(i)
Advancing in
Mathematics
Student’s Book
Form 2
Kinyua Mugo
Lucy Maina
Jared Ondera
(ii)
Published by
Longhorn Publishers (K) Ltd.,
Funzi Road, Industrial Area,
P.O. Box 18033 – 00500,
Nairobi, Kenya.
Longhorn Publishers (Uganda) Ltd.,
Plot 731, Kamwokya Area,
Mawanda Road, P.O. Box 24745,
Kampala Uganda.
Longhorn Publishers (Tanzania) Ltd.,
Kinondoni Plot No.4,
Block 37B, Kawawa Road,
P.O. Box 1237,
Dar es Salaam, Tanzania.
© Kinyua Mugo, Lucy Maina, Jared Odera. 2008
All rights reserved. No part of this publication may be reproduced, stored
in a retrieval system or transmitted in any form or by any means, electronic,
mechanical, photocopying, recording or otherwise without the prior written
permission of the publisher.
First published 2003
Reprinted 2004
Corrected 2008
Reprinted 2009, 2010, 2011, 2013(twice), 2014, 2015, 2016, 2017
Typesetting, graphics and design by Justus Mogaki.
ISBN 9966 49 583 5
Printed by Printing Services Ltd,
Factory Street, Industrial Area,
P.O Box 32197 - 00600,
Nairobi.
(iii)
Chapter 1: Cubes and cube roots .................1
Cubes.................................................................1
Finding cubes by multiplication........................1
Cubes from tables .............................................3
Cube roots .........................................................4
Cube roots of numbers by factor method..........4
Cube roots from tables......................................6
Applications of cubes and cube roots ...............7
Chapter 2: Reciprocals ..................................8
Reciprocals by division..................................... 8
Reciprocal tables............................................... 9
Computation using reciprocals .......................10
Chapter 3: Indices........................................ 11
Indices............................................................. 11
Writing numbers in index form.......................12
Laws of indices ...............................................13
Multiplication law of indices ..........................13
Division law of indices ...................................14
The zero index.................................................15
Negative indices.............................................. 15
Other integral powers (powers of powers)......16
Fractional indices............................................17
Simple equations involving indices ................19
Chapter 4: Logarithms ................................22
Standard form (Revision)................................22
Logarithms ......................................................23
Logarithms of numbers between 1 and 10...... 23
Logarithms of numbers greater than 10 ..........25
Logarithms of numbers between 0 and 1........ 25
Antilogarithms ................................................26
Using logarithms in multiplication .................28
Using logarithms in division........................... 29
Bar numbers....................................................29
Power and roots...............................................31
More combined operations ............................32
Chapter 5: Equations of straight lines .......34
The gradient of a straight line .........................34
To determine gradient of a line through
known points...............................................34
The equation of a straight line ........................37
Gradient-intercept form of the equation
of a straight line ..........................................38
Double intercept (or x-, y-intercept) form
of the equation of a straight line .................40
Parallel and perpendicular lines......................42
Chapter 6: Reection and congruence.......44
Symmetry........................................................ 44
Paper folding and cutting to make
symmetrical shapes..................................... 44
Properties of symmetrical shapes....................45
Symmetry in solids ........................................47
Reection........................................................48
Properties of reection.................................... 48
Reection in the Cartesian plane ....................50
Reection in the mirror lines y = x
and y =
–
x ....................................................50
Reection in the mirror lines x = k
and y = k...................................................... 51
Geometric deductions using reection............52
Angle bisector .................................................52
Vertically opposite angles ...............................52
An isosceles triangle .......................................53
Mediators of the sides of a triangle..............53
Chord of a circle..............................................53
Congruence .....................................................55
Congruent triangles: tests for congruence.......55
Chapter 7: Rotation .....................................58
Rotational symmetry....................................... 58
Rotational symmetry of plane gures.............58
Contents
(iv)
Rotational symmetry of solids ........................59
Rotation as a transformation ...........................60
Properties of rotation.......................................62
Rotation and congruence.................................62
Locating an image given the object, centre
and angle of rotation ...................................62
Finding the centre and angle of rotation .........63
Rotation in the Cartesian plane.......................65
Revision exercise 1 (Chapters 1 to 7)...........68
Revision exercise 1.1 ......................................68
Revision exercise 1.2 ......................................69
Revision exercise 1.3 ......................................71
Chapter 8: Similarity and enlargement .....73
Introduction .................................................... 73
Similar plane gures .......................................73
Construction of similar gures........................74
Similar solids ..................................................76
Enlargement and its properties........................77
Construction of objects and images ................78
Locating the centre of enlargement and
nding the scale factor................................ 78
Fractional scale factor .....................................80
Negative scale factor....................................... 81
Enlargement in the Cartesian plane ................81
Area scale factor .............................................83
Volume scale factor......................................... 84
Chapter 9: Pythagoras’ Theorem ...............86
The theorem ....................................................86
Proof of Pythagoras’ theorem .........................88
Using Pythagoras’ theorem .............................89
Pythagorean triples..........................................91
Using Pythagoras’ theorem in real life
situations .....................................................92
Chapter 10: Trigonometry ..........................94
Introduction..................................................... 94
Tangent of an acute angle................................95
Degrees and minutes.......................................96
Table of tangents .............................................97
Sine and cosine of an acute angle ...................98
Sine and cosine tables. ....................................99
Sines and cosines of complementary
angles. .......................................................101
Relationship between sine, cosine
and tangent................................................ 101
Trigonometrical ratios of special
angles (0°, 30°, 45°, 60°, 90°)...................102
Tangent, sine and cosine of 45° ....................102
Tangent, sine and cosine of 30° and 60° .......103
Tangent of 0° and 90° ...................................103
Sine and cosine of 0° and 90° .......................104
Logarithms of tangents, sines and cosines.... 105
Application of trigonometry .........................106
Chapter 11: Area of a triangle .................. 110
The sine formula ...........................................110
Hero’s formula .............................................. 111
Areas of triangles with equal bases and
between parallel lines................................113
Application of area of a triangle ................... 114
Chapter 12: Areas of quadrilaterals and
other polygons ...................... 116
Introduction................................................... 116
Area of a parallelogram ................................ 116
Area of a trapezium....................................... 117
Area of a regular polygon .............................119
Area of an irregular polygon......................... 120
Chapter 13: Area of part of a circle..........121
Parts of a circle..............................................121
Area of a sector .............................................121
Area of a segment .........................................122
Area of a common region between
two circles................................................. 123
Chapter 14: Surface area of solids............126
Surface area of prisms...................................126
Surface area of a pyramid .............................127
(v)
Surface area of a cone ...................................128
Frustum of a cone or pyramid....................... 129
Surface area of a sphere ................................131
Revision exercise 2 (Chapters 8 to 14).......131
Revision exercise 2.1 ....................................132
Revision exercise 2.2 ....................................133
Revision exercise 2.3 ....................................134
Chapter 15: Volume of solids
Volume of a prism.........................................136
Volume of a pyramid.....................................137
Volume of a cone...........................................138
Volume of a frustum......................................139
Volume of a sphere........................................140
Chapter 16: Quadratic expressions
and equations........................ 142
Simple algebraic expressions........................ 142
Binomial expansions..................................... 142
Binomial products (quadratic identities).......143
Quadratic expressions ...................................147
Factorising quadratic expression...................147
Perfect squares ..............................................150
Quadratic equations ......................................151
Solving quadratic equations.......................... 151
Forming quadratic equations from
given roots.................................................153
Equations leading to quadratic equations .... 155
Word problems leading to quadratic
equations ...................................................155
Chapter 17: Linear inequalities................158
Inequalities.................................................... 158
Compound statements................................... 158
Forming inequalities from word
statements..................................................159
Solution of linear inequalities in
one unknown.............................................160
Solving simultaneous inequalities.................161
Graphical representation of linear
inequalities................................................ 162
Linear inequalities in two unknowns ............163
Graphical solution of simultaneous
linear inequalities..........................................164
Chapter 18: Linear motion........................166
Linear motion................................................ 166
Displacement.................................................166
Speed............................................................. 166
Velocity .........................................................168
Acceleration ..................................................168
Distance-time graph......................................170
Velocity-time graph.......................................172
Interpretation of graphs of linear motion......174
Relative speed ...............................................175
Chapter 19: Statistics ................................ 177
What is statistics?..........................................177
Collecting and organising data .....................177
Frequency distribution table .........................178
Grouped data ................................................ 180
Presentation of data, rank order
list, frequency distribution, table
pictograph, pie chart .....................................182
Frequency polygon....................................... 187
Line graphs and trend................................... 188
Averages; The mean, mode and median ......190
Mean, mode and median of grouped data..... 193
Chapter 20: Angle properties of a circle .197
Introduction................................................... 197
Angles in the same segment..........................197
Angle in a semicircle ....................................199
Angles at the centre and the
circumference of a circle...........................200
Cyclic quadrilateral....................................... 202
Chapter 21: Vectors (1) ............................. 204
Vector and scalar quantities ..........................204
Displacement vector and rotation ................205
(vi)
Equivalent vectors.........................................205
Addition of vectors .......................................205
Scalar multiplication .....................................207
Vectors in the Cartesian plane....................... 208
Position vector ............................................. 211
Midpoints ......................................................213
Length of a vector .........................................213
Translation and displacement vectors ...........215
Revision exercises 3 (Chapters 15 to 21) .218
Revision exercise 3.1 ....................................218
Revision exercise 3.2 ....................................220
Revision exercise 3.3 ....................................223
4-Figure Mathematical tables ..................228
Logarithm base 10 ........................................228
Antilogarithms .............................................230
Sine of angles ............................................... 232
Cosine of angles ........................................... 234
Tangent of angles .........................................236
Cubes of numbers ........................................238
Reciprocals of numbers ................................240
1
In Form 1, we learnt how to nd squares and
square roots of numbers. In this chapter, we
shall learn how to nd and apply cubes and
cube roots of numbers.
Cubes
Activity 1.1
Study the number pattern 1, 8, 27, 64. State
the next ve terms. Compare your answers
with those of other members of your class and
discuss how you arrived at the answers.
From your discussion, you should have noticed
that each of the numbers can be written as a
product of three identical factors.
i.e. 1 = 1 × 1 × 1 = 1
3
8 = 2 × 2 × 2 = 2
3
27 = 3 × 3 × 3 = 3
3
64 = 4 × 4 × 4 = 4
3
Numbers such as 1, 8, 27, 64…… are known
as cubes.
Finding cubes by multiplication
The cube of a number is found by multiplying
the number by itself thrice, i.e. multiplying a
number by its square.
The cubes of the rst 10 natural numbers are:
1 × 1 × 1 = 1
3
= 1 6 × 6 × 6 = 6
3
= 216
2 × 2 × 2 = 2
3
= 8 7 × 7 × 7 = 7
3
= 343
3 × 3 × 3 = 3
3
= 27 8 × 8 × 8 = 8
3
= 512
4 × 4 × 4 = 4
3
= 64 9 × 9 × 9 = 9
3
= 729
5 × 5 × 5 = 5
3
= 125 10 × 10 × 10 = 10
3
= 1 000
Note that each of these cubes can be expressed
as a product of prime factors in power form. In
each case, the power of each factor is always a
multiple of 3. For example,
1728 = (2 × 6)
3
= (2 × 2 × 3)
3
= (2 × 2 × 3) × (2 × 2 × 3) × (2 × 2 × 3)
= 2
6
× 3
3
The two prime factors in 1 728 are 2 and 3 and their
powers are 6 and 3 respectively, both of which are
multiples of 3. Therefore 1 728 is a cube.
Example 1.1
Find the cube of 13. Use your answer to state
the cube of: (a) 1.3 (b) 0.13 (c)130
Solution
The cube of 13 can be expressed as 13
3
.
Thus, 13
3
= 13 × 13 × 13
= 169 × 13
= 2 197
(a) 1.3 ≈ 1
1.3 lies between two consecutive cubes
1 and 8.
∴ 1
3
< 1.3
3
< 2
3
⇒ 1 < 1.3
3
< 8
Since 13
3
= 2 197, then
1.3
3
= 2.197 (The decimal point is
inserted by inspection.)
(b) 0.13
3
= 0.13 × 0.13 × 0.13
= (13 × )
3
= 13
3
× ( )
3
= 2197 ×
= 0.002 197
(c) 130
3
= (13 × 10)
3
= 13 × 10 × 13 × 10 × 13 × 10
= 13
3
× 10
3
= 2 197 × 1 000
= 2 197 000
1
10
6
––
1
10
2
––
1
100
–––
CUBES AND CUBE ROOTS
1
2
3
5
–
4
7
–
3
5
–
3
5
–
3
5
–
3 × 3 × 3
5 × 5 × 5
–––––––
27
125
–––
3
5
–
3
11
7
––
3
–
3
1
4
7
11
7
––
11
7
––
11
7
––
11 × 11 × 11
7 × 7 × 7
––––––––––
1 331
343
––––
302
343
–––
3
8
–
11
13
––
3
7
–
1
2
–
= 5 × 5 × 5 × 2
3
× 2
3
× 2
3
= 5
3
× (2 × 2 × 2) × (2 × 2 × 2) ×
(2 × 2 × 2)
= 5
3
× 2
9
(ii) (3 × 5 × 7
2
)
= (3 × 5 × 7
2
)(3 × 5 × 7
2
)(3 × 5 × 7
2
)
= 3 × 3 × 3 × 5 × 5 × 5 × 7
2
× 7
2
× 7
2
= 3
3
× 5
3
× 7 × 7 × 7 × 7 × 7 × 7
= 3
3
× 5
3
× 7
6
(iii) (2xy
2
z)
3
= (2xy
2
z)(2xy
2
z)(2xy
2
z)
= 2xy
2
z × 2xy
2
z × 2xy
2
z
= 2 × 2 × 2xxxyyyyyyzzz
= 2
3
x
3
y
6
z
3
(b) All the intergers should be raised to 3.
Therefore, 2 ×2
2
× 4 × 4
2
×6
2
× 6
2
2
× 4
2
× 6 = 384
Exercise 1.1
1. Find the cube of each of the following numbers.
(a) 2.3 (b) 5.4 (c) 52.0 (d) 79
2. Evaluate:
(a) 1.5
3
(b) 4.81
3
(c) 2.31
3
(d) 5.01
3
3. Using your answers to Question 2, state the
answers to:
(a) (i) 15
3
(ii) 0.15
3
(iii) 150
3
(b) (i) 0.481
3
(ii) 48.1
3
(iii) 481
3
(c) (i) 0.231
3
(ii) 23.1
3
(iii) 231
3
(d) (i) 0.501
3
(ii) 50.1
3
(iii) 501
3
4. Evaluate:
(a) (2
3
× 3)
3
(b) (4 × 3
2
)
3
(c) (2
2
× 3 × 13
2
)
3
(d) (6
2
× 7)
3
5. Find the smallest positive integer by which
each of the following can be multiplied to
make it a cube.
(a) 2 × 3 × 7
2
(b) 2 × 3
2
(c) 3 × 7 × 11
2
(d) 5
3
× 8
2
6. Find the cubes of the following numbers.
(a) (b) (c) 2 (d) 4
We can also nd cubes of fractions and mixed
numbers as illustrated in Example 1.2 below.
Example 1.2
Find the cube of: (a) (b) 1
Solution
(a) = × ×
=
=
(b) =
= × ×
=
=
= 3
Note that:
1. To nd the cube of a fraction, obtain
the cube of numerator and denominator
separately.
2. To nd the cube of a mixed number,
(a) Write the number as an improper
fraction.
(b) Proceed as for a proper fraction.
Example 1.3 illustrates how to nd the cube
of a number written in power form.
Example 1.3
(a) Find the cube of:
(i) 5 × 2
3
(ii) 3 × 5 × 7
2
(iii) 2xy
2
z giving
your answer in the same form.
(b) What would you multiply by 2 × 4 × 6
2
to
make it cube?
Solution
(a) (i)(5 × 2
3
)
3
= (5 × 2
3
)(5 × 2
3
)(5 × 2
3
)
3
Fig. 1.1
Since 52.3 = 5.23 × 10 and
5.23
3
= 143.06 ,
then 52.3
3
= 5.23
3
× 10
3
= 143.06 × 10
3
= 143 060
Note that by multiplication,
52.3
3
= 143 055. 667
≈ 143 060 (5 s.f.).
If the number to be cubed has more than 4 s.f.,
round off the number to 4 s.f. then proceed as in
Example 1.5.
Example 1.5
Use tables to evaluate 0.586 367
3
.
Solution
0.586367 ≈ 0.5864 (4 s.f.)
∴ 0.586367
3
≈ 0.5864
3
0.5864
3
=
5.864
3
10
=
5.864
3
10
3
Fig. 1.2
( )
Main columns Differences columns
x
0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9
1.0
1.1
1.2
1.3
5.7
5.8
4
– – – – – – – – – – – – – – – – – – – – – – – –
201.2
CUBES
– – – – – – – – – –
– – – – – – – – – –
Main columns Differences columns
x
0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9
1.0
1.1
1.2
1.3
5.1
5.2
– – – – –
143.06
CUBES
– – – – – – – – – –
7. Which of the following represent cubes?
(a) 2
3
× 3 × 5 (b) 2
6
× 4
3
× 5
3
(c) 7
3
× 9
6
× 11
11
(d) 5
3
× 12
3
8. Find the smallest integral multiplier which
makes each of the following a cube.
(a) 3x
3
y
4
(b) 2a
6
× 5b
3
(c) 4x
7
y
2
× 2xy
3
× z
2
(d) 3x
10
× y × 3x
2
× y
5
Cubes from tables
Sometimes, nding cubes of large numbers
by multiplication can be cumbersome. The
alternative method of nding cubes of numbers
is by use of tables. However, tables of cubes
only give approximate values, for example, a
4-gure table gives cubes of numbers correct
to 4 s.f. , sometimes to 5 s.f. The following
examples illustrate how to nd cubes of
numbers using tables.
Example 1.4
Use 4-gure tables to nd the cube of 52.3.
Solution
52.3 ≈ 50 (1 s.f.)
∴ 52.3
3
≈ 50
3
≈ 125 000
Since 52.3 >50,
52.3
3
is just over 125 000.
We also know that,
52.3 = 5.23 × 10
∴ 52.3
3
= (5.23 × 10)
3
= 5.23
3
× 10
3
Using tables of cubes locate 5.2 in the rst
column on the left, usually headed x. Follow
the row of 5.2 across the page until you get to
the column headed 3 in the main columns.
The intersection of row 5.2 and column 3 gives
143.06, see Figure 1.1)
4
3
3
3
A
10
n
–––
3
3
3
Thus, row 5.8
give 201.2
main column 6
+
Differences column 4 gives
4
201.6
∴5.864
3
= 201.6
⇒
5.864
3
= 201.6
10
3
1 000
0.586 367
3
≈ 0.201 6
Note that:
1. To nd the cube of a number less than
1, rst express the number in the form
– where 1 ≤ A < 10 and n is a positive
integer.
2. If the number is greater than 10, express it
in the form A × 10
n
where 1 ≤ A < 10 and
n is a positive integer.
Exercise 1.2
Use 4-gure tables to nd the cubes of the
numbers in Questions 1 to 3.
1. (a) 7 (b) 29 (c) 398 (d) 1 238
2. (a) 3.891 (b) 0.817 (c) 0.002 5
(d) 0.206 388
3. (a) 6.831 9 (b) 3.999 9 (c) 80.901
(d) 288.48
4. Use tables to evaluate x
3
+ y
3
given that:
(a) x = 63.2 and y = 41.9,
(b) x = 0.842 1 and y = 0.6158,
(c) x = 9.872 and y = 4.518 7.
Cube roots
If a and b are two non-zero numbers such that
a
3
= b, we say that a is the cube root of b,
written as a = b for example,
2 × 2 × 2 = 2
3
= 8 ⇒ 2 = √8 and
–
3 ×
–
3 ×
–
3 =
–
3
3
=
–
27 ⇒
–
3 = √
–
27.
Cube roots of numbers by factor method
If a number has an exact cube root, it can be
expressed as a product of prime factors in
power form. In this case, the power(s) of the
prime factor(s) will be divisible by 3.
For example,
216 = 6 × 6 × 6
= 6
3
216 = 6
Since 6 = 2 × 3,
then, 216 = (2 × 3)
3
= 2 × 3 × 2 × 3 × 2 × 3
= 2
3
× 3
3
The powers of the prime factors 2 and 3 are
both divisible by 3.
Thus to nd the cube root of a number by
factor method, the procedure is:
(i) Express the number as a product of prime
factors in power form.
(ii) Divide the power of each factor by 3.
(iii) Multiply out the result to get the cube
root.
Example 1.6
Use factor method to nd the cube root of 2
744.
Solution
2 744 = 2 × 1 372
= 2 × 2 × 686
= 2 × 2 × 2 × 343
= 2 × 2 × 2 × 7 × 49
= 2 × 2 × 2 × 7 × 7 × 7
= 2
3
× 7
3
√2 744 = √2
3
× 7
3
= 2
3 ÷ 3
× 7
3 ÷ 3
= 2 × 7
= 14
5
–––––
3
27
1 000
–––
3
3
3
10
3
27
1 000
–––––
––
216
1 000 000
–––––––––
6
100
–––
3
216
1 000 000
3
3
2
10
––
8
1 000
–––––
1
5
–
3
3
–––––
8
1 000
3
3
3
27
1 000
–––––
3
10
––
3 × 3 × 3
10 × 10 ×10
–––––––––––
61
64
––
125
64
–––
–––––
27
1 000
3
3
1
61
64
––
125
64
–––
3
125
64
3
3
3
61
64
––––––
3
3
27
1 000
3
3
5 × 5 × 5
4 × 4 × 4
1
4
–
5
4
–
3
3
5
3
4
3
We can also nd cube roots of fractions and
mixed numbers. In general,
a = a , where a and b are whole numbers.
b
b
To nd the cube root of a mixed number, we
rst convert it into an improper fraction.
Example 1.7
Find the cube root of: (a) (b) 1
Solution
(a) =
Since = ,
then =
=
(b) 1 =
∴ =
=
=
=
=
= 1
1. To nd the cube root of a proper fraction,
nd the cube roots of the numerator and
denominator separately.
2. To nd the cube root of a mixed number:
(a) Express the number as an improper
fraction.
(b) Find the cube root of the numerator
and denominator separately.
(c) Simplify the result.
Remember that any terminating or recurring
decimal can be expressed as a fraction.
Example 1.8 illustrates the procedure of nding
the cube root of a terminating decimal.
Example 1.8
Use factor method to evaluate:
(a) 0.008 (b) 0.000 216
Solution
(a) 0.008 =
∴ 0.008 =
=
=
(b) 0.000 216 =
0.000 216=
=
= 0.06
To nd the cube root of a terminating decimal:
1. Express the decimal as a fraction whose
denominator is a power of 10 divisible
by 3.
2. Proceed as with an improper fraction.
Exercise 1.3
1. Find the cube roots of the following leaving
your answers in power form.
(a) 2
6
× 3
3
× 5
3
(b) 4
6
× 7
9
(c) 2
18
× 4
3
× 9
6
(d) 27 × 125 × 3
3
2. Use factor method to nd the cube root of:
(a) 64 (b) 225 × 15
(c) 10 648 (d) 17 576
3. Find the cube root of:
(a) 121 × 11 (b) 1 296 × 36
(c) 3 969 × 63 (d) 784 × 28
6
7. Evaluate:
(a) (b) (c)
Cube roots from tables
Some of the numbers used in every day life
are not whole numbers. Therefore approximate
methods of nding cube roots of numbers are
sometimes used. Mathematical tables give
approximate values of cube roots of numbers
between 1 and 1 000.
To nd the cube root of a number:
1. If the number lies between 1 and 1 000,
use the table of cubes and reverse the
process of nding the cube.
2. If the number is greater than 1 000,
express it in the form A × 10
n
where
1 ≤ A < 1 000 and n is a multiple of 3.
3. If the number is less than 1, express it in
the form where 1 ≤ A < 1 000 and n
is a positive integer divisible by 3.
Example 1.9
Use tables to calculate:
(a) 36 (b) 9 869 (c) 0.000 731 9
A
10
n
–––
4. Evaluate:
(a) (b)
(c) (d)
5. Find the cube root of :
(a) 0.2
3
(b) 0.003
6
(c) 0.05
12
(d) 0.225 × 0.015
6. Find the least positive integer by
which each of the following must be
multiplied to make it a perfect cube. Hence
nd the cube roots of the resulting numbers.
(a) 2
3
× 2
4
× 3
2
(b) 2
6
× 4
2
× 5
(c) 128 (d) 5
3
× 6
2
3
25 × 5
49 × 7
––––––
3
225 × 15
625 × 25
––––––––
361 × 19
121 × 11
––––––––
3
3
23
121
–––
1 ×
12
11
––
3
343a
3
b
3
–––––
3
8
x
3
––
3
8
2
a
3
b
6
x
3
––––––
3 3 3
3
3
3
3
3
3
3
3 3
67
– – – – – – – – – – – – – – – – – – – –
CUBES
Main columns Differences columns
x
0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9
1.0
1.1
1.2
1.3
3.2
3.3
3.4
35.937
– – – – – – – – – –
– – – – – – – – – –
Solution
(a) We know that 36 = 36.0
Fig. 1.3
Using tables of cubes (Fig 1.3), locate 36.0
or a number nearest to it but smaller. 36.0
does not appear in the table. The number
nearest to it is 35.937 which is in row 3.3
and column 0. Therefore 3.30 is the number
that will lead to the cube root of 36. The
difference between 36 and 35.937 is 63.
