

10. The length of the sides of a cubes is
related to the volume of the cube x
according to the formula: x = 3 V .
a) What is the volume of the cube if the
side length is 25cm. x
x
b) What is the volume of the cube if the Figure 1.19 Cube
side length is 40 cm.
11. In Figure 1.20 to the right find:
a) the surface area of a cube. 4cm

b) the volume of a cube.

c) compare the surface area and the 4cm
4cm
volume of a given a cube
Figure 1.20 Cube
12. Prove that the difference of the square of an even number is multiple of 4.
13. Show that 64 can be written as either 26 or 43.
14. Look at this number pattern.
72 = 49
672 = 4489
6672 = 444889
66672 = 44448889
This pattern continues.
a) Write down the next line of the pattern.
b) Use the pattern to work out 66666672.
15. Find three consecutive square numbers whose sum is 149.
16. Find the square root of 25x2 − 40xy + 16y2.
17. Find the square root of
18. Find the cube root of 27a3 + 54a2b + 36ab2 + 8b3.

36
UNIT
2 FURTHER ON
WORKING WITH
VARIABLES

Unit outcomes
After Completing this unit, you should be able to:
 solve life related problems using variables.
 multiply binomial by monomial and determine the product
of binomials.
 determine highest common factor of algebraic
expressions.

Introduction
By now you are well aware of the importance of variables in mathematics. In this
unit you will learn more about variables, specially you will learn about
mathematical expression, its component parts and uses of variables in formulas
and solving problems. In addition to these you will study special expressions
known as binomials and how to perform addition and multiplication on them.

37
 Grade 8 Mathematics [FURTHER ON WORKING WITH VARIABLES ]

2.1. Further on Algebraic Terms and Expressions

2.1.1 Use of variables in formula
Group Work 2.1
Discuss with your friends
1. What a variable is?
2. Find what number I am left with if
a. I start with x, double it and then subtract 6.
b. I start with x, add 4 and then square the result.
c. I start with x, take away 5, double the result and then divide
by 3.
d. I start with w, subtract x and then square the result.
e. I start with n add p, cube the result and then divide by a.
3. Translate the following word problems in to mathematical
expression.
a. Eighteen subtracted from 3.
b. The difference of -5 and 11.
c. Negative thirteen subtracted from – 10.
d. Twenty less than 32.
4. Describe each of the following sets using variables.
a. The set of odd natural numbers.
b. The solution set of 3x – 1 ≥ 4.
c. The solution set of x + 6 = 24.
d. The set of all natural number less than 10.

Definition 2.1: A variable is a symbol or letter such as x, y and z

used to represent an unknown number (value).

Example 1: Describe each of the following sets using variables.

a. The solution set of 3x – 5 > 6.
b. The solution set of 2x + 1 = 10.

38
 Grade 8 Mathematics [FURTHER ON WORKING WITH VARIABLES ]

Solution
a) 3x – 5 > 6 ……………. Given inequality
3x – 5 + 5 > 6 + 5 ……... Adding 5 from both sides
3x > 11 ………………… Simplifying
3x 11
> ………………….. Dividing both sides by 3
3 3
11
x>
3
 11
The solution set of 3x – 5 > 6 is x : x >  .
 3
b) 2x + 1 = 10 ………….. Given equation.
2x + 1 – 1 = 10 – 1 ……Subtracting 1 from both sides.
2x = 9 …….. ... Simplifying.
2x 9
= ………… Dividing both sides by 2.
2 2
9
x=
2
9 9 
The solution set of 2x + 1 = 10 is x = or S.S =  
2 2
Example 2: Find the perimeter of a rectangle in terms of its length  and width w.
Solution Let P represent the perimeter of the rectangle.
Then P =

=+w++w D C
= 2 + 2w
w w
= 2( + w)
A B


Figure 2.1

Example 3: The volume of a rectangular prism

equals the product of the numbers
h
which measures of the length, the
width and the height. Formulate the

statement using variables. w
Figure 2.2 A rectangular prism

39
 Grade 8 Mathematics [FURTHER ON WORKING WITH VARIABLES ]

Solution let  represent the length, w the width and h the height of the
prism. If V represents the volume of the prism, then
V=×w×h C
V = wh

Example 4: Express the area of a triangle in h

terms of its base ‘b’ and altitude
‘h’.
A B
Solution Let “b” represent the base and “h” b
the height of the triangle. Figure 2.3 triangle

A = bh
Example 5: The area of a trapezium (see Figure 2.4) below can be given by the
formula A = where A = area, h = height, b 1 = upper
base and b 2 = lower base. If the area is 170 cm2, height 17cm
and b 2 = 12cm then:
a) Express b 1 in terms of the b1

other variables in the

formula for A. h

b) Use the equation you

obtained to find b 1 . b2
Solution Figure 2.4 Trapezium

a) h (b 1 + b 2 ) = A …. Given equation
h (b 1 + b 2 ) = 2A ….. Multiplying A by 2
b1 + b2 = ….. Dividing both sides by h
b1 = - b 2 ….. Subtracting b 2 from both sides
-
b1 = …… Simplifying
b) For (a) above we have
-
b1 =
-
b1 =
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-
b1 =
b 1 = 8cm
Therefore the upper base (b 1 ) is 8cm.
 Check: A = h(b 1 + b 2 ) when b 1 = 8cm

170cm2

170 cm2
170 cm2 = 170 cm2 (True)

Exercise 2A
Solve each of the following word problems.
1. The perimeter of a rectangular field is 1000m. If the length is given as x,
find the width in terms of x.
2. Find
i) The perimeter of a square in terms of its side of length ″s″ unit.
ii) The area of a square interms of its side of length ″s″ units.
3. Express the volume of the cube in
Figure 2.5.
2k-1

2k-1
2k-1
Figure 2.5 Cube

4. The area of a trapezoid is given by the formula A = then give

the height h interms of its bases b 1 & b 2 .
5. A man is 8x years old now. How old he will be in:
a. 10 years time? b. 6x years time? c. 5y years time?
6. How many days (d) are there in the given number of weeks (w) below?
a. 6 weeks c. y weeks
b. 104 weeks d. 14 weeks

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2.1.2 Variables, Terms and Expressions

Activity 2.1
Discuss with your teacher before starting the lesson.
1. What do we mean by like terms? Given an example.
2. Are 7a3, 5a2 and 12a like terms? Explain.
3. What is an algebraic expression?
4. What is a monomial?
5. What is a binomial?
6. What is a trinomial?
7. What are unlike terms? Give an example.

Definition 2.2: Algebraic expressions are formed by using numbers,

letters and the operations of addition, subtraction,
multiplication, division, raising to power and taking
roots.

Some examples of algebraic expressions are:

x + 10, y – 16, 2x2 + 5x – 8, x – 92, 2x + 10, etc.
Note:
i. An algebraic expression that contains variables is called an
expression in certain variables. For examples the expression
7xy + 6z is an algebraic expression with variables x, y and z.
ii. An algebraic expression that contains no variable at all is called
constant. For example, the algebraic expression 72 - 16π is constant.
iii. The terms of an algebraic expression are parts of the expression that
are connected by plus or minus signs.

Examples 6: List the terms of the expression 5x2 – 13x + 20.

Solution: The terms of the expression 5x2 – 13x + 20 are 5x2, - 13x, and 20.

Definition 2.3: An algebraic expression in algebra which contains

one term is called a monomial.

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−2
Example 7: 8x, 13a2b2, , 18xy, 0.2a3b3 are all monomial.
3

Definition 2.4: An algebraic expression in algebra which

contains two terms is called a binomial.

Examples 8: 2x + 2y, 2a – 3b, 5p2 + 8, 3x2 + 6, n3 – 3 are all binomial.

Definition 2.5: An algebraic expression in algebra which

contains three terms is called a trinomial.

Examples 9: 4x2 + 3x + 10, 3x2 – 5x + 2, ax2 + bx + c are all trinomial.

Definition 2.6: Terms which have the same variables, with the corresponding
variables are raised to the same powers are called like terms;
other wise called unlike terms.

For example:
Like terms Unlike terms
34xy and -8xy 12xy and 6x …… Different variables.
18p2q3 and p2q3 8p2q3 and 16p3q2 …Different power
5w and 6w 10w and 20 …… Different variables.
7 and 20 14 and 10a…… Different variables.

Example 10: Which of the pairs are like terms: 80ab and 70b or 4c2d2, and
-6c2 d2.

Solution 4c2d2 and -6c2d2 are like terms but 80 ab and 70 b are unlike
terms.

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Note:
i. Constant terms with out variables, (or all constant terms) are like
terms.
ii. Only like terms can be added or subtracted to form a more
simplified expression.
iii. Adding or subtracting like terms is called combining like terms.
iv. If an algebraic expression contains two or more like terms, these
terms can be combined into a single term by using distributive
property.

Example 11: Simplify by collecting like terms.

a. 18x + 27 – 6x – 2
b. 18k – 10k – 12k + 16 + 7

Solution
a. 18x + 27 – 6x – 2
= 18x - 6x + 27 – 2 ……. Collecting like terms
= 12x + 25 ……………… Simplifying
b. 18k – 10k – 12k + 16 + 7
= 18k – 22k + 16 + 7 ….. Collecting like terms
= -4k + 23 …….. ……… Simplifying
Example 12: Simplify the following expressions
a. (6a + 9x) + (24a – 27x)
b. (10x + 15a) – (5x + 10y)
c. – (4x – 6y) – (3y + 5x) – 2x)
Solution
a. (6a + 9x) + (24a – 27x)
= 6a + 9x + 24a – 27x ……. Removing brackets
= 6a + 24a + 9x – 27x ….... Collecting like terms
= 30a – 18x ….. Simplifying
b. (10x + 15a) – (5x + 10y)

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= 10x + 15a – 5x – 10y ……. Removing brackets

= 10x – 5x + 15a – 10y ……. Collecting like terms
= 5x + 15a – 10y …… . . . . . .Simplifying
c. –(4x – 6y) – (3y + 5x) – 2x
= -4x + 6y – 3y – 5x – 2x ……Removing brackets
= -4x – 5x – 2x + 6y – 3y ……Collecting like terms
= -11x + 3y …………………. Simplifying

Group work 2.2

1. Three rods A, B and C have lengths of (x + 1) cm; (x + 2) cm and
(x – 3)cm respectively, as shown below.
x +1 x+2 x-3
(A) (B) (C)
Figure 2.6
In the figures below express the length  interms of x.
Give your answers in their simplest form.
a. A C d. B C
  A

A  A C
b. B B e. B 

A B
C C
B A
c. f. 

Figure 2.7
2. When an algebraic expression was simplified it became 2a + b.
a. Write down as many different expressions as you can which
simplify to 2a + b.
b. What is the most complex expression you can think of that
simplifies to 2a + b?
c. What is the simplest expression you can think of that simplifies to
2a + b?

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Note: An algebraic formula uses letters to represent a relationship

between quantities.

Exercise 2B
1. Explain why the terms 4x and 4x2 are not like terms.
2. Explain why the terms 14w3 and 14z3are not like terms.
3. Categorize the following expressions as a monomial, a binomial or
a trinomial.
a. 26 d. 16x2 g. 20w4 – 10w2
b. 50 bc2 e. 10a2 + 5a h. 2t – 10t4 – 10a
c. 90 + x f. 27x + i. 70z + 13z2 – 16
4. Work out the value of these algebraic expressions using the values given.
a. 2(a + 3) if a = 5
b. 4(x + y) if x = 5 and y = -3
-
c. if x = -3 and y = -2

d. if a = 3, b = 4 and c = 2
e. 2(b + c )2 – 3 (b – c)2 if b = 8 and c = -4
f. (a + b)2 + (a + c)2 if a = 2, b = 8 and c = -4
g. c (a + b)3 if a = 3, b = 5 and c = 40

Challenge Problems

5. Solve for d if d = - - if x 1 = 3, y 1 = 4
and x 2 = 12, y2 = 37.
-
6. y if x = 5 and y =
7. Collect like terms together.
a. xy + ab – cd + 2xy – ab + dc d. 5 + 2y + 3y2 – 8y – 6 + 2y2 + 3
b. 3x2 + 4x + 6 – x2 – 3x – 3 e. 6x2 – 7x + 8 – 3x2 + 5x – 10
c. 3y2 – 6x + y2 + x2 + 7x + 4x2 f. 2x2 – 3x + 8 + x2 + 4x + 4

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2.1.3 Use of Variables to Solve Problems

Activity 2.2
Discuss with your teacher
1. Prove that the sum of two even numbers is an even numbers.
2. If the perimeter of a rectangle is 120cm and the length is 8cm more than the
width, find the area.
3. The sum of three consecutive integers is 159. What are the integers?
4. The height of a ballon from the ground increases at a steady rate of x metres
in t hours. How far will the ballon rise in n hours?
In this topic you are going to use variables to solve problems involving some
unknown values and to prove a given statement.

? What is a proof?

A proof is an argument to show that a given statements is true. The argument

depends on known facts, such as definitions, postulates and proved theorems.

Example13: (Application involving consecutive integers)

The sum of two consecutive odd integers is -188. Find the integers.

Solution: Let x represent the first odd integer, hence

x + 2 represents the second odd integer.
(First integer) + (Second integer) = total ……. Write an equation in words.
x + (x + 2) = -188 …… Write a mathematical equation
x + x + 2 = -188
2x + 2 = -188
2x = -190
=
x = -95
Therefore the integer are -95 and -93.
Example14: (Application involving Ages)
The ratio of present ages of a mother and her son is 12 : 5. The mother’s
age, at the time of birth of the son was 21 years. Find their present ages.

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Solution:
Let x be the present age of the mother and y be that of her son.
x 12
Thus x:y = 12:5 or =
y 5
5x – 12y = 0 …... Equation 1
x – y = 21 …... Equation 2
From equation 2, we get x = 21 + y …… Equation 3
Substituting equation 3 in to equation 1, we will get:
5(21 + y) – 12y = 0
105 + 5y – 12y = 0
105 – 7y = 0
105 = 7y
y = 15
Thus x = 21 + y
x = 21 + 15 ⇒ x = 36
Therefore, the present ages of a mother and her son are 36 years and
15 years respectively.

Exercise 2C
Solve each of the following word problems.
1. A 10 meter piece of wire is cut in to two pieces. One piece is 2 meters
longer than the other. How long are the pieces?
2. The perimeter of a college basket ball court is 96 m and the length is 14m
more than the width. What are the dimensions?
3. Ten times the smallest of three consecutive integers is twenty two more
than three times the sum of the integers. Find the integers.
4. The surface area “S” of a sphere of radius r is given by the formula:
S = 4 πr2.
Find (i) the surface area of a sphere whose radius is 5 cm.
1
(ii) the radius of a sphere whose surface area is 17 cm2.
9
5. By what number must be 566 be divided so as to give a quotient 15 and
remainder 11?
6. I thought of a number, doubled it, then added 3. The result multiplied by
4 came to 52. What was the number I thought of ?

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7. One number is three times another, and four times the smaller added to
five times the greater amounts to 133; find them.

Challenge Problems
8. If a certain number is increased by 5, one – half of the result is three –
fifths of the excess of 61 over the number. Find the number.
9. Divide 54 in to two parts so that four times the greater equals five times
the less.
10. Prove that the sum of any 5 consecutive natural numbers is divisible by 5.

2.2 Multiplication of Binomials

In grade seven you have studied about certain properties of multiplication and
addition such as the commutative and associative properties of addition and
multiplication and the distributive of multiplication over addition. In this sub–
unit you will learn how to perform multiplication of monomial by binomial and
multiplication of binomial by binomial.

2.2.1 Multiplication of Monomial by Binomial

Activity 2.3
Discuss with your friends /partners/
1. Multiply 4a by 2ab 3. Multiply 6b by (3a + 15b)
2. Multiply 4b by (2ab + 6b) 4. Multiply 7ab by (3ab – 6a)
You begin this topic, let us look at some examples:
Example 15: Multiply 2x by 4yz
Solution:
2x × 4yz = 2 × x × 4 × y × z
= (2 × 4) (x × y × z)
= 8 xyz
Example 16: Multiply 4c2 by (16 abc – 5a2 bc)
Solution:
4c2 × (16abc – 5a2bc)
= (4c2 × 16abc) – (4c2 × 5a2bc)
= (4 × 16 × c2 × c × a × b) – (4 × 5 × c2 × c × a2 ×b)
= 64 c3 ab – 20 c3 a2b
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Example 17: Multiply 4rt (5pq – 3pq)

Solution:
4rt × (5pq – 3pq)
= (4rt × 5pq) – (4rt × 3pq)
= (4 × 5 × r × t × p × q) – (4 × 3× r × t ×p × q)
= (20rt pq – 12 rt pq)
= (20-12) rt pq
= 8pqrt

? Do you recall the properties used in examples 15, 16, and 17 above?

1. Distributive properties
Group Work 2.3
1. In Figure 2.8 below, find the area of the shaded region.
2a

b
b

2a
Figure 2.8
2. If the area of a rectangle is found by multiplying the length times
the width, express the area of the rectangle in Figure 2.9 in two ways
to illustrate the distributive property for a(b + c).
b c

b+c
Figure 2.9

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3. Express the shaded area of the rectangle in Figure 2.10 in two ways
to illustrate the distributive property for a (b – c).
b-c

b-c c
Figure 2.10
4. In Figure 2.11, find the area of each rectangle.
x 3

x R1 R3
(x + 2)

2 R2 R4

(x + 3)
Figure 2.11
Consider the rectangle in Figure 2.12 which has been divided in to two smaller
ones:
The area A of the bigger rectangle is given by: A = a(b + c).
The area of the smaller rectangles are given by A 1 = ab and A 2 = ac; but the sum
of the areas of the two smaller rectangles are given by,
A 1 + A 2 gives the area A of the bigger rectangle:
b+c

A1 A2
a a

b c
Figure 2.12 Rectangle

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That means: A = A 1 + A 2
a(b + c) = ab + ac
This suggests that a(b + c) = a × b + a × c the factor out side the bracket multiply
each number in the bracket, this process of removing the bracket in a product is
known as expansion.
Similarly consider another rectangle as in Figure 2.13 below.
Area of the shaded region = area of the bigger rectangle–area of the un shaded
region.
Therefore, a (b – c) = ab – ac.
b

a a

b-c c
Figure 2.13 Rectangle

You have seen that the above two examples on area of rectangle, this could be
generalized as in the following way:

Note: For any rational numbers a, b, and c

a. a(b + c) = ab + ac b. a (b – c) = ab – ac
These two properties are called the distributive properties of
multiplication over Addition (subtraction).

Exercise 2D
1. Expand these expressions by using the distributive properties to remove the
brackets in and then simplify.
1. 2(a + b) + 3 (a + b) 6. 7(2d + 3e) + 6 (2e – 2d)
2. 5(2a – b) + 49(a + b) 7. 3(p + 2q) + 3(5p – 2q)
3. 4(5a+c) + 2(3a – c) 8. 5(5q + 4h) + 4(h – 5q)
4. 5(4t – 3s) + 8(3t + 2s) 9. 6(p + 2q + 3r) + 2(3p – 4q + 9r)
5. 5(3z + b) + 4(b – 2z) 10. 2(a + 2b – 3c) + 3(5a – b + 4c) + 4(a + b + c)

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Challenge Problems
11. Remove the brackets and simplify.
a) (x + 1)2 + (x + 2)2 d) (x + 2)2 – (x – 4)2
b) (y – 3)2 + (y – 4)2 e) (2x + 1)2 + (3x + 2)2
c) (x – 2)2 + (x + 4)2 f) (2x – 3)2 + (5x + 4)2

2.2.2 Multiplication of Binomial by Binomial

Activity 2.4
Discuss with your friends / partners
Find the following products
1. (2x + 8) (3x – 6) 4. (2x - 10) (x – 8)
2. (5a + 4) (4a + 6) 5. (2a2 – ab) (20 + x)
3. (2x – 8) (2x + 8) 6. (3x2 + 2x – 5) (x – 1)

Sometimes you will need to multiply brackets expressions. For example (a + b) (c + d).
This means (a + b) multiplied by (c + d) or (a + b) × (c + d).
Look at the rectangles 2.14 below.
The area ‘A’ of the whole rectangle is (a + b) (c + d). It is the same as the sum of the
areas of the four rectangle so: A = A 1 + A 2 + A 3 + A 4
(a + b) (c + d) = ac + ad + bc + bd

c d

a A1 = ac A2 = ad

A3= bc A4 = bd
b

Figure 2.14 Rectangle

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Notice that each term in the first brackets is multiplied by each term in the second
brackets:
a×d
ac
(a + b) (c + d)

b×c
b×d

You can also think of the area of the rectangle as the sum the areas of two separate part
(the upper two rectangles plus the lower two rectangle) see Figure 2.15:
Thus, (a + b) (c + d) = a (c + d) + b(c + d)
Think of multiplying each term in the first bracket by the whole of the second bracket.
These are two ways of thinking about the same process. The end result is the same. This
is called multiplying out the brackets.
C+d

a Area = a(c+d)

b Area = b(c+d)

Figure 2.15 Rectangle

This process could be described as follows.

Note: If (a + b) and (c +d) are any two binomials their product

(a + b)  (c + d) is defined as:

(a + b) × (c + d) = ac + ad + bc + bd

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Example18: Multiply (4x + 4) by (3x + 8)

Solution
(4x + 4) (3x + 8) = (4x  3x) + (4x  8) + (4  3x) + (4  8)
= 12x2 + 32x + 12x + 32
= 12x2 + 44x + 32

Example 19: Multiply (2x + 10) by (3x - 6)

Solution
(2x + 10) (3x - 6) = (2x  3x) – (6  2x) + (10  3x) – (6  10)
= 6x2 - 12x + 30x – 60
= 6x2 + 18x – 60
Example20: Multiply (2x - 3) by (4x - 12)
Solution
(2x - 3) (4x - 12) = (2x  4x) – (12  2x) – (3  4x) + (3  12)
= 8x2 - 24x - 12x + 36
= 8x2 – 36x + 36
In the multiplication of two binomials such as those shown in example 20 above,
the product 2x  4x = 8x2 and -3  -12 = 36 are called end products. Similarly,
the product -12  2x = -24x and -3  4x = -12x are called cross product. Thus
the product of any two binomials could be defined as the sum of the end
products and the cross products. The sum of the cross products is written in the
middle.

Example21: Multiply (4x - 6) by (4x + 10)

end product
Solution
(4x - 6) (4x + 10) = (4x  4x) + (4x  10) – (6  4x) – (6  10)

Cross product

= 16x2 + 40x - 24x – 60

= 16x2 + 16x – 60

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Example22: Multiply (4x-10) by (6x-2)

end product
Solution:
(4x - 10) (6x -2) = (4x  6x)-(24x)-(106x)+(102)

cross product
2
= 24x -8x-60x+20
= 24x2-68x+20
Exercise 2E
Find the products of the following binomials.
1. (2x + 2y) (2x – 2y) 6. -

2. (3x + 16) (2x – 18) 7.

3. (-4x – 6) (-20x + 10) 8. - - -

4. -5 [ ] 9.

5.

Challenge Problems
10. (2x2 + 4x – 6) (x2 + 4)
11. (2x2 – 4x – 6) -
12. -

2.3 Highest Common Factors

Activity 2.5
Discuss with your teacher before starting the lesson.
1. Define and explain the following key terms:
a. Factorizing a number b. Prime factorization
2. Find the HCF of the following.
a. 72 and 220 b. 36, 48 and 72 c. 120, 150 and 200

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3. Find the HCF of the following:

a. 20xyz and 18x2z2
b. 5x3y and 10xy2
c. 3a2b2, 6a3b and 9a3b3
4. Factorize the following expressions.

c.

b. x(2b + 3) + y(2b + 3) d. -

Factorizing
This unit is devoted to the method of describing an expression is called
Factorizing. To factorize an integer means to write the integer as a product of
two or more integers. To factorize a monomial or a Binomial means to express
the monomial or Binomial as a product of two or more monomial or Binomials.
In the product 25 = 10, for example, 2 and 5 are factors of 10. In the product
(3x + 4) (2x) = 6x2 + 8x, the expressions (3x + 4) and 2x are factors of 6x2 + 8x.

Example 23: Factorize each monomial in to its linear factors with coefficient
of prime numbers.
a. 15x3 b. 25x3
Solution:
a. 15x3 = (3× 5) × (x × x × x)
= (3x) × (5x) × (x).
b. 25x = (5 × 5) × (x × x × x)
3

= (5x) × (5x) (x).

Example 24: Factorize each of the following expressions.

a. 6x2 + 12 b. 5x4 + 20x3

Solution:
a. 6x2 + 12 = 6x2 + 6  2
= 6(x2 + 2)
b. 5x4 + 20x3 = 5x3  x + 5x3  4
= 5x3(x + 4)
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Exercise 2F
Factorize each of the following expressions.
1. 2x2 + 6x 6. 3a4b – 5bc3
2. 18xy2 – 12xy3 7. 6x4yz + 15x3y2z
3. 5x3y + 10xy2 8. 8a2b3c4 – 12a3b2c3
4. 16a2b + 24ab2 9. 8xy2 + 28xyz – 4xy
5. 12ab2c3 + 16ac4 10. -10mn3 + 4m2n – 6mn2

Challenge Problems
11. 7a2b3 + 5ab2 + 3a2b 14. 16x4 – 24x3 + 32x2
12. 2a3b3 + 3a3b2 + 4a2b 15. 10x3 + 25x2 + 15x
13. -30abc + 24abc – 18a2b

Highest common factor of two integers

Group Work 2.4
Discuss with your group
For Exercise 1 – 4 factor out the HCF.
1. 15x2 + 5x
2. 5q4 – 10 q5
3. y(5y + 1) – 9(5y + 1)
4. 5x (x – 4) – 2(x – 4)

You begin the study of factorization by factoring integers. The number 20 for example
can be factored as 120, 210, 45 or 225. The product 225 (or equivalently 225)
consists only of prime numbers and is called the prime factorization.
The highest common factor (denoted by HCF) of two or more integers is the highest
factor common to each integer. To find the highest common factor of two integers, it is
often helpful to express the numbers as a product of prime factors as shown in the next
example.
Example 25: Find the highest common factor of each pair of integers.
a. 24 and 36
b. 105 and 40

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Solution:
First find the prime factorization of each number by multiplication or by factor tree
method.
a. i. by multiplication ii. By using factor trees

Factors of 24 = 2×2 ×2× 3 24 36

Factors of 36 = 2×2 ×3× 3 2 12 2 18

2 6 2 9
2 3 3 3

The numbers 24 and 36 share two factors of 2 and one factor of 3. Therefore, the highest
common factor is 223 =12
b. i. By multiplication
Factors of 105 = 3  7 × 5
Factors of 40 = 22 2 × 5
ii. By using factor trees
105 40

35 20
3 2
10
5 7 2

2 5

Therefore, the highest common factors is 5.

Highest common factor (HCF) of two or more monomials

Example 26: Find the HCF among each group of terms.
a. 7x3, 14x2, 21x4
b. 8c2d7e, 6c3d4

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 Grade 8 Mathematics [FURTHER ON WORKING WITH VARIABLES ]

Solution:
List the factors of each term.

a) 7x3 = 7  x x x
14x = 2 
2
7  x x
21x = 3 
4
7xx xx

Therefore, the HCF is 7x2.

8c 2 d 7 e = 2 3 c 2 d 7 e 
b)  The common factors are the common powers of 2, c and d
6c 3 d 4 = 2 × 3 c 3 d 4 
appearing in both factorization to determine the HCF we will take the common
least powers. Thus
The lowest power of 2 is : 21
The lowest power of c is : c2
The lowest power of d is : d4

Therefore, the HCF is 2c2d4.

Example 27: Find the highest common factor between the terms:
3x(a + b) and 2y(a + b)
Solution
3x (a + b)
The only common factor is the binomial (a + b).
2y(a + b)
Therefore, the HCF is (a + b).