But in the differences column, the number
nearest to 63 is 67 which corresponds to
column 2.
Thus, 36.0 = 3.302.
You could conrm that our method is
correct by nding the cube of 3.302, which
you will nd to be approximately equal to 36.
(b) 9 869 = 9.869 × 1 000
9 869 = 9.869 × 10
3
= 9.869 × 10
3
In the table of cubes 9.800 is nearest to
9.869 with a difference of 69.
Thus,
9.800 is at the intersection of row 2.1 and
column 4.
Difference 69 is at the intersection of
differences column 5 and row 2.1
∴9.869 = 2.145
⇒ 9.869 × 10
3
= 9.869 × 10
3
i.e. 9 869 = 2.145 × 10 = 21.45
7
Applications of cubes and cube roots
When solving problems involving volumes,
we usually calculate cubes of numbers. It is
often necessary to reverse the process of nding
volume, say to nd the length of a cube given
its volume.
This process calls for nding the cube root of a
number. You handled such problems in Form 1.
Exercise 1.5
1. A solid cube has its edges measuring 3.25
cm long. Calculate the volume of the cube.
2. Calculate the volume of a solid cuboid of
sides 19.2 cm long, 8.8 cm wide and 5 cm
high. Use your result to nd the size of a
solid cube with the same volume.
3. A cubic container has a volume of 7 893 cm
3
.
Find the height of a cuboid of the same
volume if its base is 21.3 cm by 14.5 cm.
4. A cylindrical metal bar of height 12 cm and
radius 3.5 cm is melted down and recast
into a cubic block. Find the length of the
side of the cube, giving your answer to one
decimal place.
5. Find the length of a cube whose volume
equals that of a cylinder of height 8.6 cm
and radius 3.8 cm.
6. A cubic block of wood has sides 8 cm long.
A cylindrical hole of radius 3.6 cm is drilled
through the block. Find the volume of the
part of the block that remains.
3
3
3
3
–––––––––
731.9
1 000 000
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
731.9
100
–––––––
3
(c) 0.000 731 9 =
=
In the tables of cubes, the number nearest to
731.9 is 731.4 with a difference of 5.
∴Row 9.0 give 731.4
Column 1
Difference 5 corresponds to column 2
731.9 = 9.012
0.000 731 9 =
731.9
√1 000 000
=
9.012
= 0.090 12
100
Exercise 1.4
1. Each of the numbers given in this question
lies between two consecutive perfect cubes.
In each case, state the two cubes and hence
use tables to evaluate their cube roots.
(a) 5.831 (b) 85.76 (c) 985.3
(d) 503.77
2. Use tables to evaluate:
(a) 98.36 (b) 628.5 (c) 1 977
(d) 4.247 8 × 10
4
3. Use tables to evaluate:
(a) 0.062 16 (b) 0.004 32
(c) 0.618 79 (d) 16.8 × 10
6
4. Simplify:
(a) 216a
3
b
3
(b) 24
3
× 10
6
(c) 12
3
a
12
b
3
(d) 13
3
x
6
y
3
5. (a) Given that x
3
= 158, nd x.
(b) If a = 8 and b = 14, evaluate:
(i) a
2
+ b
2
(ii) b
2
– a
2
(iii) 3a
3
b
2
(iv)
3
a
3
+ b
3
8
In Form 1, we learnt that if the product of
two numbers equals 1, each number is said
to be the multiplication inverse of the other.
For example, , and are multiplication
inverses of 4, 8 and 12 respectively. These
multiplication inverses are also known as
reciprocals.
Reciprocals by division
Since the product of a number and its reciprocal
is 1, we nd the reciprocal of any number by
dividing 1 by the number.
Example 2.1
Find the reciprocals of: (a) (b) 9 (c) 3
Solution
(a) Reciprocal of = 1 ÷
= 1 × (Division by a fraction
means multiplication by
its inverse.)
=
Note that × = 1.
∴ is the reciprocal of .
(b) Reciprocal of 9 means 1 ÷ 9
1 ÷ 9 =
∴ is the reciprocal of 9.
(c) Reciprocal of 3 means 1 ÷ 3
1 ÷ 3 = 1 ÷
= 1
× ( is the multiplication
inverse of )
=
∴ the reciprocal of 3 or is .
Generally if a and b are two non-zero numbers,
and ab = 1, then a is the reciprocal of b and vice
versa.
i.e. ab = 1
⇒ a =
and b =
Thus the reciprocal of b is , that of a is and
that of is .
Reciprocals may be given in fraction or decimal
form as the question may demand.
Exercise 2.1
1. State the reciprocals of:
(a) (b) (c)
2. Given that x = 3, y = and z = , evaluate:
(a) The reciprocals of x, y and z.
(b) (c) (d) (e)
(f) (g) (h)
3. Evaluate z + + 1.5 given that z = 2.
4. Evaluate the following:
(a) 3 × (b) 1 ×
(c) 1 + 2 ×
1
3
–
2
5. Find the value of x if = + +
6. Find the reciprocals of:
(a) (b)
(
√125
)
2
(c) – + 2
1
4
–
1
12
––
2
3
–
1
2
–
2
3
–
2
3
–
3
2
–
3
2
–
2
3
–
3
2
–
3
2
–
2
3
–
1
9
–
1
9
–
1
2
–
1
2
–
1
2
–
7
2
–
2
7
–
2
7
–
–
7
2
–
2
7
–
1
b
–
1
a
–
a
b
–
b
a
–
3
7
–
5
8
–
7
20
––
2
9
–
5
6
–
x
y
–
y
x
–
y
z
–
x
z
–
1
x + y
––––
1
y + z
––––
1
x + z
––––
2
3
–
1
2
–
1
2
–
5
6
–
4 ÷
4
1
1
2
–
1
8
–
1
x
–
1
3
–
1
4
–
1
5
–
1
z
–
1
3
–
1
4
–
3
14
––
1
1
–
3
7
1
1
4
1
2
–
–
7
2
–
2
7
–
1
2
2
1
b
–
1
a
–
RECIPROCALS
2
9
7. Find the reciprocals of:
(a) 2.8 (b) 0.28 (c) 28 (d) 280
giving your answers correct to 4 s.f.
Comment on your answers.
Reciprocal tables
The table of reciprocals is used in exactly
the same way as other tables except that the
difference must be subtracted. This is because
the reciprocal of a number becomes smaller as
the number becomes larger, for example, the
reciprocal of 4 is = 0.25, while that of 8 is
= 0.125.
Thus < ⇒ 8 > 4.
Generally, reciprocal tables cater for numbers
between 1 and 10 only. Thus, to nd the
reciprocal of a number greater than 10, or less
than 1, rst express the number in the form A ×
10
n
or
A
10
n
–––
where 1 ≤ A < 10 and n is an integer,
i.e. the number must rst be written in standard
form.
Since the reciprocal of 1 is = 1, and that of
10 is = 0.1, the reciprocal of any number
between 1 and 10 must lie between 0.1 and 1.
Example 2.2
Use tables to evaluate the reciprocal of 14.76.
Solution
Express 14.76 in the form A × 10
n
.
Thus, 14.76 = 1.476 × 10
1
Now, using the reciprocal tables, locate row
1.4 in the left hand column. Follow this row
until you are in the column headed 7. The
intersection of row 1.4 and column 7 gives
number 0.6803.
Continue in the same row to the differences
column 6 where you will nd a difference of 29.
Subtract 29 from 0.6803 to get 0.6774, which
will lead to the reciprocal of 14.76.
This example is best arranged as follows.
Row 1.4 0.680 3
Column 7 –
Differences column 6 29
0.677 4
= 0.677 4
= ×
= 0.677 4 ×
Thus, = 0.067 74.
Example 2.3
Use tables to nd the reciprocal of 0.072 34.
Solution
0.072 34 =
=
Now, using tables,
Row 7.2 column 3 give 0.138 3
Differences column 4 gives - 1
0.138 2
∴
= 0.138 2
Since 0.0723 4 = or 7.234 ×
then = or ×
=
= 0.138 2 × 100
= 13.82
1
8
–
1
4
–
1
8
–
1
4
–
1
1
–
1
10
––
1
1.476 × 10
1
–––––––––
1
1.476
–––––
1
1.476
–––––
1
10
1
–––
1
10
––
7. 234
100
–––––
1
7.234
–––––
7.234
100
–––––
1
0.072 34
–––––––
1 × 100
7.234
––––––
1
7.234
100
–––––
Fig. 2.1
1
14.76
–––––
1
100
–––
1
7.234
–––––
1
1
100
––
Main columns Subtract mean difference
x 0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9
1.0
1.1
1.2
1.3
1.4
1.5
29
– – – – – – – – – – – – – – – – – – – – – – – – – – –
0.6803
– – – – – – – –
– – – – – – – –
CUBES
7. 234
10
2
–––––
10
3. (a) 38.17 (b) 72.46 (c) 1 300
(d) 19.76 (e) 64.52 (f) 93.51
(g) 435.7 (h) 98 631
4. Find the reciprocal of 28.36. Without any
further use of tables state the reciprocals of
(a) 2.836 (b) 0.283 6 (c) 283.6
(d) 2 836
5. Evaluate:
1 000
12 × 16.39
Computation using reciprocals
Example 2.4 below illustrates how to use
reciprocals in computation.
Example 2.4
Find the value of x given that = +
Solution
In this example, evaluate the given reciprocals
separately, then combine by addition.
Thus: =
= ×
= 0.762 3 ×
= 0.076 23
=
= ×
1
10
––
= 0.117 43
i.e. = 0.117 43
x =
= 8.516
Exercise 2.3
Use tables to solve the questions.
1. Given that p = 2.382, nd the value of
p + .
2. If = + , nd the value of x.
3. If v = 17.24, r = 16.41 and = – , nd
the value of u.
4. Evaluate:
(a) + (b) –
5. If = + and a = 131 when b = 129,
nd the value of c.
6. Evaluate:
(a) ÷ 0.008 (b) 8.06 × 60 ÷ 108
(c) 88 ÷ 0.054 (d) 1 ÷ +
7. Evaluate using tables:
8. Evaluate:
(a) 1 ÷ √127.5 (b)
9. If x = 788 when b = 88, evaluate.
.
1
x
–
1
0.117 43
–––––––
1
p
–
1
14.64
–––––
1
x
–
1
8.73
––––
1
u
–
1
r
–
1
v
–
1
4 – √2
–––––
1
4 + √5
–––––
1
130
–––
1
√7 – 2
–––––
1
√107 + 3
––––––––
3
a
2
–
1
b
–
1
c
–
1
7.934
–––––
1
7.81
––––
3
x – b
x + b
–––––
1
10
––
1
13.12
–––––
1
10
––
1
1.312 × 10
––––––––
1
1.312
–––––
3
√
1
0.013
–––––
+
1
0.0147
––––––
3
√
100
7.85
––––
Exercise 2.2
Use tables to nd the reciprocals of the numbers
in Questions 1 to 3.
1. (a) 8.317 (b) 1.732 (c) 6.452
(d) 1.111
2. (a) 0.873 1 (b) 0.008 317 (c) 0.084 547
(d) 0.064 52 (e) 0.744 44 (f) 0.333 33
1
13.12
–––––
1
24.27
––––– .
1
x
–
= 0.412 0 ×
= 0.041 20
⇒ = 0.041 20
∴ + = 0.076 23 + 0.041 2
1
24.27
–––––
1
2.427 × 10
––––––––
1
2.427
–––––
1
24.17
–––––
1
24.27
–––––
1
13.12
–––––
1
10
––
11
1
2
–
1
2
–
1
2
–
1
2
–
2
3
–
2
3
–
2
3
–
2
3
–
2
3
–
(b) In 4
5
, 4 is the base and 5 is the index.
∴ 4
5
= 4 × 4 × 4 × 4 × 4
5 factors
(c) In 2
4
, 4 is the index which means that 2
multiplied together.
∴2
4
= 2 × 2 × 2 × 2
= 16
Exercise 3.1
1. Express each of the following in index
form.
(a) 3 × 3 (b) 4 × 4 × 4 × 4
(c) 2 × 2 × 2 × 2 × 2 (d) 6 × 6 × 6 × 6
(e) 9 × 9 × 9 × 9 × 9 × 9
(f) 11 × 11 × 11 × 11 × 11
(g) 13 × 13 × … × 13
12 factors
2. Write the following in expanded form.
(a) 2
3
(b) 3
4
(c) 3
7
(d) 4
3
(e) 7
3
(f) 5
6
(g) 15
6
(h) 12
3
3. Copy and complete.
(a) 4
2
= 16. Thus 16 is the second power of
4.
(b) 3
3
= __. Thus __ is the third power of 3.
(c) 6
3
= __. Thus __ is the __ power of __.
4. Which is greater, 5
3
or 3
5
?
5. How many factors are there in:
(a) 2 (b) 2
4
(c) 2
10
?
6. Write in a shorter form using index notation.
(a) × × × (b) × × × ×
Indices
In Form 1, we learnt that:
2
3
means 2 × 2 × 2
3 factors
3
5
means 3 × 3 × 3 × 3 × 3 , and so on.
5 factors
Thus, a
4
is a × a × a × a .
4 factors
The raised numeral is called an index (plural:
indices). It is also refered to as a power or
exponent.
For example, 64 can be written as 64 = 4 × 4 × 4 = 4
3
.
In this case, 3 is called the index and 4 is called
the base. We say that 64 is the third power of 4.
4
3
is the index form (to base 4) of 64.
Generally,
If n = a
x
, then
n = a × a × … × a
x factors
where x is a positive integer, a
x
is the index
form of n, where a is the base and x is the
index or power.
Example 3.1
(a) Express 5 × 5 × 5 × 5 × 5 × 5 in index form.
(b) Write 4
5
in expanded form.
(c) Evaluate 2
4
.
Solution
(a) In 5 × 5 × 5 × 5 × 5 × 5,
6 factors
∴5 × 5 × 5 × 5 × 5 × 5 = 5
6
{
{
{
{
{
{
{
INDICES
3
5 is the base and
6 is the index
12
(c) b × b × b × b × b × b
(d) m × m × m × m
(e) t × t × … × t
20 factors
Writing numbers in index form
We have seen that we can write 4 as 2
2
, 25 as
5
2
, and so on. Thus it should be possible to
write any given number in index form in the
simplest form as long as the number has factors.
For example, 16 can be factorised using a factor
tree as shown below.
16 16 is divisible by 2 or 16
2 8 8 is divisible by 2 4 4
2 4 4 is divisible by 2 2 2 2 2
2 2
∴ 16 = 2 × 8
= 2 × 2 × 4
= 2 × 2 × 2 × 2
same factors
= 2
4
(simplest form)
Note that 2
4
is considered to be a simpler form
since in 4
2
, 4 is not a prime number.
Example 3.2
Write each of the following in its simplest index
form.
(a) 81 (b) 96 (c) 5 × c × c × 5 × c × 5
Solution
(a) 81
9 9
3 3 3 3
(b) 96 96 = 2 × 2 × 2 × 2 × 2 × 3
8 12
2 4 2 6 = 2
5
× 3
2 2
2 3
(c) 5 × c × c × 5 × c × 5
= 5 × 5 × 5 × c × c × c
same same
factors factors
= 5
3
× c
3
= 5
3
c
3
Exercise 3.2
1. Write each of the following in index form
using the specied base.
(a) 25 (base 5) (b) 64 (base 4)
(c) 49 (base 7) (d) 1 000 (base 10)
2. Write each of the following in its simplest
index form.
(a) 2 × a × a × a
(b) 3 × y × y
(c) h × h × h × 7 × h × 21
(d) 3 × b × b × a × b × b × b
(e) 3 × a × 3 × a × a × a
3. Write each of the following in its simplest
index form.
(a) 2 (b) 8 (c) 32
(d) 16 (e) 64 (f) 128
4. Evaluate:
(a) 2
2
× 3
3
(b) 2
4
× 3
2
(c) 4
2
× 3
2
(d) 5
2
× 2
2
(e) 2
3
× 10
4
(f) 4
3
× 10
5
5. Evaluate the following if a = 2.
(a) 3a
2
(b) 5a
3
(c) 9a
2
(d) 6a
2
(e) 16a
2
(f) 2a
3
Also 16 = 4 × 4
same factors
= 4
2
81 = 9
2
(This is not the
simplest form since 9 is
not a prime number.)
∴ 81 = 3 × 3 × 3 × 3
4 factors
= 3
4
(This is the
simplest
index form.)
same single
factors factor
(Rearranging the
factor)
{
{
{
{
{
{
{
{
13
= 10 × 10 × 10 × 10 × 10 × 10 × 10
(2 + 5) factors
= 10
(2 + 5)
= 10
7
(d) 2
3
× 2
7
= (2 × 2 × 2) × (2 × 2 × 2 × 2 × 2 × 2 × 2)
= 2
(3 + 7)
= 2
10
When two numbers, written in index form
with a common base, are multiplied, the
indices are added while the base remains
unaltered for example,
a
x
× a
y
= a
(x + y)
If in a multiplication there is more than one
letter to be multiplied, they must be multiplied
separately because each represents a different
value.
Example 3.4
Simplify 4x
3
y
3
× 5x
4
y
5.
Solution
4x
3
y
3
× 5x
4
y
5
= 4 × 5 × x
3
× x
4
× y
3
× y
5
(Rearranging the factors)
= 20 × x
(3 + 4)
× y
(3 + 5)
= 20 × x
7
× y
8
= 20x
7
y
8
Squaring an expression simply means
multiplying the expression by itself, for
example,
(a
5
)
2
= a
5
× a
5
= a
(5 + 5)
(Multiplication law)
= a
(5 × 2)
= a
10
Thus, in order to square an algebraic expression,
square the coefcient and double the indices of
the letters.
6. If n is a whole number, nd the smallest
value of n for which a
n
is greater than 35
given that:
(a) a = 2 (b) a = 3 (c) a = 4
7. Express 2 × 2 × 2 × 3 × 3 × 5 × 5 × 5 × 5 in
index form.
8. Express 72 and 108 as products of powers
of 2 and 3.
Laws of indices
When numbers are in index form, they are
easier to write and also to use in calculations.
The following laws of indices are used in
calculations involving multiplication, division,
power and roots.
Multiplication of numbers in index form
Recall that,
2
3
= 2 × 2 × 2 and 2
5
= 2 × 2 × 2 × 2 × 2.
3 factors 5 factors
Therefore, 2
3
× 2
5
= (2 × 2 × 2) × (2 × 2 × 2 × 2 × 2)
3 factors 5 factors
= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
(3 + 5) same factors
= 2
8
∴ 2
3
× 2
5
= 2
(3 + 5)
= 2
8
Example 3.3
Simplify each of the following giving your
answer in index form.
(a) 10
2
× 10
5
(b) 2
3
× 2
7
Solution
(a) 10
2
× 10
5
= (10 × 10) × (10 × 10 × 10 × 10 × 10)
2 factors 5 factors
{
{
{
{
{
{
{
{
14
Example 3.5
Find the square of 4x
4
y
3
.
Solution
(4x
4
y
3
)
2
= 4
2
× x
4 × 2
× y
3 × 2
= 16x
8
y
6
Or (4x
4
y
3
)
2
= 4x
4
y
3
× 4x
4
y
3
= 4 × 4 × x
4
×
x
4
×
y
3
×
y
3
= 16x
8
y
6
(Adding indices of terms
with common bases.)
Exercise 3.3
1. Simplify:
(a) a
2
× a
4
(b) n
5
× n
7
(c) p
3
× p
5
(d) 5
8
× 5
5
(e) p
3
× p
4
× p
5
(f) z
7
× z
12
× z
(g) t
3
× t
7
× t
2
2. Evaluate the following leaving your
answers in index form.
(a) 2 × 2
3
(b) 3
2
× 3
3
(c) 3
2
× 7
2
× 3 (d) 10
3
× 10
5
3. Simplify the following.
(a) 4x
7
y
2
× 2xy
3
z
2
(b) 2x
4
× 5x
3
(c) 6x
2
y × 3x
3
y
5
(d) 2a
2
b × 4a
3
b
2
4. Simplify:
(a) (x
3
)
2
(b) (a
2
b
3
)
2
(c) (5a
3
bc
2
)
2
(d) (9a
5
b
5
c
2
)
2
Division of numbers in index form
In factor form, 2
5
= 2 × 2 × 2 × 2 × 2
and 2
3
= 2 × 2 × 2
Thus, 2
5
÷ 2
3
=
2
5
2
3
5 factors
=
2 × 2 × 2 × 2 × 2
2 × 2 × 2
3 factors
= 2 × 2
(5 – 3) factors
= 2
2
∴2
5
÷ 2
3
= 2
(5 – 3)
= 2
2
In this example, the power of 2 in the answer
is the difference between that of the numerator
and that of the denominator.
Example 3.6
Simplify each of the following giving your
answer in the simplest index form.
(a) 3
7
÷ 3
4
(b) x
8
÷ x
3
Solution
7 factors
(a) 3
7
÷ 3
4
=
3
7
=
3 × 3 × 3 × 3 × 3 × 3 × 3
3
4
3 × 3 × 3 × 3
4 factors
= 3 × 3 × 3
(7 – 4) factors
= 3
(7 – 4)
= 3
3
8 factors
(a) x
8
÷ x
3
=
x
8
=
x × x × x × x × x × x × x × x
x
3
x × x × x
3 factors
= x × x × x × x × x
(8 – 3) factors
= x
(8 – 3)
= x
5
If there is more than one letter to be divided,
they must be divided separately as they
represent different values.
{
{
{
{
{
{
{
{
{
15
Solution
12x
4
y
5
÷ 3x
3
y
2
=
12x
4
y
5
3x
3
y
2
=
4x
4
y
5
(Dividing the
x
3
y
2
coefcients)
=
4xy
5
(Dividing by x
3
)
y
2
= 4xy
3
(Dividing by y
2
)
∴ 12x
4
y
5
÷ 3x
3
y
2
= 4xy
3
When two numbers, written in index form
with a common base are divided, the indices
are subtracted, i.e. a
x
÷ a
y
= a
(x – y)
.
Exercise 3.4
1. Use the division law of indices to evaluate
(a) 2
6
÷ 2
3
(b) 2
8
÷ 2
5
(c)
2
18
÷ 2
15
(d) 4
19
÷ 4
18
(e) 2
13
÷ 2
8
(f)
2
9
÷ 2
4
(g) 2
10
÷ 2
8
(h) 5
3
÷ 5
2. Simplify:
(a)
4
6
(b)
4
7
(c)
10
9
(d)
h
11
4
4
4
3
10
7
h
(e)
a
8
(f)
g
12
(g)
g
6
(h)
p
5
a
3
g
9
g p
(i)
h
9
(j)
a
6
h
7
a
4
3. Simplify the following.
(a) 8a
4
÷ 4a
2
(b)
x
12
y
5
x
7
y
4
(c) 4x
2
y
5
÷ 2xy
3
(d)
35a
7
b
12
c
3
5a
5
b
4
c
2
(e) 14p
9
q
6
r
2
÷ 2pq
4. If A = 27x
4
y
3
z
4
and B = 3x
2
yz
2
, nd:
(a) AB (b) A ÷ B (c) A
2
(d) B
2
.
The zero index
Consider 3
4
÷ 3
4
.
Using the division law of indices,
3
4
÷ 3
4
= 3
(4 – 4)
= 3
0
.
Using factor form,
3
4
÷ 3
4
=
3
4
=
3 × 3 × 3 × 3 = 1
3
4
3 × 3 × 3 × 3
∴ 3
0
= 1
Similarly,
x
3
÷ x
3
= x
(3 – 3)
= x
0
and
x × x × x
= 1
x × x × x
∴ x
0
= 1
Any non-zero number raised to power zero,
equals to 1 i.e. a
0
= 1 for all values of a.
Negative indices
To divide x
5
by x
2
, we simply subtract the
indices, for example,
x
5
÷ x
2
= x
5 – 2
= x
3
.
Now consider x
2
÷ x
5
.
x
2
÷ x
5
=
x × x
x × x × x × x × x
subtract indices cancel by factors
x
2 – 5
1
x
3
x
–3
1 1 1 1
1 1 1 1
Example 3.7
Simplify 12x
4
y
5
÷ 3x
3
y
2
.
16
Any number raised to a negative power is
the same as the reciprocal of the equivalent
positive power of the same number, for
example,
a
–x
=
1
and not
–
a
x
, provided a ≠ 0.
a
x
Example 3.8
Simplify the following giving answers with
positive indices.
(a) x
5
÷ x
8
(b) x
3
÷ x
–4
Solution
(a) x
5
÷ x
8
= x
5 – 8
= x
–3
=
1
x
3
(b) x
3
÷ x
–4
= x
3 – (–4)
= x
3 + 4
= x
7
Or
x
3
÷ x
–4
=
x
3
=
x
–4
= x
3
× x
4
= x
3 + 4
= x
7
Other integral powers (Powers of powers)
Consider (2
2
)
3
.
This means that ‘2 squared is cubed’.