Factorizing out the highest common factor

Factorization process is the reverse of multiplication process. Both processes use
the distributive property: ab + ac = a (b + c)

Example 28: Multiply: 5y(y2 + 3y + 1)

= 5y (y2) + 5y(3y) + 5y(1)
= 5y3 + 15y2 + 5y

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 Grade 8 Mathematics [FURTHER ON WORKING WITH VARIABLES ]

Factor: 5y3 + 15y2 + 5y

= 5y(y2) + 5y(3y) + 5y(1)

= 5y (y2 + 3y + 1)
Example 29: Find the highest common factors
a. 6x2 + 3x b. 15y3 + 12y4 c. 9a4b – 18a5b + 27a6b
Solution a. The HCF of 6x2 + 3x is 3x … Observe that 3x is a common factor.
6x2 + 3x = (3x  2x) + (3x  1) … Write each term as the product
of 3x and another factor.
= 3x (2x + 1) …… Use the distributive property to
factor out the HCF.
2
Therefore, the HCF of 6x + 3x is 3x.

 Check: 3x (2x + 1) = 6x2 + 3x

b. The HCF of 15y3 + 12y4 is 3y3 …Observe that 3y3 is a common factor.
15y3 + 12y4 = (3y3  5) + (3y3  4y) …. Write each term as the product of
3y3 and another factor.
= 3y3(5 + 4y) ……Use the distributive property to factor out
the HCF.
Therefore, the HCF of 15y + 12y is 3y3.
3 4

c. 9a4b – 18a5b + 27a6b is 9a4b …Observe that 9a4b is a common factor.

= (9a4b  1) – (9a4b  2a) + (9a4b  3a2) …… Write each term as the
product of 9a4b and another factor.
= 9a4b(1 – 2a + 3a2) …… Use the distributive property to factor out the
HCF.
Therefore the HCF of 9a b - 18a5b + 27a6b is 9a4b.
4

Factorizing, out a binomial factor

The distributive property may also be used to factor out a common factor that
consists of more than one term such as a binomial as shown in the next example.

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 Grade 8 Mathematics [FURTHER ON WORKING WITH VARIABLES ]

Example 30: Factor out the highest common factor:

2x (5x + 3) – 5(5x + 3)
Solution
2x(5x + 3) – 5(5x + 3) …….. The Highest common factor is the binomial
5x + 3
= (5x + 3)  (2x) – (5x + 3)  (5) ……… Write each term as the product
of (5x + 3) and another factor.
= (5x + 3) (2x – 5) ….. Use the distributive property to factor out the
HCF.

Check: (5x + 3) (2x – 5) = (5x + 3) (2x) + (5x + 3) (-5)

= 2x (5x + 3) – 5(5x + 3)

Exercise 2G
1. Find the highest common factor among each group of terms.
a. -8xy and 20y e. 3x2y, 6xy2 and 9xyz
b. 20xyz and 15yz2 f. 15a3b2 and 20ab3c
c. 6x and 3x2 g. 6ab4c2 and 12a2b3cd
d. 2ab, 6abc and 4a2c
2. Find the highest common factor of the pairs of the terms given below.
a. (2a – b) and 3(2a – b) e. 21x (x + 3) and 7x2(x + 3)
b. 7(x – y) and 9(x – y) f. 5y3(y – 2) and -20y(y – 2)
c. 14(3x + 1)2 and 7(3x + 1)
d. a2(x + y) and a3(x + y)2
3. Factor out the highest common factor.
a. 13(a + 6) – 4b (a + 6) d. 4(x + 5)2 + 5x(x + 5) – (x + 5)
b. 7(x2 + 2) – y(x2 + 2) e. 6(z – 1)3 + 7z(z – 1)2 – (z – 1)
c. 8x(y2 – 2) + (y2 – 2) f. x4 – 4x

Challenge Problems
4. Factor by grouping: 3ax + 12a + 2bx + 8b

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Summary For Unit 2

1. A variable is a symbol or letter used to represent an unspecified value in
expression.
2. An algebraic expression is a collection of variables and constant under
algebraic operations of addition or subtraction. For example, y + 10 and
2t – 2 × 8 are algebraic expressions.
The symbols used to show the four basic operations of addition,
subtraction, multiplication and division are summarized in Table 2.1

Table 2.1
Operation Symbols Translation
Addition x+y  Sum of x and y
 x plus y
 y added to x
 y more than x
 x increased by y
 the total of x and y
Subtraction x–y  difference of x and y
 x minus y
 y subtracted from x
 x decreased by y
 y less than x
Multiplication x  y, x(y), xy  product of x and y
 x times y
 x multiplied by y
Division x ÷ y, ,  Quotient of x and y
 x divided by y
 y divided into x
 ratio of x and y
 x over y
3. For any rational numbers a, b and c
a. a(b + c) = ab + ac
b. a(b – c) = ab – ac
These two properties are called the distributive properties.

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 Grade 8 Mathematics [FURTHER ON WORKING WITH VARIABLES ]

4. If (a + b) and (c + d) are any two binomials whose product (a + b) (c + d) is

defined as (a + b) (c + d) = ac + ad + bc + bd.
5. The highest common factor (HCF) of numbers is the greatest number which
is a common factor of the numbers.
The procedures of one of the ways to find the HCF is given below:
1) List the factors of the numbers.
2) Find the common factors of the numbers.
3) Determine the highest common factors of these common factors.

Miscellaneous Exercise 2
I. State whether each statement is true or false for all positive integers x, y, z and w.
1. If a number y has z positive integer factors, then y and 2z integer
factors.
2. If 2 is a factor of y and 3 is a factor of y, then 6 is a factor of y.
3. If y has exactly 2 positive integer factors, then y is a prime numbers.
4. If y has exactly 3 positive integer factors, then y is a square.
5. If y has exactly 4 positive integer factors, then y is a cube.
6. If x is a factor of y and y is a factor z, then x is a factor of z.

II. Choose the correct answer from the given four alternatives.
7. A triangle with sides 6, 8 and 10 has the same perimeter as a square
with sides of length ?
a. 6 b. 4 c. 8 d. 12
8. If x + y = 10 and x – y = 6, what is the value of x3 – y3?
a. 604 b. 504 c. 520 d. -520
9. If ab + 5a + 3b + 15 = 24 and a + 3 = 6, then b + 5 = ?
a. 5 b. 50 c.4 d. 12
2 2 2
10. If ab = 5 and a + b = 25, then (a + b) = ?
a. 35 b. 20 c. 15 d. 30
11. If n is an integer, what is the sum of the next three consecutive even
integers greater than 2n?
a. 6n + 12 b. 6n + 10 c. 6n + 4 d. 6n + 8

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Grade 8 Mathematics [FURTHER ON WORKING WITH VARIABLES ]

12. One of the following equation is false.

a. A = bh for h = c. P = 2( + w) for -w

-
b. A = 2s2 + 4sh for h = d. A = bh for h =

13. If x = 6 and y = 2, then what is the value of 3x2 – 4 + 8.

b. c. d.
14. Find the value of y, if y = x2 – 6 and x = 7.
a. 49 b. 7 c. 43 d. 45
15. If x = 2 and y = 3, then what is the value of yx + xy  y + x?
a. 9 b. -29 c. 29 d. 18
a
a+
16. If a = 4 and b = 7, then what is the value of b?
a
a−
b
a. 8 b. 1 c. d.

III. Work out Problems

17. Simplify each of the following expressions.
a. (x3 + 2x – 3) – (x2 – 2x + 4)
b. 2x(3x + 4) – 3(x + 5)
c. x (y2 + 5xy) + 2xy(3x – 2y )
d. 2(a2b2 – 4a3b3) – 8(ab2 – 3a2b2)
18. Express the volume of this cube.

x-1

x-1
x-1
x-1

Figure 2.16 cube

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Grade 8 Mathematics [FURTHER ON WORKING WITH VARIABLES ]

19. Find the surface area of this cube.

3x-y

3x-y
3x-y
Figure 2.17 cube

20. Prove that the sum of five consecutive natural number is even.
21. Prove that 6(n + 6) – (2n + 3) is odd numbers for all n
22. Multiply the expressions.
a. (7x + y) (7x – y) f. (5a – 4b) (2a – b)
b. (5k + 3t) (5k + 3t) g. (5x – 3)
c. (7x – 3y) (3x – 8y) h. (2h + 2·7) (2h - 2·7)
d. (5z + 3)(z2 + 4z – 1) i. (k – 3)3

e. - j. (k + 3)3
23. Find the highest common factor for each expression.
a. 12x2 – 6x e. 4x (3x – y) + 5(3x – y)
b. 8x(x – 2) -2(x – 2) f. 2 (5x + 9) + 8x(5x + 9)
c. 8(y + 5) + 9y(y + 5) g. 8q9 + 24q3
d. y(5y + 1) – 8 (5y + 1)
24. Find three consecutive numbers whose sum shall equal 45.
25. Find three consecutive numbers such that twice the greatest added to three
times the least amount to 34.
26. Find two numbers whose sum is 36 and whose difference is 10.
27. If a is one factor of x, what is the other factor?
28. Find the value of (x+5) (x+2) + (x−3) (x −4) in its simplest form. What is the
numerical value when x = -6?
29. Simplify (x + 2) (x + 10) − (x − 5) (x − 4). Find the numerical value of this
expression when x = -3.

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UNIT
3 LINEAR EQUATIONs
AND
INEQUALITIES

Unit outcomes
After Completing this unit, you Should be able to:
 understand the concept equations and inequalities.
 develop your skills on rearranging and solving linear
equations and inequalities.
 apply the rule of transformation of equations and
inequalities for solving problems.
 draw a line through the origin whose equation is given.

Introduction
In this unit you will expand the knowledge you already have on solving linear
equations and inequalities by employing the very important properties known as
the associative property and distributive property of multiplication over addition
and apply these to solve problems from real life. More over you will learn how
to set up a coordinate plane and drawing straight lines using their equation.

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 Grade 8 Mathematics [LINEAR EQUATION AND INEQULITIES ]

3.1 Further on Solutions of Linear Equations

Group Work 3.1
Discuss with your friends/partners.
1. Solve the following linear equations using equivalent transformation.
a. 6x – 8 = 26 c. 5x – 17 - 2x = 6x -1 –x
b. 14x +6x = 64 d. 5x – 8 = -8 +3x –x
2. Solve the following linear equations.
a. 7(x-1) -x = 3- 5x + 3 (4x - 3)
b. 0.60x + 3.6 = 0.40(x + 12)
c. 8(x + 2) +4x + 3 = 5x + 4 + 5 (x + 1)
d. 8y – (5y – 9) = -160
e. 3(x + 7) + (x -8)6 = 600
3. Do you recall the four basic transformation rules of linear equations.
Explain.

3.1.1 Solution of Linear Equations Involving Brackets

Activity 3.1
Discuss with your friends/partners.
Solve the following linear equations.
a. 4(2x + 3) =3(x + 8) d. 3(6t +7 )= 5 (4t + 7)
b. 6(5x – 7) = 4(3x + 7) e. 7 (9d – 5) = 12 (5d -6)
c. 4(8y + 3) = 6(7y + 5 ) f. 10x – (2x + 3) = 21

To solve an equation containing brackets such as 5(4x + 6) = 50 –(2x + 10), you

transform it into an equaivalent equation that does not have brackets. To do this
it is necessary to remember the following rules.

Note: For rational numbers a, b and c,

a) a + (b + c) = a + b + c
b) a - (b + c) = a – b – c
c) a(b + c) = ab + ac
d) a(b – c) = ab – ac

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 Grade 8 Mathematics [LINEAR EQUATION AND INEQULITIES ]
Example 1: Solve: x – 2( x -1) = 1 – 4(x + 1) Using the above rules.
Solution x – 2 (x – 1) = 1 -4 (x +1 ) …Given equation
x – 2x + 2 = 1 – 4x – 4 ……. Removing brackets
x – 2x +4 x = 1 - 4 – 2………Collecting like terms
3x = -5 …………….Simplifying
………..……Dividing both sides by 3
x= ………..…… x is solved.
 Check: For x =

1- 4

1+

(True)

Example 2: Solve 4(x – 1) +3 (x + 2) = 5 (x – 4) using the above rules.

Solution 4(x -1 ) + 3(x + 2) = 5(x – 4) ……Given equation
4x- 4 + 3x + 6 = 5x - 20 ………Removing brackets
4x + 3x – 5x = -20 + 4 – 6…..…..Collecting like terms
2x = -22 ………………Simplifying
……………...Dividing both sides by 2
x = - 11 ………..……X is solved
 Check: For x = -11
4(-11 – 1) +3 (-11 + 2) 5(-11 – 4)

4(-12) + 3 (-9) 5(-15)

- 48 – 27 -75
- 75 = - 75 (True)

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 Grade 8 Mathematics [LINEAR EQUATION AND INEQULITIES ]

Exercise 3A
1. Solve each of the following equations, and check your answer in the
original equations.
a. 7x – 2x + 6 = 9x - 32 e. 8x + 4 = 3x – 4
b. 21 – 6x = 10 – 4x f. 2x + 3 = 7x + 9
c. 2x – 16 = 16 – 2x g. 5x -17 = 2x + 4
d. 8 – 4y = 10 – 10y h. 4x + 9= 3x + 17
2. Solve each of the following equations, and check your answer in the
original equations.
a. 7- (x + 1) = 9 –(2x -1) e. 4(8y +3 ) = 6 (7y + 5 )
b. 3y + 70 + 3 ( y – 1) = 2(2y + 6) f. 8(2k – 6) = 5(3k – 7)
c. 5(1 – 2x) -3(4 + 4x) = 0 g. 5(2a +1) +3 (3a – 4) = 4(3a – 6)
d. 3 – 2(2x + 1) = x + 17

Challenge problems
3. Solve the equation 8x + 10 -2x = 12 +6x -2.
4. solve the equation -16( 2x - 8)- (18x – 6) = -12 + 2(6x – 6).
5. Solve the equation (8x – 4) (6x + 4) = (4x +3) (12x – 1).
6. Solve for x in each of the following equations:
a. m(x + n) = n
b. x( a + b) = b(c – x)
c. mx = n(m + x)

3.1.2 Solution of Linear Equations Involving Fractions

Group work 3.2
Discuss with your friends/partners.
1. Work out
c. 2 e. 1

d. 3

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 Grade 8 Mathematics [LINEAR EQUATION AND INEQULITIES ]

2. Work out.
b. 4 c. 6 d. 7
3. Work out.
b. 2 c. 21 d. 3
4. Work out
a. 3 b. 36 c. 4 d. 2
5. In a school, of the students are girls. What fraction of the students
are boys?
6. A box containing tomatoes has a total weight of 5 kg.
The empty box has a weight of 1 kg. what is the weight of the
tomatoes?
7. A machine takes 5 minutes to produce a special type of container.
How long would the machine take to produce 15 container?

From grade 5 and 6 mathematics lesson you have learnt about addition,
subtraction, multiplication and division of fractions. All of these are shown on
the following discussion.

Adding fractions
It is easy to add fractions when the denominators (bottom) are the same:
Easy to add: what about this?

Denominators are the same Denominators are different

Adding fractions with the same denominator

= Add the numerators (top)

Add them over the same denominators, ( bottom)

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 Grade 8 Mathematics [LINEAR EQUATION AND INEQULITIES ]
Example 3:
Adding fractions with different denominators

First find equivalent fractions to these ones which have the same denominator
(bottom):
Fractions equivalent to Fractions equivalent to
333
418 99
99

Equivalent to Equivalent to

So

Note: To add fractions, find equivalent fractions that have the same
denominator or (bottom).

Subtracting fractions
It is easy to subtract fractions when the denominators (bottom) are the same:
Easy to subtract: What about this?

Denomintaors are the same Denominators are different.

Example 4: (Subtracting fraction with different denominators)
work out

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 Grade 8 Mathematics [LINEAR EQUATION AND INEQULITIES ]
Solution: Find equivalent fractions to these ones which have the same
denominator (bottom). An easy way is to change both denominator
to 36 because 9 4 = 36 is LCM of the denominators.

×9
×4
= =
×9
×4

Equivalent to
Equivalent to
So

Note: To subtract fractions, find equivalent fractions that have the same
denominator (bottom).

Multiplying fractions
To multiply two fractions, multiply the numerators together and multiply the
denominators together.
For example,
Multiply the numerators (top)
Multiply the denominators
(bottom)

You can simplify this to ( by dividing the top and bottom of by 10).
50 7 35
Therefore, × =
18 10 18
Dividing fractions
To divide fractions, invert or take the reciprocal of the dividing fraction (turn it
upside down) and multiply by the divisor.
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 Grade 8 Mathematics [LINEAR EQUATION AND INEQULITIES ]
For example
Chang the ″
Sign in to change the fraction you are dividing by up side down.
a ″×″ sign This is called inverting the fraction.

Now let us consider linear equations having fractional coefficients.

Example 5: Solve

Solution: …..Given equation

The LCM of the denominators is 3  10 = 30 since 3 and 10 do not have any
common factors.
Therefore, multiplying both sides by 30.
30

30 ……..By the distributive property

10(x + 1) +3(x -1) = 360 ………Simplifying
10x + 10 + 3x – 3 = 360 ………Removing brackets
13x + 7 = 360 ………Collecting like terms
13x + 7- 7 = 360 – 7……Subtracting 7 from both sides
13x = 353………Simplifying
Dividing both sides by 13
x =
The solution set is

 Check:
353 353
+1 −1
13 13 ?
+1 12
3 10 =

12
12

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 Grade 8 Mathematics [LINEAR EQUATION AND INEQULITIES ]

12
12
12 = 12 (True)
Example 6: Solve

Solution: The LCM of the denominators is 24.

24 ….Multiplying both sides by 24.
24 …Distributive property
7 = 3x + 4 ………….Removing brackets
7 – 4 = 3x + 4 – 4……….Subtracting 4 from both sides
3 = 3x ………….Simplifying
Or Dividing both sides by 3
x=
The solution set is .

Example 7: Solve =16.

Solution: The LCM of the denominators is 12.

12 …Multiplying both sides by 12.
12 +12 +12 12×16……Distributive property
6(x + 1) + 4(x +2) +3(x + 3) = 12×16…….Simplifying
6x +6 + 4x + 8 + 3x + 9 = 192……….Removing brackets
13x + 23 -23 = 192-23...Subtracting 23 from both sides
13x = 169 ………….Simplifying
………..Dividing both sides by 13
x = 13
The solution set is .
 Check: 16

16

7+5+4 16
16 = 16 ( True)
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 Grade 8 Mathematics [LINEAR EQUATION AND INEQULITIES ]

Example 8: Solve (x + 7) -

Solution: The LCM of the denominators 3 and is 6.

(x + 7) - (x + 1) = 4 ………Given equation

6 = 6×4 ……Multiply both sides by 6.

6 -6 = 6×4 …Distributive property

2(x + 7) -3(x + 1) = 24 …Simplifying
2x + 14 -3x – 3 = 24 …Removing brackets
2x – 3x + 14 – 3 = 24
-x + 11 = 24 ….Collecting like terms
-x + 11-11 = 24-11...Subtracting 11 from both sides
-x = 13 ……………….Simplifying
…………Dividing both sides by -1.
x = -13
The solution set is
 Check: 4

-2 + 6 4
4 = 4 (True)
Exercise 3B
1. Solve each of the following equations.
a.

b. e.

c. f.

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 Grade 8 Mathematics [LINEAR EQUATION AND INEQULITIES ]
2. Solve each of the following equations and check your answer in each
case by inserting the solution in original equation.
a. f.

b. g.

c. h.

d.

e.

Challenge Problems
3. Solve the following equations.
2
a. 12 – c.

b. d. 0.78 –

3.1.3 Solve Word Problems Using Linear Equations

Group work 3.3
1. Two complementary angles are drawn such that one angle is 10°
more than seven times the other angle. Find the measure of each
angle.
2. A certain number of two digits is three times the sum of its digits, and
if 45 be added to it the digits will be reversed; find the number.

Mathematical problems can be expressed in different ways. Common ways of

expressing mathematical problems are verbal or words and formulas or open
statements. In this sub-unit you will learn how to translate verbal problems to
formulas or mathematical expressions so that you can solve it easily. It is
important to translate world problems to open statements because it will be clear
and concise.
Although there is no one definite procedure which will insure success to translate
word problems to open statements to solve it, the following steps will help to
develop the skill.
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 Grade 8 Mathematics [LINEAR EQUATION AND INEQULITIES ]
Table 3.1 Problem – solving Flow chart for word problems
Step 1 Read the problem carefully. • Familiarize yourself with the problem.
Identify the unknown, and if possible
estimate the answer.
Step 2 Assign lables to unknown • Identify the unknown quantity or
quantities quantities. Let x represent one of the
unknowns. Draw a picture and write
down relevant formulas.
Step 3 Develop a verbal models • Write an equation in words.

Step 4 • Replace the verbal model with a

Write a mathematical equation
mathematical equation using x or
other variable.
Step 5 • Solve for the variable using steps for
Solve the equation
solving linear equations.
Step 6 • Once you have obtained a numerical
Interpet the results and write value for the variable, recall what it
the final answer in words
represents in the context of the
problems.
Step 7 Check all answers back into
the original statements of
the problems

Example 9: The sum of a number and negative ten is negative fifteen. Find the
number.
Solution: Step 1: Read the problem
Let x represent the unknown number Step 2: Label the unknown
(a number) + ( -10) = -15 Step 3: Develop a verbal model
x + (-10) = -15 Step 4: Write the equation
x + (-10) + 10 = -15 +10 Step 5: Solve for x
x = -5 Step 6: Write the final answer in
Therefore, the number is -5. words.

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 Grade 8 Mathematics [LINEAR EQUATION AND INEQULITIES ]
Example 10: (Applications involving sales Tax)
A video game is purchased for a total of Birr 48.15 including
sales tax. If the tax rate is 7%. Find the original price of the video
game before sales tax is added.

Solution: Step 1: Read the problem

Let x represent the price of the video game.
0.07x represents the amount of sales tax. Step 2: Label variables
+ = Step 3: Write a verbal equation
x + 0.07x = Birr 48.15 Step 4: Write a mathematical
equation
1.07x = 48 .15 Step 5: Solve for x multiply by
100 to clear decimals
100( 1.07x) = 100(48.15)
107x = 4815
= Step 6: Divide both sides by 107
x=
x = 45 Step 7: Interpret the results and
write the answer in words.
Therefore, the original price was Birr 45.
Example 11: (Applications involving consecutive integers)
Find three consecutive even numbers which add 792.
Solution: Step 1: Read the problem
Let the smallest even number be x. Step 2: label the unknow
Then the other even numbers are (x+2) and (x+4) Step 3: Develop a verbal
model
Because they are consecutive even numbers.
x+(x+2) +(x+4) = 792 Step 4: Write the equation
3x + 6 = 792
3x = 786
x = 262
The three even numbers are 262, 264 and 266. Step 5: Write the final answer in

Word

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 Grade 8 Mathematics [LINEAR EQUATION AND INEQULITIES ]
 Check
262 + 264 +266 = 792 Step 6: Check
Example 12: (Applications involving ages)
The sum of the ages of a man and his wife is 96 years. The man is
6 years older than his wife. How old is his wife?

Solution: let m = the man’s age and w = the wife’s age.

⇒ m + w= 96 …………….Translated equation (1)
⇒ m = 6 + w ……………...Translated equation (2)
⇒ (6 + w) + w = 96 ……... Substituting equation (2) into 1
⇒ 2w + 6 = 96 ……………Collecting like terms
⇒ 2w = 96 – 6 ……………Subtracting 6
⇒ 2w = 90 ………………..Collecting like terms
⇒ w = 45 …………………Divided both sides by 2
Therefore, the age of his wife is 45 years old.

Exercise 3C
Solve each problem by forming an equation.
1. The sum of three consecutive numbers is 276. Find the numbers.
2. The sum of three consecutive odd number is 177. Find the numbers.
3. Find three consecutive even numbers which add 1524.
4. When a number is doubled and then added to 13, the result is 38. Find the
number.
5. Two angles of an isosceles triangle are x and (x+10). Find two possible
values of x.
6. A man is 32 years older than his son. Ten years ago he was three times as
old as his son. Find the present age of each.
Challenge Problems
7. A shop –keeper buys 20kg of sugar at Birr y per kg. He sells 16kgs at
Birr per kg and the rest at Birr ( y + 1) per kg. what is his profit.
8. A grocer buys x kg of potatoes at Birr 1.50 per kg and y kg of onions at
Birr 2.25 per kg. how much money does he pay in Birr?
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9. If P is the smallest of four consecutive even integers, what is their sum
interms of P?
10. The sum of a certain number and a second number is -42. The first
number minus the second number is 52. Find the numbers.

3.2 Further on Linear Inequalities

Activity 3.2
Discuss with your friends/partners.
1. Can you recall the definition of linear inequality?
2. Discuss the four rules of transformation of linear inequalities using examples
and discuss the result with your teacher.
3. Solve the following linear inequalities.
a. 4x -16< 12, x c. 20 − -18, x  +

< -4(x -5), x  + d. 0.5 (x -8) ≤10 + x

From grade 6 and 7 mathematics lesson you have learnt about to solve linear
inequalities in one variable based on the given domain.

Example 13: (Solving an inequality)

Solve the inequality -3x +8 ≤ 22.
Solution: -3x +8 ≤ 22………….Given inequalities
-3x +8 - 8 ≤ 22 – 8 ……..Subtracting 8 from both sides
-3x ≤ 14 ……Simplifying
…Dividing both sides by -3; reverse the inequality
sign
x≥ …….Simplifying
Therefore, the solution set is .
Example 14: (Solving an inequality)
Solve the inequality 24x – 3 < 4x + 10, x  .
Solution: 24x – 3 < 4x +10 x ………….Given inequalities
24x -3 +3 < (4x +10)+3 ……….Adding 3 from both sides
24x < 4x+13………………….Simplifying
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 Grade 8 Mathematics [LINEAR EQUATION AND INEQULITIES ]
24x – 4x < 4x – 4x +13………Subtracting 4x from both sides
20x < 13……………..Simplifying
………Dividing both sides by 20
x< …………Simplify
Therefore, the solution set is
Example 15: (Solving an inequality)
Solve the inequality 4x -6 > 10, x .

Solution: 4x – 6 > 10 ………….Given inequalities

(4x -6) + 6 >10 + 6 ………Adding 6 from both sides
4x > 16 …………Simplifying
…………..Dividing both sides by 4
x>4
The solution of the inequality is x > 4.
Therefore, the solution set is .

Example 16: (Solving an inequality)

Solve the inequality .