Thus, (2
2
)
3
= 2
2
× 2
2
× 2
2
= 2
2 + 2 + 2
= 2
6
But 2 + 2 + 2 = 2 × 3
So, in evaluating (2
2
)
3
, we have in fact
multiplied the indices together.
Similarly, (x
5
)
4
= x
5
× x
5
× x
5
× x
5
= x
5 + 5 + 5 + 5
= x
5 × 4
= x
20
When a number, written in index form is
raised to another power, the indices are
multiplied, for example,
(a
x
)
y
= a
x × y
= a
xy
.
Now consider (2 × 3)
3
.
(2 × 3)
3
means ‘(2 × 3) cubed’.
Thus, (2 × 3)
3
= (2 × 3) × (2 × 3) × (2 × 3)
= 2 × 2 × 2 × 3 × 3 × 3
(Regrouping like factors)
= 2
3
× 3
3
.
Similarly, (a × b)
3
= a
3
b
3
.
All the numerals which are multiplied together
in a bracket are raised to the power of that
bracket, for example,
(a × b)
x
= a
x
b
x
.
( )
3
means ‘ cubed’
( )
3
= × ×
=
= 2
3
3
3
Similarly, ( )
4
= × × ×
=
=
a
4
b
4
All the numerals which are divided in a
bracket are raised to the power of the bracket,
for example,
a
x
=
a
x
.
b b
x
2
3
–
2
3
–
2 × 2 × 2
3 × 3 × 3
–––––––
a
b
–
a
b
–
a
b
–
a
b
–
a
b
–
a × a × a × a
b × b × b × b
–––––––––––
Thus, x
–3
=
1
x
3
(Since we are doing the same
thing but using different methods,
the results must be equal.)
x
4
x
3
1
2
3
–
2
3
–
2
3
–
2
3
–
17
3
5
–
p
q
–
3
7
–
x
y
–
a
b
–
1
8
–
m
n
l
m
x
y
–
x
2
y
––
n
2
m
––
ab
2
c
–––
3
4
–
1
6
2
––
1
z
8
––
1
y
4
––
1
a
3
––
1
2
9
––
1
2x
5
–––
1
2
–
1
2
–
1
2
–
1
2
–
9. Write the following using positive indices.
(a) x
–6
(b) y
–3
(c) 2
–1
(d) p
–8
(e) q
–4
(f) 3
4
x
–2
10. Find the value of each of the following.
(a) x
–1
if x = 11 (b) z
0
if z = 3
(c) x
–5
if x = 2 (d) p
–2
if p = 9
(e) x
–2
if x = 12
11. If x is greater than 1, which is greater, x or
x
2
?
12. If 0 < x < 1, which is greater, x or x
2
?
13. If 0 < x < 1, arrange x
2
, x
4
, x
3
in an
ascending order.
Fractional indices
Note that if n is any positive integer, a
n
is the
short form for
a × a × … × a
n factors
Fractional powers obey the same laws of
indices as integral powers. What then is the
meaning of:
(i) 4 (ii) 8 (iii) 8 ?
Consider
4 × 4 = 4
+
(Addition law of indices.)
= 4
1
= 4
∴ 4 is the number which when multiplied by
itself gives 4.
Also we know that √4 × √4 = 2 × 2 = 4
∴ 4 = √4
Similarly, a × a = a
+
= a
1
= a
and √a × √a = a.
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
Exercise 3.5
1. Simplify:
(a) 2
0
(b) x
1
(c) 100
0
(d) (4x)
1
(e) (xy)
0
(f) 3x
0
(g) 5
2
x
0
y
0
(h) (3a)
0
+ 3 (i) 4a
0
– 4
2. Write the following with positive indices
and then simplify.
(a) a
–4
× a (b) m
12
× m
–2
÷ m
–5
(c) 2x
–5
× x (d) x ×
1
x
–2
3. Work out the following.
(a)
(
x
4
)
2
× x
2
(b)
(
a
5
)
3
× a
–5
(c)
(
y
7
)
2
÷ y
14
(d)
(
a
3
)
4
× a
–12
(e)
(
x
–2
)
6
× x
13
(f) a
8
×
(
a
3
)
2
4. Simplify:
(a) (3 × 5)
2
(b) (2ab)
3
(c) (72xy)
0
(d)
(
x
2
y
2
)
3
5. Simplify, giving your answers in positive
index form.
(a)
(
)
2
(b)
(
)
5
(c)
(
)
–
2
(d)
(
)
3
(e)
(
)
–
3
(f)
(
)
–
1
(g)
(
)
3
(h)
(
)
–
2
(i)
(
)
7
6. Simplify:
(a)
(
)
3
(b)
(
)
–2
(c)
(
x
3
)
3
÷ x
5
(d)
(
)
5
7. Evaluate:
(a) 6
0
(b) 1 000
0
(c) 10
9
× 10
–8
(d) 7
3
÷ 7
4
(e)
(
)
–2
(f) (2 × 5)
–1
(g) (3
3
)
2
(h)
(
4
2
)
0
8. Write the following with negative indices.
(a) (b) (c)
(d) (e) (f)
1
2
–
2
3
–
1
3
–
{
18
Example 3.10
Evaluate 16
1
2
–
1
3
–
1
3
–
1
3
–
1
3
–
1
3
–
1
3
–
1
3
–
1
3
–
1
3
–
1
3
–
1
3
–
1
3
–
1
3
–
2
3
–
1
3
–
1
3
–
2
3
–
1
3
–
1
3
–
2
3
–
1
3
–
1
3
–
1
3
–
2
3
–
2
3
–
3
3
1
2
–
1
2
–
1
3
–
1
3
–
1
3
–
1
3
–
1
3
–
1
3
–
1
3
–
1
3
–
3
1
3
–
1
2
–
3
3
33
n
3 3
1
3
–
2
3
–
3
1
2
–
1
2
–
1
2
–
4
–
1
n
1
2
–
∴a = √a for all positive values of a.
Consider
8 × 8 × 8 = 8
+ +
= 8
1
= 8
∴ 8 is the number which when multiplied by
itself three times gives 8.
n
Also √8 × √8 × √8 = 2 × 2 × 2 = 8
∴ 8 =
3
√8
Similarly, a
3
= a × a × a
= a
+ +
= a
1
or a
3
= a
1
Since
a
3
= a
× 3
and
3
√a
3
= a (This means that a cube
root of a number cubed gives the number itself.)
∴ a = √a for all values of a.
We have seen that a = √a and a = √a.
Likewise, √a means the nth root of a.
n is called the order of the root. So in
general, a = √a, where n is a positive integer.
Consider
8 = 8 × 8 since = +
= 8
2
since 8
2
= 8
× 2
= 8
= √8
2
since 8 = √8
But 8 = 2
3
∴
3
(2)
3
= 2
√
Hence,
3
2
8
√
= 2
2
= 4
Similarly, a = a
2
= √a
2
3
4
Since a can also be written as (a
2
) , then
a = √a
2
= √a
2
.
Example 3.9
Find the square root of 16x
4
y
2
:
(a) by factor method, and
(b) using laws of indices.
Solution
(a) 16x
4
y
2
= 2 × 2 × 2 × 2 × x × x × x × x × y × y
(Expanded form)
= 2 × 2 × 2 × 2 × x × x × x × x × y × y
(Pairing off factors)
∴ √16x
4
y
2
= 2 × 2 × x × x × y
(Picking one out of every
two like factors.)
= 2
2
× x
2
× y
= 4x
2
y
(b) √16x
4
y
2
=
(
16x
4
y
2
)
(Since a = √a.)
= 16 (x
4
) (y
2
) (Raising each
of the factors to power .)
= 4x
2
y (Since (a
m
)
n
= a
mn
.)
Note that method (b) is quicker.
For any root, order n, we simply nd the n
th
root of the coefcient and divide the indices
of the letters by the order n of the root, for
example,
√81a
4
b
12
= 3ab
3
.
19
In general, if a
x
= a
y
, then x = y.
Similarly, if a
x
= b
x
, then a = b, provided
both the bases are positive numbers.
To solve equations involving indices, follow
the procedure below.
1. Express both sides of the equation with
a common base and simplify as far as
possible to reduce to one term on LHS
and one term on RHS.
2. If the variable is in the exponent, equate
the indices and solve the resulting
equation.
3. If the variable is in the base, ensure that
the powers are the same. Equate the bases
and solve the resulting equation.
Example 3.11
Solve the equation.
(a) 81
n
= 3 (b) 9
(a – 3)
× 81
(1 – a)
= 27
–a
1
3
–
1
2
–
16
25
––
1
2
–
1
4
–
2
5
–
1
81
––
2
3
–
3
2
–
3
2
–
1
2
–
1
2
–
1
2
–
4
3
–
9
16
––
1
2
–
1
4
–
49
16
––
1
2
–
1
5
–
2
3
–
3
4
–
1
2
–
5
3
–
3
4
–
1
4
–
1
4
–
1
4
–
1
4
–
1
4
3
4
–
1
4
1
4
–
1
4
1
4
–
1
4
1
4
–
4
444
a
b
–
1
b
–
1
5
–
10. Simplify the following.
(a) √9x
2
y
4
(b) √36n
4
m
6
(c) √3
4
x
8
y
6
z
2
(d) √64s
4
t
10
Simple equations involving indices
Sometimes we may be required to solve
equations involving indices.
Consider 2
x
= 4. What value of x makes the
equation true?
2
x
= 4 is the same as
2
x
= 2
2
(Expressing RHS in index form.)
The base on the LHS is equal to the base on the
RHS. Since LHS = RHS, the indices must also
be equal.
∴ if 2
x
= 2
2
, then
x = 2
b
b
b
a
b
–
1
b
–
1
4
–
3
4
–
3
4
–
1
4
–
3
4
–
4
3
–
1
3
–
1
27
––
1
2
–
1
3
–
–
1
2
–
2
3
–
–
Solution
16 = 16
3
since × 3 =
= 16 × 16 × 16
= 16 × 16 × 16 since 16 = 16
= 2 × 2 × 2 = 8
Also 16 = 16
3
since 3 × =
= 16 × 16 × 16
= 2
4
× 2
4
× 2
4
= 2
4
×
×
2
4
×
×
2
4
×
= 2 × 2 × 2 = 8
In this example, it is easier to take the root rst
since it is exact.
In general
x = x
a
= √x
a
or
x = x
a
= √x
a
(best when √x is exact).
Exercise 3.6
Evaluate the expressions in Questions 1-8.
1. (a) 8 (b) 9 (c) (d)
2. (a) 27 (b) 9 (c) 32 (d) 81
3. (a) 16
–
(b) 8
–
(c)
–
(d) 6
–
(e)
–
4. 8 + 8
0
– 2
–
2
5. 81 –
–1
– 27
0
6. +
–
2
+ 16 7. 4 × 3
–2
÷ 8
–
8. (x – 1) + (x + 7) + 3x
0
when x = 9.
9. Express as a single power of 2.
8 × 64 × 4
–3
20
Example 3.14
Solve for x in 9
x + 1
+ 3
2x + 1
= 36
Solution
9
x + 1
+ 3
2x + 1
= 36
Expressing in index form where possible.
3
2(x + 1)
+ 3
2x + 1
= 4 × 3
2
3
2x + 2
+ 3
2x + 1
= 4 × 3
2
3
2x + 1
× 3
1
+ 3
2x + 1
= 4 × 3
2
3
2x + 1
× (3
1
+ 1) = 4 × 3
2
(Factoring out
3
2x+1
.)
⇒ 3
2x + 1
= 3
2
(Dividing both sides by
the common factor 4)
⇒ 2x + 1 = 2 (Equating indices.)
∴ x =
Exercise 3.7
Solve the equations in Questions 1 to 16.
1. 2
x
= 32 2. 4
x + 1
= 32
Expressing RHS in index form with the
same base as LHS.
27
–
a
= (3
3
)
–
a
= 3
–
3a
Since LHS = RHS
3
–
2a – 2
= 3
–
3a
⇒
–
2a – 2 =
–
3a
∴ a = 2
Example 3.12
Solve the equation y
4
= 81.
Solution
RHS = 81
= 3
4
LHS = y
4
∴ y
4
= 3
4
Since indices are the same, then the bases must
be equal.
∴ y = 3
Since the power 4 is even, then y could also be
equal to the addition inverse of 3, i.e.
–
3
1
2
–
Solution
(a) 81
n
= 3
LHS = 81
n
= (3
4
)
n
(Expressing LHS in index
form.)
= 3
4n
(Since (a
x
)
y
= a
xy
.)
RHS = 3 = 3
1
∴ 3
4n
= 3
1
⇒ 4n = 1 (Equating the indices since LHS
and RHS have a common base.)
n =
(b) 9
(a – 3)
× 81
(1 – a)
= 27
–a
LHS = 9
(a – 3)
× 81
(1 – a)
Expressing in index form and simplifying.
⇒LHS = 3
2(a – 3)
× 3
4(1 – a)
= 3
(2a – 6) +
3
(4 – 4a)
= 3
–2a – 2
1
4
–
Example 3.13
Solve for x and y in 3
x
× 3
y
= 1 …………(i)
and 2
2x – y
= 64 ………(ii)
Solution
In equation (i), 3
x
× 3
y
= 1
3
x
× 3
y
= 3
0
(Expressing RHS as a
power
of 3)
⇒ 3
x + y
= 3
0
In equation (ii), 2
2x – y
= 64
⇒ 2
2x – y
= 2
6
x + y = 0 +
2x – y = 6
3x = 6
∴ x = 2
If x = 2, x + y = 0 becomes
2 + y = 0
y
=
–
2
Thus, the simultaneous solution is x = 2, y =
–
2.
(Equating the indices
and solving
the equations
simultaneously.)
21
29. 16
(3 + n)
× 2
(1 + n)
=
(
)
(1 – n)
30. 25
2n
÷ 5
n
= 5
6
31. 5
x
× 6
4
= 180
2
32. 6
2x + 1
= 2
2x + 1
33. 8 × 2
2x – 1
= 16 × 2
x – 1
34. If
(
)
m
× 81
–n
= 243, express m in terms
of n.
Solve for x and y in Questions 35 to 38.
35. 2
2x + y
= 8 and 3
x – y
= 1
36. 3
5x – 2y
= 243 and 3
2x – y
= 3
37. 5
x
× 5
2y
= 25 and 3
2x
÷ 3
y
= 81
38. 3
2x – y
= 27 and 4
x
÷ 16
y
= 1
1
2
–
1
27
––
3. 3x
3
= 24 4. 3
x + 1
= 9
2
5. 4
x + 1
= 8
x
6. 2x = 18
7. 3
2x – 5
= 3
7
8. 8
x – 1
= 32
x
9. 3
2x – 5
= 27 10. 81
n
= 1
11. 81
n
= √3 12. 5
n
=
13. 5
n
= 14. (0.01)
n
= 10
15. 0.01
n
= √10 16. 128
x
= 2
Solve for n in Questions 17 to 26.
17. y
n
× y
n
= 1 18. y
n
× y
n
=
19. x
3
÷ x
3
= x
n
20. x
18
÷ x
6
= x
n
21. x
3
÷ x
n
= x
–6
22. x
n
× x
6
= x
4
23. x
n
× x
n
= x
16
24. x
4
÷ x
12
= x
n
25. x
n
× x
6
= x
–6
26. x
n
× x
n
= x
1
5√5
–––
1
y
–
1
25
––
Solve the equations in Questions 27 to 33.
27. 9
x
× 3
(2x – 1)
= 3
15
28. 9
(x – )
= 27
( – x)
1
2
–
1
2
–
3
4
–
22
Standard form (Revision)
Consider a number such as 6 520.
it can be written down as 6.52 × 1 000
i.e. 6 520 = 6.52 × 1 000
= 6.52 ×10
3
, which means 6.52 ×10
3
is said to be the standard form of 6 520.
Similarly, the number 0.000 15 can be written
as
0.000 15 =
1.5
10 000
=
1.5
10
4
= 1.5 ×
1
10
4
= 1.5 × 10
–
4
1.5 ×10
–
4
is said to be the standard form of
0.000 15.
A number is in standard form if it is expressed
in the form a × 10
n
, where a is a number
between 1 and 10, and n is an integer.
Example 4.1
(i) Express 78 956 340 in standard form,
correct to 4 s.f.
(ii) Calculate 7.15 × 10
2
× 4 × 10
3
giving your
answer in standard form.
(iii) Evaluate (1.2 ×10
9
) ÷ (4.8 × 10
–
4
.)
Solution
(i) 78 956 340 = 7.895 634 × 10 000 000
= 7.895 634 × 10
7
= 7.896 × 10
7
(4 s.f)
(ii) 7.15 × 10
2
× 4 × 10
3
= 7.15 × 4 × 10
2
×10
3
(Regrouping)
= 28.60 × 10
5
= 2.86 × 10 × 10
5
= 2.86 × 10
6
(Adding powers of 10.)
(iii) (1.2 × 10
9
) ÷ (4.8 × 10
–
4
)
=
1.2 × 10
9
4.8 × 10
–
4
=
12 × 10
9
48 × 10
–
4
= × 10
9 –
–
4
= × 10
13
= 0.25 × 10
13
=
× 10
13
= 2.5 × 10
–
1
× 10
13
= 2.5 × 10
12
Exercise 4.1
1. Write the following in the ordinary form.
(a) 1.2 × 10
1
(b) 4.275 × 10
0
(c) 8.863 × 10
3
(d) 6.2 ×10
3
(e) 9.578 × 10
–
2
(f) 3.94 ×10
–
3
(g) 7.859 × 10
–
1
(h) 3.65 × 10
2
2. Express the following numbers in standard
form.
(a) 74.3 (b) 65.2 (c) 86.25
1
4
–
1
4
–
2.5
10
1
–––
(Multiplying numerator
and denominator each
by 10.)
(Subtracting powers
of 10.)
(Adding powers
of 10.)
LOGARITHMS
4
23
3. Write the following in the ordinary form.
(a) 2 × 10
–
1
(b) 0.3 × 10
2
(c) 0.02 × 10
–
2
(d) 0.158 × 10
–
3
4. Express the following numbers in standard
form.
(a) 0.2 (b) 0.03
(c) 0.001 5 (d) 0.105 02
(e) 0.000 526 8 (f) 0.000 045 8
(g) 0.003 84 (h) 0.001 056
(i) 0.468 9 (j) 0.003 97
(k) 0.001 25 (l) 0.000 000 914 8
Logarithms
If a number is expressed in index form with 10
as the base, the index (or power) is called the
common logarithm of the number or simply
the logarithm of the number.
Consider the number 100., 100 = 10
2
⇒ 2 is the logarithm of 100 to the base 10.
This is written in short as
log
10
100 = 2
base number logarithm
Note: Common logarithms are usually written
without indicating the base, for example, log
100 means the same as log
10
100.
100 000 000 = 10
8
∴ log 10
8
= 8
Likewise,
0.000 001 =
1
= 10
–
6
10
6
∴ log 10
–
6
=
–
6
Table 4.1 below shows logarithms of some
powers of 10.
Number Power of 10 Logarithm
1 1 = 10
–
6
–
6
1 000 000 10
6
1 1 = 10
–
5
–
5
100 000 10
5
1 1 = 10
–
4
–
4
10 000 10
4
1 1 = 10
–
3
–
3
1 000 10
3
1 1 = 10
–
2
–
2
100 10
2
1 1 = 10
–
1
–
1
10 10
1
1 1 = 10
0
0
10 1 = 10
1
1
100 1 = 10
2
2
1 000 1 = 10
3
3
10 000 1 = 10
4
4
100 000 1 = 10
5
5
1 000 000 1 = 10
6
6
Table 4.1
In general, the logarithm of any power of 10
is equal to the power or index, for example,
log 10
n
= n and log 10
–
n
=
–
n.
Logarithms of numbers between 1 and 10
We know that 1 = 10
0
. Thus, log 1 = 0.
Also, 10 = 10
1
. Hence, log 10 = 1.
From this we observe that if a number lies
between 1 and 10, its logarithm must lie
between 0 and 1. Such a number is not an exact
power of 10. Its logarithm, therefore, can only
be approximated.
(d) 97.38 (e) 112.2 (f) 432.6
(g) 413 (h) 982 (i) 724.9
(j) 324.8 (k) 5 401 (l) 3 096
(m) 23 847 (n) 92 652 (o) 827 400
24
the column headed 5 in the main columns
(Table 4.3).
Table 4.3
The number at the intersection of the row and
the column is 5 119.
Continue across the page along the same row
until you get to the differences column headed 4.
The number at this intersection is 5.
The number 5 119 will lead to the logarithm of
3.25. But we require the logarithm of 3.254.
∴ Add the difference 5 to 5 119 to get 5 124
∴ Log 3.253 8 = 0.512 4 (since 3.254 lies
between 1 and 10)
Example 4.4
Find the logarithm of 5.034 356.
Solution
5.034 356 ≈ 5.034 (4 s.f. since tables cater
for only 4 s.f.)
Row 5.0 and main columns 3 give 7 016.
Differences column 4 gives 3.
∴ Total = 7 016 + 3 = 7 019.
Hence log 5.034 356 = 0.7 019 since 5.034 lies
between 1 and 10.
Exercise 4.2
Use 4-gure tables to nd the logarithm of
1. (a) 5.372 (b) 4.185 (c) 7.321
2. (a) 3.712 (b) 4.713 (c) 8.495
3. (a) 7.430 (b) 2.49 (c) 6.213
4. (a) 3.161 4 (b) 3.214 3 (c) 4.111 6
Mathematical tables provide logarithms of
numbers between 1 and 9.999. Many tables
give these values correct to 4 signicant gures,
hence the name ‘4-gure tables.
The following examples illustrate how to nd
logarithms of numbers using logarithm tables.
Example 4.2
Find the logarithm of 3.25 using the table of
logarithms.
Solution
Look for 3.2 in the rst column and move
across the page along the row of 3.2 until you
get to the column headed 5 in the main columns
(Table 4.2).
Table 4.2
The number at the intersection of row 3.2 and
column 5 is 5 119, and will lead to the logarithm
of 3.25.
Since 3.25 lies between 1 and 10, its logarithm
must lie between 0 and 1.
∴ Log
10
3.25 = 0.511 9 The d.p. is located by
inspection according to the size of the number.
⇒ 10
0.5119
= 3.25
Example 4.3
Use tables to nd the logarithm of 3.253 8.
Solution
3.253 8 ≈ 3.254 (4 s.f. since tables cater for
only 4 s.f.)
Look for 3.2 in the rst column and move
across the page along the row until you get to
Main columns Differences columns
x 0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9
ADD
3.2
– – – – – – – – – –
– – – – – – – – –
5119
– – – – – – – – –
Main columns Differences columns
x 0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9
ADD
32
– – – – – – – – – – –
– – – – – – – – –
5119
– – – – – – – – –
– – – – – –
– – – – – – – – – – – – – – – –
5
25
Logarithms of numbers greater than 10
Recall that
log 10 = 1
log 100 = 2
log 1 000 = 3, and so on.
Therefore, if a number is greater than 10,
its logarithm must be greater than 1. The
logarithm must, therefore, contain an integral
part (whole number) and a decimal part. The
integral part is called the characteristic and the
decimal part the mantissa.
The logarithm of a number greater than 10 is
obtained as follows.
1. Write the number in the form, a ×10
n
, for
example, log 535.7 = 5.357 × 10
2
.
2. Using tables, nd the logarithm of a, for
example, log 5.357 = 0.729 0
3. By inspection, state the logarithm of 10
n
,
for example log 10
2
= 2.
4. Add the powers to get logarithm of the
number, for example 10
0.729 0
× 10
2
= 10
2.729 0
.
∴log 535.7 = 2.729 0.
Example 4.5
Find the logarithm of:
(a) 20.73 (b) 30 550
Solution
Number Standard As power Logarithm
form of 10 of number
from tables Characteristic
(a) 20.73 2.073 × 10
1
10
0.316 6
× 10
1
= 10
1 + 0.316 6
= 10
1.316 6
1.316 6
from tables
(b) 30 550 3.055 × 10
4
10
0.485 0
× 10
4
= 10
0.485 0 + 4
= 10
4
.485 0
4.485 0
Exercise 4.3
1. Express the following in standard form:
(a) 973 (b) 48.6 (c) 5 000
(d) 5 270 000 (e) 4 850 (f) 72 536
2. State the characteristic of the logarithm of:
(a) 15 (b) 4 800 (c) 600
(d) 2 (e) 93 000 000 (f) 39.8
(g) 1 250 (h) 23.4 (i) 735.6
3. Use 4-gure tables to nd the logarithm of:
(a) 3.16 (b) 31.6 (c) 316
(d) 3 160 (e) 31 600 (f) 316 000
What do you notice about the logarithms of
these numbers?
4. Find the logarithm of:
(a) 247 (b) 3 600 (c) 10 900
(d) 15.7 (e) 9 402 (f) 15 455
5. Write the following numbers as powers
of 10.
(a) 7 863 (b) 12.41 (c) 428.45
(d) 98 370 (e) 91.93 (f) 108 341
(g) 128 (h) 25.997 (i) 368 701
Logarithms of numbers between 0 and 1
Consider a number such as 0.064 8.