Solution: ……….Given inequalities

12 ……Multiply both sides by 12 to clear
fractions.
12 …….Apply the distributive
property
-3x + 2 ≤ 24 + 8x ………….Simplifying
-3x – 8x + 2 ≤ 24 + 8x -8x ………Subtracting 8x from both sides
-11x + 2 ≤ 24 …………Collecting like terms
-11x + 2 – 2 ≤ 24 – 2 ……….Subtracting 2 from both sides
-11x ≤ 22 ……………Simplifying
………….Dividing both sides by -11. Reverse
the inequality sign.
x ≥ -2
There fore, the solution set is {-2, -1, 0, 1, 2, 3, …}.
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Example 17: (Solving an inequality)
Solve the inequality 3x – 2(2x − 7) ≤ 2(3 + x) − 4 , x.
Solution:
3x − 2 (2x − 7) ≤ 2(3 + x ) − 4 …Given inequalities
3x – 4x +14 ≤ 6 + 2x − 4 ………Removing brackets
-x + 14 ≤ 2x + 2 ………………...Simplifying
-x – 2x +14 ≤ 2x − 2x +2 ……….Subtracting 2x from both sides
-3x + 14 ≤ 2 ……………………..Simplifying.
-3x + 14 – 14 ≤ 2 – 14…………...Subtracting 14 from both sides
-3x ≤ − 12 ……………………….Simplifying
…………….Dividing both sides by -3. Reverse
the inequality sign
x≥4
The solution of the inequality is x ≥ 4.
Therefore, the solution set is = .
Exercise 3D
1. Solve the following inequalities:
a. e.
b. f. 9 +
c. 8x – 5 > 13 – x g.
d. 4x + 6 > 3x + 3
2. Solve each of the following linear inequality in the given domain.
a. 4 - x > x − 8, x e. 5x + 6 ≤ 3x + 20, x
b. 4y – 6 < (28 – 2y), y f. > y
c. < -8 (x – 6), x +
g. 6x ≥16 +2x -4, x
d. -2 (12 – 2x) ≥ 3x - 24, x +
h. 10x + 12 ≤ 6x + 40, x
3. Eight times a number increased by 4 times the number is less than 36. What is
the number?
4. If five times a whole number increased by 3 is less than 13, then find the
solution set.
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Challenge Problems
5. Solve each of the following linear inequalities:
a. 3 (x + 2) – (2x – 7)≤ (5x -1) -2(x + 6)
b. 6 – 8(y + 3) +5y > 5y –(2y -5) +13
c. -2 - ≤
d. -0. 703 < 0.122 -2 .472
e. 3.88 – 1.335t ≥5.66

3.3 Cartesian Coordinate System

3.3.1 The Four Quadrants of the Cartesian Coordinate Plane
Group work 3.4

1. Write down the coordinates of all the 6 • •

points marked red in Figure 3.1 5
4• • • •
to the right . 3
2
1 • •
0 1 2 3 4 5 6 7 8 9
Figure 3.1

2. Write the coordinates of the points

A, B, C, D, E, F, G and H shown in A
3
Figures 3.2 to the right. 2 B
1 C
-5 -4 -3 -2 -1-10 1 2 3 4 5
F
-2 D
H - 3
G E

Figure 3.2

3. Name the quadrant in which the point of p(x, y) lines when:

a. x > 0 , y > 0 c. x > 0, y < 0
b. x < 0, y>0 d. x <0, y <0

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 Grade 8 Mathematics [LINEAR EQUATION AND INEQULITIES ]
For determing the position of a point on a plane you have to draw two mutually
perpendicular number lines. The horizontal line is called the X –axis, while the
vertical line is called the Y-axis. These two axes together set up a plane called
the Cartesian coordinate planes. The point of intersection of these two axis is
called the origin. On a suitably chosen scale, points representing numbers on the
X-axis are called X-coordinates or abscissa, while chose on the y-axis are
called Y-coordinates or ordinat. The x-coordinate to the right of the y-axis are
positive, while those to the left are negative. The y- coordinates above and below
the X-axis are positive and negative respectively. Let XOX' and YOY' be the
X-axis and the Y-axis respectively and let P be any point in the given plane.
For determing the coordinates of the point P, you draw lines through P parallel to
the coordinate axis, meeting the X-axis in M and the y-axis in N.
y
P(a,b)
N
b
x
M a 0

y′
Figure 3.3

y
The two axes divide the given plane
into four quadrants. Starting from the
positive direction of the X-axis and 2nd quadrant 1st quadrant
moving the anticlockwise (counter (0, 0)
x
clockwise) direction, the quadrants
which you come across are called the 3rd quadrant 4th quadrant

first, the second, the third and the

fourth quadrants respectively.
Figure 3.4

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Note: i. In the first quadrant all points have a positive abscissa and a
positive ordinate.
ii. In the second quadrant all points have a negative abscissa and
a positive ordinate.
iii. In the third quadrant all points have a negative abscissa and a
negative ordinate.
iv. In the fourth quadrant all points have a positive abscissa and a
negative ordinate.

EXERCISE 3E
1. Draw a pair of coordinate axes, and plot the point associated with each of
the following ordered pair of numbers.
A(-3, 4) D (0, -3) G (0, 6)
B(4, 6) E (-3, -2) H (2.5, 3)
C(4, -3) F (-5, 6) I (-2, 4.5)
2. Based on the given Figure 3.5 to y
the right answer the following
questions. 3 P
T 2 A
a. Write the coordinates of the point A, B, 1
S Q x
P, S, N and T. 0 1 2 3
-3 C-2 -1 -1 B
b. Which point has the coordinates (-1, -2)? D -2 M
N -3
c. Which coordinate of the points Q is
zero?
Figure 3.5
d. Which coordinate of the points D and M
is the same?
e. To which axis is the line DM parallel?
f. To which axis is the line AT parallel?
g. If F is any point on the line AT, state its
y-coordinate.
h. To which axis is the line PQ parallel?
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Challenge Problems
3. Answer the following:
a. On which axis does the point A(0,6) lie?
b. In which quadrant does the point B(-3, -6) lie?
c. Write the coordinates of the point of intersection of the x-axis and
y- axis.

3.3.2 Coordinates and Straight Lines

Group Work 3.5
1. Write down the equations of the y
lines marked (a) to (d) in the given
3 a
Figure 3.6 to the right. 2
1 b
x
-3 -2 -1 0-1 1 2 3 c
-2
-3 d

Figure 3.6
2. Write down the equations of the y
lines marked (a) to (d) in the given
a b c d
Figure 3.7 to the right. 3
2
3. Draw the graphs of the following 1
0 x
equations on the same coordinate -3 -2 -1 -1 1 2 3
-2
system: -3

a. y = x c. y = 4x
b. y = -x d. y = -4x Figure 3.7

4. True or false. If the statement is false, rewrite it to be true.

a. The line x = 30 is horizontal. b. The line y = -24 is horizontal.
5. True or false. If the statement is false, rewrite it to be true.
a. A line parallel to the y – axis is vertical.
b. A line perpendicular to the x – axis is vertical.
For exercise 6 – 9, identify the equation as representing a vertical line or
a horizontal line.
6. 2x + 7 =10 7. 9 = 3 + 4y 8. -3y + 2 = 9 9. 7 = -2x – 5
10. Write an equation representing the x – axis.
11. write an equation representing the y – axis.
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Graph of an equation of the form x = a (a ∈ )
The graph of the equation x= a (a , a≠0) is a
line parallel to the y-axis and at a distance of a unit from it.

Note: i. If a > 0, then the line lies to the y

right of the y-axis.

x=-a

x=0

x=a
ii. If a <0, then the line lies to the left
of the y-axis. (-a, 0)
0
(0, 0) (a, 0)
x
iii. The graph of the equation x= 0 is
the y-axis. a unit a unit

Figure 3.8

Example 18: Draw the graphs of the following straight lines.

a. x = 6 b. x = -6
Solution: First by drawing tables of values for x, and y in which x-is constant
and following this you plot these points and realize that the points lie
vertical line.

a.. x 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6
y -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8

b. x -6 -6 -6 -6 -6 -6 -6 -6 -6 -6 -6
y -5 -4 -3 -2 -1 0 1 2 3 4 5

8
7
6
x = -6 5 x=6
4
3
2
1 x
-6 -5 -4 -3 -2 -1 -1 0 1 2 3 4 5 6
-2
-3
-4
-5
-6

Figure 3.9

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Graph of an equation of the form y = b (b  )

The graph of the equation y= b (b , b≠0) y
is the line parallel to the x-axis and at a
distance of b from it. (0,b) y=b

Note: i. If b >0, then the line lies above the 0

(0,0)
y=0 x
x-axis. y = -b
(0,-b)
ii. If b < 0, then the line lies below the
x-axis.
iii. The graph of the equation y= 0 is Figure 3.10
the x-axis.

Example 19: Draw the graphs of y = 4.

Solution: First by drawing tables of values for x, and y in which y is constant
and following this you plot these points and realize that the points
lie horizontal line.
x -4 -3 -2 -1 0 1 2 3 4
y 4 4 4 4 4 4 4 4 4
y

5 y=4
4
3
2
1
-5-4-3 -2 -1-10 1 2 3 4 5
x
-2
-3
-4
-5

Figure 3.11
Graph of an equation of the form y = mx (m and m ≠0)
In grade 6 and 7 mathematics lesson we discussed about y = kx, where y is
directly proportional to x, with constant of proportionality k. For example y = 4x
where y is directly proportional to x with constant of poroportionality 4.
Similarly how to draw the graph of y= mx, (m ), look at the following
examples.
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Example 20: Draw the graphs of y = 5x. y
Solution:
y=5x
Step i: Choose some values for x, for
8
example let x= -2,-1, 0, 1 and 2. 7
6
Step ii: Put these values of x into the 5
4
equation y = 5x: 3
2
When x= -2: y =5(-2) = -10 1
x
-5 -4-3 -2-1-1 0 1 2 3 4 5
When x = -1: y = 5(-1) = -5 -2
When x = 0: y = 5(0) = 0 -3
-4
-5
When x = 1: y = 5(1) = 5 -6
-7
When x = 2: y = 5(2) = 10 -8

Step iii: Write these pairs of values in a table.

x -2 -1 0 1 2
y -10 -5 0 5 10

Step iv: Plot the points (-2, -10), (-1, -5), (0, 0) (1, 5) and (2, 10) and join them
to get a straight line. Figure 3.12
Step v: Lable the line y = 5x.
Example 21: Draw the graphs of y = -5x.
Solution:
Step i: Choose some values for x, for example let x= -2, -1, 0, 1 and 2.
Step ii: Put these values of x into the equation y= -5x y
When x= -2: y = -5(-2) =10
9
When x = -1: y= -5(-1) = 5 8
7
6
When x = 0: y = -5(0)=0 5
4
3
When x= 2: y = -5(2) = -10 2
1
-5 -4 -3 -2 -1-10 1 2 3 4 5 x
Step iii: Write these pairs of values in a table. -2
-3
x -2 -1 0 1 2 -4
-5
-6
y 10 5 0 -5 -10 -7
-8
-9 y=-5x
Step iv: Plot the points (-2, 10), (-1, 5), (0, 0)
(1, -5) and (2, -10) and join them to get Figure 3.13
a straight line.
Step v: Label the line y=-5x

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3.3.3. The Slope “m” Of Straight Line
Activity 3.3
Discuss with your friends.
1. What is a slope?
2. What is the slope of a line parallel to the y-axis?
3. What is the slope of a horizontal line?
4. What is the slope of a line parallel to the x – axis?
5. What is the slope of a line that rises from left to right?
6. What is the slope of a line that falls from left to right?
7. a. Draw a line with a negative slope.
b. Draw a line with a positive slope.
c. Draw a line with an undefined slope.
d. Draw a line with a slope of zero.

From your every day experience, you might be familiar with the idea of slope. In
this sub – topic you learnt how to calculate the slope of a line by dividing the
change in the y – value by change in the x – value, where the y – value is the
vertical height gained or lost and the x – value is the horizontal distance
travelled.
change in y − value
Slope =
change in x − value y
In Figure 3.14 to the right, B
Q(x2,y2)
consider a line drawn through the
points P(x 1 , y1 ) and Q(x 2 , y2 ).
From P to Q the change in the x P(x1,y1)
coordinate is (x 2 – x 1 ) and the
change in the y coordinate is x
(y2 – y 1 ). By definition, the slope A
of the line AB is given by:
y 2 − y1
;x2 ≠ x2 Figure 3.14
x 2 − x1

Note: If we denote the slope of a line by the letter “m”.

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Definition 3.1: If x1 ≠ x2 the slope of the line through the points

(x1, y1) and (x2, y2) is the ratio:
Change in y − value
Slope = m =
change in x − value
y 2 − y1
=
x 2 − x1

Group work 3.6

Discuss with your friends (partners).
1. In Figure 3.15 below, determine the slope of the roof.
B

4m
12 m C
A

Figure 3.15
2. State the slope of the straight line that contains the points p(1, -1) and
Q(8, 10).
3. Find the slope of a line segment through points (-7, 2) and (8,6).
4. Find the slope of each line.
a) y = 4 b) x = 7

Example 22: Find the slope of the line passing through the point P(-4, 2) and
Q(8, -4).
y 2 − y1 − 4 − 2 − 6 −1
Solution: Slope = m = = = =
x 2 − x 1 8 − (−4) 12 2
1 1
Therefore, − is the coefficient of x in the line equation y = − x .
2 2

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Example 23: Find the slope of the line passing through each of the following
pairs of points.
−1 −1
a) P(4, -6) and Q(10, -6) b) P  ,−4  and Q ,4 
 4   4 
Solution:
y 2 − y 1 −6 − (−6) − 6 + 6 0
a. m = = = = =0
x 2 − x1 10 − 4 6 6
y 2 − y1 4 − (−4) 8
b. m = = = undefined
x 2 − x1 −1  −1  0
− 
4  4 

Note: i. The horizontal line has a slope of 0.

ii. The vertical line has no slope (not defined).

Example 24: Draw the graphs of the following equations on the same
Cartesian coordinate plane.
a. y = b. y = -3x c. y = 4x d. y =
Solution: First to draw the graph of the equation to calculated some ordered
pairs that belongs to each equation shown in the table below.
x -3 -2 -1 0 1 2 3
y= 0

y = -3x 9 6 3 0 -3 -6 -9
y = 4x -12 -8 -4 0 4 8 12
y= -2 0 2

y
y=-3x y=4x

9
8
7
6
5
4
3
2
1
-5 -4 -3 -2 -1 0 1 2 3 4 5
x
-1
-2
-3
-4
-5
y= x -6
-7
-8
-9

y= x
Figure 3.16
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From the above graphs, you can generalize that:
i. All orderd pairs, that satisfy each linear equation of the form y = mx
(m , m≠0) lies on a straight lines that pass through the origin.
ii. The equation of the line y = mx, m is called the slope of the line, and the
graph passes the 1st and 3rd quadrants if m > 0, and the graph passes
through the 2nd and 4th quadrants if m< 0.

EXERCISE 3F
1. Draw the graphs of the following equations on the same coordinate
system:
a. y= -6x b. y = 6x c. y = d. y =
2. Draw the graphs of the following equations on the same coordinate
system:
a. y + 4x = 0 c. x = 3 e. 2x – y = 0
b. 2y = 5x d. x + 4 = 0 f. -
3. Complete the following tables for drawing the graph of y =
x 1 6 3
y
(x, y)

Challenge Problems
4. Point (3, 2) lies on the line ax+2y = 10. Find a .
5. Point (m, 5) lies on the line given by the equation 5x – y = 20. Find m.
6. Draw and complete a table of values for the graphs y=2x -1 and y= x- 2
7. a. Show that the choice of an ordered pair to use as (x 1 , y1 ) does not affect
the slope of the line through (2, 3) and (-3, 5).
y 2 − y1 y1 − y 2
b. Show that =
x 2 − x1 x1 − x 2
For Exercise 8 – 11 fined the slope of the line that passes through the
two points.
 − 2 1 8 −5
8. P  ,  and Q , 
 7 3 7 6 
9. A  ,  and B , 
1 3 1 -4
 2 5 4 5 
10. C (0, 24) and D (30, 0)
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11. E 0,  and F 0, 

5 9
 7  26 
12. Find the slope between the points
A (a + b, 4m – n) and B (a – b, m + 2n)
13. Find the slope between the points
C(3c – d, s + t) and D (c – 2d, s – t)
14. Write the equation of the line which has the given slope “m” and which
passes through the given point.
3 3
a. (2, 10) and m = - 4 b. (4, -4) and m = c. (0, 0) and m =
2 5
15. State the slope and y-intercept of the line 2x + y + 1 = 0.
16. Find the slope and y-intercept of y − y o = m(x − x o ) where x o and y o are
constants.
17. Find the slope and y-intercept of each line:
a. (x + 2) (x + 3) = (x − 2) (x − 3) + y
b. x = mu + b
18. State the slope and y-intercept of each linear equations.
a. 6(x + y) = 3(x − y)
b. 2(x + y) = 5(y + 1)
c. 5x + 10y − 20 = 0
19. Write the slope-intercept equation of the line that passes through (2,5)
and (-1,3).

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Summary For unit 3

1. You can transform an equation into an equivalent equation that does not
have brackets. To do this it is necessary to remember the following rules.
a. a + (b + c) = a + b + c c. a (b + c) = ab + ac
b. a – (b + c) = a – b – c d. a ( b – c) = ab – ac
2. The following rules are used to transform a given equation to an equivalent
equation.
a. For all rational numbers a, b and c:
If a =b then a + c = b + c and a – c = b – c. that is, the same number may
be added to both sides and the same number maybe subtracted from both
sides without affecting the equality.
b. For all rational numbers a, b and c where c≠ 0:
If a = b then ac = bc and . That is both sides may be multiplied by the
same non-zero number and both sides may be divided by the same non-
zero number without affecting the equality.
3. To solve word problems, the following steps will help you to develop the
skill. The steps are:
a. Read the problem carefully, and make certain that you understand the
meanings of all words.
b. Read the problem a second time to get an overview of the situation being
described and to determine the known facts as well as what is to be found.
c. Sketch any figure, diagram or chart (if any) that might be helpful in
analyzing the problem.
d. Choose a variable to represent an unknown quantity in the problem.
e. Form an equation containing the variable which translates the conditions
of the problem.
f. Solve the equation.
g. Check all answer back into the original statements of the problem.
4. The following rules are used to transform a given inequality to an
equivalent inequality.
a. For all rational numbers a ,b and c, if a < b then a + c < b + c or
a - c < b - c. That is, if the same number is added to or subtracted from
both sides of an inequality, the direction of the inequality remains
unchanged.
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Grade 8 Mathematics [LINEAR EQUATION AND INEQULITIES ]
b. For all rational numbers a, b and c
i. If a < b and c > 0, then ac < bc or That is, if both sides of an
inequality are multiplied or divided by the same positive number, the
direction of the inequality is unchanged.
ii. If a < b and c < 0, then ac > bc or . That is, if both sides are
multipled or divided by the same negative number, the direction of the
inequality is reversed.
5. The two axes divide the given plane into four quadrants. Starting from the
positive direction of the X-axis and moving the anticlockwise direction, the
quadrants which you come across are called the first, the second, the third
and the fourth quadrants respectively.
y

ii. quadrant i. quadrant

x<0 x>0
y>0 y>0

x
iii. quadrant iv. quadrant
x<0 x>0
y<0 y<0

Figure 3.17
6. If x 1 ≠ x 2 the slope of the line through the points (x 1 , y 1 ) and (x 2 , y 2 ) is the ratio:
Slope = m =

=
7. All orderd pairs, that satisfy each linear equation of the form
y = mx (m , m≠ 0) lies on a straight lines that pass through the origin.
8. The equation of the line y = mx, m is called the slope of the line, and the graph
passes the 1st and 3rd quadrants if m > 0, and the graph passes through the 2nd
and 4th quadrants if m< 0.

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 Grade 8 Mathematics [LINEAR EQUATION AND INEQULITIES ]

Miscellaneous Exercise 3
I. Write true for the correct statements and false for the incorrect ones.
1. For any rational numbers a, b and c, then a(b + c) = ab + ac.
2. If any rational number a > 0, ax +b > 0, then the solution set is
3. If any rational number a < 0, ax +b > 0, then the solution set is
.
4. The equation of the line y = 4x, 4 is the slope of the line and the graph
passes the 2nd and 3rd quadrants, since 4 > 0.
5. The equation of the line y= 4x + 6 that pass through the origin of
coordinates.
6. The graphs of the equation y= b (b  , b ≠ 0), if b > 0 then the equation of
the line lies above the x-axis.
7. The graph of the equation x= a (a  , a ≠ 0), if a < 0 then the equation of
the lines to right of the y-axis.

II. Choose the correct answer from the given alternatives

8. In one of the following linear equations does pass through the origin?
a. y = c. y =

b. y = -3x + d. y = 2x - 6
9. The solution set of the equation is:
b. c. d.
10. The solution set of the equation 2x +3 (5 -3x) =7 (5- 3) is:
a. {5} b. c. d.
11. If , (x ≠0), then which of the following is the correct value of
x?
c. d.
12. If x is a natural number, then what is the solution set of the inequality
0.2x - ≤ 0.1x?
b. φ c. d.

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 Grade 8 Mathematics [LINEAR EQUATION AND INEQULITIES ]
13. Which one of the following equations has no solution in the set of
integers ?
a. 6x + 4 = 10 c. 9 – 12x = 3
b. 8x +2 = 4x – 6 d.
14. What is the solution set of the inequality 20 (4x -6) ≤80 in the set of
positive integers?
c. φ d.
15. The sum of the ages of a boy and his sister is 32 years. The boy is 6 years
older than his sister. How old is his sister?
a. 15 b. 19 c. 14 d. 13

III. Work out problems

16. Solve each of the following linear equation by the rules of
transformation.
a. 4x + 36 = 86 – 8x
b. 12x – 8+ 2x -17 = 3x – 4 – 8 +74
c. 4(2x -10) = 70 + 6x
d. 20 – 2x = 62 (x – 3)
e. 2(6y – 18) -102 = 78 – 18(y+2)
f. 7(x + 26 + 2x) = 5(x + 7)
17. Sove each of the following equations.
c.
18. Solve each of the following linear inequalities by the rules of transformation.
a. 6x – 2 < 22 c. 8x – 44 < 12(x -7) e.
b. -9 ≤ 3x + 12 d. 8(x -3) ≥15x -10 f. 6(2+6x) ≥ 10x -12
19. (word problems)
a. The sum of three consecutive odd integers is 129. Find the integers.
b. Two of the angles in a triangle are complementary. The third angle is
twice the measure of one of the complementany angles. What is the
measure of each of the angles?
c. Abebe is 12 years old and his sister Aster is 2 years old. In how many
years will Abebe be exactly twice as old as Aster?
20. Draw the graphs of the equations y = and y = - on the same coordinate
plane. Name their point of intersection as p. State the coordinate of the
point p.
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 Grade 8 Mathematics [LINEAR EQUATION AND INEQULITIES ]

21. Find the equation of the line with y-intercept (0,8) and slope .
22. Find the slope and y-intercept of y = 10x −
23. Find the slopes of the lines containing these points.
a) (4,-3) and (6, -4)
b) and

c) and
24. Find the slope of the line x = -24.
25. Find a and b, if the points P(6,0) and Q (3,2) lie on the graph of ax + by = 12.
26. Points P(3,0) and Q(-3,4) are on the line ax + by = 6. Find the values of
a and b.
27. Point (a,a) lies on the graph of the equation 3y = 2x − 4. Find the value of a.
28. Find anequation of the line containing (3,-4) and having slope -2. If this line
contains the points (a,8) and (5,b), find a and b.

100
UNIT
SIMILAR FIGURES
4

Unit outcomes
After Completing this unit, you should be able to:
 know the concept of similar figures and related
terminologies.
 understand the condition for triangles to be similar.
 apply tests to check whether two given triangles are
similar or not.

Introduction
You may see the map of Ethiopia either in smaller or larger size, but have you
asked yourself about the difference and likness of these maps? In geometery this
concept is described by “similarity of plane figures” and you learn this concept
here in this unit. You begin this by studying similarity of triangles and how to
compare their areas and perimeters.

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 Grade 8 Mathematics [SIMILAR FIGURES]

4.1 Similar Plane Figures

Activity 4.1
Discuss with your teacher
1. Which of the following maps are similar?

(a) (b) (c) (d)

Figure 4.1
2. Which of the following pictures or polygons are similar?

(a) (b) (c) (d)

Figure 4.2

3. Which of the following polygons are similar?

D 8cm u 8cm T
C
P 2cm Q
6cm 6cm
6cm

Polygon I
6cm

Polygon II Polygon III 2cm

2cm

A B
8cm R S S R
8cm 2cm

Figure 4.3
Similar geometric Figures are figures which have exactly the same shape. See
Figure 4.4, each pair of figures are similar.

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 Grade 8 Mathematics [SIMILAR FIGURES]

Small
Larger
Larger rectangle Rectangle Small
triangle
triangle

(a)
Figure. 4.4 Geometric Figures (b)

Therefore, geometric figures having the same shape, equal corresponding angles
and corresponding sides are proportional are called similar figures.

4.1.1 Illustration and Definition of Similar Figures

Group Work 4.1
1. A square and a rectangle have corresponding angles congruent.

W C
D

W W

A B
W
Figure. 4.5
Are they similar? Why?

2. A square and a rhombus have the corresponding sides proportional.

x C S
D

t t
x x
P R

A B t t
x

Are they similar? Why? Figure. 4.6 Q

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 Grade 8 Mathematics [SIMILAR FIGURES]
3. Which members of these families of shape are similar:
a) squares d) circles g) regular hexagons
b) rectangles e) equilateral triangles h) trapeziums
c) Parallelograms f) isosceles triangles
4. Which members of these families of solid shape are similar:
a) cubes c) spheres e) pyramids
b) cuboids d) tetrahedrons f) cones

Definition 4.1: Two polygons are similar, if:

i. their corresponding sides are proportional.
ii. their corresponding angles are congruent (equal).

Example1: Which of the following polygons are similar? Which are not? state
the reason.

Square Rectangle
Rectangle Rhombus

Figure 4.7 Polygons

Solution: From the given Figure 4.7 above
a. The square and the rectangle are not similar because their corresponding
angles are congruent but their corresponding sides are not proportional.
b. The square and the rhombus are not similar because their corresponding
sides are proportional but their corresponding angles are not congruent.
c. The rectangle and the rhombus are not similar because their
corresponding angles are not congruent and corresponding sides are not
proportional.

Example2: Tell whether each pairs of in Figures 4.8 is similar or not.

(a) (b)
Figure 4.8 Polygons

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 Grade 8 Mathematics [SIMILAR FIGURES]

Solution:
a. Paris of the quadrilaterals have the same shape and angles but have not
the same size. Therefore, they are not similar.
b. Paris of the quadrilaterals have not the same shape. Therefore, they are
not similar.

Exercise 4A
Which of the following figures are always similar?
a. Any two circles. e. Any two squares.
b. Any two line segments. f. Any two rectangles.
c. Any two quadrilaterals. g. Any two equilateral triangles.
d. Any two isosceles triangles.

4.1.2 Scale Factors and Proportionality

Activity 4.2
Discuss with your friends.
1. Have you observed what they do in the film studio?
2. How do you see films in the cinema house?
3. How are the pictures enlarged on the cinema screen?
4. What is meant by scale factor?
5. What is meant by proportional sides of similar figures?

To discuss Activity 4.2, it is important to study central enlargement (central

stretching). Central enlargement which is either increases or decreases the size
of figures with out affecting their shapes.

Under an enlargement
1. Lines and their images are parallel.
2. Angles remain the same.
3. All lengths are increased or decreased in the same ratio.

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Grade 8 Mathematics [SIMILAR FIGURES]
Positive enlargement
In Figure 4.9 triangle A 1 B 1 C 1 is the image of triangle ABC under enlargement.
O is the centre of enlargement and the lines AA 1 , CC 1 and BB 1 when produced
must all pass through O.

B1

C1
Image

B
C Object

Centre of enlargement
A1 A O

Figure 4.9 Positive enlargement

The enlargement in Figure 4.9 is called a positive enlargement, because both

the object and its image are on the same side of the centre of enlargement. Also
the image is further from O than the object.

Note: - AB is parallel to A1B1.

- BC is parallel to B1C1.
- AC is parallel to A1C1.
- Angle BAC≅ angle B1 A1 C1.
- Angle ABC ≅ angle A1 B1 C1.
- Angle BCA≅angle B1C1 A1.

"In Figure 4.9 above the object and the image are similar why?"
?
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 Grade 8 Mathematics [SIMILAR FIGURES]

Example3: Given the rectangle, ABCD and A as the centre of enlargement.

Draw the image AB'C'D' after enlargement of each side of ABCD
twice.
B’ B’ C’
C’

B C C
B B C

A D A D’ A D’
D D

Figure. 4.10
Solution:
Point A is the centre of enlargement and is fixed. A' is at A. Since each sides of
A'B'C'D' enlarged twice of each side of ABCD,
AB AD AC DC 1
= = = = or A'B'=2AB, A'D'=2AD A'C'=2AC and,
A' B' A' D' A' C' D' C' 2
D'C' = 2DC. The number 2 in this equation is called the constant of
proportionality or scale factor.

? "Can you define a scale factor based on example 3 above?"

Definition 4.2: Scale factor – the ratio of corresponding sides usually

expressed numerically so that:
length of line segment on the enlargement
Scale factor=
length of line segment on the original

Notation of scale factor = K

Example 4: The vertices of triangle ABC have co-ordinates A(2,1), B(4,1) and
C(3,4). Find the co-ordinates of triangle A 1 B 1 C 1 after an
enlargement, scale factor 2, with centre at O.
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 Grade 8 Mathematics [SIMILAR FIGURES]
Y

8 C1
7
6
5
4 C
3
2 A1
B1
1 A B
0 1 2 3 4 5 6 7 8 X

Figure. 4.11
Solution:
In Figure 4.11 you can see the object, triangle ABC and its image under
enlargement, triangle A 1 B 1 C 1 with co-ordinates: A 1 (4,2), B 1 (8,2) C 1 (6,8).

Example 5: Give the shape PQRS and point O. Draw the image P'Q'R'S' after
enlargement of each side of PQRS twice.
P’
P

P O Q
O• Q’

Q R
S
R’
S R S’
Figure 4.12

Solution: Join O to P, Q, R and S. Since each sides of P' Q' R' S' enlarged
twice of each side of PQRS.
OP OQ OR OS 1
= = = = or
OP' OQ' OR' OS' 2
OP'=2OP, OQ'= 2OQ, OR'=2OR and OS'=2OS
Hence, the number 2 is called the constant of proportionality or scale factor.
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 Grade 8 Mathematics [SIMILAR FIGURES]

Can you define a central enlargement (central stretching) in your own

? word?

Definition 4.3: A mapping which transform a figure following

the steps given below is called central enlargement.

Step i: Mark any point O. This point O is called center of the central
enlargement.
Step ii: Fix a number K. This number K is the constant of proportionality.
Step iii: Determine the image of each point A such as A' such that A'O=KAO.
Step iv: The image of point O is itself.
According to the definition 4.3 when you find the image of a plane figure,
i. If K > 1, the image figure is larger than the object figure.
ii. If 0 < K < 1, the image figure is smaller than the object figure.
iii. If K=1, the image figure is congruent to the object figure.

Example 6: Enlarge triangle PQR by scale factor 3 and O is the centre of

enlargement as shown below.

P O Q
Figure 4.13

Solution:
R′
Copy triangle PQR and the point O inside the
triangle. R
Step i: Make point P', R' and Q' on
O
OP, OQ and OR such that P Q
P'O = 3PO, Q′O = 3QO and P′ Q′
Figure 4.14
R′O = 3RO see Figure 4.14 to the
right.
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 Grade 8 Mathematics [SIMILAR FIGURES]
Stepii: Join the points P',Q', R' with
line segment to obtain P'Q'R'
(which is the required Figure).

Exercise 4B
1. Draw the image of the shape KLMN K •C
after an enlargement by scale factor
N
1
with center O. Label the image
2
K' L' M' N'.
M L

Figure 4.15

L N
2. Work out the scale factor of the
enlargement that takes in Figure
A C
4.16, triangle ABC on the
triangle LMN.

M
B

Figure 4.16
3. Copy the Figure 4.17 below.
With O as centre, draw the O
image of the shaded shape after
enlargement by:
1
a. scale factor
4
3
b. scale factor
4 Figure 4.17

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 Grade 8 Mathematics [SIMILAR FIGURES]

4.2 Similar Triangles

4.2.1 Introduction to Similar Triangles
Group Work 4.2
1. Consider Figure 4.18 below:
D M

40 20 x 12

0 0 0 0
54 62 54 62
E y F N L
24
Figure 4.18
a. Are ∆DEF similar to ∆MNL? Why?
b. If ∆DEF similar to ∆MNL then find the value of X and Y.

2. Consider Figure 4.19 below:

A P

10c 10b 6c 6b

B C Q 6a R
10a
Figure 4.19

a. ∆ABC is similar to ∆PQR. Explain the reason.

b. ∆ABC is not similar to ∆PRQ. Explain the reason.
3. ∆XYZ is given such that ∆DEF similar ∆XYZ. Find XY and YZ when the scale
factor from ∆XYZ to ∆DEF is 6 and DE=7, EF=12 and XZ=36.

You have define similar polygon in section 4.1.1. You also know that any
polygon could be dived into triangles by drawing the diagonals of the polygon.
Thus the definition you gave for similar polygons could be used to define similar
triangles.

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Grade 8 Mathematics [SIMILAR FIGURES]

A D
Definition 4.4: ∆ABC is
similar to ∆DEF, if 4c 4b
6c 6b
i. their corresponding sides
are proportional.
B C E F
ii. their corresponding angles 6a 4a
Figure. 4.20
are congruent.

That is symbolically:
∆ABC~∆DEF if and only if
1. ∠A≡∠D
2. ∠B≡∠E corresponding angles are congruent
3. ∠C≡∠F

4.
AB BC
=
DE EF

5. corresponding sides are proportional

BC AC
=
EF DF

6.
AB AC
=
DE DF
or the above three facts 4, 5 and 6 can be summarized as
AB BC AC
= = =K
DE EF DF

Note: From definition (4.4) of similar triangles it is obvious that similarity is

a transitive relation. That is: If ∆ABC~ ∆DEF and ∆DEF ~∆XYZ, then
∆ABC~∆XYZ.

Example 7: Let ∆ABC~∆DEF. As shown in Figure 4.21 below. Find

a. m(∠F) b. m(∠E) c. the length of BC.

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Grade 8 Mathematics [SIMILAR FIGURES]

C F
0
100
a 5

0 0
A 30 30
16 B 12 E
D

Figure 4.21

Solution: Given ∆ABC ∼ ∆DEF …. By definition 4.4

a. ∠A≡∠D if and only if m (∠A) = m (∠D) = 30°
∠B ≡ ∠E ....... Marked angels
Therefore, ∠C ≡ ∠ F if and only if m(∠C) = m (∠F) = 100°
Therefore, m (∠F) = 100°
b. m(∠A) + m(∠B) + m (∠C) = 180° ...Angle sum theorem
30° + m (∠B) +1000 = 180° ............. Substitution
m (∠B) = 180°- 130° = 50°
Therefore, m (∠E) = 50°.
c. Given ∆ ABC~∆DEF
AB BC AC
implies = = ................ Definition 4.4
DE EF DF
16 a
= .......................... Using the 1st and 2nd proportions
12 5
12a = 80 ........................ Cross multiplication
80
a= .........................Dividing both sides by 12
12
80
Therefore, the length of BC= unit.
12

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Grade 8 Mathematics [SIMILAR FIGURES]
Example 8: In Figure 4.22 below, show that ∆ABC and ∆LMN are similar.
A
L
0
55
5 cm 10 cm
2cm 4cm

0
0 75
0 50
75
M B 7.5 cm C
3 cm N

Figure 4.22

Solution: You begin by finding the unknown angels in the triangles. You
know the size of two angles in each triangle and you also know that
the sum of the angles of a triangle is 1800. Therefore, it is easy to
calculate the size of the unknown angles.
In ∆ABC, m(∠ABC)+m(∠BCA)+m(∠CAB)=180°.... Why?
⇒ 750+500+m(∠CAB)=1800 .......................... Substitution
⇒ m(∠CAB)=1800-1250
⇒ m(∠CAB)=550
In ∆LMN, m(∠LMN)+m(∠MNL)+m(∠NLM)=180°....Why?
⇒ 750+m(∠MNL)+550=1800 ....................... Substitution
⇒ m(∠MNL)=1800-1300
⇒ m (∠MNL) =500
The corresponding angles are equal, to show the corresponding
sides are in the same ratio. Let us check whether the corresponding
sides are proportional or not. In short let us check.
∆ABC~∆LMN
AB BC AC
⇒ = = =K (constant of proportionality)
LM MN LN
5cm 7.5cm 10cm
= = = 2.5
2cm 3cm 4cm

114
Grade 8 Mathematics [SIMILAR FIGURES]

The corresponding sides are proportional with constant of proportionality

equals 2.5. Therefore ∆ABC~∆LMN….By definition 4.4.

Example 9: In Figure 4.23, ∆ABC~∆XYC. A

3
If CX=6, CY=5, AX= 3 and
X
AB= 8, find XY. 8
6

B Y 5 C

Figure 4.23

Solution:
By definition of similar triangles, we have:
AB BC AC
= = =K (constant proportionality)
XY YC XC
AB AC
Thus = .............. Using the 1st and 3rd proportions
XY XC
8 AX + XC
⇒ = since AX+XC=3+6=9
XY XC
8 9
⇒ =
XY 6

⇒ 9XY = 48 .......................... Cross multiplication

48
⇒ XY = ..............................Dividing both sides by 9.
9

Example 10: If ∆ABC≡∆DEF, then

a. Is ∆ABC~∆DEF?
b. Justify your answer.

Solution:
a. Yes
b. Suppose ∆ABC ≡ ∆DEF are as shown in Figure 4.24 below:

115
 Grade 8 Mathematics [SIMILAR FIGURES]
B
D E

F A C

Figure 4.24
Then i. ∠ABC≡∠DEF
∠BCA≡∠EFD and
∠CAB≡∠FDE
Hence corresponding angles are congruent.
ii. AB ≡ DE , BC ≡ EF and AC ≡ DF implies
AB=DE, BC=EF, and AC=DF. Thus
AB BC AC
= = =1
DE EF DF
Hence the corresponding sides are proportional. Since (from (i) and (ii))the
corresponding sides are proportional with constant of proportionality 1, and
also the corresponding angles are congruent, then the triangles are similar by
definition 4.4. From example 10 above you can make the following
generalization.

Note: Congruence is a similarity where the constant of proportionality is 1.

This general fact is equivalent of the statement given below. If two
triangles are congruent, then they are similar.

Exercise 4C
1. If ∆ABC~ ∆B'A'C', what are the pairs of corresponding angles and the pairs
of corresponding sides?
2. If ∆ABC~∆A'B'C' and AC=20cm, A'C'=15cm, B'C'=12cm and A'B'=9cm,
find the lengths of the other sides of ∆ABC.

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 Grade 8 Mathematics [SIMILAR FIGURES]

3. The sides of a triangle are 4cm, 6cm, and a cm respectively. The corres-
ponding sides of a triangle similar to the first triangle are b cm, 12 cm and
8 cm respectively. What are the lengths a and b?
4. Are two similar triangles necessarily congruent? Why?
5. What is the length of the image of a 20cm long segment after central
1
stretching with a scale factor ?
2
6. If ∆DEF~∆KLM such that DE =(2x + 2)cm, DF = (5x − 7)cm, KL = 2cm,
KM = 3cm and EF= 10cm, then find LM. Z
7. In Figure 4.25 if ∆XYZ~∆WYP, P
express d in terms of a, b and c. a
d

X b W C Y
Figure 4.25

8. In Figure 4.26 below, if∆ABC~∆XBZ with XB=6cm, BZ=5cm, CX=8cm

and AC=7cm. C
What is the length of 8 cm

a. BC? b. XZ? 7 cm X
6 cm

A B
Z
Figure. 4.26

Challenge Problems A
9. Write down a pair of similar triangles B

in Figure 4.27 to the right. Find CD 15 18

10
and AC, if AE is parallel to BD.
30 C
E D
Figure. 4.27

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 Grade 8 Mathematics [SIMILAR FIGURES]
4.2.2 Tests for Similarity of Triangles (SSS, SAS and AA)
Activity 4.3
Discuss with your teacher before starting the lesson.
1. Can you apply AA, SAS and SSS similarity theorems to decide whether a given
triangles are similar or not?
2. Which of the following is (are) always correct?
a. Congruent by SAS means similar by SAS.
b. Similar by SAS means congruent by SAS.
c. Congruent by SSS means similar by SSS.
d. Similar by SSS means congruent by SSS.

"But to decide whether two triangles are similar or not, it is necessary to

? know all the six facts stated in the definition 4.4?"
To prove similarity of triangles, using the definition of similarity means checking all
the six conditions required by the definition. This is long and tiresome. Hence we
want to have the minimum requirements which will guarantee us that the triangle
are similar, i.e all the six conditions are satisfied. These short cut techniques are
given as similarity theorems.

In this section you will see similarity theorems as you did see in grade 6
mathematics lessons congruence theorems for congruency of triangles.

Theorem 4.1: (AA-Similarity theorem)

If two angles of one triangle are congruent to the corresponding
two angles of another triangle, then the two triangles are similar.

V
Example 11: In Figure 4.28 below
∠R≅∠V and show that ∆RSW~∆VSB. 0
45
W

S
B
0
45
Figure 4.28
R
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 Grade 8 Mathematics [SIMILAR FIGURES]

Proof:
Statements Reasons
1. ∠R≅∠V 1. Degree measures are equal
2. ∠RSW≅∠VSB 2. Vertical opposite angles
3. ∆RSW~∆VSB 3. AA similarity theorem

Example 12: In ∆ABC and ∆XYZ if ∠ABC ≅ ∠XYZ, ∠ACB≅ ∠XZY,

AB=8cm, AC=10cm and XY=4cm as shown in Figure 4.29
below; find the length of XZ .
A X

B C Y Z
Figure 4.29

Solution: since ∠B ≅ ∠Y and ∠C ≅ ∠Z, we can say that

∆ABC ~ ∆XYZ by AA similarity theorem. Then,
AB AC
= ....................... Definition of similar triangle.
XY XZ
8 10
Thus: = ..................Substitution
4 XZ
8XZ= 4 × 10 ...................Cross multiplication
40
XZ= = 5 cm
8

Theorem 4.2: (SAS-Similarity theorem)

If two sides of one triangle are proportional to the corresponding
two sides of another triangle and their included angles are also
congruent, then the two triangles are similar.

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 Grade 8 Mathematics [SIMILAR FIGURES]
Example 13: In Figure 4.30 below, find DE.
\

24
A 12
B

10 18
E
16

C
Figure 4.30
Solution:
1. ∠ABC≅∠EBD 1. Vertical opposite angles
AB 12 2 BC 16 2 2. The ratio of the lengths of the
2. = = and = =
EB 18 3 BD 24 3 corresponding sides are equal
3. ∆ABC~∆EBD 3. SAS similarity theorem
CA 2 4. Corresponding sides of similar triangles
4. =
DE 3 are proportional.
10 2 5. Substitution
5. =
DE 3
6. 2DE=30 6. Cross-product property
7. DE = 15 cm 7. solve for DE

Example 14: In Figure 4.31 below ∆ABC and ∆DEF, are given where,
m(∠A) = m(∠D) = 550, AB = 30 cm, AC = 100cm,
DE = 15cm and DF = 50cm.
a. Are the two triangles similar?
b. Justify your answer.

B
E

30 cm
15 cm

0 0
55 55
F
A 100 cm C D 50 cm
Figure 4.31

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 Grade 8 Mathematics [SIMILAR FIGURES]

Solution:
a. Yes
b. Suppose ∆ABC and ∆DEF are as shown in Figure 4.31 then,
AB 30cm
= =2
DE 15cm
AC 100cm
= =2
DF 50cm
AB AC
= =2
DE DF

Hence two sides of ∆ABC are proportional to two corresponding sides of ∆DEF.
Furthermore, m(∠A) = m(∠D) = 550, which shows that m(∠A) = m(∠D). Thus
the included angles between the proportional sides of ∆ABC and ∆DEF are
congruent. Therefore ∆ABC~∆DEF by SAS similarity theorem.

Theorem 4.3: (SSS-Similarity theorem)

If the three sides of one triangle are in proportion to the three sides
of another triangle, then the two triangles are similar.

Example 15: Based on the given Figure 4.32 below decide whether the two
triangles are similar or not. Write the correspondence.

C E

8 cm 10 cm 16 cm 20 cm

F D
A 14 cm B 28 cm

Figure 4.32

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 Grade 8 Mathematics [SIMILAR FIGURES]

Solution:
AC 8cm 1
= =
FE 16cm 2
While finding proportional sides don't forget to
CB 10cm 1
= = compare the smallest with the smallest and the
ED 20cm 2
AB 14cm 1 largest with the largest sides.
= =
FD 28cm 2
AC CB BA 1
Hence = = = or the sides are proportional.
FE ED DF 2
Therefore ∆ABC~∆FDE ............................. By SSS similarity theorem.
From this you can conclude that: ∠A≡∠F, ∠B≡∠D and ∠ C≡∠E.

Exercise 4D
1. If ∆ABC~∆XYZ and AC=10cm, AB=8cm and XY=4cm, find the length of
XZ .
2. Prove that any two equilateral triangles are similar.
3. In Figure 4.33 below determine the length x of the unknown side of
∆ABC, if ∆ABC~∆DEF.
x cm E
A B

5 cm 10 cm 12 cm
6 cm

D F
C Figure 4.33 10 cm

4. In Figure 4.34 below C C

∠CAB≡∠CDE,
AC=4cm, DC=5cm and 4 cm 5 cm
DE=7cm. Determine the
length of sides AB of B D E
A 7 cm
∆ABC. Figure 4.34

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 Grade 8 Mathematics [SIMILAR FIGURES]

5. In Figure 4.35 of ∆ABC, AC = 20cm, C

AB = 16cm, BC = 24cm.If D is a point

on AC with CD = 15cm and, E is a
point on BC with CE = 18cm, then:
a. Show that ∆DEC~∆ABC. D E
b. How long is DE?
6. For any triangles ABC, if ∠A≡∠B, then
A B
show that ∆ABC~∆BAC. Figure 4.35

7. Show that the corresponding altitudes of similar triangles ABC and PQR
have the same ratio as two corresponding sides (See Figure 4.36).
B
Q

P R
A C S
D
Figure 4.36
Challenge Problems C
8. For the plane Figure 4.37 below
BE and AD are altitude of ∆ABC. D
E
prove that
a. ∆ADC~∆BEC F
b. ∆AFE~∆BFD
A B
Figure 4.37

9. In Figure 4.38 to the right E is

G
any point on AB and G is any D C
point on DC. If AB // DC ,
prove that:
DG DF F
a. ∆DGF~∆BEF. b. = .
EB FB A
E B
Figure. 4.38
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 Grade 8 Mathematics [SIMILAR FIGURES]
10. In Figure 4.39 to the right determine the C
length AB of ∆ABC.
5cm

7cm
D E
4cm
A B
Figure 4.39

4.2.3 Perimeter and Area of Similar Triangles

Group work 4.3

1. Find the ratio of the areas of two
similar triangles:
5
a. if the ratio of their corresponding sides is .
4
10
b. if the ratio of their perimeters is .
9
2. For the given Figure 4.40 below, find 8cm

a. the area .
b. the perimeter. 6cm 5cm

Figure 4.40

In lower grades you have seen how to find the perimeter and area of some
special plane figures such as triangles, rectangles, squares, parallelograms
and trapeziums. In the proceeding section of this unit you have been dealing
with the areas and perimeters of similar plane figures. The perimeters and areas
of similar plane figures have very interesting relations to their corresponding
sides. You can compare the ratios of perimeters or that of the areas of similar
polygons with out actually calculating the exact values of the perimeters or the
areas. Look at the following example to help you clearly see these relations.

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 Grade 8 Mathematics [SIMILAR FIGURES]

a b c
Example 16: In Figure 4.41 below if ∆ABC ~ ∆XYZ with = = .
x y z
Determine the relationship between:
a. The altitudes of the two triangles
b. The perimeters of the two triangles.
c. The areas of the triangles.
C Z

a y x
b h2
h1

A B X Y
D W
c z
Figure. 4.41
Solution:
∆ABC ~ ∆XYZ. let the constant of proportionality between their corresponding
sides be K, i.e.
a
=k implies a = kx
x
b
= k implies b=ky
y
c
= k implies c=kz
z
h1
= k implies h 1 =kh 2
h2
a. Let CD be the altitude of ∆ABC from vertex C on AB and ZW be the
altitude of ∆XYZ from vertex Z on XY
Then ∠CDB ≅ ∠ZWY …. Both are right angles.
∠B ≅ ∠Y …… Corresponding angles or similar triangles.
Therefore, ∆CDB ∼ ∆ZWY … By AA similarity theorem.
CD CB
Thus = ……. Definition of similar triangles.
ZW ZY
h1 a
⇒ = ……. Substitution
h2 x
h a
⇒ 1 = k …….. Since = k proportional sides
h2 x
h1
=k implies h 1 = kh 2
h2
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 Grade 8 Mathematics [SIMILAR FIGURES]
b. P(∆ABC)=a+b+c
=kx+ky+kz
=k(x+y+z)
and p(∆xyz) = x+y+z
P( ∆ABC) K(x + y + z)
then = =k
P( ∆XYZ) x+y+z

Hence the ratio of the perimeters of the two similar triangles is “k” which is
equal to the ratio of the lengths of any pair of corresponding sides.
1
c. a(∆ABC) = c.h 1
2
1
= (kzh 1 )
2
1
= (kz.kh 2 )
2
1
and a(∆xyz)= zh 2
2
1
kz.kh 2
a (ΔABC) 2
then = = k2
a (ΔXYZ) 1
zh 2
2
Hence the ratio of the areas of the two similar triangles is k2, the square of
the ratio of the lengths of any pair of corresponding sides. The above
examples will lead us to the following two important generalization which
could be stated as theorems.

Theorem 4.4: If the ratios of the corresponding sides of two

similar polygons is k, then the ratio of their
P 1 S1
perimeters is given by = =k.
P 2 S2

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Grade 8 Mathematics [SIMILAR FIGURES]

Theorem 4.5: If the ratios of the corresponding sides of two

S1
similar polygons is = k, then the ratio of
S2
2

A1  S1 
their areas, is given by: =   = k 2 .
A 2  S2 

Example 17: Find the ratio of the areas of two similar triangles,
5
a. If the ratio of the corresponding sides is .
4
10
b. If the ratio of their perimeters is .
9
Solution:
Let A 1 , A 2 be areas of two similar triangles, P 1 , P 2 be the perimeters of the
two triangles and S 1 , S 2 be their corresponding sides.
2

A S 
a. 1 =  1  ..........................Theorem 4.5
A2  S 2 
2

A1  5 
=   ........................... Substitution
A2  4 
A1 25
Therefore, = .
A 2 16
2
 
A1  P1 
b. =  ................................... Theorem 4.4 and 4.5
A2  P 
 2 
2

 
A1  10 
= ................................. Substitution
A 2  9 
 
A 100
Therefore, 1 = .
A2 81

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 Grade 8 Mathematics [SIMILAR FIGURES]
Example 18: The areas of two similar polygons are 80cm and 5cm2. If a side 2

of the smaller polygon is 2cm, find the corresponding sides of the

larger polygons.
Solution:
Let A 1 and A 2 be areas of the two polygons and S 1 , S 2 be their
corresponding sides, then
2
 S1  A
  = 1 .................................................. Theorem 4.5
 S1  A2
2
 S1  80
  = ................................................... Substitution
 S1  5
S 21 80
=
22 5
2
S1
= 16
4
= 64
2
s 1

S1 × S1 = 8 × 8
S1 = 8cm
Therefore, the corresponding sides of the larger polygon is 8cm.

Example 19: The sum of the perimeters of two similar polygon is 18cm. The
ratios of the corresponding sides is 4:5. Find the perimeter of
each polygon.
Solution:
Let S 1 and S 2 be the lengths of the corresponding sides of the polygon and
P 1 and P 2 be their perimeters.
P 1 +P 2 =18
P 1 =18-P 2 .................. Solve for P 1
P 1 S1
Thus = ...............Theorem 4.4
P 2 S2
18 − P 2 4
= .................... Substitution
P2 5
4P 2 =90-5P 2 .................... Cross multiplication
9P 2 =90
P 2 =10 ............................ Divides both sides by 9.
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 Grade 8 Mathematics [SIMILAR FIGURES]

Therefore, when P 2 =10

P 1 +P 2 =18
P 1 +10=18
P 1 =8

Exercise 4E
1. In two similar triangles, find the ratio of:
a. corresponding sides, if the areas are 50cm2 and 98cm2.
b. the perimeter, if the areas are 50cm2 and 16cm2.
2. Two triangles are similar. The length of a side of one of the triangles is
6 times that of the corresponding sides of the other. Find the ratios of the
perimeters and the area of the triangles.
3. The sides of a polygon have lengths 5, 7, 8, 11 and 19 cm. The perimeter
of a similar polygon is 75cm. Find the lengths of the sides of larger
polygon.
4. A side of a regular six – sided polygon is 8cm long. The perimeter of a
similar polygon is 60cm. What is the length of a side of the larger
polygon?
5. The ratio of the sides of two similar polygon is 3:2. The area of the
smaller polygon is 24cm2. What is the area of the larger polygon?
6. Two trapeziums are similar. The area of one of the trapeziums is 4 times
that of the other. Determine the ratios of the perimeters and the corres-
ponding side lengths of the trapeziums.
7. Two triangles are similar. The length of a side of one of the triangles is 4
times that of the corresponding side of the other. Determine the ratios of
the perimeters and the areas of the polygon.

Challenge Problems
8. Two pentagons are similar. The area of one of the pentagons is 9 times
that of the other. Determine the ratios of the lengths of the corresponding
sides and the perimeters of the pentagons.

129
Grade 8 Mathematics [SIMILAR FIGURES]
9. Two triangles are similar. The length of a side of one of the triangles
2 times that of corresponding sides of the other. The area of the smaller
triangle is 25sq.cm. Find the area of the larger triangle.
10. The lengths of the sides of a quadrilateral are 5cm, 6cm, 8cm and 11cm.
The perimeter of a similar quadrilateral is 20cm. Find the lengths of the
sides of the second quadrilateral.
11. The picture represents a man made pool surrounded by a park. The two
quadrilateral are similar and the area of the pool is 1600 sq.cm. What is
the area of the park of A'B' is four times the length of AB?

A’ B’
A Park B

Pool

D C
D’ C’
Figure 4.42

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Grade 8 Mathematics [SIMILAR FIGURES]

Summary For Unit 4

1. Similar geometric figures are figures which have the same shape.
2. Two polygons are similar if:
i. their corresponding sides are proportional.
ii. their corresponding angles are congruent.
3. Under an enlargement
a. Lines and their images are parallel.
b. Angles remain the same.
c. All lengths are increased or decreased in the same ratio. .
4. Scale factor- the ratio of corresponding sides usually expressed numerically
so that:
length of line on the enl arg ement
scale factor =
length of line on the original
5. ∆ABC similar to ∆DEF if:
i. their corresponding sides are proportional.
ii. their corresponding angles are congruent.
6. AA Similarity theorems
If two angles of one triangle are congruent to the corresponding two angles
of another triangle, then the two triangles are similar.
T P

R S L M

Figure 4.43
In the above Figure 4.43 you have:
• ∠T ≡∠P if and only if m(∠T)=m(∠P)
• ∠R ≡∠L if and only if m(∠R=m(∠L)
• ∆TRS∼∆PLM .............................. by AA similarity.

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Grade 8 Mathematics [SIMILAR FIGURES]
7. SAS Similarity theorem
If two sides of one triangle are proportional to the corresponding two sides
of another triangle and their included angles are also congruent, then the
two triangles are similar. A Q

B C R S

Figure 4.44

• ∠A≡∠Q if and only if m(∠A)=m(∠Q)

AB AC
• =
QR QS
Therefore ∆ABC~∆QRS ................................. by SAS similarity.
8. SSS Similarity theorem
If the three sides of one triangle are in proportion to the three sides of
another triangle, then the two triangles are similar.
A Q

B C R S

Figure 4.45

In the above Figure 4.45 you have:

AB BC AC
= =
QR RS QS

Therefore, ∆ABC~∆QRS ...................... SSS Similarity.