0.064 8 = 6.48 × 10
–
2
From tables, log 6.48 = 0.8116
∴0.064 8 = 10
0.811 6
× 10
–
2
= 10
–
2 + 0.811 6
Thus, log 0.064 8 =
–
2 + 0.8116
Negative characteristic Positive mantissa
Such logarithm as
–
2 + 0.811 6 is usually
written as 2.811 6 to show that only 2 is
negative.
Mantissa
26
4. Copy and complete Table 4.4.
Number Standard Product of Logarithm
form powers of 10
2.073 2.073 × 10
0
10
0.316 6
× 10
0
0.316 6
0.207 3 2.073 × 10
–
1
10
0.316 6
× 10
–
1
1.316 6
0.020 73
0.002 073
0.000 207 3
0.000 020 73
Table 4.4
What do you notice about the logarithms of
the numbers in Table 4-4?
Give a reason for your observation.
5. Express the following as powers of 10.
(a) 0.053 46 (b) 0.896 3
(c) 0.996 3 (d) 0.873 6
(e) 0.004 89 (f) 0.000 462 1
Antilogarithms
Given the logarithm of a number, the
corresponding number can be obtained by
using Tables of antilogarithms. This means
reversing the process of nding the logarithm of
the number.
Example 4.7
Find the value of 10
0.707 6
.
Solution
Since the power of 10 represents the logarithm
of a number, then the logarithm of the required
number is 0.707 6.
∴ log x = 0.707 6 where x is the required
number.
Since 0.707 6 lies between 0 and 1, x must lie
between 1 and 10.
Using antilogarithm tables (Table 4.5), look for
0.70 down the rst column.
–
–
–
–
2.811 6 is read as ‘bar two point eight one one
six’.
Notice that if a number lies between zero and
1, its logarithm consists of two parts, a negative
characteristic and a positive mantissa. Such
a logarithm is written with the minus sign
above the characteristic to show that only the
characteristic is negative.
Example 4.6
Find the logarithm of 0.003 681.
Solution
Number Standard As power Logarithm
form of 10
0.003 681 3.681 × 10
–
3
10
0.565 9
× 10
–
3
= 10
–
3 + 0.565 9
= 10
3.565 9
3.565 9
Note that logarithms of negative numbers do
not exist, for example, log (
–
814 5) does not
exist.
Exercise 4.4
1. Express the following in standard form.
(a) 0.091 (b) 0.008 1
(c) 0.673 (d) 0.002 8
2. State the characteristic of the logarithm of.
(a) 0.48 (b) 0.008 914
(c) 0.000 003 482 (d) 0.069 35
3. Work out the logarithm of:
(a) 0.763 815 (b) 0.008 413
(c) 0.104 4 (d) 0.000987
(e) 0.342 (f) 0.048 3
(g) 0.007 6 (h) 0.902
(i) 0.045 034 (j) 0.401 3
(k) 0.007 137 9 (l) 4.03 × 10
–
5
(m) 6.104 × 10
–6
(n) 4.28 × 10
–
10
27
Antilogarithms
Table 4.5
Move across the page along row 0.70 until you
get to the column headed 7 in the main columns.
The number at the intersection of the row and
the column is 5 093. Note it down.
Continue across the page along the same row
until you get to the differences column headed
6. The number at this intersection is 7. This
7 must be added to 5 093 and we get 5 100.
Therefore, the number whose logarithm is
0.707 6 is 5.1.
i.e. log x = 0.707 6
∴x = 5.1
Example 4.8
Use antilogarithm tables to nd the number
whose logarithm is:
(a) 2.713 6 (b) 3.873 8
Solution
(a) Let a be the number whose logarithm is
2.713 6.
∴log a = 2.713 6
⇒ a = 10
2.713 6
= 10
2
× 10
0.713 6
Now using antilogarithm tables,
Row 0.71 and column 3 give 5 164.
Differences column 6 gives 7.
The total is 5 164 + 7 = 5 171.
∴ 10
0.713 6
= 5.171
Thus, a = 5.171 × 10
2
= 517.1.
(b) Let b be the number whose logarithm is
3.873 8.
∴log b = 3.873 8
⇒ b = 10
3.873 8
= 10
–
3
× 10
0.873 8
Using antilogarithm tables,
Row 0.87 and column 3 give 7 464.
Differences column 8 gives 14.
The total is 7 464 + 14 = 747 8.
∴ 10
0.873 8
= 7.478
Thus, b = 7.478 × 10
–3
= 0.007 478.
Exercise 4.5
Use antilogarithm tables to nd the numbers
whose logarithms are given in Questions 1 to 5.
1. (a) 0.481 5 (b) 2.307 1 (c) 1.811 4
2. (a) 3.042 1 (b) 2.116 3 (c) 0.947 2
3. (a) 0.832 1 (b) 4.673 9 (c) 1.816 4
4. (a) 1.621 3 (b) 1.814 4 (c) 3.411 8
5. (a) 1.651 1 (b) 2.495 3 (c) 4.688 3
Evaluate the following:
6. (a) 10
0.483 5
(b) 10
1.704 0
(c) 10
2.780 1
7. (a) 10
1.905 2
(b) 10
2.525 5
(c) 10
3.660
3
8. (a) 10
4.199 8
(b) 10
2.071 6
(c) 10
1.077 9
Find the numbers whose logarithms are given in
Questions 9 to 11.
9. (a)
–
1 + 0.592 2 (b)
–
2 + 0.8645
10. (a)
–
1 + 0.477 1 (b)
–
2 + 0.608 5
11. (a)
–
4 + 0.908 5 (b)
–
2 + 0.301 6
Main columns Differences columns
x 0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9
ADD
0.70
– – – – – – – – – – – – – –
– – – – – – – – –
5 093
– – – – – – – – –
– – – – – – – – –
– – – – – – – – – – – – – – – –
7
–
–
–
– ––
– ––
– ––
– ––
–
28
Using logarithms in multiplication
When multiplying numbers, we could use
logarithms. Usually, logarithms are used to
avoid long and cumbersome multiplications,
and when accuracy is required only to a limited
number of signicant gures.
The knowledge acquired from writing a number
to power 10 and realisation that the index is the
logarithm of the number makes it possible to
apply common logarithms in multiplication.
Using the multiplication law of indices, we
get the logarithm of the product and nd its
antilogarithm.
Let a = 10
x
and b = 10
y
Then a × b = 10
x
× 10
y
= 10
x + y
Now, log a = x, log b = y, and
log (a × b) = log 10
x + y
= x + y
Thus, log (a × b) = log a + log b
So, to get the product a × b, get the sum of
logs and nd the antilog.
Example 4.9
Use logarithms to evaluate:
(a) 3.742 × 41.68
(b) 91.85 × 3.467 × 125.3
Solution
(a) Number Standard form Logarithm
3.742 3.742 ×10
0
0.5731 +
41.68 4.168 × 10
1
1.6199
10
2.193 0
2.1930
= 10
2
× 10
0.193
= 10
2
× 1.559
156 ∼ 10
2
× 1.560
∴ 3.742 × 41.68 = 156 correct to 3 s.f.
(b) Number Standard form Logarithm
91.85 9.185 × 10
1
1.963 0
3.467 3.467 × 10
0
0.540 0 +
125.3 1.253 × 10
2
2.097 9
10
4.600 9
4.600 9
= 10
4
× 10
0.600 9
39 890 = 10
4
× 3.989
∴ 91.85 × 3.467 × 125.3 = 39 890
To nd the product of two numbers, add their
logarithms. Then nd the antilogarithm of
the sum, i.e. log (x × y) = log x + log y since
if log x = a, and log y = b, then
xy = 10
a
× 10
b
= 10
(a + b)
.
Then nd the antilogarithm of (a + b).
Exercise 4.6
Use 4-gure logarithm tables to work out the
following.
1. (a) 4.8 × 6.4 (b) 23.66 × 14.21
2. (a) 3.2 × 17.41 (b) 16.82 × 72.13
3. (a) 49.2 × 2.345 (b) 3.081 × 17.26
4. (a) 42.44 × 53.81 (b) 17.6 × 2.48
5. (a) 3.481 × 2.673 (b) 23.12 × 42.67
6. (a) 3.731 × 21.42 (b) 74.01 × 2.335
7. (a) 6.71 × 42.62 (b) 9.92 × 4.59
8. (a) 25.9 × 96.3 (b) 451.9 × 637.9
9. 2.041 × 5.763 × 241.2
10. 8.743 × 763.1 × 42.68
11. 563.7 × 26.14 × 3.211
12. 546.6 × 9 748 × 4.385
29
Using logarithms in division
To divide one number by another, we subtract
the logarithm of the denominator from the
logarithm of the numerator (division law of
indices), then we nd the antilogarithm of
the difference, i.e.
10
x
÷ 10
y
= 10
x – y
,
then log (x ÷ y) = log x – log y.
Example 4.10
Use 4-gure logarithm tables to nd the value
of 35.82 ÷ 14.21.
Solution
Number Standard form Logarithm
35.82 3.582 × 10
1
1.554 1 –
14.21 1 421 × 10
1
1.152 6
10
0.401 5
0.401 5
2.521 = 2.521 × 10
0
∴35.82 ÷ 14.21 = 2.521
Exercise 4.7
Use 4-gure logarithm tables to nd the value of
the following.
1. (a) 24.2 ÷ 16.1 (b) 327 ÷ 1.53
2. (a) 48.12 ÷ 6.123 (b) 78.12 ÷ 12.64
3. (a) 812.1 ÷ 16.52 (b) 218.4 ÷ 17.3
4. (a) 512 ÷ 6.43 (b) 17.21 ÷ 14.18
5. (a) 653.1 ÷ 37.24 (b) 518.6 ÷ 173.4
6. (a) 83.41 ÷ 15.76 (b) 92.86 ÷ 68.93
Bar numbers
Earlier in this chapter, we wrote the logarithms
of numbers between 0 and 1 with a minus sign
above the characteristic. When written in this
form, the logarithms may then be referred to as
bar numbers.
Adding bar numbers
Example 4.11
Simplify: (a) 2.89 + 5.47
(b) 3.76 + 4.85
Solution
(a) 2.89 + 5.47
= 2 + 0.89
5 + 0.47
7 + 1.36
= 7 + 1 + 0.36
= 6 + 0.36
= 6.36
(b) 3.76 + 4.85 = 3 + 0.76
4 + 0.85
1 + 1.61
= 2.61
Subtracting bar numbers
Example 4.12
Simplify: (a) 4.21 – 1.73 (b) 2.36 – 6.87
Solution
–
1
+
1
(a) 4.21 – 1.73 = 4 + 0.21 –
1 + 0.73
–
6 + 0.48
= 6.48
–
1
+
1
(b) 2.36 – 6.87 = 2 + 0.36 –
6 + 0.87
3 + 0.49
= 3.49
– –
––
(Write the numbers
vertically, separating the
integral parts from the
decimal parts and add.)
(Simplify the integral
parts, i.e.
–
7 + 1 =
–
6.)
(Write as a single number in bar
form.)
(Borrow 1 from
–
4 to leave
–
5.)
(
–
5 – 1 =
–
6)
(Borrow 1 from
–
2 to leave
–
3.)
(i.e.
–
3 – (
–
6)
=
–
3 + 6 = 3)
(Subtract
–
6 from
–
3.)
– –
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
– –
+
+
30
Exercise 4.8
Calculate the following.
1. (a) 1.2 + 3.7 (b) 3.2 + 2.4 (c) 2.3 + 3.4
2. (a) 1.3 + 2.8 (b) 4.8 + 2.8 (c) 1.6 + 2.7
3. (a) 3.6 – 2.4 (b) 2.7 – 3.5 (c) 4.1 – 2.7
4. (a) 1.6 – 2.3 (b) 3.4 – 4.2 (c) 2.3 – 1.6
Division of bar numbers
Example 4.13
Evaluate: 7.42 ÷ 2
Solution
To divide a bar number, the characteristic must
be exactly divisible by the divisor. If not, we
subtract the least positive number which makes
it so.
Thus 7.42 ÷ 2 = (
–
7 + 0.42)÷ 2
= (
–
7 – 1 +1 + 0.42 ) ÷ 2
(Subtract 1 from the
characteristic and add to the
mantissa so that the value of
the number does not change.)
=
–
8 + 1.42
2
=
–
8 + 1.42
2 2
=
–
4 + 0.71
= 4.71
Multiplication of bar numbers
Example 4.14
Simpliy 3.6 ×
–
2.
Solution
3.6 ×
–
2 = (
–
3 + 0.6 ) ×
–
2
=
–
3 ×
–
2 + 0.6 ×
–
2
= 6 +
–
1.2
= 4.8
– –
– –
– –
–
–
–
–
–
–
––
–
–
–
–
–
–
–
–
–
–
–
–
––
Exercise 4.9
Evaluate the following.
1. (a) 2.12 × 8 (b) 4.78 ×
–
2
2. (a) 5.264 ÷ 2 (b) 2.714 ÷ 3
3. (a) 4.62 ×
–
3 (b) 2.913 ×
–
4
4. (a) 3.722 2 ÷
–
2 (b) 6.199 7 ÷ 4
Using bar numbers in multiplication
and division
Example 4.15
Calculate: (a) 42.8 × 0.003 251
(b) 0.356 ÷ 0.015 38
Solution
(a) Number Logarithm
42.8 1.6 31 4
+
0.003 251 3.5 12 0
0.139 1 1.1 434
∴42.8 × 0.003 251 = 0.139 1
(b) Number Logarithm
0.356 1.551 4
–
0.015 38 2.186 9
23.15 1.364 5
∴0.356 ÷ 0.015 38 = 23.15
Example 4.16
Calculate: 2.61 × 21.83 × 0.073
61.72 × 11.73
Solution
Number Logarithm
2.61 0.416 6
21.83 1.339 1 +
0.073 2.863 3
Numerator. 0.619 0
61.72 1.790 4
+
11.73 1.069 3
Denominator. 2.859 7
0.005 745 3.759 3
Evaluate
logarithm of
numerator
–
–
–
–
–
–
–
Evaluate logarithm
of denominator
and subtract log of
denominator from
log of numerator
31
∴2.61 × 21.83 × 0.073 = 0.005 745
61.72 × 11.73
Exercise 4.10
Use logarithm tables to calculate:
1. 543.2 × 0.562 1
2. 8.743 × 763.1 × 0.426 8
3. 456.6 × 974.8 × 0.003 485
4. 563.7 × 26.14 × 0.032 1
5. 342.2 ÷ 4.244 6. 0.621 3 ÷ 2.842
7. 0.834 1 ÷ 15.76 8. 596.6 ÷ 2 432
9. Calculate the following.
(a) PQ (b) Q ÷ P (c) P ÷ Q
if P = 37.32 and Q = 46.54.
10. 5.4 × 0.006 4 11. 0.48
0.086 0.73 × 0.92
12. 0.872 × 3 150 × 0.034 5
0.56 × 0.93
Powers and roots
Powers
Raising a number to a power corresponds to the
multiplication of the logarithm of the number
by the index.
Example 4.17
Calculate 29.11
2
correct to 4 s.f.
Solution
Method 1 Method 2
Number Logarithm Number Logarithm
29.11 1.464 0 + Or 29.11
2
1.464 0 × 2
29.11 1.464 0 847.2 2.928 0
847.2 2.928 0
∴ 29.11
2
= 847.2
Example 4.18
Find the value of (0.849 5)
4
.
Solution
Method 1 Method 2
Number Logarithm Number Logarithm
0.849 5 1.929 2 0.849 5
4
1.929 2 × 4
0.849 5 1.929 2 0.521 0 1.716 8
0.849 5 1.929 2
0.849 5 1.929 2
0.521 0 1.716 8
∴ 0.849 5
4
= 0.521 0
Note: Method 1 is only suitable for small
powers, such as squares and cubes. However,
method 2 can be used for any other including
fractional powers.
In general, log x
n
= n × log x, written simply
as n log x for all values of n.
Roots
The inverse process to raising a number to a
power is taking the corresponding root.
To nd the root of a number, we divide the
logarithm of the number by the order of the
root.
Example 4.19
Calculate the cube root of 35.64.
Solution
Remember
√35.64 = 35.64
∴ log 35.64 = log 35.64
Number Logarithm
35.64 1.551 9 ×
3.291 0.517 3
∴ √35.64 = 3.291 (correct to 4 s.f.)
3
1
3
–
1
3
–
–
–
–
–
–
–
–
1
3
–
(Multiplying by is
the same as dividing
by 3.)
1
3
–
1
3
–
1
3
–
3
32
Solution
Number Logarithm
√48 700 4.687 5 ×
= 2.343 8 +
8.93 0.950 9
3.294 7
3.142
2
0.497 2 × 2
= 0.994 4 +
5.67 0.753 6
1.748 0
35.22 1.546 7
∴
√48 700 × 8.93
= 35.22 (4 s.f)
3.142
2
× 5.67
Example 4.22
Evaluate using logarithm tables.
√0.843 2 – √0.752 6
√0.843 2 + √0.752 6
Solution
Remember that logarithms are not used for
addition or subtraction.
1. Calculate √0.843 2
Number Logarithm
√0.843 2 1.925 9 ÷ 2
0.918 1 1.962 9
2. Calculate √0.752 6
Number Logarithm
√0.752 6 1.876 5 ÷ 3
0.909 5 1.958 8
3. Now, calculate the difference in the
numerator, and the sum in the denominator.
√0.843 2 – √0.752 6
=
0.918 1 – 0.909 5
√0.843 2 + √0.7526 0.918 1 + 0.909 5
=
0.008 6
1.827 6
n
1
n
–
1
n
–
2
3
–
4
3
4
3
3
4
3
1
2
–
Evaluate and
simplify logarithm
of the numerator.
Evaluate
and simplify
logarithm of the
denominator
and subtract it
from the log of
numerator.
3
1
3
–
–
–
–
–
3
3
3
3
√
3
In general, log √x = log x
= log x
Exercise 4.11
Use logarithms to evaluate.
1. (a) 17.3
2
(b) 1.312
5
(c) 2.312
4
2. (a) 2.812
3
(b) 1.517
4
(c) 2.723
5
3. (a) 17.41 (b) 82.7 (c) 2.971
2
4. (a) 34
3
(b) 2.245
5
(c) √1.62
5. (a) √55 (b) √22.5 (c) √181.2
6. (a) √28.41 (b) √3 100 (c) √891 0
7. (a) √3.765 (b) √786.5 (c) √25
More combined operations
Example 4.20
Find the value of
14.73 × 22.41 .
82.3
Solution
Number Logarithm
14.73 1.168 2 +
22.41 1.350 4
2.518 6 –
82.3 1.915 4
0.603 2 ÷ 2
2.003 0.301 6
∴
14.73 × 22.41
= 2.003
√ 82.3
Example 4.21
Calculate the value of √48 700 × 8.93
3.142
2
× 5.67
33
Number Logarithm
0.008 6 3.934 5 –
1.827 6 0.262 0
0.004 704 3.672 5
∴
√0.843 2 – √0.752 6
= 0.004 704.
√0843 2 + √0.752 6
Exercise 4.12
1. Use logarithms to evaluate the following
expressions.
(a) 22.67 – 11.43 (b) 0.003
22.67 + 11.43 √ 0.017 26
(c) √0.828 (d) 0.671
3
× 0.042
2
√0.061
(e) √1.1 × 14.23
2
(f) 8.072 × √0.743 2
39.67 0.824
2
(g) 17.26 (h) 82.41 × 76.62
√ 43.81 √ 7.389
(i) 4.12 × 71.93 (j) 48√592
1.458 × 82.44 9.761
2. Given that log 3 = 0.477 1 and
log 5 = 0.699 0, without further use of
logarithm tables, nd:
(a) log 3
2
(b) log 5
3
(c) log 15
(d) log 45 (e) log 30 (f) log 50
3. If log x = 2.634 8 and log y = 3.516 3, nd
the logarithm of:
(a) xy (b) x ÷ y (c) x
2
y
(d) xy
2
(e) √xy (f) √
4. Find the logarithms of the following numbers
to the indicated base (For example: 49 = 7
2
.
∴log
7
49 = 2).
(a) 81 to base 9 (b) 64 to base 4
(c) 25 to base 5 (d) 16 to base 2
(e) 27 to base 3 (f) 216 to base 6
x
y
–
3
1
3
–
3
–
–
5
(
)
3
3
34
The gradient of a straight line
A bus conductor wishes to move some luggage
to the roof rack of his bus. To get onto the roof,
he uses an extendable ladder which he places in
position AC, as shown in Figure 5.1.
Fig. 5.1
The height BC of the bus is 2.8 m and the
distance AB is 2 m. For every one metre that
the conductor moves horizontally, what is the
corresponding vertical distance?
This may be expressed as:
Vertical distance
=
2.8
= 1.4
Horizontal distance
2
Suppose that the conductor extends the ladder
further so that it is in position DC (dotted).
Will he nd the climb to the top of the bus any
easier? Why?
If the ladder is in position DC, and given that
BD = 2.5 m, what is the corresponding vertical
distance for every one metre that the conductor
moves horizontally?
You should nd that the ratio
Vertical distance
is now 1.12.
Horizontal distance
The ratio
vertical distance
is a measure
horizontal distance
of steepness or slope. It is known as gradient.
Gradient measures how steep an incline is.
Notice, therefore, that gradient is higher for a
steeper incline.
To determine gradient of a line through
known points
In Fig. 5.2, OC is a straight line through the
origin.
Fig. 5.2
When moving from O to C, every 1 vertical
step made results into horizontal movement
equal to 2 steps.
Thus the gradient of line OC is
Vertical distance
=
1
2
Horizontal distance
4
3
2
1
–
1
1 2 3 4 5 6
x
y
– – – – – – – – – – –
– – – – –
– – – – – –
– – – – – – – – – – –
– – – – – –
A(2 , 1)
C(6 , 3)
B(4 , 2)
O
ILLUSTRATION
A
D
C
B
EQUATIONS OF STRAIGHT LINES
5
35
y
x
– – – – – – – – –
– – – – – – – – – – – – – – – –
Increase
Increase
0
Positive
gradient
Stop
Start
Fig. 5.5(a)
Fig. 5.4
Solution
Using points A(4 ,0) and B(6 ,4), we get
Gradient of line AB =
change in y-coordinate
change in x-coordinate
=
4 – 0
=
4
= 2
6 – 4 2
A is at (4 , 0) and C is at (0 , 7). When moving
from C to A, line AC slopes downwards.
The change in the y–coordinate is 0 – 7 =
–
7,
and the change in the x–coordinate is 4 – 0 =
4.
∴ gradient of line AC =
–
7
=
–
1.75.
4
Note that:
1. If an increase in the x-coordinate causes an
increase in the y-coordinate (Fig. 5.5(a)),
i.e. the line slopes upwards from left to
right, the gradient is positive.
x
y
0
×
×
B(6, 4)
A
C
4
7
y
2
y
1
x
y
– – – – – – – – –
– – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – –
0
x
2
– x
1
y
2
– y
1
P(x
1
, y
1
)
Q(x
2
, y
2
)
x
1
x
2
Using any two points on the line, say A(2 , 1)
and C(6 , 3), we have
Vertical distance = 3 – 1
Horizontal distance = 6 – 2....
Gradient =
3 – 1
=
2
=
1
6 – 2 4 2
For any line passing through points P(x
1
, y
1
)
and Q(x
2
, y
2
) in Fig. 5.3,
Fig. 5-3
Gradient =
Vertical distance
Horizontal distance
=
change in y-coordinate
corresponding change in x-coordinate
=
y
2
– y
1
x
2
– x
1
Note:
1. Gradient measures the steepness of a line.
2. When the coordinates of any two points on
a line are known, the gradient can be found
without drawing the graph. In this case,
Gradient =
y
2
– y
1
x
2
– x
1
or gradient =
y
1
– y
2
x
1
– x
2
.
Example 5.1
Determine the gradients of lines AB and AC in
Figure 5.4.
36
2. If an increase in the x-coordinate causes a
decrease in the y-coordinate (Fig 5-5 (b)),
i.e. the line slopes downwards from left
to right, the gradient is negative.
Fig. 5-5 (b)
Example 5.2
Find the gradients of lines DE and EF.
Fig. 5.6
Solution
Moving from D to E, the
change in the x-coordinate is 4 –
–
1 = 5,
change in the y-coordinates is 5 – 5 = 0.
∴ gradient of line DE is
0
= 0.
5
Moving from F to E, the
change in the x-coordinate is 4 – 4 = 0,
change in the y-coordinate is 5 –
–
2 = 7
∴ gradient of line EF is
7
0
The value of
7
is undened. See note below.
0
Thus the gradient of line EF is undened
Note:
6
= 2 means that 2 (quotient) multiplied by
3
3 (divisor) equals 6 (dividend).
Similarly, 0 = 0 means that 0 (quotient)
7
multiplied by 7 (divisor) equals 0 (dividend),
which is true.
But
7
= 0 would mean 0 (quotient) multiplied
0
by 0 (divisor) equals 7 (dividend), which is not
true.
Likewise, if x is any number,
x
= 0 has no
meaning.
0
Thus,
0
= 0 while
x
is undened.
x
0
We notice that:
1. If, for an increase in the x-coordinate,
there is no change in the y-coordinate,
then
the line is horizontal meaning that the
gradient is zero.
2. If there is no change in the x-coordinate
while there is an increase in the
y-coordinate, then the line is vertical
and the gradient is undened.