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 Grade 8 Mathematics [SIMILAR FIGURES]

9. In Figure 4.46, If ∆ABC~∆DEF with constant of proportionality k, then

a. P(∆ABC)=KP(∆DEF)
b. a(∆ABC)=K2× a(∆DEF)
c. AB = KDE, B E

BC = KEF
AC = KDE

A C D F

Figure 4.46

Miscellaneous Exercise 4
I Write true for the correct statements and false for the incorrect one.
1. AB ≡ CD if and only if AB=CD.
2. ∠ABC≅∠DEF if and only if m(∠ABC)=m(∠DEF).
3. All rhombuses are similar.
4. All congruent polygons are similar.
5. All Isosceles triangles are similar.
6. Any two equilateral triangles are similar.

II Choose the correct answer from the given alternatives.

7. In Figure 4.47 given below PQ ⊥ QT , ST ⊥ QT and P, R and S are on the same
line. If ∆PQR~∆STR, then which of the following similarity theorems
supports your answer?
a. SAS Theorem P

b. AA Theorem
c. SSS Theorem
T
d. None Q
R

Figure 4.47 S

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 Grade 8 Mathematics [SIMILAR FIGURES]
AB BC AC
8. Given ∆ABC and ∆DEF, if = = , then which of the following
DE EF DF
postulates or theorem shows that ∆ABC is similar to ∆DEF?
a. AA similarity theorem. c. SAS similarity postulate.
b. AAA similarity theorem. d. SSS similarity theorem.
9. Which of the following plane figures are not necessarily similar to each other?
a. two equilateral triangles.
b. two isosceles triangles.
c. two circles.
d. two squares.
10. Which of the following is different in meaning from ∆ADF~∆LMN?
a. ∆DFA~∆NML c. ∆AFD~∆LMN
b. ∆FAD~∆MNL d. ∆DAF~∆NLM
11. In Figure 4.48 below MN//YZ. If XN = 10cm, NZ = 5cm and MY = 4cm,
then what is the length of XY ?
a. 9cm Z
5cm
b. 15 cm
N
c. 12 cm
d. 18 cm 10cm

X
M 4cm Y
Figure 4.48

III Work out problems

12. Rectangle ABCD is similar to rectangle PQRS. Given that AB=14cm,
BC=8cm and PQ=21 cm, calculate the length of QR.
13. A football field measures 100m by 72m. A school marks a football field
similar in shape to a full size foot ball field but only 30m long. What is its
width?

134
 Grade 8 Mathematics [SIMILAR FIGURES]

14. In Figure 4.49 below ∠QRP≅∠XYZ and show that ∆QRP~∆XYZ.

Z
P

8cm
6cm

Q R X
3cm 4cm Y
Figure 4.49
A
15. The sides of a polygon are 3cm, D
5cm, 6cm, 8cm and 10cm. The
perimeter of a similar polygons is
O
48cm. Find the sides of the second
polygon.
16. In Figure 4.50 to the right DC // AB,
C
AO BO
prove that = Figure 4.50 B
OD OC

17. A piece of wood is cut as shown in Figure 4.51 below. The external and
internal edge of the wood are similar quadrilaterals:
I 6m H
D 2cm C

b a 6m

A
c B
F G
Figure 4.51

i.e. ABCD~ FGHI. The lengths of the sides are indicated on the figure. How
long are the internal edges marked as a, b, and C?

135
UNIT
5 CIRCLES
Unit outcomes
After Completing this unit, you should be able to:
 have a better understanding of circles.
 realize the relationship between lines and circles.
 apply basic facts about central and inscribed angles and
angles formed by intersecting chords to compute their
measures.

Introduction
In the previous grades you had learnt about circle and its parts like its center,
radius and diameter. Now in this unit you will learn about the positional
relationship of a circle and lines followed by chords and angles formed inside a
circle and how to compute their degree measures of such angles.

5.1 Further On Circles

Activity 5.1
Discussed with your teacher orally.
Define and show by drawing the following key terms:
a. circles c. diameter e. circumference of a circle
b. radius d. chord

136
 Grade 8 Mathematics [CIRCLES]

Now in this lesson you will discuss more about parts of a circle i.e minor arc
and major arc, sector and segment of a circle, tangent and secant of a circle
and center of a circle by construction.

Parts of a circle
Group Work 5.1
1. a. Draw a circle of radius 4cm.
b. Draw a diameter in your circle. The diameter divides the circle in to two
semicircles.
c. Colour the two semicircles indifferent colours.
d. Draw a minor arc in your circle and label your minor arc.
e. Draw a major arc in your circle and label your major arc.
2. A circle has a diameter of 6cm.
a. write down the length of the radius of the circle.
b. Draw the circle.
c. Draw a chord in the circle.

Definition 5.1: The set of points on a circle (part of a circle) contained

in one of the two half-planes determined by the line
through any two distinct points of a circle is called an
arc of a circle.
C

A B
O

P
Figure 5.1 Circle

137
 Grade 8 Mathematics [CIRCLES ]

The center of the circle is O and PO is the radius. The part of the circle
determined by the line through points A and B is an arc of the circle. In Figure
5.1 above arc ACB is denoted by ACB or arc APB is denoted by APB.
P

A. Classification of Arcs
i. Semi-circle: Is half of a circle whose A
O
B

end points are the end points of a diameter

of the circle and measure is 1800.
C

Figure 5.2 APB and ACB are a semi-circles.

ii. Minor arc: is the part of a circle •Y

which is less than a semi-circle.
O
A B

•
X

Figure 5.3 AX , BY, AY and BX are a minor arcs

iii. Major arc: is the part of a circle which is greater than a semi-circle.
X
•

A B
O

•
Y

Fig. 5.4 AXBY and BYAX are major arcs.

138
 Grade 8 Mathematics [CIRCLES]

Example 1: In Figure 5.5 below determine whether the arc is a minor arc,
a major arc or a semicircle of a circle O with diameters AD and BE .
B
a. AFB e. CDE F•
C
b. ABD f. BCD
A
c. BED g. AED
d. CAE h. ABC O
D
Solution:
a. minor arc e. minor arc
b. semi-circle f. minor arc E
c. major arc g. semi-circle Figure 5.5 Circle
d. major arc h. minor arc
B. Sector and segments of a circle
Y
Definition 5.2: A sector of a circle is the
region bounded by two
O
radii and an intercepted
arc of the circle.
r r

A B
X
Figure 5.6 Circle

In Figure 5.6 above, the shaded region AOB and the unshaded region AYB are
sectors of the circle.

X
Definition 5.3: A segment of a circle is the
A
region bounded by •a chord B

and the arc of the circle.

O

Y
Figure 5.7 Circle

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 Grade 8 Mathematics [CIRCLES ]
In Figure 5.7 above the shaded region AXB and the unshaded region AYB are
segments of the circle.

C. Positional relations between a circle and a line

A circle and a line may be related in one of the following three ways.
1. The line may not intersect the circle at all.


Figure 5.8 Circle

2. The line may intersect the circle at exactly one point.
P Q
• n

Figure 5.9 Circle

3. The line may intersect the circle at two points

A B m

Figure 5.10 Circle

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 Grade 8 Mathematics [CIRCLES]

Definition 5.4: a. If a line intersect a circle at exactly one point,

then the line is called a tangent of the circle.
b. The point at which it intersect the circles is
called point of tangency.
c. If a line intersect a circle at two points then the
line is called a secant of the circle.
P

X Y

A B

Figure 5.11 Circle

In Figure 5.11 above P is the point of tangency, XY is a tangent to the circle O

and AB is a secant line to the circle O.

D. Construction
To find the center of a circle by construction the
following steps is important:
Step i : Draw a circle by using coins
Figure 5.12 circle

Step ii : Draw a chord AB A B

Figure 5.13 circle

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 Grade 8 Mathematics [CIRCLES ]

Step iii: Construct the perpendicular bisector of AB

A B

Figurer 5.14 circle

Step iv: Draw another chord CD .

A B

D
Figure 5.15 circle

Step v: Construct the perpendicular

A B
bisector of CD .
C

Figure 5.16 circle

B

Step vi: The perpendicular bisectors of AB

A
and CD intersect at O, the centre of O C

the circle.

D
Figure 5.17 The required circle

142
Grade 8 Mathematics [CIRCLES]

Exercise 5A
1. Write true for the correct statements and false for the incorrect ones of
each of the following.
a. A secant of a circle contains chord of the circle.
b. A secant of a circle always contains diameter of the circle.
c. A tangent to a circle contains an interior point of the circle.
d. A tangent to a circle can pass through the center of the circle.
2. In Figure 5.18 below A is an interior point of circle O. B is on the circle
and C is an exterior point. Write correct for the true statements and false
for the incorrect ones of each of the following.
a. You can draw a secant line through point C.
b. You can draw a secant line through point B.
O B
c. You can draw a tangent line through point A.
d. You can draw tangent line through point C. A C

Figure 5.18 Circle

3. Consider the following Figure 5.19 to complete the blank space.

a. is tangent to circle O.
b. is secant to circle O. A
D E
c. is tangent to circle Q.
d. is secant to circle Q.
O Q
e. is the common chord to circle O and Q. P

B C F

Figure 5.19 Circle

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 Grade 8 Mathematics [CIRCLES ]

5.2 Angles in the Circle

Now in this lesson you will discuss more about central angle, inscribed angle,
Angles formed by two intersecting chords and cyclic quadrilaterals.

5.2.1 Central Angle and Inscribed Angle

Group Work 5.2

1. What is central angle?
2. What is inscribed angle?
3. Explain the relationship between the measure of the inscribed angle
and measure of the arc subtends it.
4. In the given Figure 5.20 below m(∠CAO)=30° and m(∠CBO)=40°. Find
m(∠ACB) and m(∠AOB).
C

A B

Figure 5.20
5. If in Figure 5.21 arc BD is two times the arc AC, find ∠ BAD.
D

800
B

A
C
Figure 5.21
6. O is the center of the circle. The straight line AOB is parallel to DC.
Calculate the values of a, b and c.
B
O c

A 46° b
a C

D Figure 5.22

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 Grade 8 Mathematics [CIRCLES]

Definition 5.5: A Central angle of a circle is an angle whose vertex is

the center of the circle and whose sides are radii of
the circle. C
A B
r r
O

Figure 5.23 ∠AOB is a central

Note: 1. ACB is said to be intercepted by ∠AOB and ∠AOB is said to

be subtended by ACB.
2. The chord AB is said to subtend ACB and ACB is said to be
subtended by chord AB.
3. Chord AB subtends ∠AOB.
4. The measure of the central∠AOB is equal to the measure of
the intercepted ACB. i.e m (∠AOB) = ACB.

Fact:- If the measure of the central angle is double or halved, the length of the
intercepted arc is also doubled or halved. Thus you can say that the
length of an arc is directly proportional to the measure of the central
angle subtended by it. Hence you can use this fact to determine the
degree measure of an arc by the central angle under consideration.

Definition 5.6: An inscribed angle is an angle whose vertex is on a

circle and whose sides contain chords of the circle.
P

A B
Figure 5.24
C

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 Grade 8 Mathematics [CIRCLES ]
In Figure 5.24 above ∠APB is inscribed angle of circle O. we say ∠APB is
inscribed in ACB and ACB subtends ∠APB.
Note: In Figure 5.25 below, the relationship between the measure of the central
angle and inscribed angle by the same arc is given as follows:
1. The measure of the inscribed angle is D
half of the measure of central angle. •
2. The measure of the inscribed angle is A C
half of the measure of the arc subtends
it.
O
m(∠ABC) = m(∠AOC)
m (∠ABC) = m(ADC)
B
Figure 5.25

3. In Figure 5.26 to the right the

relationship of inscribed angles B C
subtended by the same arc is D
i.e m (∠ABE)=m(∠ACE)=m(∠ADE)
= m (AXE)

A E
•
X
Figure. 5.26

Examples 2: In Figure 5.27 to the right, O is the center of the circle. If

m(∠AOC)=520, find B

m (∠ABC) and m (AC).

Solution:
m (∠AOC) = m (AC)= 520 and C
A O
1
m (∠ABC) = m (AC)
2
1
= (520)
2 A
= 260 Figure 5.27
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 Grade 8 Mathematics [CIRCLES]

Examples 3: In Figure 5.28 to the right, O is the

B
center of the circle, m (∠ABC) = 650, and
m (∠AOE) =700, find m (∠CFE).
A
O

Solution:
F
m(∠AOE)=700 .................... Given
m(∠ABC)=650 .................... Given C E
1
m(∠ABC)= m (AFEC)
2 Figure 5.28

1
650= m (AFEC)
2
m (AFEC)=1300
m (∠AOE) = m (AFE)=70
Thus m (EC) = m(AFEC) − m (AFE)
=1300 − 700
=600
1
Therefore, m (∠CFE) = m (EC)
2
1
= (600)
2
= 300

Examples 4: In Figure 5.29 to the right, O is the center of the circle,

m (∠AQB) = 350 R
Find a. m (∠AOB)
b. m (∠APB)
O Q
c. m (∠ARB)
350
P
A

B
Figure 5.29

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 Grade 8 Mathematics [CIRCLES ]
Solution:
m (∠AOB) =m (AB)
1
a. m(∠AQB)= (m∠AOB)
2
⇒m(∠AOB)=2m(∠AQB)
=2(350)
=700
1
b. m(∠APB)= m (∠AOB)
2
1
= (700)=350
2
1
c. m(∠ARB)= m(∠AOB)
2
1
= (700)=350
2
A
Examples 5: In Figure 5.30 to the right, find the x 0 B
32
values of the variables.
y
240 C
Solution: D
1 Figure 5.30
m (∠ABD)= m (AD)
2
⇒ m(AD)=2m(∠ABD)
=2320
=640
1
m(∠ACD)= m (AD)
2
1
y= (640)
2
y=320
1
m(∠BDC) = m (BC)
2

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 Grade 8 Mathematics [CIRCLES]

⇒ m(BC) =2m(∠BDC)
= 2240
= 480
1
m(∠BAC) = m(BC)
2
1
x = (480)
2
x = 240
5.2.2 Theorems on Angles in A Circle
You are already familiar with central angles and inscribed angles of a circle.
Under this sub-section you will see some interesting result in connection with
central and inscribed angles of a circle.
It is well know that the measure of a central angle is equal to the measure of the
intercepted arc. But, a central angle is not the only kind of angle that can
intercept an arc.

Theorem 5.1: The measure of an inscribed angle is equal to

one half of the measure of its intercepted arc.

This important theorem proved in three cases. But here you can consider only the
first case.
Proof: Given an inscribed angle ABC with sides BC passing through the
center O.

O C
B
x
A

Figure 5.31

1
We want to show that: m(∠ABC)= m(AXC)
2
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 Grade 8 Mathematics [CIRCLES ]

Statements Reasons

1. Draw OA 1. Through two points there is

exactly one line.
2. ∆BOA is isosceles 2. OB ≡ OA (radii).
3. ∠OBA≅∠OAB 3. Base angles of isosceles triangle
4. m(∠OBA)+m(∠OAB)=m(∠AOC) 4. ∠AOC is supplementary to
∠OBA, m(∠OAB), +
m(∠OBA) + m(∠BOA) = 180o.
1 5. since ∠BAO≅∠ABO.
5. m(∠ABO)= m(∠AOC)
2
6. central angle AOC intercepts
6. m(AXC)=m(∠AOC)
AXC.
7. m(∠ABC)=m(∠ABO) 7. Naming the same angle.
1 8. Substitution in step 5 .
8. m(∠ABC) = m(AXC)
2

Example 6: In Figure 5.32 below, O is the centre of a circle.

m(∠QPT)=540 and m(∠TSQ)=210. P
Find: m (∠ROS).
540

Solution: Q
1 T
m(∠RPS) = m (RS)
2 O
210

R
⇒ m(RS) =2m(∠RPS)
S
=2(540)
=1080 Figure 5.32
Thus m(∠ROS)=m(RS)=108°.

Theorem 5.2: In a circle, inscribed angles subtended by the

same arc are congruent.
[[

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 Grade 8 Mathematics [CIRCLES]

Proof: Given circle O, inscribed angles B and D A

subtended by the same arc AC.
We want to show that: m(∠ABC)=m(∠ADC) B
O

D
Figure 5.33

Statements Reasons

1 1. Theorem 5.1
1. m(∠ABC)= m (AC)
2
1 2. Theorem 5.1
2. m(∠ADC) = m (AC)
2

3. m(∠ABC)=m(∠ADC) 3. Substitution

Examples 7: In Figure 5.34 to the right, B C

0 0
m(∠CPD) =120 , m(∠PCD) = 30
find m(∠A).
P

A D
Solution:
Figure 5.34
m(∠CPD)=120 .......... Given
m(∠PCD)=30............ Given
m(∠CPD)+m(∠PCD)+m(CDP)=180 … why?
120+30+m (∠CDP)=180
m(∠CDP)=180-150
=30
Therefore, m(∠CDP)=30.
Hence m(∠CDB)=m(∠CAB)=30................... Theorem 5.2

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Exercise 5B B
1. In Figure 5.35 to the right, O is the
center of the circle. If m(∠ABC)=300,
D O
CB // OA and CO and AB intersect at D, C

find m(∠ADC).

2. In Figure 5.36 to the right, if m(AHB) = 100o A

Figure 5.35
o
and m(DIC) = 80 m(∠BAC) = 50°
and FG is tangent to the circles at C,
then find each of the following.
a. m(∠BDC) P
b. m(∠ACD) H
A
c. m(∠AEB)
B
Y E
D •X
• •
F I
C •
G
Figure 5.36

3. In Figure 5.37 below, O is the centre of the circle PA and PB are tangents
to the circle at A and B, respectively. If m(∠ACB)=1150 ,then find:
a. m(AB)
A
b. m(ACB)

P O
C

Figure 5.37

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Grade 8 Mathematics [CIRCLES]

4. In Figure 5.38 to the right, O is the M

center of the circle. If m(∠B)=1400,

What is m(∠AMC)?
O
•
A C
0
140

B
Challenge Problems Figure 5.38

5. In Figure 5.39 to the right, O is the C

center of the circle, m(∠ABC) = 800 and

B
m(∠AED) = 200 then what is
m(∠DOC)? O

D
E A

Figure 5.39

B y
• C
6. In Figure 5.40 to the right, O is the
center of the circle and m (BYC) = 400,
m(AXD)=1200 and m(∠ADB) =500. A
O
What is m(∠DOC) and m(DC)?

x•
D
Figure 5.40
B

7. In Figure 5.41 given to the right , O is the

center of the circle and m(∠BOC) = 1200.
What is m(∠ADC)? O
D
C

Figure 5.41
A
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 Grade 8 Mathematics [CIRCLES ]
5.2.3 Angles Formed by Two Intersecting Chords
Activity 5.2
Discuss with your friends.
1. In Figure 5.42 given to the right, find m(∠x).
P
Q

x 1060
1740

R S

Figure 5.42
D
2. In Figure 5.43 given to the right, can
you derived a formula m(∠a) and
m(∠b) a C
A b

B
Figure 5.43

Theorem 5.3: The measure of an angle formed by two chords

intersecting inside a circle is half the sum of
the measures of the arcs subtending the angle
and its vertical opposite angle.

Proof: Given AB and CD intersecting at P inside a circle

We want to show that: B
C P •X
1 1
m (∠BPD) = m ( AYC) + m (DXB) D
2 2 Y•
•Z
A
F

Figure 5.44

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 Grade 8 Mathematics [CIRCLES]

Statements Reasons
1. Draw a line through A such that AF CD 1. construction

2. m(∠ BPD) = m (∠BAF) 2. corresponding angles

formed by two parallel
lines and a transversal line
1 3. Theorem 5.1
3. m (∠ BAF) = m (BDF)
2
4. m (∠ AYC) = m (DZF) 4. Why ?
1 5. Why?
5. m (∠ BPD) = m ( BDF)
2
1 1 6. Why?
6. m(∠BPD) = m ( BXD)+ m (DZF)
2 2
1 1 7. Substitution
7. m (∠BPD)= m (AYC) + m (BXD)
2 2

Examples 8: In Figure 5.45 given to the right, find B

the value of β, if m (AB) = 82° and 820

m (DC) = 46° A
β
x
•
O C
Solution:
To find β, simply by theorem 5.3 460
D
m (∠AXB) =
Figure 5.45
=

= = 64°
B
m (∠AXC) = 180°. . . . . . (Why)? 820

m (∠AXC) = m (∠AXB) + m (∠BXC) β

A x
180° = 64° + m(∠BXC) •
O C
116° = m(∠BXC)
460
Therefore, β = 116° D
Figure 5.46

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 Grade 8 Mathematics [CIRCLES ]
Examples 9: In Figure 5.47 given to the
A
right, find the value of a and b.
D
450
Solution: a
b B
1 145 0 O
m(∠a) = [m(DC)+m(AB)] .................. Theorem 5.3
2
1 C
= (145 0 +45 0 )
2 Figure 5.47
1
= (1900)
2
= 950 and
m(∠a)+ m(∠b) = 1800 .................. Angle sum theorem
950+ m(∠b) = 1800
m(∠b) = 850

Exercise 5C
B
1. Find the values of the variables in Figure A
550
5.48 to the right. a
b C
0 O
165
2. Find the values of the variables in Figure
5.49 below. D
80 B
A Figure 5.48
a x
z y
C

120
D B
Figure 5.49 70o

C
3. Find the m (AB) and m (DC) in O•
A
Figure 5.50 to the right.
56o
D
Figure 5.50

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 Grade 8 Mathematics [CIRCLES]

5.2.4 Cyclic Quadrilaterals

Group Work 5.3
Discuss with your friends/ partners.
1. What is a cyclic quadrilateral?
2. Based on Figure 5.51 answer A
the following questions. B
a. What is the sum of the measure
D
angles A and C?
b. What is the sum of the measure angles
C
B and D?
Figure 5.51
c. Is ABCD is a cyclic quadrilateral?
3. Find the sizes of the other three angles in A B
the cyclic quadrilateral, if AB // DC. 700 x

z y
D C
Figure 5.52

C
4. Angle ∠ECD =800. Explain why AEDB is a cyclic
quadrilateral. Calculate the size of angle ∠EDA. 300

O
D
E

A Figure 5.53 B

C
5. In Figure 5.54 below ABOD is acyclic quadrilateral
57°
and O is the center of the circle. Find x ,y, z and w,
if DO // AB. O
x z
D w B
y
A
Figure 5.54

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 Grade 8 Mathematics [CIRCLES ]
Draw a circle and mark four points A, B, C and D on it. Draw quadrilateral
ABCD as shown in Figure 5.55. This quadrilateral has been given a special name
called cyclic quadrilateral.

Definition 5.7: A quadrilateral inscribed in A D

a circles is called cyclic

quadrilateral.
B C

Figure 5.55 ABCD is cyclic quadrilateral

Property of Cyclic Quadrilateral

i. In Figure 5.55 above m(∠A)+m(∠c)=180°.
ii. Similarly m(∠B)+m(∠D)=180°.

Theorem 5.4: In cyclic quadrilateral, opposite angles are

supplementary.

Given: ABCD is a quadrilateral inscribed in

D
a circle. C

We want to show that:

i. m(∠A)+m(∠C)=1800
A B
ii. m(∠B)+m(∠D)=1800
Figure 5.56
Proof
Statements Reasons

1. ∠DAB and ∠DCB are opposite angles 1. Given

1 1 2. Theorem 5.1
2. m(∠DAB)= m (DCB) and m(∠DCB)= m (DAB)
2 2

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 Grade 8 Mathematics [CIRCLES]

1 3. By addition
3. m(∠DAB)+m(∠DCB)= [m(DCB)+m(DAB)]
2
property

1
4. m(∠DAB)+m(∠DCB)= (3600) 4. Degree measure
2
of a circle

5. Supplementary
5. m(∠A)+m(∠C)=1800

? Can you prove similarly m(∠B)+m(∠D)=1800?

C
y
Examples 10: In Figure 5.57 to the right, ABCD
B X
is a cyclic quadrilateral E is on
CD. If m(∠C)=1100, find x and 0
110
0
D
y. A 83

Solution:
m(∠BAD + m(∠BCD)=1800 ............Theorem 5.4 •E

m (∠ y) + 1100 = 1800 Figure 5.57

Therefore, y=700
m(∠ADE) + m(∠CDA) = 1800 ........straight angle and CE is a ray.
830 + m(∠CDA) = 1800
Therefore, m(∠CDA)=970
m(∠CBA) + m(∠CDA) = 1800 .........Theorem 5.4
m(∠x) + 970=1800
Therefore, x=830
A B
4x
Examples 11: ABCD is an inscribed quadrilateral
.O
as shown in Figure 5.58. Find the
m(∠BAD) and m(∠BCD). 5x
D C

Figure 5.58
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 Grade 8 Mathematics [CIRCLES ]
Solution:
m(∠DAB) + m(∠DCB)=1800 ................... Theorem 5.4
4x+5x =1800 .................................... Substitution
9x = 1800
x = 200
Hence m(∠DAB) = 4(200) = 800 and m(∠DCB) = 5(200) = 1000.

Exercise 5D A
0
1. In Figure 5.59 to the right,A,B,C,D and E are 100
E
points on the circle. If m(<A)=1000, find: B

a. m(∠C)
b. m(∠D) D
C
Figure 5.59

2. In Figure 5.60 to the right, ABCD is a cyclic A

trapezium where AB // CD. If m(∠A) = 950, 950
then find the measure of the other three D
B
angles.

C
Figure 5.60

3. Consider the quadrilateral ABCD. Is it a 760

A
0 C
cyclic quadrilateral? 110
D
Figure 5.61

4. In Figure 5.62 to the right, ABCD is A B

6x
an inscribed quadrilateral. Find the
measure of ∠BAD and ∠BCD.

4x
D C
Figure 5.62

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Grade 8 Mathematics [CIRCLES]

Challenge Problems C
5. In Figure 5.63 to the right, find 42 0

a. m(∠ABC) B
b. m(∠ADC) D 0
69
c. m(∠PAB) A
P T

Figure 5.63

Summary for Unit 5

1. A circle is the set of all points on a plane that are equidistant from a given
point, called the centre of the circle.
2. A Chord of a circle is a segment whose end points are on the circle.
3. A diameter of a circle is any chord that passes through the center and
denoted by d.
4. A radius of a circle is a segment that has the center as one end point and a
point on the circle as the other end point, and denoted by r.
5. The perimeter of a circle is called its circumference.
6. An arc is part of the circumference of a circle.
Arcs are classified in the following three ways. .
a. Semi-circle: an arc whose end points are also end points of a diameter of
a circle.
b. Minor arc: is the part of a circle less than a semi circle.
c. Major arc: is the part of a circle greater than a semi-circle.
7. A sector of a circle is the region bounded by two radii and an arc of the
circle.
8. A segment: of a circle is the region bounded by a chord and the arc of the
circle.
9. A central angle of a circle is an angle whose vertex is the center of the circle
and whose sides are radii of the circle.
10. An inscribed angle is an angle whose vertex is on a circle and whose sides
contain chords of the circle.

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 Grade 8 Mathematics [CIRCLES ]
11. A quadrilateral inscribed in a circles is called cyclic quadrilateral.
12. The measure of an inscribed angle is equal to one half of the measure of its
intercepted arc.
13. The measure of an angle formed by two chords intersecting inside a circle is
half the sum of the measure of the arcs subtending the angle and its vertical
opposite angle.
14. The sum of the opposite angles of cyclic quadrilateral is supplementary

Miscellaneous Exercise 5
I. Write true for the correct statements and false for the incorrect ones.
1. Opposite angles of an inscribed quadrilateral are supplementary.
2. A central angle is not measured by its intercepted arc.
3. An angle inscribed in the same or equal arcs are equal.
4. A tangent to a circle can pass through the center of the circle.
5. If the measure of the central angle is double, then the length of the intercepted
arc is also double.