Exercise 5.1
1. For each of the following pairs of points,
nd the change in the x-coordinate and the
corresponding change in the y-coordinate.
Hence, nd the gradients of the lines
passing through them.
(a) (0 , 2), (3 , 4) (b) (0 , 2), (5 , 0)
(c) (
–
2 ,
–
2), (2 , 0) (d) (
–
1 ,
–
2), (1 , 8)
(e) (
–
1 , 2), (3 ,
–
2) (f) (0 ,
–
2), (0 , 3)
(g) (2 ,
–
8) (
–
2 , 8) (h) (
–
2 , 0), (3 , 0)
2. Find the gradient of the line which passes
through each of the following pairs of
points.
y
x
– – – – – – – – –
– – – – – – – – – – – – – – – –
Increase
Decrease
0
Negative
gradient
Stop
Start
×
×
×
0
D(
–
1 , 5)
E(4 , 5)
F(4 ,
–
2)
y
x
37
(a) (3 , 5), (9 , 8) (b) (2 , 5), (4 , 10)
(c) (7 , 3), (0 , 0) (d) (1 , 5), (7 , 2)
(e) (0 , 4), (4 , 0) (f) (
–
2 , 3), (5 , 5)
(g) (
–
7 , 3), (8 ,
–
2) (h) (
–
3 ,
–
4), (3 ,
–
4)
(i) (
–
1, 4), (
–
3,
–
1) (j) (3 ,
–
1) , (3, 1)
3. Find the gradients of the lines l
1
to l
5
in
Fig. 5.7.
Let us now see how to obtain the equation of a
straight line.
(a) Given the gradient and a point on
the line
Example 5.3
A straight line with gradient 3 passes through
point A(3 ,
–
4). Find the equation of the line.
Solution
Fig. 5.8 is a sketch of the line.
The point B(x , y) represents any general point
on the line.
Using points A and B, we have,
gradient =
y –
–
4
x – 3
=
y + 4
x – 3
1
2
–
1
2
–
Fig. 5.7
4. In each of the following cases, the
coordinates of a point and the gradient of a
line through the point are given. State the
coordinates of two other points on the line.
(a) (3 , 1), 3 (b) (4 , 5),
(c) (
–
2 , 3),
–
1 (d) (5 , 5),
–
(e) (
–
4 , 3), undened (f) (
–
4, 3), 0
The equation of a straight line
In Form 1, you dealt with equations of the form
y = 3, x =
–
2, y = 3x – 4, 2y + 3x = 2, and so
on.
These are called equations of straight lines,
or linear equations, because their graphs are
straight lines.
Fig. 5.8
Since gradient = 3,
Then,
y + 4
= 3
x – 3
⇔ 3(x – 3) = y + 4
⇔ 3x – 9 = y + 4
⇔ 3x – y = 13 (This is the equation of
the line.)
In general,
the equation of a straight line, of gradient
m, which passes through a point (a, b) is
given by
y – b
= m.
x – a
0
A(3 ,
–
4)
B(x , y)
–
4
– – – – – – – – – – – –
– – – – – – – – – – – –
×
×
×
y
x
x
y
5
4
3
2
1
0
–
1
–
2
–
2
–
1
1 2 3 4
l
3
l
4
l
2
l
5
l
1
38
(b) Given two points on the line
Example 5.4
Find the equation of a straight line which
passes through points A(3 , 7) and B (6 , 1).
Solution
Fig. 5-9 is a sketch of the line.
Exercise 5.2
1. In each of the following cases, the gradient
of a line and a point on the line are given.
Find the equation of the line.
(a) 3, (3 , 1) (b) , (4 , 5)
(c)
–
1, (
–
2 , 3) (d)
–
, (5 , 5)
(e) 0, (
–
4 , 3) (f) Undened, (
–
4 , 3)
(g) , (
–
3 , 0) (h)
–
, (0 , 2)
2. Find the equation of the line that passes
through the following points
(a) (
–
1 , 1), (3 , 2) (b) (7 , 2) (4 , 3)
(c) (2 , 5), (0 , 5) (d) ( 5 ,
–
2), (6 , 2)
(e) (6 , 3), (
–
6 , 2) (f) (2 ,
–
5), (2 , 3)
(g) ( , ), ( , )
(h) (0.5 , 0.3), (
–
0.2 ,
–
0.7)
3. (a) Find the equation of the straight line
which passes through the points (0, 7)
and (7, 0).
(b) Show that the equation of the straight
line which passes through (0, a) and
(a, 0) is x + y = a.
4. A triangle has vertices A (
–
2 , 0), B (
–
1 , 3)
and C(2 , 3). Find the equations of the sides
of the triangle.
5. Two lines, l
1
and l
2
, both pass through the
point (4, k).
(a) If l
1
passes through the point (5 ,
–
3), and
has a gradient
–
1 , nd the value of k.
(b) If l
2
passes through (
–
14 , 0), nd its
equation.
Gradient-intercept form of the equation
of a straight line
So far, we have written the equations of straight
lines in the form ax + by = c where a, b and c
are constants. This is called the general form
of the equation of a straight line.
Fig. 5-9
Using points A and B,
gradient =
1 – 7
=
–
6
=
–
2.
6 – 3 3
Point C(x, y) is any general point on line AB.
Using points A and C,
gradient =
y – 7
.
x – 3
(We could equally well use points B and C.)
Since ACB is a straight line, the two values of
the gradient are equal.)
∴
y – 7
x – 3
=
–
2
⇔
–
2(x – 3) = y – 7
⇔
–
2x + 6 = y – 7
⇔ 2x + y = 13 (This is the equation of
the line.)
In general,
the equation of a straight line which passes
through points (a , b) and (c , d) is given by
y – b
=
d – b
or
y – d
=
d – b .
x – a c – a x – c c – a
2
3
–
1
2
–
1
5
–
1
4
–
1
3
–
1
3
–
1
4
–
2
3
–
1
3
–
B(6 , 1)
1
y
x
7
0
3
6
×
×
×
A(3 , 7)
C(x , y)
39
Activity 5.1
Rewrite each of the equations x – 2y =
–
6,
2x + 3y = 6 and 4x + 2y = 5 in the form
y = mx + c .
Find any two convenient points on each line
and draw the lines on a squared paper. Calculate
the gradient of each line.
Copy and complete Table 5.1 below.
Equation The form m c Gradient y-intercept
of line y = mx + c
x – 2y =
–
6 y = x + 3 3 3
2x + 3y = 6
4x + 2y = 5
Table 5.1
When we write the equation of a line in the
form y = mx + c, what do the constants m and c
represent?
y = mx + c is known as the gradient-
intercept form of the equation of a straight
line. m represents the gradient and c is the
y-intercept of the line.
When we write the equation of a line in the
gradient-intercept form, we can easily tell the
gradient and y-intercept of the line and hence
sketch it, if need be.
Example 5.6
Find the gradient and y-intercept of the line
whose equation is 4x – 3y – 9 = 0. Sketch the
line.
Solution
4x – 3y – 9 = 0 is equivalent to y = x – 3.
Comparing with y = mx + c gives
gradient, m = ,
y-intercept, c =
–
3.
Fig. 5.11 is a sketch of the line.
Fig. 5.10
At what value does the line 3x – y = 2 cut the
y-axis? At y = –2.
This value is referred to as the y-intercept.
Consider the following example.
Example 5.5
Find the gradient of the line 3x – y = 2 and
draw the line on squared paper.
Solution
To nd the gradient of the line, we need to rst
nd any two points on the line.
We write the equation 3x – y = 2 as y = 3x – 2,
choose any two convenient values of x and nd
corresponding values of y.
For example, when x = 0, y = 3 × 0 – 2 =
–
2,
∴ point (0,
–
2) lies on the line.
When x = 2, y = 3 × 2 – 2 = 4,
∴ point (2,4) lies on the line.
Thus gradient of the line is
4 –
–
2 = 6
= 3 .
2 – 0 2
Fig. 5.10 shows the line 3x – y = 2.
y
4
3
2
1
0
–
1
–
2
1 2 4
3x – y = 2
–
1
–
3
3
x
×
×
1
2
–
4
3
–
4
3
–
1
2
–
40
Fig. 5.12
Double intercept (or x-, y-intercept) form
of the equation of a straight line
Consider the following example.
Example 5.7
Find the equation of the line which passes
through points (5 , 0) and (0 , 9). Sketch the
line.
Solution
The gradient of the line is
9 – 0
=
9
=
–
9
0 – 5
–
5
5
If (x , y) is a general point on the line,
gradient = y – 0
x – 5
∴ the equation of the line is
y
=
–
9
x – 5 5
⇔ 5y =
–
9 x + 45
⇔ 9x + 5y = 45
⇔ x + y = 1
⇔ x + y = 1
Fig. 5.13 is a sketch of the line.
Fig. 5.11
Note that for an increase of 3 units in the
x-coordinate, the increase in the y-coordinate
is 4 units. Hence, point (3 , 1) is on the line.
Exercise 5.3
1. In each of the following cases, determine
the gradient and y-intercept by writing the
equation in the form y = mx + c. Sketch the
line.
(a) 5x = 2y (b) y – 3x – 1 = 0
(c) 2x + y = 3 (d) 4x – 2y + 3 = 0
(e) 2x + 3y = 3 (f) 5 = 5x – 2y
(g) 2x + 3y = 6 (h) 8 – 7x – 4y = 0
2. Find the y-intercepts of the lines with the
given gradients and passing through the
given points.
(a) 3, (2 , 6) (b) , (
–
2 , 3)
(c)
–
2, (7 , 4) (d)
–
, (2 , 4)
(e) 0, (
–
3 ,
–
2) (f) Undened, (1, 3)
3. Write down, in the gradient-intercept form,
the equations of lines (a), (b), (c) and (d) in
Fig. 5.12.
y
x
4
3
2
1
0
–
1
–
2
–
3
–
2
–
1
1 2 3
(a)
(c)
(b)
(d)
9
45
––
1
5
–
1
9
–
5
45
––
1
4
–
5
3
–
4x – 3y – 9 = 0
×
(3 , 1)
y
1
0
-1
-2
-3
1 2 3
41
Fig. 5-13
Note that the RHS of the equation obtained in
Example 5.7 equals 1 and that on the LHS, x
and y are respectively divided by 5 and 9. Here,
5 and 9 are respectively the x- and y-values of
the points (5 , 0) and (0 , 9).
At what values does the line cut the x- and y-
axes? These values are respectively called the
x- and y-intercepts of the line.
Activity 5.2
Find the equation of the line that passes through
each of the pairs of points (1, 0) and (0 , 3),
(
–
1 , 0) and (0 , 5), (
–
3 , 0) and (0 ,
–
4), (5, 0)
and (0,
–
2), writing each equation in the
form
x + y
= 1. Sketch each line.
a b
Copy and complete Table 5.2 about your lines.
Line through Equation x + y = 1 a b x-int. y-int.
points: a b
(1 , 0) and (0 , 3)
(
–
1 , 0) and (0 , 5)
(
–
3 , 0) and (0 ,
–
4)
(5 , 0) and (0 ,
–
2)
Table 5.2
When we write the equation of a line in the form
x + y
= 1,
a b
what do the constants a and b stand for?
x + y
= 1 is called the double intercept or
a b
x-, y-intercept form of the equation of a
straight line. a represents the x-intercept and
b represents the y-intercept of the line.
If the equation of a straight line is written in the
double intercept form, we can easily tell the x-
and y-intercepts of the line and hence sketch it
if necessary.
Exercise 5.4
1. Find, in the form y = mx + c, the equation of
each of the lines whose x- and y-intercepts
are given below.
(a) x-intercept 3, y-intercept 4
(b) x-intercept 3, y-intercept
–
1
(c) x-intercept
–
2, y-intercept 5
(d) x-intercept
–
4, y-intercept
–
2.5
2. Write the equations of lines l
1
to l
4
in
Fig. 5.14 in the form x + y = 1.
a b
x
9
0
y
×
×
x + y = 1
1
5
–
1
9
–
5
Fig. 5.14
3. Rewrite the following equations of straight
lines in the double intercept form.
(a) y = 3x + 7 (b) 7 – 2x = 4y
(c) 4y + x – 8 = 0 (d) y + x + = 0
(e) y – 15 = x (f)
–
10 (x + 3) = 0.5y
3
2
–
2
3
–
1
6
–
1
3
–
2
5
–
6
4
2
0
–
2
–
4
–
6
–
4
–
2
2 4 6 8
x
y
l
2
l
1
l
3
l
4
42
You should have noticed that:
Parallel lines have the same gradient. Hence,
when the equations of parallel lines are
written in the form y = mx + c, they have the
same value of m.
Activity 5.4
Now consider Fig. 5.16.
4. Find the coordinates of the points where
each of the following lines cuts the x- and
y- axes.
(a) y = 5x – 4 (b) 2y + x = 7
(c) 5x + 6y – 4 = 0 (d) 2y – 8 = 7x
5. Find the equations of the lines passing
through each of the following pairs of
points, writing your answers in double
intercept form.
(a) (0 , 4) and (
–
1 ,
–
2) (b) (1 , 0) and (3 , 2)
(c) (3 , 7) and (5 , 10) (d) (
–
3 , 3) and (3 , 1)
Parallel and perpendicular lines
Activity 5.3
Consider Fig. 5.15.
Which sets of lines are parallel?
Calculate the gradients of all the lines. What do
you notice about the gradients of the parallel
lines?
x
y
4
3
2
1
–
1
0
–
2
–
3
–
2
–
1
1 2 3
l
3
l
6
l
5
l
1
l
7
l
2
l
4
–
3
l
5
l
3
x
3
2
1
0
–
1
–
2
–
2
–
1
1 2 3 4
4
y
l
2
l
1
l
4
l
6
Fig. 5.15
Fig 5.16.
Which pairs of lines are perpendicular?
Calculate the gradients of all the lines.
Use your results to complete Table 5.3 about
the pairs of perpendicular lines.
Gradient of Product of gradients
Pair 1
st
line (m
1
) 2
nd
line (m
2
)
(m
1
m
2
)
Table 5.3
You should notice, from your table, that
m
1
m
2
=
–
1, i.e.
The product of the gradients of perpendicular
lines is
–
1.
43
2
5
–
2
5
–
2
5
–
2
5
–
2
5
–
∴
y – 2
=
x – 5
⇔ 5y – 10 = 2x – 10
⇔ 5y = 2x
⇔ y = x (This is the equation
of the required line.)
(b) The required line is perpendicular to
5y – 2x = 10. If its gradient is m, then
m × =
–
1
m =
–
.
If (x, y ) is a general point on this line,
gradient =
y – 2 .
x – 5
∴
y – 2
=
x – 5
⇔ 2y – 4 =
–
5x + 25
⇔ 2y =
–
5x + 29
⇔ y =
–
x + (This is the
equation of the
required line.)
Exercise 5.5
1. Without drawing, determine which of the
following pairs of lines are parallel.
(a) 2y + x = 7, 2y = 20 – x
(b) 6y = 4 – 5x, 5y + 6x = 2
(c) x + y + = 0, x + y + = 0
(d) 2y – 7x = 8, 14x – 17 = 4y
(e) y =
–
x + 5 and the line through(
–
3 , 3)
and (0 , 1).
(f) A line through (
–
3 , 1) and (1 , 3) and
another line through (
–
1 , 3 ) and (1 , 5).
2. Without drawing, determine which of the
following pairs of lines are perpendicular.
(a) y = 2x + 5, 2y + x = 3
(b) 2x – y = 7, x + y = 5
(c) 3y = 2x + 1, 2y + 3x – 5 = 0
(d) 7x – 2y =
–
2, 14y – 4x =
–
1
(e) y = x – 2 and the line through (8 , 10)
and (2 , 2).
(f) A line through (2 , 2) and (10 , 8) and
another through (5 , 6) and (8 , 2).
3. In each of the following cases determine the
equation of the line through the given point
and parallel to the given line.
(a) (5 , 3), y = 2x + 3
(b) (0 , 0), 2x + y =
–
3
(c) (
–
3 , 2), 2(y – 2x) = 3
(d) (3 , 0), 2x + 5y = 6
4. In each of the cases in Question 3, determine
the equation of the line through the given
point and perpendicular to the given line.
5.
ABCD is a rectangle with A as the point (
–
3 , 1).
(a)
If AB is parallel to the line 3y – x = 4,
nd the equation of line AB.
(b) Find the equation of line AD.
(c) If C has coordinates (2 , 6), nd the
equations of lines BC and CD.
(d) Find the coordinates of points B and D.
(e) Find the equations of the diagonals of
the rectangle.
(f) Calculate the length of a diagonal of the
rectangle. (Hint: Draw the rectangle on
a squared paper).
5
2
–
29
2
––
2
5
–
5
2
–
2
5
–
5
2
–
2
5
–
5
2
–
2
3
–
3
4
–
2
5
–
5
2
–
Example 5.8
Find the equation of the line through (5 , 2)
which is (a) parallel, (b) perpendicular to the
line 5y – 2x = 10.
Solution
The line 5y – 2x = 10 can be written as y = x + 2.
∴ its gradient is .
(a) The gradient of the line through (5 , 2)
is also since this line is parallel to
5y – 2x = 10.
If (x, y) is a general point on this line,
gradient =
y – 2 .
x – 5
−5
2
44
Fig. 6.2 shows some symmetrical geometric
shapes with the lines of symmetry drawn as
broken lines.
(a) an isosceles triangle (b) a rectangle
(c) a kite
Fig. 6.2
How many lines of symmetry has:
(a) an equilateral triangle, (b) a square,
(c) a rhombus, (d) a circle?
Paper folding and cutting to make
symmetrical shapes
Activity 6.1
Fold a rectangular piece of paper once so that
the corners coincide. Cut off a corner as shown
in Fig. 6.3.
– – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – –
– – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – –
corner
cut
Fig. 6.3
♣
♠
Symmetry
What do the pictures in Fig. 6.1 have in
common?
(a) (b)
(c) (d)
Fig. 6.1
The left hand side of ech picture is exactly the
same shape and size as the right hand side.
This property is known as symmetry and the
pictures are said to be symmetrical.
Choose a picture from Fig. 6.1 and make a
tracing of it. Draw a line which will divide
the picture into two identical parts. This line is
called a line of symmetry and the gure is said
to have line symmetry.
Place the edge of a mirror along the line of
symmetry. What do you notice?
A line which divides a shape into two equal
parts i.e. one part is a mirror image of the
other, is called a line of symmetry. Each
part is a reection of the other.
A shape which has one or more lines of
symmetry is said to be symmetrical about
the line(s).
REFLECTION AND CONGRUENCE
6
45
Unfold the corner which you have cut off.
What shape do you get?
What can you say about:
(i) the shape and angles of your gure,
(ii) the two parts of the shape on each side of
the fold,
(iii) the distances of the corresponding points
on opposite sides of the fold line?
What line does the fold line represent?
Activity 6.2
Fold a rectangular sheet of paper as shown in
Fig. 6.4 ensuring that the corners coincide. Cut
off the corner as shown.
Fig. 6.4
Unfold the corner which you have cut off.
What shape do you get?
What can you say about the sides and angles
of your gure?
Place the edge of a mirror on the fold lines
in turn. What do you notice?
How many lines of symmetry does your gure
have?
Since one half of a symmetrical shape is a
mirror image, i.e. a reection of the other,
line symmetry is sometimes called mirror
symmetry, reection symmetry or bilateral
symmetry (bilateral means two-sided).
Properties of symmetrical shapes
From Activities 6.1 and 6.2, you should have
discovered that a line of symmetry divides a
shape into two parts with the properties that:
(i) The corresponding sides of the two
parts are equal;
(ii) The corresponding angles of the two
parts are equal; and
(iii) The corresponding points on the two
parts are the same distance away from
the line of symmetry.
For example, in Fig. 6.5, m is the line of
symmetry of the given shape.
The left-hand side is the mirror image of the
right-hand side. Hence,
∠a′ = ∠a (Corresponding angles)
A′B′ = AB (Corresponding sides)
OB′ = OB (Corresponding distances).
Fig. 6.5
Exercise 6.1
1. Fold a rectangular sheet of paper as shown
in Fig. 6.6 and cut off the corner as shown.
(Note that the corners do not coincide).
Is the fold line a line of symmetry?
Fig. 6.6
2. Fold a rectangular piece of paper as shown
in Fig. 6.7. Cut off the corner as shown and
unfold it. (Note that on folding the second
time, the corners do not coincide).
– – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – –
corner
cut
– – – – – – – – – – – – –
– – – – – – – – – – – – – – – – –
O
C′
C
m
A′
B′
B
A
a′
a
cut
46
Draw a diagram to show how you would cut
the corner to obtain a regular octagon.
5. Which of the shapes in Fig. 6.10 have
lines of symmetry and how many lines of
symmetry does each have?
Copy each shape and on the copy draw the
lines of symmetry.
– – – – – – – – – – – – – – – –
cut
Fig. 6.7
(a) How many lines of symmetry does your
gure have?
(b) Do all the folds represent lines of
symmetry? If not, what are they?
3. Fold a piece of paper as in Fig. 6.8 ensuring
that the corners coincide. Cut as shown and
unfold the part indicated with an arrow.
Fig. 6.8
(a) What is the shape of the gure obtained?
(b) How many lines of symmetry does it
have?
4. Fold a sheet of paper twice as in Fig. 6.4
and then fold again as in Fig. 6.9. Cut as
indicated and unfold the corner that you
have cut off. (Note the equal angles.)
Fig. 6.9
(a) What shape do you obtain?
(b) Are the fold lines the only lines of
symmetry of your gure? If not, where
are the others?
How many lines of symmetry does the gure
have?
cut
–– – – – – –
– – – – – – – – – –
cut
Fig. 6.10
(a) (b)
(g) (h)
(i) (j)
(k) (l)
(e) (f)
(c) (d)
DIAGRAM
(to come)
cut
– – – – – – – – –
– – – – – – – – – – – – – – – – –
47
(c) (d)
6. Draw a line segment PQ on a piece of paper.
Does PQ have a line of symmetry? Fold the
paper so that the fold is a line of symmetry
of PQ. What is the size of the angles
between the fold and PQ? What can you
say about the distances of P and Q from any
point on the line of symmetry?
Symmetry in solids
Activity 6.3
Carefully cut an orange in half. Place the cut
surface of one half against a shiny plane surface
(an ordinary mirror will do).
What do you see? The result is slightly false.
Why?
Place the other half against the shiny surface.
Do the two ‘whole’oranges that you see appear
identical?
If you cut an orange in two so that the two
parts are identical, the cut is called a plane of
symmetry.
Note: A rubber ball will give more accurate
results if used in place of an orange because it is
a more perfect sphere.
How many planes of symmetry does a sphere
have?
Exercise 6.2
1. Make a list of objects which have planes of
symmetry in your classroom.
2. What parts of your body are mirror images
of each other?
3. How many planes of symmetry has:
(a) a cube, (b) a cuboid,
(c) a cylinder, (d) a cone,
(e) a regular tetrahedron,
(f) a prism with an isosceles triangle
cross-section,
(g) a prism with an equilateral triangle
cross-section,
(h) a prism with a scalene triangle cross-
section?
4. If you place the following solids against a
plane mirror, what new solid do you see?
(a) A cube
(b) A rectangular block
(c) A cylinder (on one of its plane surfaces)
(d) A hemisphere (on its plane surface)
(e) A square-based pyramid (on its base)
5. How many planes of symmetry does each of
the solids in Fig. 6.11 have?
– – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – –
– – – – – –
– – – – – –
– – – – – –
– – – – – –
– – – –
– – – –
– – – –
– – – –
(a)
(b)
Fig. 6.11
48
REFLECTION
AIDS KILLS
AIDS KILLS
Fig. 6.13
Reection
We have already seen that the two parts of a
shape on opposite sides of a line of symmetry,
are mirror images of each other.
Now consider looking at yourself in a mirror.
Do you see yourself as others see you?
In what ways does your image differ from
yourself?
The gure being reected is called the object
and its reection the image.
Now answer the following questions.
1. If you raise your right arm, which arm is
raised in the mirror?
2. Which is taller, you or your image?
3. Imagine a line joining the tip of your nose
to its image in the mirror. What angle
does this line make with the mirror?
4. If you stand 3 m infront of the mirror,
where does your image appear to be?
5. If you walk towards the mirror, what
happens to your image?
Now look at the pictures in Fig. 6.12(a) and
(b) each of which is a picture of a car and its
reection.
(a)
(b)
Fig. 6.12
In Fig. 6.12(a), it is not possible to tell which
is the reection of the other. Why? However,
we can easily see which the reection is in Fig.
6.12(b). Why? Which letters in Fig. 6.12(b)
remain the same when reected? What can you
say about their lines of symmetry?
Do your observations apply also to Fig 6.13?
The process or act of reecting an object is
a transformation.
In Mathematics, a transformation is said to
be a change in the position or dimensions
(or both) of a shape. The figure being
transformed is referred to as the object and
the gure which results after transformation
as the image.
You are going to encounter other
transformations later.
Under reection, an object and its image have
reection symmetry and the mirror acts as the
line or plane of symmetry.