II. Choose the correct answer from the given four alternatives
6. In Figure 5.64 to the right, O is the center
A
of the circle. What is the value of x?
4x O
a. 360 c. 100
b. 600 d. 180 5x C

Figure 5.64

7. In Figure 5.65 below, OA and OB are radii of circle O. Which of the

following statement is true? A
a. AB = OA
b. AB>OA O 600
c. AB<OA
B
d. None
Figure 5.65
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 Grade 8 Mathematics [CIRCLES]

8. The measure of the opposite angles of a cyclic quadrilateral are in the ratio
2:3. What is the measure of the largest of these angles?
a. 270 b. 1200 c. 600 d. 1080
III. Workout problems 800
9. In Figure 5.66 to the right, lines D C

AB and CD are parallel and E

m (AB) = 1000 and m(CD) = 800. A B

What is m(∠AED)? 1000

Figure 5.66

10. In Figure 5.67 below, find the value of the measure ∠a.
a

A O C

B
Figure 5.67

11. Construct the circle through A, B and C where AB=9cm, AC = 4cm and
BC = 4cm.
B
12. In Figure 5.68 given below;
a. If m (∠AOC) =1400, find m(∠ABC) and m(∠ADC).
O
b. If m(∠ABC) = 600, find m(∠AOC) and m(∠ADC).
c. If m(∠AOC) = 2000 find m(∠ABC). A C
0
d. If m(∠ABC) = 80 , find m(∠OAC). D
e. If m(∠OCA) = 200, find m (∠ADC). Figure 5.68
13. ABCD is a quadrilateral inscribed in a circle BC = CD. AB is parallel to DC
and m (∠DBC) = 500. Find m (∠ADB). Q

14. O is the center of the circle.

Calculate the size of angle ∠QSR P 560 O R

S
Figure 5.69
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 Grade 8 Mathematics [CIRCLES ]
n
15. For the given circle O is its center and the m D
two secant lines m and n are parallel.
A 60°
Find x.
O
120° x C
16. In Figure 5.71 below is a chord and O
is the center of the circle. Calculate the B
size of m (∠TPO). P
Figure 5.70

O 70° B

D
T
Figure 5.71 A O

17. O is the center in Figure 5.72 to the right, B 158°

Angle ABC = 158°. C
Find (a) reflex angle AOC (c) angle ADC Figure 5.72
(b) angle AOC
18. Find all the unknown angles in the figure in
D C
which // and angle ACD = 24°. β 24°
A x
α θ
19. Given that angle BFC = 58°. B

Angle BXG = 48° and angle CBF = 22°.

Find (i) ∠BGX (iv) ∠BCG Figure 5.73
(ii) ∠BGF (iv) ∠BFG
(iii) ∠BCF

G F E
H
x 58°
48°
D
22° C
A
B
Figure 5.74
164
UNIT
6 INTRODUCTION
TO
PROBABILITY

Unit outcomes
After Completing this unit, you should be able to:
 understand the concept of certain, uncertain and
impossible outcomes.
 know specific facts about event, sample space and
probability of simple events.

Introduction
When you buy a lottery ticket you cannot be 100% sure to win. Some things can
occur by chance or things what you expected may not occur at all. The occurrence
or non-occurrence of these things are studied in mathematics by the theory of
probability. So in this unit you will learn the simple and introductory concepts of
probability.

165
 Grade 8 Mathematics [INTRODUCTION TO PROBABILITY ]

Figure 6.1

This morning there is a chance of heavy rain with the possibility of thunder. In
the afternoon the rain will die away and it is likely that the sun will break
through the clouds, probably towards evening.

Weather forecasts are made by studying weather data and using a branch of
mathematics called probability.
Probability uses numbers to represent how likely or unlikely it is that an event
such as 'a thunderstorm' will happen.
Probability is used by governments, scientists, economists, medical researchers
and many other people to predict what is likely to happen in the future by
studying what has already happened.

Group work 6.1

Discuss with your group.
1. Find out your classmates' favorite and least favorite school subjects.
Ask them what they like and dislike about each.
2. Also find out if they like or dislike going to school overall. Find out why
or why not.
3. Make a chart presenting your findings to the class.

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 Grade 8 Mathematics [INTRODUCTION TO PROBABILITY ]

6.1 The Concept of Probability

Consider a two sided coin. For convenience let the two faces (sides) of the coin
be called head (H) and tail (T). If we toss (flip) the coin, the experience shows
that the possible outcomes are H (head) or T (tail) but not both. Similarly discuss
example 1 with your group.

Example 1
a. Tossing a coin.
b. Tossing two coins.
c. Tossing three coins.
d. Tossing four coins.
e. Rolling a die. Figure 6.2

Definition 6.1: A process of an observation is often called an experiment.

Definition 6.2: The set of all possible out comes of an experiment is called
a sample space/the possibility set/ of the experiment and
denoted by S.
Definition 6.3: A subset of the sample space of an experiment is called
an event and denoted by E.

To illustrate the above definition /explanation/ consider the following summary

of probability experiments.
Experiment Sample space
a. Toss a coin. {Head, tail} ={H, T}
b. Toss two coins. {HH, HT, TH, TT} a) Showing
head
c. Toss three coins. {HHH, HHT, HTH, HTT,
THH, THT, TTH, TTT}
d. Roll a die. {1, 2, 3, 4, 5, 6}
e. Choose an English {a, e, i, o, u}
vowel.
f. Answer a true – false {True, False} b) Showing tail
question.
Figure 6.3 coins
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 Grade 8 Mathematics [INTRODUCTION TO PROBABILITY ]
Example2: suppose an experiment is rolling a die.
a. List the elements of the sample spaces.
b. List the elements of the set of the event "the number shown
is prime".
c. List the elements of the set of the event "the number shown
is odd".
d. List the elements of the set of the event "the number shown
is even".

Solution
a. The sample space are
b. The prime number is =E
c. The odd number is =E
d. The even number is =E Figure 6.4 Die

Activity 6.1
1. Identify the following events has certain or impossible events.
a. You will grow to be 30 centimeters tall.
b. You will live to be 240 years old.
c. You will die.
d. A newly born baby will be a girl.
2. Give two examples of events that you think
a. are impossible.
b. are certain.
3. Locate each of the following situations on the probability scale.
a. You will have match home work to night.
b. Ababy born today was a girl.
c. The local meteorologist predicts a 40% chance that it will rain
tomorrow.
d. If will snow in your town in August.

Definition 6.4: If the probability of an event is one, then it is called

certain event.

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 Grade 8 Mathematics [INTRODUCTION TO PROBABILITY ]
Example 3: a. Night will follow day.
b. December following November next year.
c. The next person to come into the room will be right handed.

Definition 6.5: If the probability of an event is 0, then it is called

impossible out comes.

Example 4: a. When water boils it changes to milk.

b. Two lines intersect at four points.

c. You will grow to be 50 meters tall.

d. It will rain to morrow.

Definition 6.6: All probabilities must have a value greater than or

equal to 0 and less than or equal to 1 or 0 ≤ P(E) ≤ 1.
This can be shown on a probability scale:
0
½ 1

Impossible unlikely even likely certain

chance
Figure 6.5 Probability scale

Exercise 6A

1. Draw a 0 to 1 probability scale and mark on it the probability that:

a. the sea will disapper.

b. you will buy a new pair of shoes soon.

c. the sun will not rise next week.

169
 Grade 8 Mathematics [INTRODUCTION TO PROBABILITY ]

d. a member of the class will be late tomorrow.

e. a newely born baby will be a boy.

f. you will watch TV sometime to night.

g. a coin thrown in the air will land heads up.

h. a coin thrown in the air will land tails up.

2. Given two examples of events that you think

a. are impossible d. are likely
b. are unlikely e. are certain
c. have about an even chance

6.2 Probability of Simple Events

Group work 6.2
1. A fair six sided die is rolled. What is the probability of getting.
a. a number 4 f. a number 1 or a number 2
b. an odd number g. an even number
c. a multiple of 3 h. 3 or more
d. less than 5 i. a prime number
e. more than 6
Figure 6.6 die
2. Suppose an experiment is rolling a die:
a. List the elements of the sample spaces.
b. List the elements of the set of the event "the sum of the number
is prime.
c. List the elements of the set of the event "the sum of the number
is even".
d. List the elements of the set of the event "the sum of the number
is 12".
e. List the elements of the set of the event "the sum of the number
is 9".

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Grade 8 Mathematics [INTRODUCTION TO PROBABILITY ]
3. Nine playing cards are numbered 2 to 10. A card is selected from them
at random. Calculate the probability that the card will be

9 10
a. an odd number. 7 8
6
b. a multiple of 4.

5
4
2 3
Figure 6.7 Playing Cards

4. What is the probability of an event that is certain occur?

5. What is the probability of an event that cannot occur?

Historical Note
The first book written on the subject of
probability was the Book on Games of
Chance by Jerome Cardano. He was an
Italian physician and mathematician who
lived in the 16th century.

Figure 6.8 Jerome Cardan

Definition 6.7: (probability of an event )

In words: The probability of an event is the ratio of the number of
successful out comes in the event to the total number of possible out
comes in the sample space.

In symbols: P (event) =

or P(E) =
Assuming that the out comes are all equally likely.

Example 5: Find the probability of getting a number 5 or 6 when a fair die

is rolled.
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 Grade 8 Mathematics [INTRODUCTION TO PROBABILITY ]
Solution:
These are two successful out comes: 5 or 6 and there are six possible out
comes: 1, 2, 3, 4, 5, 6 or n(E) = 2 and n (S) = 6.
Thus P(5 or 6) = = =

Note: The word fair means that each number has an equal chance of
turning up: the outcomes are equally likely.

Example 6: Find the probability of drawing a card with a prime number on it

from a deck of cards numbered l to 20.
Solution:
1 2 3 4 5 6 7 8 9 10 
The sample spaces are  
11 12 13 14 15 16 17 18 19 20 

The events are

Hence P(prime number) = = =

Example 7: A bag contains 8 discs of which 4 are red, 3 are blue and 1 is
yellow. Calculate the probability that when one disc is drawn
from the bag it will be
a. red c. blue
b. yellow d. yellow or blue

Solution: There are 8 discs altogether so the total

number of possible out comes is 8 Figure 6.9

a. P (red) = = =

b. P(yellow) = =

c. P(blue) = =

d. P(yellow or blue) = = =

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 Grade 8 Mathematics [INTRODUCTION TO PROBABILITY ]
Example 8: You randomly draw a slip of paper from a box containing 4 slips.
A red slip, a black slip, a white slip and a pink slip.

a. How many possible outcomes are there?

b. What is the probability of each outcomes?
Solution: a. There are 4 possible outcomes.
b. Since each slip is equality likely to be drawn, the probability of
each outcomes is .
Example 9: A letter is chosen at random from the word PROBABILITY.
Work out the probability that it will be:
a. R b. Y c. B d. I or A e. B or I
Solution: Total number of possible outcomes is 11.
a. P(R) = =

b. P(Y) = =

c. P(B) = =

d. P(I or A) = =

e. P( B or I) = =

Note: An outcome is said to occurre at random if each out comes is

equally likely to occur.

Example 10: Six slips of paper one labeled with the letters of “POTATO”. The
slips are shuffled in a hat and you randomly draw one slip. What
is the probability that the slip you draw:
a. the letter T?
b. either the letter O or the letter A?
c. the letter Z?
d. one of the letters P, O. T or A?

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 Grade 8 Mathematics [INTRODUCTION TO PROBABILITY ]

Solution: a. P(t)= =

b. P (O,A)= =

c. P(Z) = =

d. P (P, O, T or A) =

Example 11: A bowl of fruit contains 3 apples, 4 bananas, 2 lemon and

1 orange. Abebe takes one piece of fruit without looking. What
is the probability that he takes.
a. an apple c. a lemon
b. a banana d. an orange
Write each answer in three ways,
that means
i. as a fraction
ii. as a decimal and
iii. as a percentage Figure 6.10

Solution: A bowl of fruit contains = 3 apples + 4 bananas + 2 lemon + 1

orange = 10
a. P (Lemon) = = .

Therefore i. probability of an apple as a fraction is .

ii. probability of an apple as a decimal is 0.3.
iii. probability of an apple as a percentage is % = 30%.
b. P (banana) = = .

Therefore, i. probability of an apple as a fraction is .

ii. probability of an apple as a decimal is 0.4.
iii. probability of an apple as a percentage is 40%.
c. P (lemon) = = .

Therefore, i. probability of a lemon as a fraction is .

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 Grade 8 Mathematics [INTRODUCTION TO PROBABILITY ]
ii. probability of a lemon as a decimal is 0.2.
iii. probability of a lemon as percentage is 20%.
d. P(an orange) = = .

Therefore , i. probability of an orange as a fraction is .

ii. probability of an orange as a decimal is 0.1.
iii. probability of an orange as a percentage is 10%.

Example 12: Two dice are rolled. State the probability of each event.
a. The sum is 7.
b. The sum is 13.
c. The sum is less than 13.

Figure 6.11 Dice

Solution: There are 6 numbers on each die. The sample space has 36 or 62 out
comes.
Sample space

Second die
First Die 1 2 3 4 5 6
1 (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
2 (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
3 (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
4 (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
5 (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
6 (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)
a. The out comes with sum 7 are:
{(1, 6), (2, 5), (3, 4), (4, 3) (5, 2), (6, 1)}.
P(sum =7) = = = .
b. There are 0 out comes with sum 13.
⇒ P(sum = 13) = .
c. All 36 out comes have a sum less than 13.
P(sum < 13) = .

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 Grade 8 Mathematics [INTRODUCTION TO PROBABILITY ]

Exercise 6B
1. A counting number less than 30 is chosen at random. What is the probability
that the number chosen:
a. is a multiple of 4? c. is a cube number?
b. is a square? d. is a prime?
2. A jar contains 2 orange, 5 blue, 3 red and 4 yellow
marbles. A marble is drawn at random from the jar.
Find each probability.
a. p (orange)
b. p (red)
c. p ( blue)
d. p (green) Figure 6.12

3. Two dice are thrown at the same time. State the probability of each event:
a. The sum is 5
b. The sum is 9
c. The sum is 12

4. A game is played with two spinners. You multiply the two numbers on the
spinners land to get the score
Spinner A Spinner B

2
3 1
1 3 4
2

This score is 2 4 = 8
Figure 6.13 Spinner

a. Copy and complete the table to show all the possible scores.
One score has been done for you.
Spinner B
× 1 2 3 4
Spinner A

1
2 8
3
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 Grade 8 Mathematics [INTRODUCTION TO PROBABILITY ]
b. Work out the probability of getting a score of 6.
c. Work out the probability of getting a score that is an odd number.
5. This spinner is spun. What is the probability of getting:
a. a number 1
b. an odd number
1 3
2

The three-sided spinner has landed on two

Figure 6.14

6. Nine counters numbered 2 to 10 are put in a bag. One counter is selected at

random.
What is the probability of getting a counter with
a. a number 5 d. a perfect square number
b. an odd number e. a multiple of 3?
c. a prime number
7. A hundred raffle tickets are sold. Raman buys 8 tickets, Susan 5 tickets and
Aster 12 tickets. What is the probability that the first prize will be won by
a. one of these three c. Susan
b. Raman d. Somebody other than Aster?
Write each answer in three ways:
a. as a fraction c. as a percentage
b. as a decimal and
5.

Challenge Problem
8. When three dice are thrown at the same time what is the probability that the
sum of the number of dots on the top faces will be 6?
9. A number is selected at random from 1 to 100. State the probability that:
a. The number is odd c. the number is even or divisible by 5
b. the number is divisible by 5 d. the number is divisible by 5 or 3.

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 Grade 8 Mathematics [INTRODUCTION TO PROBABILITY ]

Summary for Unit 6

1. A process of an observation is often called an experiment.
2. The set of all possible outcomes of an experiment is called a sample space
/the possibility set/ of the experiment and denoted by S.
3. A subset of the sample space of an experiment is called an event and
denoted by E.
4. An event which is certain to happen has a probability of 1.
5. An event which cannot happen is (impossible) outcomes has a probability
of 0.
6. The probability that an event will happen is calculated by:
Probability of an event = .
7. The probability of an event happening is always greater than or equal to 0
(impossible) and less than or equal to 1 (certain). This can be written as
0 ≤ probability an event ≤ 1.

Miscellaneous Exercise 6
I. Write true for the correct statements and false for the incorrect one
1. The probability of an event that is certain to occur is 1.
2. The probability of an event that is an impossible out comes to occur is 0.
3. The probability of getting a sum of 7 or 11 by rolling two dice is
4. If the set of all possible outcomes is equal to an event then the probability
of an event is 1.
5. Suppose that two dice are tossed, and then the probability of the sum 1 is
also 1.

II. Choose the correct answer from the given alternatives

6. If 2 fair dice are tossed, what is the probability that the sum of the
number of dots on the top faces will be 7?
c.
b. d.

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 Grade 8 Mathematics [INTRODUCTION TO PROBABILITY ]
7. A bag contains 6 white balls, 4 red balls and 5 black balls. If a ball is
drawn from the bag at a random, then which of the following is true.
a. Probability of a white ball is .
b. Probability of red or a black ball is .
c. Probability of not getting a black ball is .
d. All are true

8. A pairs of fair dice is tossed. What is the probability of not getting a sum
5 or 9?
b. c. d.
9. Which of the following is true about a probability scale?
a. Probability of unlikely between 0 and .
b. Probability of even chance is .
c. Probability of likely between and 1.
d. All are true
III. Work out problems
10. A letter of the English alphabet is chosen at random. Calculate the
probability that the letter so chosen be:
a. vowel
b. Precedes m and is a vowels
c. follows m and is a vowels
11. If three coins are thrown. What is the probability of obtaining.
a. all heads c. at least one heads
b. all tails d. at least two heads
12. A letter is chosen at random from the words ETHIOPIA MATHS. What
is the probability that the letter E is chosen?
13. One letter is chosen at random from the word ISOSCELES. What is the
probability of choosing.
a. the letter C c. a vowel
b. the letter E d. a consonant
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UNIT
7 GEOMETRY AND
MEASUREMENTS

Unit outcomes
After Completing this unit, you Should be able to:
 understand basic concepts about right angled triangles.
 apply some important theorems on right angled triangles.
 know basic principles of trigonometric ratios.
 know different types of pyramid and common parts of
them.

Introduction
In this unit you will in detail learn about the basic properties of right angled
triangles, by using two theorems on this triangle. You will also learns about a
new concept that is very important in the field of mathematics known as
trigonometric ratios and their real life application to solve simple problems. In
addition to this you will also learn a solid objects known as pyramids and cone
and their basic parts.

7.1 Theorems on the Right Angled Triangle

In your earlier classes you have learnt many things about triangles. By now, you
do have relatively efficient knowledge on some of the properties of triangles in
general. In this sub topic we will give special attention to the properties of right
angled triangles and the theorems related to them.
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 Grade 8 Mathematics [GEOMETRY AND MEASUREMENT]
Right angled triangles have special properties as compared to other types of
triangles. Due to their special nature they have interesting properties to deal with.
There are some theorems and their converses that deal with the properties of right
angled triangles.

Group Work 7.1

1. Name the altitudes drawn from the right angle to the hypotenuse of
the given right angled triangles.
a) b) c)
Q A R B
I

R S H G
J
C
Figure 7.1
p

2.
C In Figure 7.2 To the left of the unknown
quantities
∆ ABC ∆ ADC ∆ BDC
a Hypotenuse
b
h leg
leg
α β
A e B
D f
c
Figure 7.2
3. In Figure 7.2 above, find three similar triangles.
4. In Figure 7.2 above ∆ CAB ∼ ∆ DAC. Why?

Historical note
There are no known records of the exact date
or place of Euclid's birth, and little is known
about his personal life. Euclid is often referred
to as the "Father of Geometry." He wrote the
most enduring mathematical work of all time,
the Elements, a13volume work. The
Elements cover plane geometry, arithmetic
and number theory, irrational numbers, and
solid geometry.
Figure 7.3 Euclid
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 Grade 8 Mathematics [GEOMETRY AND MEASUREMENT]
7.1.1 Euclid's Theorem and Its Converse
In Figure 7.2 above the altitude CD of ∆ABC divides the triangle in to two right
angled triangles: ∆ADC and ∆BDC. You can identify three right angled triangles
(∆ABC, ∆ADC and ∆BDC). If you consider the side correspondence of the three
triangles as indicated in Table 7.1 below, it is possible to show a similarity
between the triangles.
Table 7.1
∆ ABC ∆ ADC ∆ BDC
Hypotenuse c b a
leg a h f
leg b e h

The AA similarity theorem could be used to show that:

1. ∆ DBC ~ ∆ DCA
2. ∆ ABC ~ ∆ CBD
3. ∆ CAB ~ ∆ DAC
From similarity (2) you get the following proportion:
AB BC
=
CB BD
c a
⇒ =
a f
⇒ a 2 = cf
and from similarity (3) you get the following proportion:
CA AB
=
DA AC
b c
⇒ =
e b
b 2 = ec
These relations are known as Euclid's Theorem.
From the above discussion, you can state the Euclid's theorem and its converse.

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 Grade 8 Mathematics [GEOMETRY AND MEASUREMENT]

Theorem 7.1: (Euclid's Theorem)

In a right angled triangle with an altitude to the hypotenuse, the
square of the length of each leg of the triangle is equal to the
product of the hypotenuse and the length of the adjacent
segment into which the altitude divides the hypotenuse:
Symbolically: 1. (BC)2 = AB × BD C
Or a = c × f
2
b a
2. (AC) =AB × DA
2

Or b2 = c × e A e B
cD f
Figure 7.4

Example1: In Figure 7.5 to the right,

∆ABC is a right angled C

triangle with CD the a b

altitude on the hypotenuse.

B A
Determine the lengths of 12 cD 3
Figure 7.5
AC and BC if
AD= 3cm and DB= 12cm.
Solution:
(AC)2= (AB) × (AD) .............. Euclid's Theorem
(AC)2= (15cm) × (3cm) = 45cm2… AB = BD +AD

AC= 45 cm 2

AC = 3 5cm
(BC)2 = (AB) × (BD) ............ Euclid's Theorem
(BC)2= (15cm) × (12cm)
( BC) = 180cm 2

BC = 6 5cm

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Grade 8 Mathematics [GEOMETRY AND MEASUREMENT]

Theorem 7.2: (Converse of Euclid's Theorem)

In a triangle if the square of each shorter side of the triangle is
equal to the product of the length of the longest side of the
triangle and the adjacent segment into which the altitude to
the longest side divides this side, then the triangle is right
angled:
Symbolically: 1. a2 = cf and
2. b2= ce if and only if ∆ABC is right angled.
C

b a
h

A e f B
cD
Figure 7.6

You can combine the theorem of Euclid’s and its converse as

follows:

∆ABC with side lengths a, b, c and h the length of the altitude

to the longest side and e, f the lengths of the segments into
which the altitude divide the longest side and adjacent to the
sides with lengths a and b respectively is right angled if and
only if a2=cf and b2 =ce.
Symbolically, ∆ABC is right angled.
If and only if a2 = cf and if and only if b2 = ce.
C

b a

h
A e f B
cD
Figure 7.7

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 Grade 8 Mathematics [GEOMETRY AND MEASUREMENT]
Example 2: In Figure 7.8 to the right,
C
AD = 4cm, DB = 12cm,
b a
AC = 8cm and BC = 8 3cm
and m (∠ADC) = 900. A B
e cD f
Is ∆ABC a right angled? Figure 7.8
Solution:

a. (BC)2= (BD) × (BA) ........... Theorem 7.1

(8 3cm) 2 = (12cm) × (BA)
192cm2= (12cm) (BA)
192cm 2
BA =
12cm
BA = 16cm
Thus (AB) × (DB) = (16cm) × (12cm)
= 192cm2
Hence (BC)2 = (BD)× (BA)
b. (AC)2 = (8cm)2 = 64cm2 = b2
(AD) × (AB) = (16cm) × (4cm)
= 64cm2 = ce
Hence b2 = ec
Therefore from (a) and (b) and by theorem 7.2, ∆ABC is a right angled
triangle, where the right angle is at C.

Example 3: In Figure 7.9 below, AC = 3 13cm, BC = 2 13cm,

AB = 13cm and DB=4cm. Is ∆ ABC a right angled?
C

b a

A e f B
cD
Figure 7.9

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Grade 8 Mathematics [GEOMETRY AND MEASUREMENT]
Solution:

a. (BC)2 = (BD) ×(AB) ........... Theorem 7.1

Now (BC) 2 = (2 13cm) 2 = 4 × 13 = 52cm 2 =a 2
(AB)×(BD)=(13cm) × (4cm)=52cm2=fc
Therefore, a2=fc
b. (AC)2=(AD) × (AB) . . . . . . . Theorem 7.1
( )
Now (AC)2= 3 13cm 2 = 9 × 13 = 117cm 2 =b 2
(AB) × (AD) = 13cm × 9cm = 117cm 2 = ec
Therefore, b2 = ec
Therefore, from (a) and (b) above and by theorem 7.2 ABC is a right
angled triangle.

Exercise 7A
1. In Figure 7.10 to the right, ∆ ACB is C

a right triangle with the right angle at

C and CD ⊥ AB where D is on AB .
A B
Find the lengths of AC and 6cm D 12cm
BC, if AD = 6cm and DB = 12cm. Figure 7.10

2. In Figure 7.11 below, ∆ABC is a right triangle. m(∠ABC)= 900, BD is

the altitude to the hypotenuse AC of ∆ABC. Find the values of the
variables.
A
C
4cm
D 8 cm
y cm
9cm
z cm D
A B x cm
9 cm
(a) x cm

Figure 7.11 B
y cm C
(b)
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Grade 8 Mathematics [GEOMETRY AND MEASUREMENT]
3. In Figure 7.12 to the right, ∆ABC is right angled at B, B
BD ⊥ AC, BE = BC, BE = 6cm, AC = 12cm.
Find a.
b.
c.
A C
E D
12 cm
Figure 7.12
C
4. In Figure 7.13, AD = 3.2 cm,
DB = 1.8 cm AC = 4cm and
BC = 3 cm. Is ∆ABC a right angled?

A B
D
Figure 7.13

Challenge problems
5. In Figure 7.14 below, ABC is a semicircle with center at O. BD ⊥ AC
such that BD = 8cm and BC = 10cm.
B
Find a.
b.
c. •
A D C
d. O
Figure 7.14

6. In Figure 7.15 below, O, is the center of the semicircle ABC.

BD ⊥ AC, DO = 3cm and BD = 6cm. Find the radius of the circle.

A •
D O C

Figure 7.15

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 Grade 8 Mathematics [GEOMETRY AND MEASUREMENT]
7.1.2 The Pythagoras' Theorem and Its Converse

Historical note
Early writers agree that Pythagoras was born on
Samos the Greek island in the eastern Aegean
Sea. Pythagoras was a Greek religious leader and
a philosopher who made developments in
astronomy, mathematics, and music theories.
Figure 7.16 Pythagoras

Group work 7.2

1. Verify the Pythagorean property by counting the small squares in the
diagrams.

(a) Figure 7.17 (b)

2. State whether or not a triangle with sides of the given length is a right triangle.
a. 3m, 5m, 7m d. 10cm, 24 cm, 26 cm
b. 10m, 30m, 32m e. 20mm, 21mm, 29mm
c. 9cm, 12cm, 15cm f. 7km, 11km, 13km
3. Pythagorean triples consist of three whole numbers a, b and c which obey the
rule: a2 + b2 = c2
a. when a = 1 and b = 2, find the value of c.
b. when a = 3 and b = 4, find the value of c.
4. Pythagoras' Theorem states that a2 + b2 = c2 for the sides a, b and c of a right –
angled triangle. When a = 5, b = 12 then c = 13.
Find three more sets of rational numbers for a, b and c which satisfy
Pythagoras' Theorem.

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 Grade 8 Mathematics [GEOMETRY AND MEASUREMENT]

Theorem 7.3 (Pythagoras' Theorem)

If a right angled triangle has legs of lengths a and b and

hypotenuse of length c, then a2+b2=c2.
C

a
b

A B
c1 D c c2

Figure 7.18

Let ∆ ABC be right angled triangled the right angle at C as shown above:
Given: ∆ ACB is a right triangle and CD ⊥ AB.
We want to show that : a2+b2 = c2.
Proof:
Statements Reasons
2
1. a = c2  c 1. Euclid's Theorem
2. b2= c 1  c 2. Euclid's Theorem
3. a2+b2= (c 2 × c) + (c 1 × c) 3. Adding step 1 and 2
4. a2+b2=c(c 1 +c 2 ) 4. Taking c as a common factor
5. a2+b2=c(c) 6. Since c 1 +c 2 =c
6. a2+b2=c2 5. Proved
Example 4: If a right angle triangle ABC has legs of lengths a= 3cm and
b=4cm. What is the length of its hypotenuse?
Solution: Let c be the length of the hypotenuse
a2+b2=c2 . . . . Pythagoras' Theorem
(3cm)2+(4cm)2 = c2
9cm2+16cm2 = c2
25cm2 = c2
c = 25cm 2
c = 5cm
Therefore, the hypotenuse is 5 cm long.
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 Grade 8 Mathematics [GEOMETRY AND MEASUREMENT]
Example 5: If a right angle triangle ABC has leg of length a=24cm and the
hypotenuse c=25cm. Find the required leg.
Solution: If b is the length of the required leg, then
a2+b2 = c2 ........................ Pythagoras' Theorem
(24cm)2+b2 = (25cm)2
576cm2+b2 = 625cm2
b2 = (625-576)cm2
b2 = 49cm2
b= 49cm 2
b = 7cm
Therefore, the other leg is 7cm long.

The converse of the Pythagoras' Theorem is stated as follows:

Theorem 7.4 (Converse of Pythagoras' Theorem)

If the lengths of the sides of C

ABC are a, b and c where b a

a2+b2=c2 then the triangle is
right angled. The right angle is A c B
opposite the side of length c. Figure 7.19

C
The Pythagoras' Theorem and its
converse can be summarized as b a

follows respectively. A B
c
In ∆ABC with a, b lengths of Figure 7.20

the shorter sides and c the

Length of the longest side, then ∆ ABC is right angled if and
only if a2+b2=c2. Using the Figure above: ∆ ABC is right
angled, if and only if a2+b2=c2 .

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 Grade 8 Mathematics [GEOMETRY AND MEASUREMENT]
Example 6: In Figure 7.21 below. Is ∆ABC is a right-angled?
Solution: C
i.. a2= (12cm)2=144cm2
b = 5cm a = 12cm
ii. b2=(5cm)2=25cm2
ii. c2=(13cm)2=169cm2 A B
c = 13cm
Therefore, a2+b2=169cm2 and
Figure 7.21
c2=169cm2
Hence ∆ABC is right angled, the right angle at C….. converse of Pythagoras
theorem.

Example 7: In Figure 7.22 below. Is ∆ABC is a right-angled?

Solution:
A
i) a2= (8cm)2=64cm2
ii) b2= (12cm)2=144cm2
C = 15cm
iii) c2= (15cm)2= 225cm2 b = 12cm
Therefore, a2+b2 = (64+144) cm2
208cm2 and c2= 225 cm2
C B
Therefore, 208cm ≠ 225cm
2 2 a = 8cm
Figure 7.22
Therefore, ∆ ABC is not a right-angled.

Exercise 7B
1. In each of the following Figures ∆ABC is a right angled at C. Find the
unknown lengths of sides.
A
C A

6 cm x cm 5 cm y cm
10 cm

4 cm

B
A B C B C 24 cm
x cm 3 cm
(a) (b) (c)
Figure. 7. 23

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 Grade 8 Mathematics [GEOMETRY AND MEASUREMENT]
2. In Figure 7.24 to the right, ABCD is a A B
rectangle with length and width 6 cm and
4 cm respectively. What is the length of the
diagonal AC?
D C
Figure 7.24

3. Find the height of an isosceles triangle with two congruent sides of length
37cm and the base of length 24cm.
4. Abebe and Almaz run 8km east and then 5km north. How far were they
from their starting point?
5. A mother Zebra leaves the rest of the herd to go in search of water. She
travels due south for 0.9km and, then due east for 1.2km. How far is she
from the rest of the herd?

Figure 7.25

6. In Figure 7.26 below ∆ABC is an equilateral triangle. AD ⊥ BC and

AB= 20 cm.
A
Find: a. AD b. BD c. DC
Hint: AD bisects both ∠BAC and BC.

B C
D

Figure 7.26

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 Grade 8 Mathematics [GEOMETRY AND MEASUREMENT]
7. In Figure 7.27 to the right ABCD is a square A B
and DB the diagonal of the square BD= 6 2
cm. Find the length of side of the square.

D C
Figure 2.27

8. In Figure 7.28 to the right, Find the length of: A

3 cm D B
a. BE
b. DF 5 cm
E
c. EF F
8 cm
d. Is ∆CEF is a right-angled?
10 cm
e. Is ∆ADF is a right-angled?
C
Figure 2.28

9. The right-angled triangle ABC has sides 3cm, 4cm and 5cm. Squares have
been drawn on each of its sides. D

a. Find the number of cm squares in:

A E
i. the square CBFG I
ii. the square ACHI
H C B
iii. the square BADE
b. Add your answers for (a) (i) and (a) G F
(ii) above. Figure 7.29

c. State whether or not a triangle with sides of the given lengths is a right
triangle.

Using the theorems for calculations

Solving problems using the Euclid's and Pythagoras’ Theorems. You are now
well aware of the two theorems, their converses and their applications in
determining whether a given triangles is right angled triangle. You can now
summarize, the Euclid's and the Pythagoras’ theorems as follows:

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 Grade 8 Mathematics [GEOMETRY AND MEASUREMENT]
Given: a right angled triangle ABC as shown
C
in the figure to the right and CD is altitude to
the hypotenuse. Let a, b and c be the side b a

opposite to the angles A, B and C respectively.

If AD = c 1 and DB=c 2 , then A c1 D c2 B
c
a. a2 = c × c 2 Figure 7.30
………..Euclid's Theorem
b2 = c × c 1
b. a2+b2 = c2 ............. Pythagoras' Theorem
C
Example 8: In Figure 7.31 to the right, if
b a
DB = 8cm and AD = 4 cm
then find the lengths of:
A B
c1 D c2
c
a. AB b. BC c. AC d. DC Figure 7.31

Solution:
a. AB = AD + DB . . . . Definition of line segment.
AB = 4cm + 8cm
AB = 12cm
Hence c = 12cm
b. (BC)2 = (BD) × (BA) ...................... Euclid's Theorem
(BC)2 = (8cm) × (12cm)
(BC)2 = 96cm2
BC = 4 6 cm
c. (AC)2=(AD) × (AB) ...................... Euclid's Theorem
(AC)2=(4cm) × (12cm)
(AC)2 = 48cm2
AC = 48cm 2
AC = 4 3cm

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 Grade 8 Mathematics [GEOMETRY AND MEASUREMENT]
2 2 2
d. (DC) + (BD) =(BC) ...................... Pythagorans' Theorem
(DC)2 + (8cm)2 = (4 6cm) 2
(DC)2 + 64cm2 = 96cm2
(DC)2 = (96-64) cm2
(DC)2 = 32 cm2
DC = 32cm 2
DC = 4 2cm
Example 9: In Figure 7.32 below, find the unknown (marked) length.
(AC)2 = (CD) × (CB) ....... Euclid's Theorem A
2
x = 1216
x
x2 = 192 unit square 8 h

x = 192 B C
4 D 12
x = 8 3 unit C
Figure 7.32
Therefore, the value of x= 8 3 unit.
(AD)2 + (DC)2 = (AC)2 ............................ Pythagoras' Theorem
h2 + (12)2 = (8 3 )2
h2 + 144 = 192
h2 = 192 − 144
h2 = 48
h = 48
h = 4 3 unit.

Exercise 7C A
4
1. In Figure 7.33, find x, a and b. D
a
x
8

B C
b
Figure 7.33
2. If p and q are positive integers such that p > q. Prove that p2-q2, 2pq and
p2+q2 can be taken as the lengths of the sides of a right-angled triangled.

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 Grade 8 Mathematics [GEOMETRY AND MEASUREMENT]
3. How long is an altitude of an equilateral Z
x
triangle if a side of the triangle is: W
a. 6cm long? b. a cm long? 5 6
h
4. In Figure 7.34 to the right, find x, y and h.
X Y
y
Figure 7.34
7.2 Introduction to Trigonometry
7.2.1 The Trigonometric Ratios

Activity 7.1
Discuss with your teacher
1. In Figure 7.35 below given a right angled triangle ABC
a. What is C
i. the opposite side to angle α?
β
ii. the adjacent side to angle α?
iii. the hypotenuse of ∆ ABC?
b. What is
i. the opposite side to β? α
A B
ii. the adjacent side to β? Figure 7. 35
iii. the hypotenuse of ∆ ABC?
2. In Figure 7.35 given a right angled triangle ABC:
a. In terms of the lengths AB, BC, AC, write sin α and sin β.
b. In terms of the lengths AB, BC, AC, write cos α and cos β.

The word trigonometric is derived from two Greek words trigono meaning a
triangle and metron meaning measurement. Then the word trigonometry literally
means the branch of mathematics which deals with the measurement of triangles.
The sine, the cosine and tangent are some of the trigonometric functions.
In this sub unit you are mainly dealing with trigonometric ratios. These are the
ratios of two sides of a right angled triangle.

? What are trigonometric ratios?

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 Grade 8 Mathematics [GEOMETRY AND MEASUREMENT]
Before defining them let us consider the
following Figure 7.36
Y
•
M3

M2

M1

α •
O
X1 X2 X3 X
Figure 7. 36

In Figure 7.36 OY and OX are rays that make an acute angle X1M1, X 2 M 2 and

X 3 M 3 are any three segments each from OY perpendicular to OX . It is obvious

to show that ∆OX 1 M 1 ~∆OX 2 M 2 ~∆OX 3 M 3 (by AA Similarity Theorem). Then
X1 M1 X 2 M 2 X 3 M 3
i. = = this ratio is called the sine of ∠XOY which is
OM 1 OM 2 OM 3

abbreviated as:
X1 M1 X 2 M 2 X 3 M 3
Sin (∠XOY)= = = = sin α, (sine ≅ sin).
OM 1 OM 2 OM 3
OX1 OX 2 OX 3
ii. = = this ratio is called the cosine of ∠XOY which is
OM 1 OM 2 OM 3

abbreviated as:
OX 1 OX 2 OX 3
Cos (∠XOY) = = = = cos α, (cosine ≅ cos).
OM 1 OM 2 OM 3

X1 M1 X 2 M 2 X 3 M 3
iii. = = this ratio is called the tangent of (∠XOY)
OX1 OX 2 OX 3

which is abbreviated as:

X1 M1 X 2 M 2 X 3 M 3
tan (∠XOY) = = = = tan α, (tangent ≅ tan).
OX1 OX 2 OX 3

Note: The sine, cosine and tangent are trigonometric ratio depends on
the measure of the angle but not on the size of the triangle.

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 Grade 8 Mathematics [GEOMETRY AND MEASUREMENT]
In Figure 7.37 to the right, in a right- A
triangle ABC, if ∠C is the right-angle,

adjacent
then AB is the hypotenuse, BC is the hypotenuse

side opposite to ∠A and AC is the side

adjacent to ∠A. B C
Opposite
Figure 7.37

Definition 7.1: If ∆ ABC is right-angled at C, then

length of the side opposite to ∠ A BC
a. sine ∠ A = =
length of hypotenuse AB

length of the side adjacent to < A AC

b. Cosine ∠ A = =
length of hypotenuse AB
length of the side opposite to ∠ A BC
c. Tangent ∠ A = =
length of the side adjacent to ∠A AC

Note: i. Sine of ∠ A, cosine of ∠ A and Tangent of ∠ A are respectively

abbreviated as sin ∠A, cos ∠A and tan ∠A.
ii. The lengths of the opposite side, adjacent side and hypotenuse
are denoted by the abbreviations "opp.", "adj." and "hyp."
Respectively.

Example 10: Use Figure 7.38 to state the value of each ratio.
B
a. sin A
7 x
b. cos A
c. tan A
A C
4
Figure 7.38

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 Grade 8 Mathematics [GEOMETRY AND MEASUREMENT]
Solution:
(AC)2 + (BC)2 = (AB)2 ................. Pythagoras’s Theorem
42 + x2 = 72
x2 = 49 − 16
x2 = 33
x= 33
opp. 33
a. Sin A = =
hyp. 7
adj. 4
b. Cos A = =
hyp. 7

opp. 33
c. tan A = =
adj.. 4

Exercise 7D
1. Use Figure 7.39 at the right to state the value of each ratio
a. sin θ A α B
θ
b. cos θ
c. tan θ 7
24
d. sin α
C
e. cos α
Figure 7.39
f. tan α
2. Use Figure 7.40 at the right to find
the value of each ratio. B

a. Find the value of x β

b. sin β
17
c. cos β x

d. tan β
e. sin θ θ
A C
f. cos θ 15
Figure 7. 40
g. tan θ
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 Grade 8 Mathematics [GEOMETRY AND MEASUREMENT]

Challenge Problems A

3. In Figure 7.41 at the right to state the β

value of each ratio.
z
a. sin β d. sin α y

b. cos β e. cos α
c. tan β f. tan α
α
C x B
Figure 7.41
4. Use Figure 7.42 at the right to describe each ratio.
sin β
a.
cos β C r
B
α
cos β
b.
sin β
S
sin α t
c.
cos α β

cos α Figure 7. 42
d. A
sin α

7.2.2 The Values of Sine, Cosine and tangent for 30°, 45° and 60°
The following class activity will help you to find the trigonometric values of the
special angle 45°.
Activity 7.2
Discuss with your friends/ parents
Consider the isosceles right angle triangle in Figure 7.43.
B
a. Calculate the length of the hypotenuse AB.
b. Are the measure angles A and B equal?
c. Which side is opposite to angle A? 1
d. Which side is adjacent to angle B?
e. What is the measure of angle A?
f. What is the measure of angle B? A
1 C
g. Find sin ∠A, cos ∠A and tan ∠A. Figure 7.43

h. Find sin ∠B, cos B and tan ∠ B.

i. Compare the result ( value) of g and h.

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 Grade 8 Mathematics [GEOMETRY AND MEASUREMENT]
From activity 7.3 you have found the values of sin 450 , cos 450 and tan 450. In an
isosceles right triangle, the two legs are equal in length. Also, the angles opposite
the legs are equal in measure. Since
B
0
m(∠A)+m(∠B)+m(∠c) = 180 and
45°
m(∠c) = 900 c= 2
1
m(∠A)+m(∠B) = 900
Since m(∠A)=m(∠B) each has the measure 450. 45°
A C
In Figure 7.44, each legs is 1 unit long. From the 1
Figure 7.44
Pythagorean property:
c2 = 12 + 12
c2 = 2
c= 2
opp. 1 2
Hence sin 450 = = = .... Why?
hyp. 2 2
adj. 1 2
cos 450= = = .... Why?
hyp. 2 2
opp.
tan 450= =1
adj.
Example 11: In Figure 7.45, find the values of x and y. B

Solution: y x
tan 45° = opp.
adj.
x 450
A
1= 3 C
3 Figure 7.45
x=3
sin 45° = opp.
adj.
1 3
=
2 y
y=3 2
The following Activity will help to find the trigonometric values of the special
angles 30° and 60°.
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 Grade 8 Mathematics [GEOMETRY AND MEASUREMENT]

Activity 7.3
Discuss with your friends/ partner
Consider the equilateral triangle ABC with side 2 units long as shown in Figure 7.46
below. C

a. Calculate the length of AD .

300 300
b. Calculate the length of DC . 2 2
c. Find sin 30°, cos 30° & tan 30°.
d. Find sin 60°, cos 60° and tan 60°.
600 600
e. Compare the results of sin 30° and cos 60°. A
D
B
f. Compare the results of cos 30° and sin 60°.
Figure 7.46
g. Compare the results of tan 30° and tan 60°.

C
From activity 7.3 you have attempted to find the
values of sin 300 , cos300, tan 300, sin 600,
300 300
cos600 and tan 600. Consider the equilateral 2 2
triangle in Figure 7.47 with side 2 units. The hh
altitude DC bisects ∠C as well as side AB. Hence
m(∠ACD)=300 and AD = 1 unit….. (Why)? 600 600
A B
D
2 2 2
(AD) + (DC) = (AC) ... Pythagorean Theorem Figure 7.47

in ∆ADC.
12 + h2 = 22
h2 + 1 = 4
h2 = 3
h = 3 units.
Now in the right-angled triangle ADC
C
opp. 1
Hence, sin 300= =
hyp. 2
300
adj. 3
cos 300= =
hyp. 2 2
opp. 1 3
tan 300= = =
adj. 3 3
opp. 3 A 600 D
sin 600= = 1
hyp. 2
Figure 7.48
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 Grade 8 Mathematics [GEOMETRY AND MEASUREMENT]
adj. 1
cos 600= =
hyp. 2
opp. 3
tan 600= = = 3
adj. 1

Example 12: In Figure 7.49, find the values of x and y.

Solution:
opp. 5
sin 300= = x 5
hyp. x
1 5
= 300
2 x y
x=10 units Figure 7.49
0opp. 5
tan 30 = =
adj. y
3 5
=
3 y
3 y = 15
15
y= = 5 3 units
3
Example 13: In Figure 7.50, find the values B
of x and y. 600
4
Solution: x
opp. y
sin 600= =
hyp. 4 A C
y
3 y Figure 7.50
=
2 4
2y = 4 3
y = 2 3 unit.
adj. x
cos 600 = =
hyp. 4
1 x
=
2 4
x = 2 unit.

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 Grade 8 Mathematics [GEOMETRY AND MEASUREMENT]
Example 14: A tree casts a 60 meter shadow and makes an angle of 300 with
the ground. How tall is the tree?
B

300
A
C
60m
Figure 7.51
Solution: Let Figure 7.51 represent the given problems.
opp.
tan ∠A=
adj.
h
tan 300=
60 m
h = 60m tan 30 0
3
h = 60m ×
3
h = 20 3 meter
Therefore, the height of the tree is 20 3 meters.
D C
Example 15: The diagonal of a rectangle is 20cm
long, and makes an angle of 300 20 cm
with one of the sides. Find the
lengths of the sides of the rectangle. A 300 B
Figure 7.52
Solution: Let Figure 7.52 represent the given problems
opp.
sin 300 =
hyp.
BC
sin 300=
20
BC = 20sin 300
1
BC = 20 ×
2
BC = 10cm

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 Grade 8 Mathematics [GEOMETRY AND MEASUREMENT]
adj.
cos300 =
hyp.
AB
cos300 =
20 cm
3 AB
=
2 20 cm
AB = 10 3cm
Therefore, the lengths of the sides of the rectangles are 10cm and 10 3 cm.
Example 16: When the angle of elevation of the sun is 450, a building casts
a shadow 30m long. How high is the building?
Solution: Let Figure 7.53 represent sun

the given problem C

opp.
tan 450=
adj. h
h
1=
30 450
A
h = 30m 30m B

Therefore, the height of the Building is 30m. Figure 7.53

Example 17: A weather balloon ascends vertically at a rate of 3.86 km/hr while
it is moving diagonally at an angle of 600 with the ground. At the
end of an hour, how fast it moves horizontally (Refer to the
Figure 7.54 below).

Solution: Let Figure 7.54 represent the given

problem and x be the horizontal speed
opp.
tan 600 =
adj.
3.86 h=3.86 km/hr
3 = km/hr
x
3x = 3.86 km/hr
3.86 600
x= km/hr x
3
3.86 Figure 7.54
x= = 6.85km/hr
0.5774

Therefore, the horizontal speed of the ballon is 6.85km/hr.

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 Grade 8 Mathematics [GEOMETRY AND MEASUREMENT]
Example 18: A ladder 20 meters long, leans against a wall and makes an angle
of 450 with the ground. How high up the wall does the ladder
reach? And how far from the wall is the foot of the ladder?
Solution: Let in Figure 7.55 represent the given problem
adj.
Cos 45° =
hyp. C
1 AB
= 20 m
2 20 m

WALL
20 = 2 AB
20 m
AB = = 10 2 meters 450
2 A
B Figure 7.55

Therefore, the foot of the ladder is 10 2 meters far from the wall.
opp.
sin 450 =
hyp.
1 BC
=
2 20 m
20 = 2 BC
20
BC = = 10 2 meters
2
Therefore, the ladder reaches at 10 2 meters
high far from the ground. CC
Example 19: At a point A, 30 meters from
the foot of a school building as shown
h
in Figure 7.56 to the right, the angle to
the top of the building C 600. What is 600
the height of the school, building? A 30 meters B
Solution: Figure 7.56
By considering ∆ ABC which is right angled you can
use trigonometric ratio.
opp.
tan 60 0 =
adj.
BC
tan 60 0 =
AB
h
3=
30m
h = 30 3meters
Therefore, the height of the school building is 30 3 meters.
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Grade 8 Mathematics [GEOMETRY AND MEASUREMENT]

Exercise 7E
1. In Figure 7.57 below find the value of x.
C A

600

5 cm x

A 300 B B
C
x 8 cm
(a) (b)
Figure 7.57

2. A ladder of length 4m leans against a vertical wall so that the base of the
ladder is 2 meters from the wall. Calculate the angle between the ladder
and the wall.
3. A ladder of length 8m rests against a wall so that the angle between the
ladder and the wall is 450. How far is the base of the ladder from the wall?
4. In Figure 7.58 below, a guide wire is
used to support a 50 meters radio
antenna so that the angle of the wire
makes with the ground 600. How far
50 meters
is the wire is anchored from the base
of the antenna?
600
x
Figure 7.58

5. In an isosceles right triangle the length of a leg is 3cm. How long is the
hypotenuse?

Challenge problems
6. How long is an altitude of an equilateral triangles, if the length of a side
of the triangle is
a. 6cm b. 4cm c. 10 cm

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 Grade 8 Mathematics [GEOMETRY AND MEASUREMENT]
0 0 0
7. In a 45 -45 -90 triangle the length of the hypotenuse is 20cm. How long
is its leg?

7.3 Solids Figures

7.3.1Pyramid

Historical note
Pyramids
The Egyptian pyramids are ancient pyramid
shaped brick work structures located in Egypt.
The shape of a pyramid is thought to be
representative of the descending rays of the
sun.

Group work 7.3

Discuss with your friends/ partners.
1. What is a pyramid?
2. Can you give a model or an example of a pyramid
3. Answer the following question based on the given Figure 7.59 below.
a. Name the vertex of the pyramid. V

b. Name the base of the pyramid.

c. Name the lateral faces of the pyramid.
d. Name the height of the pyramid.
e. Name the base edge of the pyramid. D E
C
f. Name the lateral edge of the pyramid.
A
B
Figure 7.59 Rectangular pyramid

Definition 7.2: A Pyramid is a solid figure that is formed by line segments

joining every point on the sides and every interior points of a
polygonal region with a point out side of the plane of the
polygon.

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 Grade 8 Mathematics [GEOMETRY AND MEASUREMENT]
From the group work (7.3) above you may discuss the following terminologies.
 The polygonal region ABCD is called the base of the pyramid.
 The point V outside of the plane of the polygon (base) is called the vertex
of the pyramid.
 The triangles VAB, VBC, VCD, and VDA are called lateral faces of the
pyramid (see Figure 7.59).
 AB, BC, CD and DA are the edges of the base of the pyramid (see Figure 7.59).
 VA, VB, VC and VD are lateral edge of the pyramid (see Figure 7.59).
 The altitude of a pyramid is the perpendicular distance from the vertex to
the point of the base.
 The slant height is the length of the altitude of a lateral face of the
pyramid.
 Generally look at Figure 7.60 below.

V Vertex
Lateral edge

Lateral face
height of the pyramid
D C Slant height

base
V' E
A
B
Figure 7.60 Rectangular pyramid

Figure 7.61 below show different pyramids. The shape of the base determines
the name of the pyramid.

V V
V

B
C B
B A C
A C
A D
D E
(a) Triangular pyramid (b) Quadrilateral pyramid (c) Pentagonal pyramid
Figure 7.61
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 Grade 8 Mathematics [GEOMETRY AND MEASUREMENT]

Activity 7.4
Discuss with your teacher before starting the lesson.
1. Make a list of the names of these shapes. You do not have to draw them.
Choose from: hexagonal pyramid, tetrahedron, and square pyramid.
2. What is a regular pyramid?
3. What is altitude of the pyramid?

a) b) c)
Figure 7.62
V
Special class of pyramids are known as
right pyramids. To have a right pyramid
the following condition must be satisfied:
The foot of the altitude must be at the
center of the base. In Figure 7.63 to the h
right shows a rectangular right pyramid.
The other class of right pyramids are D
known as regular pyramids. To have a C
regular pyramid, the following three A
B
conditions must be fulfilled:
Figure 7.63 Rectangular right pyramid
1. The pyramid must be a right pyramid.
2. The base of the pyramid must be a regular polygon.
In Figure 7.64 shows regular pyramids.
3. The lateral edges of a regular pyramid are all equal in length.

(a) Regular triangular pyramid (b) Regular square pyramid (c) Regular pentagonal
pyramid
Figure 7.64
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 Grade 8 Mathematics [GEOMETRY AND MEASUREMENT]

Exercise 7F T

1. In Figure 7.65 shows a square pyramid.

a. Name its vertex.
b. Name its four lateral edges. 6 cm
Q R
c. Name its four lateral faces.
W 6 cm
d. Name the height.
P S
e. Name the base. 6 cm
Figure 7.65
V
2. In Figure 7.66 to the right
a. Name the vertex of the pyramid
b. Name the lateral edge of the pyramid
c. Name the lateral faces of the pyramid.
A C

B
Figure 7.66 Pyramid
7.3.2 Cone
Group Work 7.4
Discuss with your friends.
1. What is a cone? V
2. Answer the following question based on
the given Figure 7.67 to the right.

a. Name the vertex of the cone. h
b. Name the slant height of the cone.
c. Name the base of the cone.
d. Name the altitude of the cone. O E
Figure 7.67

Definition7.3: The solid figure formed by joining all points of a circle to a

point not on the plane of the circle is called cone.

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 Grade 8 Mathematics [GEOMETRY AND MEASUREMENT]
In Figure 7.68, represent a cone.
Vertex
The original circle is called the base of V

the cone and the curved closed surface is Lateral face

its lateral surface.
Base
Plane
Figure 7.68

The point outside the plane and at which the segments from the circular region
joined is called the vertex of the cone.
The perpendicular distance from the base to the vertex is called the altitude of
the cone.

Definition7.4: a. A Right Circular cone is a V

Vertex
circular cone with the foot of
its altitude is at the center of
the base as shown in Figure Slant height

7.69 to the right.

Altitude
b. A line segment from the vertex
of a right circular cone to any Base
point of the circle is called the Figure 7.69
slant height.