Note that if an object has a line of symmetry
that is parallel to the mirror line, as with
the letters T, I and O in Fig 6.12 (b), it is
unchanged when reected. Note that it is the
same case with the letters A and I and in the
skull bones in Fig. 6.13.
In Mathematics, only the reection of two-
dimensional (plane) gures is studied. It
is easier to see and study the mathematics
involved. The mirror line (m) is usually
indicated by an arrow at each end.
Properties of reection
From the foregoing discussion, you should
have noticed that an object and its image are on
opposite sides of the mirror line, and that:
1. An object and its image have the same
shape and size.
REFLECTION
49
m
– – – – – – – – – – – – – – – – – – – – – – – – – –
Q
R
P
Q′
R′
P′
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – –
m
Q
R
P
m
m
m
m
m
(c) (d)
(a) (b)
m
(e)
(f)
A′
B
m
B′
A
•
– – – – – – – – – – – – – – – – – –
•
•
– – – – – – – – – – – – – – – – – – – – – – – –
•
Fig. 6.16
(iii) Similarly, obtain Q′ and R′, the images of
Q and R respectively.
(iv) Join P′ , Q′ and R′ to obtain the image of
∆PQR.
2. A point on the object and a corresponding
point on the image are equidistant from
the mirror line.
3. The image is laterally inverted, i.e. what
is the object’s left-hand side becomes the
image’s right-hand side and vice versa.
4. The line joining a point and its image is
perpendicular to the mirror line.
5. A point on the mirror line is an image of
itself. Such a point is said to be invariant
since its position has not changed.
Note:
We think of a mirror as two-sided so that if B is
on the same side as the image A′, then its image
B′ is on the same side as the object A (Fig.
6.14).
Fig. 6.14
Example 6.1
Draw the image of triangle PQR (Fig. 6.15)
under reection in the mirror line m.
Fig. 6.15
Solution
(i) To obtain the image of point P, draw a
perpendicular from P to the mirror line
and produce it (Fig. 6.16).
(ii) Mark off P′, the image of P, equidistant
from the mirror line as P.
Exercise 6.3
1. Make a tracing of each of the drawings in
Fig. 6.17 and construct their images under
reection in the indicated mirror line m.
Fig. 6.17
2. Fig. 6.18 shows objects and their images
under reection. Trace each of the drawings
and construct the mirror line in each case.
50
(a)
(b)
(c)
(d)
Fig. 6.18
B
A′
C
B′
A
C′
Reection in the Cartesian plane
Reection in the mirror lines x-axis (y = 0)
and y-axis (x = 0).
Example 6.2
A(2 ,4 ), B(6, 4) and C(7 , 2) are the vertices
of a triangle. Find the image of the triangle
under reection in the line: (i) x-axis, (ii)
y-axis, labelling them respectively as A′B′C′
and A′′B′′C ′′.
Solution
Fig. 6.19 shows ∆ABC and its images.
How are the x-coordinates of an object point
and its image related?
How are the y-coordinates related?
You should notice that, reection in the mirror line:
1. x-axis (y = 0) maps a point P(a , b) onto
P′(a ,
–
b).
y
x
6
4
2
0
–2
–4
–6
–
8
–6 –4 –2 2 4 6
8
×
×
×
×
– – – – – – – – – –
– – – – – – – – – – – – – – – – – – – –
– – – – – – – – – –
A
A′′
A′
B′′
C′
B
B′
Fig. 6.19
×
C
×
×
×
C′′
×
x
y
5
–
4
–
3
–
2
–
1
0
1 2 3 4
4
3
2
1
–
1
–
2
–
3
–
4
–
5
y = 1
x =
–
1
A′
A′′
B′′
C
B
C′
B′
C′′
2. y-axis (x = 0) maps a point P(a , b) onto
P′(
–
a , b).
Reection in the mirror lines x = k and y = k
Example 6.3
Find the images of ∆ABC with vertices A(
–
1 ,
–
2),
B(1 , 5) and C(2 , 3) under reection in the
mirror lines (i) x =
–
1, and (ii) y = 1, labelling
them as ∆A′B′C′ and ∆A′′B′′C′′ respectively.
Solution
∆ABC and its images are shown in Fig. 6.20.
Fig. 6.20
51
State the relationship between the x-coordinates
of an object point and its image. Do likewise
for the y-coordinates.
Note that, reection in the mirror line:
1. x = k maps a point P(a , b) onto
P′(2k – a, b).
2. y = k maps a point P(a , b) onto
P′(a , 2k – b).
Reection in the mirror lines y = x and
y =
–
x
Example 6.4
A(
–
1,2), B(1,5) and C(3,4) are the vertices
of a triangle. Find the images of the triangle
when it is reected in the mirror lines: (i) y
= x, and (ii) y =
–
x, labelling them as A'B'C'
and A"B"C" respectively.
Solution
Fig. 6.21 shows ∆ABC and its images.
Fig. 6.21
How are the x-coordinates of an object point and
its image related? How about the y-coordinates?
You should notice that:
Note that reection in the mirror line:
1. y = x maps a point P(a, b) onto P′(b, a).
2. y =
–
x maps a point P(a, b) onto P'(
–
b,
–
a).
Example 6.5
The vertices of a quadrilateral are A(2 , 0.5),
B(2 , 2), C(4 , 3.5) and D(3.5 , 1). Find the
image of the quadrilateral under reection in the
line y = 0 followed by a reection in the line y =
–
x.
Solution
‘Reection in the line y = 0 followed by a
reection in y =
–
x’ means that we rst obtain
the image under a reection in the y = 0 and
then reect this image in the y =
–
x.
This is shown in Fig. 6.22. In the gure,
A'B'C'D' is the reection of ABCD in line y = 0.
A"B"C"D" is the reection of A'B'C'D' in line
y =
–
x.
Thus the required image vertices are:
A"(0.5 ,
–
2), B"(2 ,
–
2), C"(3.5 ,
–
4) and D"(1 , 3.5).
x
y
5
–
1
31 2
0
4 5 6 7
4
3
2
1
–
1
–
2
–
3
–
4
–
5
y = 0
y =
–
x
A
A′′
B′′
C
C′
A′
C′′
D
B
D′′
B′
D′
Fig. 6.22
x
0
–
6
6
4
2
–
2
–
4
–
6
– – – – – – – – – – – – – – ––
C′′
– – – –
– – – –
y
y = x
y =
–
x
–
2
2 4
– – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – – – –
6
–
4
– – – – – – – – – – – – – – – – – – – – – – –
A′
×
B′
C′
×
C
×
B
×
×
×
B′′
A′′
×
×
A
×
52
Fig. 6.23
a + a + b + b = 180° (Angles on a straight line)
2a + 2b = 180°
2(a + b) = 180°
a + b = 90°
Thus, the two angle bisectors are
perpendicular.
Vertically opposite angles
Two lines intersect as shown in Fig. 6.24. The
broken line is a mirror line m. What can you
say about the angles marked c and d?
Fig. 6.24
c + c + B = 180
0
i.e. 2c + B = 180
0
(Angles on a straight
line.)
Also, d + d + A = 180
0
i.e. 2d + A = 180
0
(Angles on a straight
line.)
But c = d
∴ 2c + B = 180°
and 2c + A = 180°
i.e. 2c + A = 2c + B
∴ A = B
Also C = D since 2c = 2d or c = d
Thus, if two lines intersect, the vertically
opposite angles formed are equal.
Exercise 6.4
1. A quadrilateral has vertices P(4 , 2),
Q(7 , 3), R(6 , 2) and S(4 , 0). Draw,
on the same axes, the quadrilateral and its
images under reection in:
(a) the x-axis, (b) the line y = x,
(c) the y-axis, (d) the line y =
–
x
labelling the images as P′Q′R′S′,
P′′Q′′R′′S′′, P′′′Q′′′R′′′S′′′ and P
iv
Q
iv
R
iv
S
iv
respectively .
State the coordinates of each image point.
2. A(
–
4 , 1), B(
–
2 ,
–
1), C(1 , 0) are the vertices
of a triangle. Find the image of the triangle
when it is reected in the mirror line:
(a) y = 1 (b) y =
–
2
(c) x =
–
3 (d) x = 1.5
3. The vertices of a triangle are A(
–
4 , 6),
B(
–
3 , 2) and C(
–
7 , 1). Find the nal
image of the triangle under:
(a) reection in line y = 0 followed by a
reection in line y = x,
(b) reection in line y =
–
x followed by a
reection in line x = 0,
(c) reection in line y = x followed by a
reection in line y = 1,
(d) reection in line x = 1.5 followed by a
reection in the same line.
4. Which properties of an object are invariant
under reection?
Geometric deductions using reection
Angle bisector
Draw two intersecting lines on tracing paper.
Thinking of the lines as indenitely long, fold
to form lines of symmetry (mirror lines) for the
pair of lines. What do you notice about the
angles?
Did you notice that the angles marked a in
Fig. 6.23 are equal? What about the angles
marked b?
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
c
c
d
d
C
D
A
B
m
m
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – –
b
b
b
b
a
a
a
a
53
1. The three mediators of the sides of a
triangle meet at a common point. The
common point is called the circumcentre
of the triangle.
2. If the triangle was equilateral, each
mediator would pass through the vertex of
the triangle opposite the side it mediates.
Chord of a circle
Draw a circle centre O on tracing paper. Draw
a chord AB on the circle (Fig. 6.27).
Fig. 6.27
Fold the circle so that A coincides with B. Does
your fold line pass through O? What angle does
the fold line make with chord AB?
1. The perpendicular from the centre of a
circle to a chord bisects the chord.
2. Conversely, the perpendicular bisector of
a chord passes through the centre of the
circle.
Exercise 6.5
1. Draw a circle on tracing paper by drawing
round a circular object. Find the centre by
folding. Check with a pair of compasses.
2. Use reection properties to nd the angles
marked by small letters in each part of
Fig. 6.28.
(a)
An isosceles triangle
Fig. 6.25 (a) shows an isosceles triangle ABC.
Draw a similar gure on tracing paper. Fold to
form a line of symmetry for BC as in Fig. 6.25.
(b). What do you notice about angles B and C?
Label as D the point where the fold line cuts BC.
(a) (b)
Fig. 6.25
What can you say about ∠ADC?
∠A is called the vertical angle of the isosceles
triangle. Does your fold line bisect this angle?
1. The base angles of an isosceles triangle
are equal.
2. The perpendicular from the vertex of an
isosceles triangle to the base bisects the
vertical angle. It also bisects the base.
Mediators of the sides of a triangle.
Draw a triangle ABC on tracing paper. Fold
to form lines of symmetry for each of AB, BC
and CA in turn, (See Fig. 6.26). The lines of
symmetry divide the respective sides of the
triangle equally. These lines are called
mediators. What do you notice about the fold
lines?
A
B
C
A
B
C
– – – – – – – – – – – – – – –
D
Fig. 6.26
– – – – – – – – – – –– – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – – – – –
C
A
B
•
O
B
A
b
75°
c
a
54
(d) (e)
Fig. 6.28.
3. In Fig. 6.29, PQRS is a circle, centre O.
OP = 5 cm, TR = 4 cm, and ST = 2 cm.
Fig. 6.29.
(a) State the size of ∠PTO.
(b) State the lengths of the following:
(i) OR (ii) OS (iii) OT
(iv) OQ (v) QT
(c) What can you say about the products
PT × TR and QT × TS?
4. Draw a triangle ABC. With BC as the
mirror line, construct the reection image
of ABC. What special gure is formed by
the object triangle ABC combined with
its image? What is the sum of the interior
angles of this gure?
Use these facts to show that the angle sum
of a triangle is 180°.
5. A line of an object is at 35° to the line of
the mirror. What is the size of the angle
between the corresponding image line and
the mirror?
6. An object line and its image make an angle
of 90°. Describe the position of the mirror
line. Draw a diagram to illustrate this.
7. Draw the gure with vertices at (
–
1 , 1),
(2 , 3), (5 , 1) and (2 ,
–
1).
(a) What kind of gure is it?
(b) How many lines of symmetry does it
have?
(c) What are the equations of the lines of
symmetry?
8. Find the equation of the mediator of the line
segment joining each of the following pairs
of points.
(a) A(
–
2 , 4), B(5 , 4)
(b) C(
–
5 , 3), D(
–
1 , 3)
(c) E(3 , 8), F(3 , 5)
(d) G(
–
1 , 1), H(
–
1 ,
–
5)
9. Points A (
–
4 , 2), B(
–
3 ,
–
3) and C (
–
1 ,
–
1)
are the vertices of a triangle. Plot and join
them to form the triangle.
(a) What type of triangle is ABC?
(b) If points B and C are images of each
other, write down the equation of the
mirror line.
10. P(
–
1 , 3), Q(
–
3 ,
–
1) and R(3 ,
–
1) are the
vertices of a triangle.
(a) Calculate the lengths PQ, QR and PR
of the sides of the triangle.
(b) What type of triangle is ∆PQR?
(c) Find the equations of the mediators of
the sides.
(d) What are the coordinates of the point of
intersection of the mediators?
40°
e
d
b
a
c
120°
(c)
78°
h
e
12°
d
b
a
c
g
i
j
f
(b)
O
S
R
P
Q
5 cm
T
2 cm
4 cm
d
a
b
e
c
O is the centre
40°
g
f
d
a
h
i
e
c
b
O
55
Congruence
Fig. 6.30 shows two successive reections of
∆ABC in the mirror lines m
1
and m
2
.
How does each image differ from the
corresponding object? Does direction change?
Does the length of a line segment change?
Does the angle between corresponding pairs of
two lines change?
Do you agree that the size and shape do not
change?
We say that the object and its images are
congruent. But there is still an important
difference between an object and its image.
What is it?
If we trace and slide ∆ABC in Fig. 6.30, onto
which triangle will it t exactly?
If we turn over and slide the tracing of ∆ABC,
onto which triangle will it t exactly?
Two shapes are congruent if they have the
same size and shape.
If two shapes are congruent and we can slide
one so that it ts exactly on the other, the
congruence is direct.
If one shape can be tted on to the other only
after turning it over, the congruence is opposite
or indirect. Thus, under reection, the object
and the image are oppositely congruent.
B
C
A
B′
C′
A′
B′′
C′′
A′′
m
2
m
1
Fig. 6.30
B
A
C
R
P
Q
20 mm
Q
P
R
B
A
C
Congruent triangles: tests for congruence
Congruent triangles are named such that the
letters are in the same alphabetical order. This
makes it easier to indicate the corresponding
sides or the corresponding angles.
Thus, if ∆ABC is congruent to ∆PQR, then AB
corresponds to PQ, BC corresponds to QR and
CA corresponds to RP, ∠ABC corresponds to
∠ PQR, and so on.
If two or more triangles satisfy any of
the following conditions, then they are
congruent.
1. Each of the three sides of one triangle
is equal to the corresponding side of the
other triangle – abbreviated as SSS
(Fig. 6.31).
Fig. 6.31
2. Two sides and the included angle of one
triangle are equal to the corresponding
two sides and the included angle of the
other triangle (abbreviated as SAS Fig.
6.32).
Fig. 6.32
3. Two angles and a side of one triangle
are equal to the corresponding two
angles and one side of the other triangle
(abbreviated as AAS or ASA Fig. 6.33).
56
4. A right angle, a hypotenuse and another
side of one triangle are equal to the
corresponding right angle, hypotenuse
and side of the other triangle – abbreviated
as RHS (Fig. 6.34).
Fig. 6.34
Note: If, in condition (2), the angle is not the
included angle, it is not possible to conclusively
say if the triangles are congruent. In such a case,
there are two possible triangles PQR, one which
is not congruent to ABC. Hence, this (ASS) is
called the ambiguous case (Fig. 6.35).
Fig. 6.35
Example 6.6
In ∆ABC and ∆DEF, AB = EF, AC = DF and
∠A = ∠F. Are these triangles congruent?
Solution
It is helpful to make a sketch, as in Fig.6.36.
Fig. 6.36
Triangles ABC and DEF are congruent
(oppositely congruent) as the condition SAS
is satised.
Example 6.7
ABCD is a quadrilateral such that AB is
parallel to DC and AB = DC. Prove that
AD = BC, and AD // BC.
Solution
Sketch quadrilateral ABCD (Fig. 6.37. Join B
to D and mark the angles as shown .
Fig. 6.37
In ∆s ADB and CBD,
AB = CD (given),
BD = DB (common line),
and x
1
= x
2
(alternate angles).
∴ ∆ADB is congruent to ∆CBD (SAS).
In short ∆ADB ≡ ∆CBD
∴ AD = BC and y = z (congruence).
Since DB is a transversal to AD and BC, and
y = z, then AD//BC.
Q
R
P
Q
R
P
x
2
y
z
x
1
D
A
C
B
B
C
A
B
C
A
R
Q
P
E
D
F
B
C
A
R
Q
P
Or
Fig. 6.33
C
A
B
C
B
A
R
P
Q
57
Exercise 6.6
1. Fig. 6.38 shows four congruent triangles.
Fig. 6.38
(a) Which pairs of triangles are directly
congruent?
(b) Which pairs are indirectly congruent?
(c) Write down pairs of triangles which
could be object and image under
reection. Trace the gure and construct
the mirror lines for these pairs.
2. In Fig. 6.39, AB =AD, BC = CD. Show
that ∆s ABC and ACD are congruent.
Fig. 6.39
3. In Fig. 6.40, points P, Q and R are on the
circumference of the circle centre O. Show
that ∆s POQ and POR are congruent.
Fig. 6.40
4. The facts below refer to two triangles,
ABC and DEF. Do the facts show that the
triangles must be congruent?
(a) CA = FD, CB = FE, ∠B = ∠E
F
E
D
B
A
D
C
(b) AC = EF, ∠A = ∠E, ∠ C = ∠F
(c) AC = DF, ∠B = ∠D, ∠C = ∠F
(d) BC = DF, ∠B = ∠F, ∠A = ∠E
(e) ∠A = ∠D, ∠B = ∠E, ∠C = ∠F
(f) AB = AC, DE = DF, ∠A = ∠D
(g) BC = DF, AC = DE, AB = EF
(h) AB = AC, DE = DF, BC = EF
5. If ∆ABC ≡ ∆YZX, which side is equal to
XY and which angle is equal to ∠ACB?
6. In Fig. 6.41, the lines ANB and CND
bisect each other. Prove that AD = CB and
AD
||
CB.
Fig. 6.41
7. O is the centre of each of the circles in
Fig. 6.42. AOB and POQ are straight lines.
Prove that AQ = PB.
Fig. 6.42
8. In ∆ABC, AB =AC. P and Q are points on
AB and AC respectively such that PQ||BC.
Show that ∆s ABQ and ACP are congruent.
9. PQS is an isosceles triangle such that
PQ = PS. R is a point on QS such that
PR ⊥QS. Show that ∆s PQR and PSR are
congruent.
10. Two chords PQ and PR of a circle are equal
in length. The bisector of ∠QPR meets arc
QR at S. Use congruence to show that:
(a) QS = RS, and
(b) PS is perpendicular to QR.
A
N
C
B
D
B
C
A
I
G
H
L
J
K
P
O
B
Q
A
R
Q
P
130°
130°
O
58
Rotational symmetry
Rotational symmetry of plane gures
Which of the shapes in Fig. 7.1 have reection
symmetry? How many lines of symmetry are
there in each?
(a) (b)
(c) (d)
Fig. 7.1
Look at the drawings in Fig. 7.2. They also
have symmetry, though this is not the symmetry
of reection. What is there about the drawings
that makes us say that they are symmetrical?
(a) (b)
(c)
Fig. 7.2
Do you notice that in Fig.7.2 each of the
drawings is made up of congruent parts?
The drawings in Fig. 7.3 are also made up of
congruent parts, but they are not symmetrical.
How are these drawings different from those in
Fig. 7.2?
(a) (b)
(c)
Fig. 7.3
Activity 7.1
Copy Fig. 7.2(a) and then make a tracing of
your copy. Stick a pin through their centres
so that the tracing can rotate.
Now rotate the tracing until it ts exactly on
top of the copy again. Through what angle
did you rotate the tracing? How many times
does this angle divide into 360°? How many
times does the tracing t on to the copy in
one revolution?
If a gure can t onto itself when it is rotated,
it is said to have rotational symmetry.
The number of times a gure ts into itself
in one complete turn is called its order of
rotational symmetry.
In plane gures, rotational symmetry is also
ROTATION
7
59
called point symmetry. Plane gures are
considered to have point symmetry only in
cases where the order of rotational symmetry
is 2 or more.
Fig 7.2(a) has rotational symmetry of order 4
because in one complete revolution, the tracing
ts four times. What is the order of rotational
symmetry of each of the shapes in Fig. 7.2(b)
and (c)?
How many times do the shapes in Fig. 7.3 t
onto themselves in one revolution?
The shapes in Fig. 7.3 t only once in one
revolution. They are, therefore, not counted as
having rotational symmetry.
State the order of rotational symmetry of each
of the drawings in Fig. 7.4.
(a) (b)
(c) (d)
(e) (f)
Fig. 7.4
Rotational symmetry of solids
Fig. 7.5 is a drawing of a square-based pyramid
with a stiff wire AB tted vertically through the
vertex V and the centre C of the base.
If the pyramid is rotated about the stiff wire,
how many times will it t on to itself in one
revolution?
The pyramid has rotational symmetry of order 4
about its axis VC.
Fig. 7.5
An axis of symmetry of a solid is a line about
which the solid has rotational symmetry.
How many axes of symmetry does a cube have?
What is the order of rotational symmetry for
each axis?
List some plant and animal forms which have
rotational symmetry.
Exercise 7.1
1. State the order of point symmetry for each
of the shapes in Fig. 7.6.
(a) (b)
– – – – – – – – – – – – – – – –– – – – – – –
– – – – – – – – – – – – – – – –
– – – – – – – – – – –
– – – – – – – – – – – – – – –
A
stiff wire
V
C
B
•
60
•
Fig. 7.6
2. Copy and complete Table 7.1.
Number of Order of
Shape lines of rotational
symmetry symmetry
Equilateral ∆
Scalene ∆ 0 1
Square
Rectangle
Rhombus
Kite
Regular pentagon
Circle
Table 7.1
3. How many axes of symmetry does each
of the following solids have? Describe
each axis by stating the points on the solid
through which it passes. What is the order
of rotational symmetry about each axis?
(a) A cuboid
(b) A triangular prism whose cross-section
is an equilateral triangle
(c) A cone
(c) (d)
(e) (f)
(g) (h)
•
(d) A regular tetrahedron
(e) A rectangular-based pyramid
(f) A hexagonal unsharpened pencil
4. Each of the shapes in Fig.7.7 is part of a
point symmetrical gure. Copy the shapes
and complete the symmetry using the given
orders and points.
(a) (b)
(c) (d)
Fig. 7.7
5. The door handle shown in Fig. 7.8 is tted
on a square prism rod. The square prism
rod ts through a square hole in the lock.
In how many ways can the door handle be
tted in the lock?
Fig. 7.8
Rotation as a transformation
We have already seen the effects of reection
on geometrical gures. As we have seen earlier,
reection is a geometrical transformation.
Rotation is another example of a transformation.
In a reection, points (except those on the
mirror line) change position and the order
(direction) of points around a gure is reversed.
But lengths and angles are unchanged.
•
Order 3 Order 2
Hole
Handle
•
Order 3 Order 2
61
Now discover the properties of rotation by
carrying out the following activities. (In these
activities it is better to work in pairs).
Activity 7.2
Draw a triangle OAB, as shown in Fig. 7.9.
Trace the triangle using tracing paper. Name
the vertices of the tracing O′, A′ and B′ to
correspond with O, A and B.
Fig. 7.9
Place the tracing exactly on top of the original
gure. Put a pin through O and O'. Keeping
the lower sheet still, rotate the tracing anti-
clockwise about O through approximately
90°. Answer the following questions.
(a) Through what angle has each of lines OA
and OB turned?
(b) Which is longer, AB or A′B′; or are they
the same length?
(c) Have any points remained in the same
position? If so, which ones?
(d) Are the angles of the image triangle the
same as the corresponding angles of the
object triangle?
Put the tracing back exactly on top of the
original triangle. Rotate the tracing clockwise
about O through 90°. Answer the above
questions now. Have you got the same result
as before?
The point about which a gure is rotated is called
the centre of rotation and the angle through
which the gure is rotated is called the angle of
rotation.
A
O
B
The direction of rotation is important! An
anticlockwise turn is taken to be positive
while a clockwise turn is taken as negative.
Therefore, an anticlockwise turn of 90° is
called a rotation of
+
90° while a clockwise turn
of 90° is called a rotation of
–
90°.
Note: This convention is used throughout in
Mathematics, Science and Engineering. It is
only in the measurement of bearings that the
positive direction is clockwise.
Activity 7.3
Using the same figure and tracing as in
Activity 7.2, arrange the tracing to coincide
with the original triangle again.
Rotate the tracing about O through 60°.
How do you know when to stop rotating the
tracing?
You may do this by rst marking the nal
position of OA on the lower sheet before
putting the tracing on top. Fig 7.10 shows
the lower sheet with the image position of
OA drawn as broken line OA′. OA′ is called
a guide line.
Arrange the tracing to coincide with the
original triangle. Rotate the tracing about O
until OA′, on the tracing, comes on top of the
guide line.
Fig. 7.10
Has OB turned the same angle as OA?
Measure to check your answer.