Exercise 7G
1. Draw a cone and indicate:
a. the base d. the slant height
b. the lateral face e. the vertex
c. the altitude
2. What is right circular cone?
3. What is oblique circular cone?
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Summary For Unit 7

C
Given
b a

A B
c1 D c2
c

Figure 7.70

For 1-4 below refer to right triangle ABC in Figure 7.70 above.
1. Euclid's Theorem i) a2= c 2  c
ii) b2 = c1 × c
2. Converse of Euclid's Theorem.
a2= c 2 c and b2= c 1 c if and only if ∆ABC is right angled.
3. Pythagorean Theorem: a2+b2=c2. A
4. Converse of Pythagorean Theorem β
If a +b =c , then ∆ABC is right angled.
2 2 2
C
b hyp
5. Trigonometric ratio in right triangle opp
ABC where ∠C is the right angle
θ
(see Figure 7.71). C B
a adj
Figure 7.71
 is the side adjacent (adj) to angle θ.

 AC is the side opposite (opp) to angle θ.

 AB is the hypotenuse (hyp) to angle θ
opp. b opp. a
a. sin θ= = d. sin β = =
hyp. c hyp. c
adj. a adj. b
b. cos θ = = e. cosβ = =
hyp. c hyp. c
opp. b opp. a
c. tan θ= = f. tan β= =
adj. a adj. b

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Grade 8 Mathematics [GEOMETRY AND MEASUREMENT]
6. Referring to the values given in table 7.2 below.
θ Sinθ cosθ tanθ
300 1 3 3
2 2 3
450
2 2 1
2 2
600 3 1
2 3
2

7. Relationship between 300 and 600 as follows:

3
a. sin 600=cos 300=
2
1
b. cos 600=sin30 =
2
8. A Pyramid is a solid figure that is formed by line segments joining every point on
the sides and every interior points of a polygonal region with a point out side of the
plane of the polygon.
9. The solid figure formed by joining all points of a circular region to a point
outside of the plane of the circle is called a circular cone.
10. V
Vertex V Slant height
Lateral edge
h
Slant height
h
height Radius
D Lateral face
>E C r
base
A B
(a) Rectangular pyramid (b) Right circular cone
Figure 7.72

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 Grade 8 Mathematics [GEOMETRY AND MEASUREMENT]

Miscellaneous Exercise 7
I. Choose the correct answer from the given alternatives.
1. A rectangle has its sides 5cm and 12cm long. What is the length of its
diagonals?
a. 17cm b. 13 cm c. 7cm d. 12cm
2. In Figure 7.73 to the right C
m (∠ACB)=900 and CD ⊥ AB.
10cm
If CD = 10cm and BD = 8cm, then what
B A
is the length of AD ? 8cm D
Figure 7.73
a. 2 41 cm c. 41 cm
b. 4 41 cm d. cm
Y
3. In Figure 7.74 to the right, right angle
triangle XYZ is a right angled at Y 6cm
and N is the foot of the perpendicular
from Y to XZ . Given that XY= 6cm X Z
N
and XZ=10cm. What is the length of 10 cm
XN ? Figure 7.74

a. 2.4 cm c. 4.3 cm
b. 3.6 cm d. 4.8 cm
4. An electric pole casts a shadow of 24 meters long. If the tip of the shadow
is 25 meters far from the top of the pole, how high is the pole from the
ground?
a. 9 meters c. 7 meters
b. 10 meters d. 5 meters
5. Which of the following set of numbers could not be the length of sides of a
right angled triangle?
a. 0.75, 1, 1.25 c. 6, 8, 10
3
b. 1, , 2 d. 5, 12, 13
2
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Grade 8 Mathematics [GEOMETRY AND MEASUREMENT]
6. A tree 18 meters high is broken off 5 meters from the ground. How far
from the foot of the tree will the top strike the ground.
a. 12 meters b. 13 meters c. 8 meters d. 20 meters

13 m

5m

x
Figure 7.75

7. In Figure 7.76 below ∆ABC is right angled at C. if BC=5 and AB=13,

then which of the following is true?
A
12
a. sin θ =
13 θ
12
b. tan β = 13
5
5
c. cos θ =
13 β
B C
13
d. cos β = 5
5 Figure 7.76

8. In Figure 7.77 below, what is the value of x?

a. 6
b. 20
c. 10 x+4
x
d. 0

8
Figure 7.77

9. One leg of an isosceles right triangle is 3cm long. What is the length of
the hypotenuse?
a. 3cm b. 3 2 cm c. 3 3 cm d. 6 cm

216
Grade 8 Mathematics [GEOMETRY AND MEASUREMENT]
10. In Figure 7.78 below which of the following is true?
C
16
a. sin ∠A=
22
16 16
b. cos ∠B= 2
22
57
c. tan ∠B = A B
8 22
d. All are correct answer Figure 7.78

11. In Figure 7.79 below, which of the following is true about the value of
the variables? 2 B
a. x=2 3 D
x 4
b. y=6 y

c. z=4 3 A C
z
d. All are true Figure 7.79

12. In Figure 7.80 to the right What is the value of x?

a. 6 A
b. 18
c. 3 2 6

d. -3 2

B C
x
Figure 7.80
13. In Figure 7.81 below, what is the value of x?
a. 10 3 A
30
b.
3
10
30 3
c.
3
300
d. All are true B x C

Figure 7.81

217
 Grade 8 Mathematics [GEOMETRY AND MEASUREMENT]
14. Which of the following is true
Q
about given ∆PQR given in
Figure 7.82 to the right? r
a. p2+q2=r2 P
2 2 2
b. q +r =p
c. (p+q)2=r2 R P
q
d. p2+r2=q2
Figure 7.82

15. In Figure 7.83 to the right, find the B

length of the side of a rhombus whose
diagonals are of length 6 and 8 unit.
8
a. 14 units A 6 C
b. 5 units
c. 10 unit
d. 15 unit D
Figure 7.83
II. Work out Question
C
16. In Figure 7.84 to the right, CD ⊥ AB
AD = 4 cm, CD = 9cm and DB=14cm.
Is ∆ABC a right angled? 9 cm
A B
D
4 cm 14 cm
Figure 7.84

B
17. ∆ABC is a right-angled triangle as
shown in Figure 7.85 to the right. If
AD = 12cm BD = DC then find the
lengths of BD and DC . A C
12cm
D
Figure 7.85

218
Grade 8 Mathematics [GEOMETRY AND MEASUREMENT]
18. In Figure 7.86 to the right, find the
A
value of the variables.
3
D
19. A triangle has sides of lengths 16, y
x
48 and 50. Is the triangle a right-
angled triangle? B C
2x
A Figure 7.86
20. In Figure 7.87 to the right, if
AC = 12 cm, BC = 5 cm, x
CD = 11 cm, then find y 12 cm
a. b.

B 5 cm C 11 cm D
Figure 7.87
21. Let ∆ABC be an isosceles triangle
and AD be its altitude. If the length
A
of side AC = 4x + 4y, BD = 6x,
DC = 2x + 2y and AB = 12, then
find the length of:
h
a. AC d. BD
b. AD e. DC
B C
c. BC D
Figure 7.88

22. In Figure 7.89 to the right, what is C

2
the value of x, if sin B = . x + 45
3 45 - x

A B
Figure 7.89

219
Grade 8 Mathematics [GEOMETRY AND MEASUREMENT]
23. In Figure 7.90 to the right, what is F
8
the value of x, if tan∠ D = .
5
13 + x

E D
13 - x
Figure 7.90
24. In Figure 7.91 to the right, what is the
A
2
value of x, if cos c= .
5
x + 21

B C
x

Figure 7.91

25. In Figure 7.92 below, if BC = 5 and AB = 13, then find

a. sinα
b. cos α A

c. tanα α
d. sinβ 13
e. cosβ
f. tanβ
sin α sin β β
g. + B
cos α cos β 5 C
h. 2
(sinα) +(cosα) 2 Figure 7.92

i. (cosβ)2+(sinβ )2

220
Grade 8 Mathematics [GEOMETRY AND MEASUREMENT]
The Function y = x2 1.00 < x < 5.99
X 0 1 2 3 4 5 6 7 8 9
1.0 1.000 1.020 1.040 1.061 1.082 1.102 1.124 1.145 1.166 1.188
1.1 1.210 1.232 1.254 1.277 1.277 1.322 1.346 1.369 1.392 1.416
1.2 1.440 1.464 1.488 1.513 1.513 1.562 1.588 1.613 1.638 1.644
1.3 1.690 1.716 1.742 1.769 1.769 1.822 1850 1.877 1.904 1.932
1.4 1.960 1.988 2.016 2.045 2.045 2.102 2.132 2.161 2.190 2.220
1.5 2.250 2.280 2316 2.341 2.341 2.102 2.434 2.465 2.496 2.528
1.6 2.560 2.592 2.310 2.657 2.657 2.722 2.756 2.789 2.822 2.856
1.7 2.890 2.924 2.624 2.993 2.993 3.062 3.098 3.133 3.168 3.204
1.8 3.240 3.276 2.958 3.349 3.349 3.422 3.460 3.497 3.534 3.572
1.9 3.610 3.648 3.686 3.725 3.764 3.802 3.842 3.881 3.920 3.960
2.0 4.000 4.040 4.080 4.121 4.162 4.202 4.244 4.285 4.326 4.368
2.1 4.410 4.452 4.494 4.537 4.580 4.622 4.666 4.709 4.752 4.796
2.2 4.840 4.884 4.928 4.973 5.018 5.062 5.108 5.153 5.198 5.244
2.3 5.290 5.336 5.382 5.429 5.476 5.522 5.570 5.617 5.664 5.712
2.4 5.760 5.808 5.856 5.905 5.954 6.002 6.052 6.101 6.150 6.200
2.5 6.250 6.300 6.350 6.401 6.452 6.502 6.554 6.605 6.656 6.708
2.6 6.760 6.812 6.864 6.917 6.970 7.022 7.076 7.129 7.182 7.236
2.7 7.290 7.344 7.398 7.453 7.508 7.562 7.618 7.673 7.728 7.784
2.8 7.840 7.896 7.952 8.009 8.066 8.122 8.180 8.237 8.294 8.352
2.9 8.410 8.468 8.526 8.585 8.644 8.702 8.762 8.821 8.880 8.940
3.0 9.000 9.060 9.120 9.181 9.242 9.302 9.364 9.425 9.486 9.548
3.1 9.610 9.672 9.734 9.797 9.860 9.922 9.986 10.05 10.11 10.18
3.2 10.24 10.30 10.37 10.43 10.50 10.56 10.63 10.69 10.76 11.82
3.3 10.89 10.96 11.02 11.09 11.16 11.22 11.29 11.36 11.42 11.49
3.4 11.56 11.63 11.70 11.76 11.83 11.90 11.97 12.04 12.11 12.18
3.5 14.25 12.32 12.39 12.46 12.53 12.60 12.67 12.74 12.82 12.89
3.6 12.96 13.03 13.10 13.18 13.25 13.32 13.40 13.47 13.54 13.62
3.7 13.69 13.76 13.84 13.91 13.99 14.06 14.14 14.21 14.29 14.36
3.8 14.44 14.52 14.59 14.67 14.75 14.82 14.90 14.98 15.08 15.13
3.9 15.21 15.29 15.37 15.44 15.52 15.60 15.68 15.76 15.84 15.92
4.0 16.00 16.08 16.16 16.24 16.32 16.40 16.48 16.56 16.65 16.73
4.1 16.81 16.89 16.97 17.06 17.14 17.22 17.31 17.39 17.47 17.56
4.2 17.64 17.72 17.81 17.89 17.98 18.06 18.15 18.23 18.32 18.40
4.3 18.49 18.58 18.66 18.75 18.84 18.92 19.01 19.10 19.18 19.27
4.4 19.96 19.45 19.54 19.62 19.71 19.80 19.89 19.98 20.98 20.16
4.5 20.25 20.34 20.43 20.52 20.61 20.70 20.79 20.88 21.90 21.07
4.6 21.16 21.25 21.34 21.44 21.53 21.62 21.72 21.81 22.85 22.00
4.7 22.09 22.18 22.28 22.37 22.47 22.56 22.66 22.75 23.81 22.94
4.8 23.04 23.14 23.33 23.33 23.43 23.52 23.62 23.72 24.80 23.91
4.9 24.01 24.11 24.24 24.30 24.40 24.50 24.60 24.70 25.81 24.90
5.0 25.00 25.10 25.20 25.30 25.40 25.50 25.60 25.70 26.83 25.91
5.1 26.01 26.11 26.21 26.32 26.42 26.52 26.63 26.73 27.88 26.94
5.2 27.04 27.14 27.25 27.35 27.46 27.56 27.67 27.77 28.94 27.98
5.3 28.09 28.20 28.30 28.41 28.52 28.62 28.73 28.84 28.94 29.05
5.4 29.16 29.27 29.38 29.48 29.59 29.70 29.81 29.92 30.03 30.14
5.5 30.25 30.36 30.47 30.58 30.69 30.80 30.91 31.02 31.14 31.25
5.6 31.36 31.47 31.58 31.70 31.81 31.92 32.04 32.15 32.26 32.38
5.7 32.46 32.60 32.72 32.83 32.95 33.06 33.18 33.29 33.41 33.52
5.8 33.64 33.76 33.87 33.99 34.11 34.22 34.34 34.46 34.57 34.69
5.9 34.81 34.93 35.05 35.16 35.28 35.40 35.52 35.64 35.76 35.88

If you move the comma in x one digit to the right (left), then the comma in x2 must be
moved two digits to the right (left)

221
Grade 8 Mathematics [GEOMETRY AND MEASUREMENT]
The Function y = x2 6.00 < x < 9.99
X 0 1 2 3 4 5 6 7 8 9
6.0 36.00 36.12 36.24 36.36 36.48 36.60 36.72 36.84 36.97 37.09
6.1 37.21 37.33 37.45 37.58 37.70 37.82 37.95 38.07 38.19 38.32
6.2 38.44 38.56 38.69 38.81 38.94 39.06 39.19 39.31 39.44 39.56
6.3 39.69 39.82 39.94 40.07 40.20 40.32 40.45 40.58 40.70 40.83
6.4 40.96 41.09 41.22 41.34 41.47 41.60 41.73 41.86 41.99 42.12
6.5 42.25 42.38 42.51 42.64 42.77 42.90 43.03 43.16 43.30 43.43
6.6 43.56 43.69 43.82 43.96 44.09 44.22 44.36 44.49 44.62 44.47
6.7 44.89 45.02 45.16 45.29 45.43 45.56 45.70 45.63 45.63 46.10
6.8 46.24 46.38 46.51 46.65 46.79 46.92 47.06 47.20 47.20 47.47
6.9 47.61 47.75 47.89 48.02 48.16 48.30 48.44 48.58 48.58 48.86
7.0 49.00 49.14 49.28 49.42 49.56 49.70 49.84 49.98 49.98 50.27
7.1 50.41 50.55 50.69 50.84 50.98 51.12 51.27 51.41 54.41 51.70
7.2 51.84 51.98 52.13 52.27 52.42 52.56 52.71 52.85 52.85 53.14
7.3 53.29 53.44 53.58 53.73 53.88 54.02 54.17 54.32 54.32 54.61
7.4 54.76 54.91 55.06 55.20 55.35 55.50 55.65 55.80 55.80 56.10
7.5 56.25 56.40 56.55 56.70 56.85 57.00 57.15 57.30 57.30 57.61
7.6 57.76 57.91 58.06 58.22 58.37 58.52 58.68 58.83 58.83 59.14
7.7 59.29 59.44 59.60 59.75 59.91 60.06 60.22 60.37 60.37 6.68
7.8 60.84 61.00 61.15 61.31 61.47 61.62 61.78 61.94 61.94 62.25
7.9 62.41 62.57 62.73 62.88 63.04 63.20 63.36 63.52 63.52 63.84
8.0 64.00 64.16 64.32 64.48 64.64 64.80 64.96 65.12 65.29 65.45
8.1 65.61 65.77 65.93 66.10 66.26 66.42 66.59 66.75 66.91 67.08
8.2 67.24 67.40 67.57 67.73 67.90 68.06 68.23 68.39 68.56 68.72
8.3 68.29 69.06 69.22 69.39 69.56 69.72 69.89 70.06 70.22 70.39
8.4 70.56 70.73 70.90 71.06 71.23 71.40 71.57 71.74 71.91 72.01
8.5 72.25 72.42 72.59 72.76 72.93 73.10 73.27 73.44 73.62 73.79
8.6 73.96 74.13 74.30 74.48 74.65 74.82 75.00 75.17 75.34 75.52
8.7 75.69 75.86 76.04 76.21 76.39 76.56 76.74 76.91 76.09 77.26
8.8 77.44 77.62 77.79 77.97 78.15 78.32 78.50 78.68 78.85 79.03
8.9 79.21 79.39 79.57 79.74 79.92 80.10 80.28 80.46 80.46 80.82
9.0 81.00 81.18 84.36 81.54 81.72 81.90 82.08 82.26 82.45 82.63
9.1 82.81 82.99 83.17 83.36 83.54 83.72 83.91 84.09 86.12 84.46
9.2 84.64 84.82 85.01 85.19 85.38 85.56 85.75 84.93 87.96 86.30
9.3 86.49 86.68 86.86 87.05 87.24 87.42 87.61 87.80 87.96 88.17
9.4 88.36 88.55 88.74 88.92 89.11 89.30 89.49 89.68 89.87 90.06
9.5 90.25 90.44 90.63 90.82 91.01 91.20 91.39 91.58 91.78 91.97
9.6 92.16 92.35 92.54 92.74 92.93 93.12 93.32 93.51 93.70 93.90
9.7 94.09 94.28 94.67 94.67 94.87 95.06 95.26 95.45 95.65 95.84
9.8 96.04 96.24 96.43 96.63 96.83 97.02 97.22 97.42 97.61 97.81
9.9 98.01 98.21 98.41 98.60 98.80 99.00 99.20 99.40 99.60 99.80

(8.47)2= 71.74 (0.847) 2= 0.7174 = 0.463

(84.7)2 = 7174 (8.472)2 = 71.77

222
Grade 8 Mathematics [GEOMETRY AND MEASUREMENT]
Y = x3 1.00 < x < 5.99
X 0 1 2 3 4 5 6 7 8 9
1.0 1.000 1.030 1.061 1.093 1.125 1.158 1.191 1.225 1.260 1.295
1.1 1.331 1.368 1.405 1.443 1.482 1.521 1.561 1.602 1.643 1.685
1.2 1.728 1772 1.816 1.861 1.907 1.953 2.000 2.048 2.097 2.147
1.3 2.197 2.248 2.300 2.353 2.406 2.460 2.515 2.571 2.628 2.686
1.4 2.744 2.803 2.863 2.924 2.986 3.049 3.112 3.177 3.242 3.308
1.5 3.375 3.443 3.512 3.582 3.652 3724 3.796 3.870 3.944 4.020
1.6 4.096 4.173 4.252 4.331 4.411 4.492 4.574 4.657 4.742 4.827
1.7 4.913 5.000 5.088 5.178 5.268 5.359 5.452 5.545 5.640 5.735
1.8 5.832 8.930 6.029 6.128 6.230 6.332 6.435 6.539 6.645 6.751
1.9 6.859 6.968 7.078 7.189 7.301 7.415 7.530 7.645 7.762 7.881
2.0 8.000 8.121 8.242 8.365 8.490 8.615 8.742 9.870 8.999 9.129
2.1 9.261 93.394 9.528 9.664 9.800 9.938 10.08 10.22 10.36 10.50
2.2 10.65 10.79 11.94 11.09 11.24 11.39 11.54 11.70 11.85 12.01
2.3 12.17 12.33 12.49 12.65 12.81 12.98 13.14 13.31 31.48 13.65
2.4 13.82 14.00 14.17 14.35 14.53 14.71 14.89 15.07 15.25 15.44
2.5 15.63 15.81 16.00 16.19 16.39 16.58 16.78 16.97 17.17 17.37
2.6 17.58 17.78 17.98 18.19 18.40 18.61 18.82 19.03 19.25 19.47
2.7 19.68 19.90 20.12 20.35 20.57 20.08 21.02 21.25 21.48 21.72
2.8 21.95 22.19 22.43 22.67 22.91 23.15 23.39 23.64 23.89 24.14
2.9 24.39 24.64 24.90 25.15 25.41 25.67 25.93 26.20 25.46 26.73
3.0 27.00 27.27 27.54 27.82 28.09 28.37 28.65 28.93 29.22 29.50
3.1 29.79 30.08 30.37 30.66 30.96 26 31.55 31.86 32.16 32.46
3.2 32.77 33.08 33.39 33.70 34.01 34.33 34.65 34.97 35.29 35.61
3.3 35.94 36.26 36.59 36.93 37.26 37.60 37.93 38.27 38.61 38.96
3.4 39.30 39.65 40.00 40.35 40.71 41.06 41.42 41.78 42.14 42.51
3.5 42.88 43.24 43.61 43.99 44.36 44.47 45.12 45.50 45.88 46.27
3.6 46.66 47.05 47.44 47.83 48.23 48.63 49.03 49.43 49.84 50.24
3.7 50.65 51.06 51.48 51.90 52.31 52.73 53.16 53.58 54.01 54.44
3.8 54.87 55.31 55.74 56.18 56.62 57.07 57.51 57.96 58.41 58.86
3.9 59.32 59.78 60.24 60.70 61.16 61.63 62.10 62.57 63.04 63.52
4.0 64.00 64.48 64.96 65.45 65.94 66.43 66.92 67.42 67.92 68.42
4.1 68.92 69.43 69.913 70.44 70.96 71.47 71.99 72.51 73.03 73.56
4.2 74.09 74.62 75.15 75.69 76.23 76.77 77.31 77.85 78.40 79.95
4.3 79.51 80.06 80.62 81.18 81.75 82.31 82.88 83.45 84.03 84.60
4.4 85.18 85.77 86.35 86.94 87.53 88.12 88.72 89.31 89.92 90.52
4.5 91.13 91.73 92.35 92.96 93.58 94.20 94.82 95.44 96.07 96.70
4.6 97.34 97.97 98.61 99.25 99.90 100.5 101.2 101.8 102.5 103.2
4.7 103.8 104.5 105.2 105.8 106.5 107.2 107.9 108.5 109.2 109.9
4.8 110.6 111.3 112.0 112.7 113.4 114.1 114.8 115.5 116.2 116.9
4.9 117.6 118.4 119.1 119.8 120.6 121.3 122.0 122.8 123.5 124.3
5.0 125.0 125.8 126.5 127.3 128.0 128.8 129.6 130.3 131.1 131.9
5.1 132.7 133.4 134.2 135.0 135.8 136.6 137.4 138.2 139.0 139.8
5.2 140.6 141.4 142.2 143.1 143.9 144.7 145.5 146.4 147.2 148.0
5.3 148.9 149.7 150.6 151.4 152.3 153.1 154.0 154.5 155.7 156.6
5.4 157.5 158.3 159.2 160.1 161.0 161.9 162.8 163.7 164.6 165.5
5.5 166.4 167.3 168.2 169.1 170.0 171.0 171.9 172.8 173.7 174.7
5.6 175.6 176.6 177.5 178.5 179.4 180.4 181.3 182.3 183.3 184.2
5.7 185.2 186.2 187.1 188.1 189.1 190.1 191.1 192.1 193.1 194.1
5.8 195.1 196.1 197.1 198.2 199.2 200.2 201.2 202.3 203.3 204.3
5.9 205.4 206.4 207.5 208.5 209.6 210.6 211.7 212.8 213.8 214.9

If you move the comm. In x one digit to the right (left), then the comma in x3 must be
moved three digits to the right (left)

223
Grade 8 Mathematics [GEOMETRY AND MEASUREMENT]
Y = x3 6.00 < x < 9.99
X 0 1 2 3 4 5 6 7 8 9
6.0 216.0 217.1 218.2 219.3 220.3 221.4 211.7 223.6 224.8 225.9
6.1 227.0 228.1 229.2 230.0 231.5 232.6 222.5 234.9 236.0 237.2
6.2 238.3 239.5 240.6 241.8 243.0 244.1 233.7 246.5 247.7 248.9
6.3 250.0 251.2 252.4 253.6 254.8 256.0 245.3 258.5 259.7 260.9
6.4 262.1 263.4 264.6 265.8 267.1 268.3 257.3 270.8 272.1 273.4
6.5 274.6 275.9 277.2 278.4 279.7 281.0 269.6 283.6 284.9 286.2
6.6 287.5 288.8 290.1 291.4 292.8 294.1 282.3 296.7 298.1 299.4
6.7 300.8 302.1 303.5 304.8 306.2 307.5 295.4 310.3 311.7 313.0
6.8 314.4 315.8 317.2 318.6 320.0 321.4 308.9 324.2 325.7 327.1
6.9 328.5 329.9 331.4 332.8 334.3 335.7 322.8 324.2 340.1 341.5
7.0 343.0 344.5 345.9 347.4 348.9 350.4 337..2 338.6 354.9 356.4
7.1 357.9 359.4 360.9 362.5 364.0 365.5 351.9 368.6 370.1 371.7
7.2 373.2 374.8 376.4 377.9 379.5 381.1 367.1 384.2 385.8 387.4
7.3 389.0 390.6 392.2 393.8 395.4 397.1 382.7 400.3 401.9 403.6
7.4 405.2 406.9 408.5 410.2 411.5 413.5 398.7 416.8 418.5 420.2
7.5 421.9 423.6 425.3 427.0 428.7 430.4 415.2 433.8 435.5 437.2
7.6 439.0 440.7 442.5 444.2 445.9 447.7 432.1 451.2 453.0 454.8
7.7 456.5 458.3 460.1 461.9 463.7 465.5 449.5 469.1 470.9 472.7
7.8 474.6 476.4 478.2 480.0 481.9 483.7 467.3 487.4 489.3 491.2
7.9 493.0 494.9 496.8 498.7 500.6 502.5 485.6 506.3 508.2 510.1
8.0 512.0 513.9 515.8 514.8 519.7 521.7 504.4 525.6 527.5 529.5
8.1 531.4 533.4 535.4 537.4 539.4 541.3 523.6 545.3 547.3 549.4
8.2 551.4 533.4 555.4 557.4 559.5 561.5 543.3 565.6 567.7 569.7
8.3 571.8 573.9 575.9 578.0 580.1 582.2 563.6 586.4 588.5 590.6
8.4 592.7 594.8 596.9 599.1 601.2 603.4 584.3 607.6 609.8 612.0
8.5 614.1 616.3 618.5 620.7 622.8 625.0 605.5 629.4 631.6 633.8
8.6 636.1 638.3 640.5 642.7 645.0 647.2 627.2 651.7 654.0 656.2
8.7 658.5 660.8 663.1 665.3 667.6 669.9 649.5 674.5 676.8 679.2
8.8 681.5 683.8 686.1 688.5 690.8 693.2 672.2 697.9 700.2 702.6
8.9 705.0 707.3 709.7 712.1 714.5 716.9 695.5 721.7 724.2 726.6
9.0 729.0 731.4 733.9 736.3 738.8 741.2 719.3 746.1 748.6 751.1
9.1 753.6 756.1 758.6 761.0 763.6 766.1 743.7 771.1 773.6 776.2
9.2 778.7 781.2 783.8 786.3 788.9 791.5 768.6 796.6 799.2 801.8
9.3 804.4 807.0 809.6 812.2 814.8 817.4 794.0 822.7 825.3 827.9
9.4 830.6 833.2 835.9 838.6 814.2 843.9 820.0 849.3 852.0 854.7
9.5 857.4 860.1 862.8 865.5 868.3 871.0 846.6 876.5 879.2 882.0
9.6 884.7 887.5 8890.3 893.1 895.8 898.6 873.7 904.2 907.0 909.9
9.7 912.7 915.5 918.3 921.2 924.0 926.9 901.4 932.6 935.4 938.3
9.8 941.2 944.1 947.0 449.0 952.8 955.7 929.7 961.5 964.4 967.4
9.9 970.3 973.2 976.2 979. 982.1 985.1 958.6 991.0 994.0 997.0

(8.47)3 = 607.6 (0.847)3 = 0.607 3

123.5 = 4.98 3 0.1235 = 0.498

(84.7)3 = 607600 (8.472)3 = 608.0 3 123500 = 49.8

224