Has any point remained xed in this rotation?
What is the angle between AB and A′B′?
(Extend the line segments if necessary).
A
O
B
– – – – – – – – – – – – – – – – – – –
60°
A′
62
Properties of rotation
From Activities 7.2 to 7.4, you should have
noticed that under rotation:
1. All points on the object turn through the
same angle in the same direction.
2. The angle between a line and its image
equals the angle of rotation.
3. Each point and its image are the same
distance from the centre of rotation.
4. The centre of rotation is invariant i.e.
it does not change its position.
Note:
(i) A rotation is fully dened when the centre
and angle of rotation are specied.
(ii) A positive rotation through an angle
θ
is
the same as a negative rotation through an
angle of
(360°–
θ
) about the same centre.
Rotation and congruence
Look at your work of Activities 7.2 to 7.4 again.
Looking at the object and the image, what can
you say about:
(a) the sizes of corresponding angles?
(b) the lengths of corresponding sides?
(c) the orientation (i.e. the direction in which
they face)?
The answers to these questions should lead you
to see that:
Under rotation, an object and its image are
directly congruent.
Locating an image given the object,
centre and angle of rotation
Example 7.1
Fig. 7.12 shows a triangle PQR in which
PQ = 3 cm, QR = 4 cm and PR = 5 cm.
A
B
C
O
90°
D
D′
– – – – – – – – –
Activity 7.4
Draw another triangle as in Activity 7.2 and
label its vertices A, B and C.
Mark on the lower sheet a point O, which is
not on the triangle.
Rotate the tracing about O through an angle
of
–
90°. How do you do this?
You may do this by using guide lines.
In Fig. 7.11, OD has been drawn. This will be
rotated to the position OD′.
Fig. 7.11
On your lower sheet, mark the guide lines OD
and OD′. Place the tracing back on the gure
and trace OD.
Now rotate the tracing through
–
90° about O.
How do you know when to stop rotating the
tracing?
What size is the angle between a line and
its image in this rotation? (Extend the line
segments, if necessary, to answer this).
What conclusion can you draw?
Which is longer, OA or its image OA′? What
about OB and OB′, OC and OC′?
Instead of rotating through
–
90°, through what
angle would you have to rotate the tracing
in a positive direction to get into the same
position?
We say that a rotation of 270° has the same
effect as one of
–
90°.
What positive rotation has the same effect as
one of
–
150°?
What negative rotation is equivalent to a
rotation of 320°?
63
– – – – – – – – – – – – – – – – – –
A
O
– – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – –
A′
Copy the gure and locate ∆P′Q′R′, the image
of ∆PQR, under a rotation of 65° about point O.
Fig. 7.12
Solution
To locate ∆P'Q'R', proceed as follows:
(a) Join P to O. With OP as the initial line,
measure an angle of 65° anticlockwise at O
and draw a construction line OA. (see Fig.
7.13.)
(b) To obtain P′ on OA, measure OP′ = OP.
Mark the point P′.
(c) Repeat step (a) for Q and R to obtain
construction lines OB and OC respectively.
Measure OQ′ = OQ and OR′ = OR on OB
and OC to obtain points Q′ and R′.
(d) Join P′, Q′, R′ to obtain ∆P′Q′R′.
Fig. 7.13
Finding the centre and angle of rotation
Under rotation, every point of an object moves
along an arc of a circle whose centre is the
centre of rotation. Thus, if a point A is mapped
onto a point A′ by a rotation about a point
O, then AA′ is a chord of the circle centre O,
through A and A′ (Fig. 7.14).
Q
R
P
Fig. 7.14
Recall that the perpendicular bisector (mediator)
of a chord of a circle passes through the centre
of the circle.
Thus, the mediator of AA′ passes through the
centre of rotation O. This fact in locating the
centre of rotation.
Example 7.2
In Fig. 7.15, ∆A′B′C′ is the image of ∆ABC after
a rotation. Copy the gure and locate the centre
of rotation. Determine the angle of rotation.
Fig. 7.15
Solution
To locate the centre of rotation, proceed as
follows.
(a) Join A to A′ and construct the mediator of
AA′ (Fig 7.16).
(b) Join B to B' and construct the mediator of
BB'.
(c) Produce the mediators in steps (a) and (b)
so that they intersect at point O.
(d) Construct the mediator of CC'. Does this
mediator also pass through O?
Using the method of Activities 7.2 to 7.4, check
that O is actually the centre of rotation.
C
B
A
A′
B′
C′
•
O
Q′
3 cm
P′
– – – – – – – – – – – – – – – – – – – – – –
B
P
R
Q
R′
C
A
O
65°
– – – – – – – – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – –– – – – – – – – –
64
C
B
A
A′
B′
C′
O
Fig. 7.17
2. In g. 7.18, PQ is a chord of a circle centre
O. The circle is rotated about O so that P′Q′
is the image of PQ.
(a) What can you say about the length of
PQ and P′Q′?
(b) What is the perpendicular distance of
P′Q′ from O, given that PQ is a
perpendicular distance x from O?
Fig. 7.18
(c) Copy and complete the following
statement:
Equal chords of a circle are the same
_______ from the centre of the circle.
3. Draw a triangle ABC. Join the midpoint M
of AC to B as in gure 7.19
Fig. 7.19
P
S
R
Q
T
•
Q
Q′P
P′
O
Fig. 7.16
To nd the angle of rotation, join any one of
the points A, B or C to the centre of rotation O.
Also join the corresponding image point to O.
Measure the angle that is formed.
Did you nd that the angle of rotation is
–
120°?
To find the centre of rotation, draw the
mediators of the line segments formed by
joining object points to their corresponding
image points. As all the mediators pass
through the centre of rotation, it is sufcient to
nd the intersection of any two mediators.
Exercise 7.2
1. The drawing in Fig. 7.17 has rotational
symmetry of order 2.
(a) What point is the centre of rotational
symmetry?
(b) Which lines are parallel?
(c) What is the image of point S?
(d) If the points P, Q, T, and S are joined,
what kind of quadrilateral is formed?
(e) If PT = 9 cm, what is the length of RT?
(f) If ∠RST = 48°, what other angle is
48°?
A
B
C
M
65
(ii)Rotate ∆ABC through 180° about M and
label the image of B as D.
(a) Name two pairs of parallel lines in
quadrilateral ABCD.
(b) What type of quadrilateral is ABCD?
(c) Name the side that is equal in length to:
(i) AB (ii) BC.
(d) Name the angle that is equal to:
(i) ∠ABC (ii) ∠BAD.
(e) What can you say about the lengths of:
(i) MB and MD? (ii) MA and MC?
(f) Which triangle is congruent to:
(i) ∆ABC? (ii) ∆BAD?
(g) Copy and complete the following
statements:
(i) The opposite sides of a
parallelogram are _____ and _____.
(ii) The opposite angles of a
parallelogram are _____.
(iii) The diagonals of a parallelogram
_____ each other.
(iv) A diagonal _____ a parallelogram.
4. Each part of Fig. 7.20 shows an object and
its image after rotation.
(a) Trace the diagrams and nd the centres
of rotation.
(b) Find the angle of rotation in each case
giving your answer both as a positive
and a negative angle.
(i)
Fig. 7.20
Rotation in the Cartesian plane
Fig. 7.21 shows a triangle ABC and its images
after rotations with different angles of rotation.
With ∆ABC as the object and ∆A′B′C′ as the
image:
(i) What is the centre of rotation?
(ii) What is the angle of rotation?
(iii) Copy and complete Table 7.2
Centre of rotation (__ , __); Angle of rotation is ___
Object point A(1 , 3) C(2 , 1) P(a , b)
Image point B′(
–
2 , 3)
Table 7.2
With ∆ABC as the object and ∆A′′B′′C′′ as the
image:
(i) What is the centre of rotation?
(ii) What is the angle of rotation?
(iii) Copy and complete Table 7.3.
B′
D′
C′
A′
C
A
B
D
(iii)
(iv)
C
A
B
D
B′
D′
C′
A′
C
A
B
D
B′
D′
C′
A′
A′
C′
B′
C
B
A
66
Centre of rotation (__ , __); Angle of rotation is ___
Object point A(1 , 3) B(3 , 2) P(a, b)
Image point C'(1 ,
–
2)
Table 7.3
With ∆ABC as the object and ∆A′′′B′′′C′′′ as
the image
(i) What is the centre of rotation?
(ii) What is the angle of rotation?
(iii) Copy and complete Table 7.4.
Centre of rotation (__ , __); Angle of rotation is ___
Object point B(3 , 2) C(2 , 1) P(a, b)
Image point A′′′(
–
1 ,
–
3)
Table 7.4
Notice that a rotation about the origin (0, 0):
1. through 90° maps a point (a, b) onto the
point (
–
b, a)
2. through
–
90° maps a point (a, b) onto the
point (b,
–
a)
3. through 180° maps a point (a, b) onto the
point (
–
a,
–
b)
Where do rotations of 0° and 360° about the
origin map point (a, b)?
A(2 , 5), B(4 , 5), C(6 , 3) and D(3 , 2) are the
vertices of a quadrilateral. With (1 , 2) as the
centre, rotate quadrilateral ABCD through 180°.
Complete Table 7.5.
Centre of rotation ( 1 , 2); Angle of rotation is 180°
Object
A(2 , 5) B(5 , 4) C(6 , 3) D(3 , 3) P(a , b)
point
Image
C′(
–
4 , 1)
point
Table 7.5
You should notice that a rotation of 180° about
a point (1 , 2) maps a point (a, b) onto the point
(2 × 1 – a, 2 × 2 – b).
Similarly:
A rotation of 180° about (h , k) maps a point
(a, b) onto the point (2h – a, 2k – b).
Exercise 7.3
1. L(4 , 2), M(
–
1 ,
–
2) and N(3 , 0) are the
vertices of a triangle. Plot these points
on squared paper and with C(2 , 1) as the
centre, rotate LMN through an angle of 90
0
.
(a) Write down the coordinates of L′′, M′
and N′.
(b) If S is the point (2 ,
–
1), what are the
coordinates of S′?
(c) If T is the point (3 , 4), what are the
coordinates of T′?
(d) Without measuring, state the angle
between LM and L′M′.
(e) What is the path traced out by L in
moving to L′?
2. A(
–
3 , 1), B(1 , 1), C(1 ,
–
3), D(
–
3 ,
–
3) and
P(
–
1 , 3), Q(3 , 3), R(3 ,
–
1), S(
–
1 ,
–
1) are
the vertices of two squares ABCD and
PQRS. Fully describe the rotation that maps
(a) ABCD onto QRSP (this means that A is
Fig. 7.21
x
3
2
1
–
1
–
2
–
3
–
3
–
2
–
1
0 1 2 3
y
B
C
A
A′
C′
B′
B′′′
C′′′
A′′′
C′′
B′′
A′′
67
mapped onto Q, B onto R, and so on).
(b) BCD onto SPQR, and
(c) ABCD onto RSPQ.
3. Write down the images of the following
points under rotation through the given
angles and about the stated centres.
(a) Centre (0 , 0), angle 90°
(i) (4 , 5) (ii) (3 ,
–
4)
(iii) (4 ,
–
7) (iv) (
–
6 ,
–
8)
(b) Centre (0 , 0) angle
–
90°
(i) (7 , 1) (ii) (
–
3 , 6)
(iii) (4 ,
–
7) (iv) (
–
2 ,
–
3)
(c) Centre (0 , 0), angle 180°
(i) (4 , 4) (ii) (
–
3 , 2)
(iii) (0 ,
–
5) (iv) (
–
3 ,
–
4)
(d) Centre (3 , 2), angle 180°
(i) (2 , 3) (ii) (
–
5 , 3)
(iii) (
–
4 ,
–
5) (iv) (3 ,
–
1)
(e) Centre (
–
2 ,
–
5), angle 180°
(i) (6 ,
–
5) (ii) (2 , 9)
(iii) (0 , 2) (iv) (
–
3 , 6)
4. A negative quarter turn about the point
(0,
–
1) maps ∆ABC onto ∆A′B′C′ with the
vertices A′(3 , 1), B′(0 , 5), and C′(0 , 1).
Find the vertices of ∆ABC.
5. A quadrilateral has vertices A(1, 3), B(2, 5),
C(4 , 4) and D (3 , 3).
(a) Find the coordinates of the image
quadrilateral A′B′C′D′ under reection
in the line y = 2.
(b) A certain transformation maps points
A′, B′, C′ and D′ onto points A′′(
–
5 ,
–
3),
B′′(
–
7 ,
–
3), C′′(
–
8 ,
–
2) and D′′(
–
6 ,
–
1)
respectively.
Fully describe this transformation.
68
Revision exercise 1.1
1. Use multiplication to evaluate:
(a) 125
3
(b) 2.16
3
(c) 0.043
3
2. Use table of reciprocals to evaluate
1
–
4
+
15
2.471 0.015 72 243.1
3. Write the following using positive indices:
(a) 12
–
2
(b) a
–
7
(c) 2x
–
4
(d)
a
2
(e)
x
–
3
(f)
a
–
5
b
–
4
w
–
4
b
–
10
(g)
xy
–
6
3
–
4
4. Simplify:
(a) 5
4
× 5
15
(b) 7
11
× 7
20
(c) a
–
6
× a
8
(d) c
–
6
× c
–
4
(e) x
3
÷ x
–
8
(f) y
5
÷ y
–
3
(g) y
–
7
÷ y
2
(h) d
–
11
÷ d
–
13
5. (a) Express the following in standard form.
(i) 68 700 000 (ii) 0.000 000 142 7
(b) Given that a = 2.5 × 10
9
and
b = 1.5 × 10
–
7
, evaluate; (i) ab (ii)
and give your answers in standard
form.
(c) Evaluate
0.68 × 1.2 × 10
–
4
× 1 000 × 2 × 10
11
and give your answer in standard form.
6. (a) State the characteristics of the
logarithms of the following numbers.
(i) 3.5 (ii) 27.2 (iii) 4 763
(iv) 18 002 (v) 10
2.5
(vi) 0.003 4
(b) State the logarithms of the following.
(i) 5.23 (ii) 45.62 (iii) 10
3.6
(iv) 4 × 10
5
(v) 0.001 389 7
7. Find the equations of the lines which pass
through the following pairs of points.
(a) (
–
2 , 4), (0 ,
–
4) (b) (
–
6 ,
–
2), (
–
2 ,
–
4)
(c) (
–
3 ,
–
1), (1 ,
–
1) (d) (0 , 4), (
–
5 , 3)
(e) (1 , 3), (2 ,
–
9) (f) (
–
4 , 2), (
–
2 ,
–
6)
(g) (a , 5), (a ,
–
3) (h) (7 , b), (15 , b)
8. Find the equation of the line whose
gradient and y-intercept are given
respectively as:
(a) ;
–
3 (b)
–
2 ; 2 (c) 0 ; 0
(d) undened
9. Which of the shapes in Fig. R.1.1 have
lines of symmetry?
(a) (b)
(c) (d)
(e) (f)
(g) (h)
Fig. R.1.1
b
a
–
1
2
–
–
REVISION EXERCISES 1
1-7
69
10. Copy each of the diagrams in Fig. R.1.2
above and construct the images under
reection in the given mirror lines.
11. Which of the following plane gures have
rotational symmetry? State the order in
each case.
(a) Square (b) Rectangle
(c) Circle (d) Rhombus
(e) Parallelogram (f) Regular hexagon
12. ABCD is a rectangle whose vertices
are A(1, 1), B(4, 1), C(4, 3) and D(1,
3).
(a) Find the coordinates of its image after
a
–
90° rotation about the origin.
(b) The object and image are said to be
congruent. Describe their congruency.
Revision exercise 1.2
1. A cube has a volume of 13 824 cm
3
. Find
the radius of a cylinder which has the same
volume and height as the cube.
Fig. R.1.2
A
C
D
m
1
B
m
2
(a)
P
Q
S
R
m
1
m
2
(b)
E
A
C
D
B
m
(c)
(d)
S
P
Q
R
m
2. Find x if x =
1
+
1
127 11.5
.
3. Simplify
1
+
4
–
1
.
35 360 73.8
4. Simplify:
(a) 3 × 3
(b)
7
(c) ax × bx (d) (ay)
2
÷ y
–
3
(e) 3x ÷ 2x (f)
33x
4
× 9x
5
11x
12
(g)
3x
3
× 18x
4
2x
5. Simplify the following giving your answers
with positive indices:
(a) 10t
3
× 4t
–
5
(b) 3t
–
4
× 8t
–
2
( )
1
3
–
4
3
–
1
2
–
1
2
–
1
2
–
4
2
3
–
5
3
–
–
( )
( )
1
3
–
1
3
–
–
(c) 4t
8
÷ 5t
–
6
(d) 4 × ( )
–
3
(e) ( )
–3
× 3t
–4
(f)
1
x
6
(g) a
0
b
–
4
÷ b
7
1
4
–
1
2
–
1
2
–
–
70
11. (a) In Fig. R.1.4(a), quadrilateral A′B′C′D′
is the image obtained when quadrilateral
ABCD is reected. Copy the gure on
squared paper and construct the mirror
line.
6. (a) Simplify the following, giving your
answers in the same bar notation.
(i) 3.2 + 1.4 (ii) 6.3 – 1.7
(iii) 4.621 × 2 (iv) 2.881 ÷ 3
(b) Find antilogarithms of:
(i) 0.765 2 (ii) 1.358 7
(iii) 2.666 6 (iv) 0.007 8
(c) Evaluate:
(i) 10
0.3
(ii) 10
1.257 6
(iii) 10
2.912
(iv) 10
3.687 1
(v) 10
1.002 5
(vi) 10
2.163 2
7. Write each of the following as a single
logarithm.
(a) log 2 + log 4 (b) log 13 – log 6
(c) 5 log 2 – log 25 (d) 3 log 5 + 2
8. Find the equation of a line which
passes through the point (2 , 3) and is
perpendicular to y = 3x – 1, giving your
answer in the double intercept form.
9. Find the equation of the line given the
gradient and a point on the line respectively
as:
(a)
–
4
; (2 , 1) (b)
2
; (8 , 4)
3 5
(c)
–
1 ; (
–
2 ,
–
3) (d) undened ; (6 ,
–
4)
10. Use Fig. R.1.3 to answer the following.
(The gure is symmetrical about m).
(a) What is the image of point C in m?
(b) What is the image of point A in m?
(c) What is the image of GF in m?
(d) Name two points which are equidistant
from m.
(e) What is the image of ∠EGF in m?
(f) What is the image of ∆CDE in m?
– – – –
– – –
–
–
–
–
1
2
–
Fig. R.1.3
B
C
D
A
G
H
F
E
m
Fig. R.1.4
A
C
B
D
(a)
C′
D′ B′
y
(b)
7
6
4
2
0
2 4 6 8
x
A′
5
3
1
1 3 5 7
D′
C′
B′
C
D
A
B
(c)
F
F
2
A′
y
(d)
8
6
4
2
0
4 6 8
x
A
A′
2
71
Revision exercise 1.3
1. Use tables to evaluate:
23.5
3
– 4 41 1 +
1
0.007 1
2. If = + , nd f given that u =
and v = .
3. Use tables to evaluate
5 + 8
63.34 0.063 6
.
3
1
8
–
1
u
–
1
v
–
1
3
–
(b) In Fig. R.1.4(b), square A′B′C′D′ is
the image of square ABCD under
reection. Copy the gure on squared
paper and locate the mirror line.
(c) In Fig. R.1.4(c), the ag F was reected
to obtain the image F
1
. F
1
was in turn
reected to obtain F
2
. Copy the gure
on a squared paper and locate the
image F
1
and the two mirror lines.
(d) On squared paper, copy Fig. R.1.4(d)
in which triangle A′ is the image of
triangle A under reection. Locate the
mirror line and nd its equation.
12. (a) Triangle ABC has vertices at A (8 , 6),
B(4 , 10) and C(2 , 6). After a negative
quarter turn about (2 , 1), ∆ABC maps
onto ∆A′B′C′. Find the coordinates of
A′, B′ and C′.
(b) Under a certain rotation, ∆PQR whose
vertices are P(3 , 4), Q(7 , 4) and
R(7 , 6) maps onto ∆P′Q′R′ whose
vertices are P′(4 ,
–
3), Q′(4 ,
–
7) and
R′(6,
–
7). Find, by construction, the
centre and angle of rotation.
(c) Triangle PQR has vertices at P(3 , 1),
Q(7 , 4) and R(4 , 3). Show the image
of ∆PQR after a rotation of 180° about
(3 , 1).
1
f
–
( )
4. Evaluate the following:
(a) 81 (b) 32
(c)
243a
6
× 2b
3
(d)
(3a
6
)
18
(243) a
(e)
6q × q
18q
5. Solve for x:
(a) 3x =
3
(b) 2
2x + 1
= 8
3
9
–
2
(c) 6 × 4
1 – m
= 1 536 (d) (9
2x
)
3
= 81
(e) ( )
x
× (2)
1 – x
= 486 (f) 5
–
2x
= 625
(g) 4x
3
= 108 (h) 3x
–
2
– 48 = 0
6. (a) Given that log y = 3.143, log x = 2.422,
evaluate the following without further
use of tables
(i) log (xy) (ii) log( )
(iii) log x (iv) 4log y + log x
(b) Use logarithms to evaluate;
(i)
142.7 × 62.3
(ii)
4.73 × 22.41
22.84 × 17.31 82.3
(iii)
321 × 0.028 3
(iv)
3.45 + 2.62
3.04
2
– √7.56 √ 786 × 0.7
7. (a) Given that log 2 = 0.301 03 and log 3 = 0.477
12, nd without further use of tables;
(i) log 6 (ii) log (iii) log 9
(iv) log 20 (v) log 15 (vi) log 2√3
(b) Solve for x in:
(i) log
10
x = 3 (ii) log
4
x = 3
(iii) log
3
81 = x (iv) log
x
16 = 4
(c) If the logarithms of the squares of a
and b are 1.024 and 0.954 respectively,
nd the logarithm of:
(i) the product of a and b
(ii) the quotient
(iii) the quotient .
a
b
–
1
2
–
1
3
–
–
1
3
––
1
12
––
1
3
–
1
3
–
2
3
–
y
x
–
1
3
–
1
3
–
2
3
–
b
a
–
4
–
1
4
––
–
2
5
––
1
4
––
2
3
––
72
Fig. R1.6
(i) Which triangles have direct
congruence?
(ii) Which triangles have opposite
congruence?
(iii) Which pair of triangles could be
object and image under reection?
(b) In Fig. R.1.7, O is the centre of the
circle. Show that triangles ABC and
ADC are congruent.
Fig. R.1.7
12. (a) Copy ABC (Fig. R.1.8) and nd
its image under rotation through
–
80°
about the marked point P. What rotation
would map ∆A′B′C′ back onto ∆ABC?
Fig. R.1.8
(b) Triangle ABC has vertices A(1 , 3),
B(
–
2 ,
–
2) and C(4 ,
–
1).
(i) Reect ∆ABC in the line x + y = 0.
State the coordinates of A′, B′ and
C′, the images of A, B, and C.
(ii) Reect ∆A′B′C′ in the line y = 0.
State the coordinates of A′′, B′′
and C′′, the images of A′, B′ and C′.
(iii) ∆ABC can be mapped onto
A′′B′′C′′ by a rotation. Find by
construction the centre and angle
of this rotation.
1
2
3
4
5
6
•
P
A
B
C
40°
A
B
C
D
O
40°
8. (a) Find the equations of the lines parallel
to the given lines and passing through
the given points.
(i) y = 4x + 3; (0 , 2)
(ii) 2y – 3x = 4; (2 , 6)
(iii) 5y + 10x – 3 = 0; (
–
2 ,
–
3)
(iv) 4x + 5y – 6 = 0; (
–
4 , 4)
(b) Write 3y = 21 – 7x in the double
intercept form and state the x- and y-
intercepts of the line.
9. Juma sells 60 pairs of shoes per week when
he sells at K£ 20 a pair. When he raises the
price to K£ 30 a pair, he sells only 40 pairs
per week.
(a) Assuming the information can be
represented on a straight line graph,
with the price plotted on the horizontal
axis, determine the gradient-intercept
form of the equation for the demand.
(b) Using the equation, predict the number
of pairs of shoes sold if the price is
lowered to K£ 15.
10. In Fig. R.1.5 the point A(5 , 0) is reected
in the mirror line m
1
which makes an
angle of 20° with the axis. The image A
1
obtained is reected in the mirror line m
2
.
The mirror lines m
1
and m
2
make an angle
of 50° at the origin. Copy the gure and
obtain the image A
2
. Find angle AOA
2
.
11. (a) In Fig. R.1.6 there are six congruent
triangles.
Fig. R.1.5
0
2
4
6
8
x
10
y
8
6
4
2
20°
50°
A
•
m
2
m
1
73
Introduction
In real life, we encounter and talk about objects
which have the same shape. In mathematics,
such objects are said to be similar. Whenever
we look at a house plan, make a scale model of
a car, etc., we are making use of similar shapes.
Note that in every day use, the word ‘similar’
often means ‘roughly the same shape’. For
example, the leaves coming from the same
tree have roughly the same shapes, although it
may be difcult to nd two leaves which have
exactly the same shape or size.
In Mathematics, however, the word ‘similar’ is
used to mean ‘exactly the same shape’. For
example, the shape of a picture on a television
screen is exactly the same irrespective of the
size of the screen. In this case, the pictures are
similar.
Similar gures
Consider Fig. 8.1(a) and (b).
(a) (b)
Measure the lengths of the corresponding lines
BC and B′C′, CD and C′D′, DE and D′E′, DF
and D′F′, etc.
Copy and complete the following.
B′C′
=
C′D′
=
BC CD
D′E′
=
D′F′
=
DE DF
What do you notice about the ratio of
corresponding sides?
Measure the corresponding angles ∠A and ∠A′,
∠B and ∠B′, ∠C and ∠C′, etc.
What do you notice?
Notice that the ratio of the lengths of
corresponding lines is constant and that the
corresponding angles are equal.
We say that the two drawings in Fig. 8.1 are
similar.
Now consider the shapes in Fig. 8.2. Which
shapes in the gure are similar to (a)? Why are
the others not?
(a)
(b)
(c) (d)
Fig. 8.2
Fig. 8.1
The two drawings are exactly the same except
for their sizes.
B
F
C
D
E
A
B′
F′
C′
D′
E′
A′
SIMILARITY AND ENLARGEMENT
8
74
When two gures are similar,
1. the ratio of corresponding lengths is
constant, and
2. corresponding angles are equal.
Construction of similar gures
Draw a triangle with sides 4.6 cm, 3.0 cm and
3.4 cm. Draw another triangle whose sides are
1 times as long. What do you notice about the
angles of the triangles? Are the two triangles
similar?
Draw two triangles, one larger than the other,
with angles of 55°, 75° and 50°. Are the two
triangles similar?
Two triangles are similar if either:
1. the ratio of corresponding sides is
constant, or
2. the corresponding angles are equal.
Would the statement still be true if ‘triangles’ is
replaced with ‘polygons’? The answer is ‘No’:
With all polygons other than triangles, the ‘or’
must be replaced with ‘and’.
Further constructions involving similar gures
will be done later in the section covering
enlargement.
Example 8.1
Fig. 8.3 shows two triangles ABC and PQR.
Calculate the lengths BC and PQ.
Fig. 8.3
Solution
Since the corresponding angles are equal, ∆’s
ABC and PQR are similar.
∴ the ratio of corresponding sides is constant.
Thus,
AB
=
BC
=
AC
PQ QR PR
i.e.
24
=
a
=
18
r 7 6
⇒
a
=
18
i.e.
a
= 3
7 6 7
∴ a = 21 i.e. BC = 21 cm
Also,
24
=
18
i.e.
24
= 3
r 6 r
⇒ 3r = 24
∴ r = 8 i.e. PQ = 8 cm
Note:
1. The symbol ~ is used for similarity so that
∆ABC ~ ∆PQR is read as ‘Triangle ABC is
similar to triangle PQR’.
2. Since the two triangles in Fig. 8.3 are alike
in everything except size, it can be said that
∆PQR is a scale drawing of ∆ABC with a
scale 1 : 3.
Example 8.2
Determine whether the hexagons in Fig. 8.4 are
similar. State your reasons.
(a)
(b)
Fig. 8.4
Solution
In Fig. 8.4(a), each side is 2 cm. In (b), each
side is 1.4 cm.
1
2
–
24 cm
75°
18 cm
B
A
C
59°
46°
a
7 cm
6 cm
Q
R
P
75°
46°
59°
r
75
∴ Ratio of corresponding sides is
1.4
= 0.7
(a constant).
2
In (a), there are two angles of 100°, and four
angles of 130° each. Each angle in (b) is 120°.
∴ not all corresponding angles are equal.
Although the ratio of corresponding sides is
constant, not all corresponding angles are equal.
Hence, the two hexagons are not similar.
Two polygons are similar if;
1. the ratio of corresponding sides is
constant, and
2. the corresponding angles are equal.
Exercise 8.1
1. State if, and why, the pairs of shapes in
Fig. 8.5 are similar.
(a)
(b)
(c)
10 cm
16 cm
20 cm
5 cm
10 cm
8 cm
12 cm
8 cm
9 cm
6 cm
(d)
(e)
Fig. 8.5
2. The triangles in each pair in Fig. 8.6 are
similar. Find x.
(a)
(b)
(c)
(d)
Fig. 8.6
6 cm
9 cm
12 cm
8 cm
15 cm
10 cm
P
R
T
N
A
24 cm
20 cm
12 cm
18 cm
39 cm
24 cm
18 cm
20 cm
15 cm
x
4 cm
3 cm
5 cm
12 cm
x
26 cm
3 cm
3 cm
3 cm
3 cm
3 cm
5 cm
5 cm
5 cm
5 cm
5 cm
20 cm
20 cm
x
10 cm
10 cm
3 cm
5 cm
3 cm
7 cm
12 cm
x
28 cm
76
3. Show that the two triangles in Fig. 8.7 are
similar. Hence calculate AC and PQ.
Fig. 8.7
4. Which two triangles in Fig. 8.8 are similar?
State the reason. If AB = 6 cm, BC = 4 cm
and DE = 9 cm, calculate BD.
Fig. 8.8
5. Construct two triangles ABC and PQR
with sides 3, 3, 5 cm and 12, 12, 20 cm
respectively. Measure all the angles. Are
triangles ABC and PQR similar?
6. The vertices of three right-angled triangles
are given below:
A(3 , 3), B(4 , 5), C(3 , 5);
P(1 , 3), Q(1 , 5), R(2 , 4);
X(
–
2 , 3), Y(1 ,
–
1), Z(
–
2 ,
–
1).
Which two triangles are similar?
7. Measure the length and breadth of this text
book. Measure also the length and breadth
of the top of your desk. Are the two shapes
similar? If not, make a scale drawing of a
shape that is similar to the top of your desk.
8. A photograph which measures 27 cm by
15 cm is mounted on a piece of card so as
to leave a border 2 cm wide all the way
round. Is the shape of the card similar to
that of the photograph? Give reasons for
your answer.
9. In Fig. 8.9, identify two similar triangles.
Use the similar triangles to calculate x and y.
Fig. 8.9
10. Lulu used similar triangles to nd the
distance across a river. To construct the
triangles he made the measurements shown
in Fig. 8.10. Find the distance across the
river.
Fig. 8.10
Similar solids
Two solids are similar if:
1. the ratio of the lengths of their
corresponding sides is constant, and
2. the corresponding angles are equal.
Example 8.3
A jewel box, of length 30 cm, is similar to a
matchbox. If the matchbox is 5 cm long, 3.5
cm wide and 1.5 cm high, nd the breadth and
height of the jewel box.
Q
R
20 cm
P
10 cm
109°
22°
A
109°
15 cm
B
49°
C
12 cm
36 m
32 m
x
– – – – – – – – – – – – – –
24 m
E
D
A
B
C
4 cm
50°
10 cm
8 cm
50°
A
y
C
B
x
D
77
Solution
The ratio of the lengths of the jewel box and the
matchbox is
Length of jewel box = 30 cm = 6
Length of match box 5 cm
This means that each edge of the jewel box is
6 times the length of the corresponding edge of
the matchbox.
width of jewel box = 6 3.5 cm = 21 cm,
and height of jewel box = 6 1.5 cm = 9 cm.
Exercise 8.2
1. A scale model of a double-decker bus is
7.0 cm high and 15.4 cm long. If the bus is
4.2 m high, how long is it?
2. A cuboid has a height of 15 cm. It is similar
to another cuboid which is 9 cm long, 5 cm
wide and 10 cm high. Calculate the area of
the base of the larger cuboid.
3. A water tank is in the shape of a cylinder
radius 2 m and height 3 m. A similar tank
has radius 1.5 m. Calculate the height of
the smaller tank.
4. Write down the dimensions of any two
cubes. Are the two cubes similar? Are all
cubes similar? Are all cuboids similar?
5. A designer has two models of a particular
car. The rst model is 15 cm long, 7.5 cm
wide and 5 cm high. The second model is
3.75 cm long, 1.70 cm wide and 1.25 cm
high. He says that the two are ‘accurate
scale models’. Explain whether or not his
claim could be true.
Enlargement and its properties
Compare the drawings in Fig. 8.11. They are
exactly alike except for the size. We say that
the larger drawing (b) is an enlargement of the
smaller one (a).
(c) (b)
Fig. 8.11
If we regard (a) as the object, then (b) is
the image under the transformation of
enlargement.
Fig. 8.12 shows a triangle ABC and its image
ABC. Lines AA', BB' and CC', produced, meet
at a common point O. The point O is called
the centre of enlargement. ∆s ABC and ABC
are similar. What is the ratio of lengths of
corresponding sides?
Fig. 8.12
By measuring, determine the value of
. This
is called the scale factor of enlargement.
Name other ratios that are equal to the scale
factor.
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
C
B
A
C′
B′
A′
O
OA′
OA
(a)
(b)
78
The scale factor of enlargement is dened
as the ratio
Linear size of image
, or
Corresponding linear size of object
Distance of an image point from the
centre of enlargement
Distance of a corresponding object point
from the centre of enlargement
Note: This particular scale factor is often
called the linear scale factor because it is the
ratio of the lengths of two line segments.
Construction of objects and images
Example 8.4
Draw any triangle XYZ. Taking a point
O outside the triangle as the centre of
enlargement, and with a scale factor 2,
construct the image triangle X′Y′Z′.
Solution
The following is the procedure of constructing
the image.
(a) Draw triangle XYZ and choose a point O
outside the triangle.
(b) Draw construction lines OX, OY and OZ,
and produce them (Fig. 8.13).
(c) Measure OX, OY and OZ. Calculate the
corresponding lengths OX′, OY′ and OZ′;
and mark off the image points.
From the denition of scale factor, it follows
that image distance = k × object distance,
where k is the scale factor. Thus,
OX′ = 3OX, OY′ = 3OY and OZ′ = 3OZ.
(d) Join points X′, Y′, Z′ to obtain the image
triangle.
Fig. 8.13
Under enlargement with a scale factor
greater than 1:
1. the object and the image are on the same
side of the centre of enlargement,
2. the image is larger than the object, and
3. any line on the image is parallel to the
corresponding line on the object.
Locating the centre of enlargement and
nding the scale factor
Example 8.5
Fig. 8.14 shows a quadrilateral and its image
under a certain enlargement.
Fig. 8.14
(a) Locate the centre of enlargement.
(b) Find the scale factor of enlargement.
(c) Given that a line measures 5 cm, nd
the length of its image under the same
enlargement.
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – ––
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
Z
Y
X
Y′
Z′
X′
O
D
C
B
A
A′
B′
C′
D′
79
Solution
(a) To locate the centre of enlargement O,
join A to A′, B to B′, C to C′ and D to D′
and produce the lines (Fig. 8.15). The
point where the lines meet is the centre of
enlargement.
Fig. 8.15
(b) Measure the distances OA, OA’; OB, OB’;
etc. and calculate the ratios , , etc.
The scale factor of enlargement is 2.
(c) Length of image
= length of object × scale factor
= 5 × 2 = 10 cm.
Note:
1. Given an object and its image, it is sufcient
to construct only two lines in order to
obtain the centre of enlargement.
2. An enlargement is fully described if the
centre and scale factor of enlargement
are given.
Exercise 8.3
1. Copy each of the shapes in Fig. 8.16. Using
the centre of enlargement indicated as X,
enlarge each of the shapes with a scale
factor of 3.
(a) (b)
(c) (d)
Fig. 8.16
2. In Fig. 8.17, PQRS is an enlargement of
PTUV.
(a) Which point is the centre of
enlargement?
(b) By measurement, nd the scale factor
of enlargement.
Fig. 8.17
3. In Fig. 8.18, ∆OA′B′ is the enlargement of
∆OAB.
(a) Which point is the centre of
enlargement?
(b) Find the scale factor of enlargement.
(c) Calculate the lengths A′B′ and AA′.
Fig. 8.18
T
U
P
V
S
Q
R
•
O
•
X
•
X
•
X
•
X
O is the centre
of circle
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
D
C
B
A
O
A′
B′
C′
D′
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
B′
B
A′
A
O
4 cm
5 cm
4 cm
3 cm
OA′
OA
OB′
OB
,
,
80
1
4
–
4. The vertices of an object and its image after
an enlargement are A(
–
1 , 2), B(1 , 4),
C(2 , 2) and A′(
–
1 ,
–
2), B′(5 , 4), C′(8 ,
–
2)
respectively.
Draw these shapes on squared paper.
Hence, nd the centre and scale factor of
enlargement.
5. On squared paper, copy points A, B, C, D
and F as they are in Fig. 8.19
Given that ∆DEF is an enlargement of
∆ABC, nd the coordinates of E.
What is the centre of enlargement?
Fig. 8.19
Fractional scale factor
If in Fig. 8.11, (b) is as the object, (a)
would be the image under enlargement. In this
case, the image is smaller than the object, so
that the transformation is actually a diminuation.
The scale factor here is a fraction since the size
of the image is smaller than the size of the object.
Example 8.6
Construct any triangle ABC. Choose a point O
outside, and a bit far from the triangle. With O
as the centre of enlargement, and with a scale
factor , construct the image triangle A′B′C′.
Solution
The procedure is the same as that of Example
8.4.
The only difference is that we multiply the
object distance by (a fraction) so as to get the
image distance,
i.e. Image distance = × Object distance.
The construction is shown in Fig. 8.20.
Fig. 8.20
Under enlargement with a scale factor that
is a proper fraction:
1. the object and the image are on the same
side of the centre of enlargement.
2. the image is smaller than the object and
lies between the object and the centre of
enlargement.
3. any line on the image is parallel to the
corresponding line on the object.
Exercise 8.4
1. Copy Fig. 8.21 and locate the image of the
ag under enlargement with centre O and
scale factor (a) (b) 2.
Fig. 8.21
1
2
–
1
2
–
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
A′
A
B′
B
C′
C
O
1
2
–
x
y
5
4
3
2
1
0
6
1 2 3 4 5 6
×
×
×
×
×
A
B
C
D
F
•
O
81
2. A′(1 , 4), B′(
–
1 , 2), C′(2 , 2) are the
vertices of the image of a triangle with
vertices A(1 , 2), B(
–
5 ,
–
4), C(4 ,
–
4) under
a certain transformation.
Draw the two triangles on squared paper
and fully describe the transformation.
3. In Fig. 8.22, P′Q′ is an enlargement of PQ
with centre O.
Fig. 8.22
(a) Find the scale factor of enlargement.
(b) Calculate the length of P′Q′.
4. Under enlargement of a polygon, what
happens to the following if the scale factor
is (i) 3, (ii) ?
(a) The lengths of corresponding sides
(b) Corresponding angles
(c) Shape
(d) Direction of lines
Negative scale factor
Consider Fig. 8.23, in which the centre of
enlargement is O and both images of ag
ABCD are 1.5 times as large as the object.
Fig. 8.23
A′B′C′D′ is the image of ABCD under
enlargement with centre O and scale factor 1.5.
A′′B′′C′′D′′ is also an image of ABCD under
enlargement with centre O. How is it different
from A′B′C′D′?
OA = 1.2 cm and OA′′ = 1.8 cm. OA and OA′′
are on opposite sides of O. If we mark A′′OA
as a number line with O as zero and A as 1.2,
then A′′ is
–
1.8 (Fig. 8.24).
Fig. 8.24
We say that the scale factor is
–
1.5 because
–
1.5 × 1.2 =
–
1.8.
Therefore, OA′′ =
–
1.5 OA.
Under enlargement with a negative scale
factor:
1. the object and the image are on opposite
sides of the centre of enlargement,
2. the object is larger or smaller than the
object depending on whether the scale
factor is greater than 1 and negative or a
negative proper fraction,
3. any line on the image is parallel to the
corresponding line on the object, but the
image is inverted relative to the object.
Enlargement in the Cartesian plane
Activity 8.1
On squared paper, draw a quadrilateral with
vertices A(0 , 3), B(2 , 3), C(3 , 1) and D(3 ,
–
2).
Copy and complete Table 8.1 for the coordinates
of A′B′C′D′ and P′, the images of points A,
B, C, D and a general point P(a , b) under
enlargement with centre the origin and the
given scale factors.
P′
P
Q′
Q
O
2 cm
3 cm
5 cm
1
3
–
–
1.8
A′′ O A
1.2
0
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
B′
B
O
D′′
B′′
D′
D
C′
C
A′
A′′
A
C′′
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
82
2. If the scale factor of enlargement if
1 or
–
1, the object and the image have the
same size.
Exercise 8.5
1. Make copies of the shapes in Fig. 8.25.
Using the points marked X as centres,
enlarge the shapes with scale factor
(a)
–
2 (b)
–
(c) .
(a) (b)
(c)
Fig. 8.25
2. In Fig. 8.26, rectangle PQRS is an
enlargement of rectangle ABCD with centre
O.
Fig. 8.26
(a) Find the scale factor of enlargement.
(b) Which point is the image of point
(i) A; (ii) B; (iii) C?
(c) Find the length of the diagonal of
rectangle PQRS given that the length of
rectangle ABCD is 15 cm.
•
•
•
B
A
C
X
X
A
X
C
B
1
2
–
1
2
–
1
2
–
1
2
–
Scale factor A′ B′ C′ D′ P′
2 (0 , 6) (2a , 2b)
(1.5 ,
–
1)
–
2 (
–
6 ,
–
2)
–
(
–
1 ,
–
1.5)
Table 8.1
An enlargement with centre (0 , 0) and scale
factor k maps a point P(a , b) onto P′(ka , kb).
Activity 8.2
Repeat Activity 8.1, but use (2 , 1) as the
centre of enlargement.
Copy and complete Table 8.2 for the images
of A, B, C and D.
Scale factor A′ B′ C′ D′
2 (
–
2 , 5)
–
2
–
Table 8.2
What is the image of a point P(a , b) under
enlargement with centre (2 , 1) and each of the
given scale factors?
You should notice that it is not possible to
generalise for P (a , b) when the centre is not
the origin. When the centre of enlargement
is not the origin, we must carry out complete
construction in order to obtain the required
image.
Notice that:
1. In enlargement, an object, the centre
and the corresponding image point are
collinear.
1
3
–
1
3
–
O
BA
C
D
Q
S
P
R
12 cm
8 cm
83
(d) If rectangle ABCD is the image, what is
the scale factor of enlargement?
3. Points A(
–
2 ,
–
1), B(1 ,
–
1), C(1 ,
–
4)
and D(
–
2 ,
–
4) are vertices of a square.
Without drawing, state the coordinates
of the vertices of the image square under
enlargement with centre origin and scale
factor
(a)
–
4 (b)
–
2 (c)
–
1 (d)
–
4. In a scale model of a building, a door which
is actually 2 m high is represented as having
a height of 2 cm.
(a) What is the scale of this model?
(b) Calculate the actual length of a wall
which is represented as being 5.2 cm
long.
5. When fully inated, two balls have radii of
10.5 cm and 14 cm respectively. They are
deated such that their diameters decrease
in the same ratio. Calculate the diameter
of the smaller ball when the radius of the
larger ball is 10 cm.
6. A tree is 6 m high. When photographing it,
a camera forms an inverted image 1.5 cm
high on the lm. The lm is then processed
and printed to form a picture in which
the tree is 6 cm high. Calculate the scale
factors for the two separate stages.
Area scale factor
Activity 8.3
Draw a rectangle. Choose a point O anywhere
outside the triangle. With O as the centre,
enlarge the rectangle with scale factor 2.
What are the dimensions of the image rectangle?
Calculate the areas of the two rectangles and
the ratio
Area of image rectangle
Area of object rectangle
1
4
–
1
2
–
3
2
–
3
2
–
9
4
–
9
4
–
This ratio is called the area scale factor.
How does this ratio compare with the linear
scale factor?
Repeat your enlargement but with scale factors
3,
–
2 and . How do the new area scale factors
compare with the corresponding linear scale
factors?
Hence,
Area scale factor = (linear scale factor)
2
.
Example 8.7
The ratio of the corresponding sides of two
similar triangles is . If the area of the smaller
triangle is 6 cm
2
, nd the area of the larger
triangle.
Solution
Since the two triangles are similar, the larger
one is an enlargement of the smaller one.
∴ linear scale factor = ratio of corresponding
sides
=
Hence, area scale factor =
(
)
2
=
Thus, area of larger triangle
= area of smaller triangle × area scale factor
= 6 × cm
2
= 13.5 cm
2
Exercise 8.6
1. A rectangle whose area is 18 cm
2
is
enlarged with linear scale factor 3. What is
the area of the image rectangle?
2. A pair of corresponding sides of two similar
triangles are 5 cm and 8 cm long.
(a) What is the area scale factor?
(b) If the larger triangle has an area of
256 cm
2
, what is the area of the smaller
triangle?
3
2
–
84
1
3
–
3. The ratio of the areas of two circles is 16 : 25.
(a) What is the ratio of their radii?
(b) If the smaller circle has a diameter of
28 cm, nd the diameter of the larger
circle.
4. Two photographs are printed from the same
negative. The area of one is 36 times that
of the other. If the smaller photograph
measures 2.5 cm by 2 cm, what are the
dimensions of the larger one?
5. A lady found that a carpet with an area of
13.5 m
2
tted exactly on the oor of a room
4.5 m long. If she moved the carpet to a
similar room which is 1.5 m longer, how
much oor area remained uncovered?
6. An architect made a model of a building
to a scale of 1 cm : 2 m. A oor of the
building is represented on the model with
an area of 12 cm
2
. What would be the
corresponding area on another model of the
same building whose scale is 2 cm : 1 m?
Volume scale factor
Fig. 8.27 shows a cuboid and its enlargement
with scale factor 2. What are the dimensions of
the image? Calculate the volumes of the two
cuboids and the ratio
Volume of image cuboid
.
Volume of object cuboid
Fig. 8.27
This ratio is called the volume scale factor.
How does this ratio compare with the linear
scale factor?
If the scale factor of enlargement is 3, what are
the dimesions and volume of the image?
How does the volume scale factor compare with
the linear scale factor now?
If the scale factor of enlargement is , what is
the volume scale factor?
Volume scale factor = (linear scale factor)
3
Example 8.8
A solid metal cone with a base radius of 7 cm
has a mass of 1.76 kg. What mass would a
similar cone made of the same metal and with
base radius 10.5 cm have?
Solution
The ratio of radii is 10.5 : 7 = 3 : 2
i.e. linear scale factor =
∴ volume scale factor =
(
)
3
=
Hence, volume of the larger cone
= × volume of smaller cone,
and mass of larger cone = × 1.76 kg
= 5.94 kg
Note: For solids of the same material, the ratio
of their masses is equal to the ratio of their
volumes.
Exercise 8.7
1. The volume scale factor of two similar
solids is 27. What are their linear and area
scale factors?
2. A cylinder with a base radius r has a volume
of 77 cm
3
. What is the volume of a similar
cylinder with base radius:
(a) 2r (b) r?
3
2
–
3
2
–
27
8
––
27
8
––
27
8
––
1
2
–
4 cm
5 cm
8 cm
85
3. Two similar jugs have capacities of 2 litres
and litre. If the height of the larger jug is
30 cm, nd the height of the smaller one.
4. Two similar cones made of the same wood
have masses 4 kg and 0.5 kg respectively.
If the base area of the smaller cone is
38.5 cm
2
, nd the base area of the larger one?
5. A concrete model of a commemoration
monument is 50 cm high and has a mass
30 kg.
(a) What mass will the full size monument
of height 9 m have if it is also made of
concrete?
(b) If the surface area of the model is
1
4
–
5 000 cm
2
, what is the surface area of
the full size monument?
6. Fig. 8.28 is a drawing of a model of a
swimming pool. Find the capacity of the
actual pool, in litres, given that its length is
27 m.
Fig. 8.28
– – – – – – – – –
– – – – – –
45 cm
22 cm
18 cm
15 cm
3 cm
1.5 cm
– – – – – – – – –
– – – – – – – – –
– – – – – – – – – – – – – – –
– – – – – – – – – – – – – – –
86
The Theorem
Given the lengths of two sides of a right-angled
triangle, how would you nd the length of the
third side?
A triangle with sides measuring 3, 4 and 5
units is right-angled. However, this is only one
isolated case and does not provide an answer
to the question above. The following activities
will help us establish the relationship between
the lengths of the sides and hence answer the
question.
Activity 9.1
Consider a oor that is tiled with tiles of the
same size, each is a right-angled isosceles
triangle, as in Fig. 9.1.
Fig. 9.1
1. Make a copy of this figure on a larger
scale.
2. Pick any of the small triangular tiles such as
A and shade the three squares standing on
its sides. The squares on the shorter sides
are equal (each comprising of two triangular
tiles). How many tiles make the square on
the longest side (called the hypotenuse)?
3. Shade a triangle composed of two triangular
tiles such as B. This triangle is similar to
the previous one. Count the number of tiles
that make the squares on its sides.
4. Continue shading triangles of the same
shape but of increasing size and record your
results as in Table 9.1.
No. of tiles
No. of tiles making the square on
making
1
st
short 2
nd
short Longest
the triangle
side side side
a b c
1 2 2 4
2
3
Table 9.1
What is the relationship between the values
a, b and c? Is this fact true if the right-angled
triangle is not isosceles? Verify this by carrying
out Activities 9.2 and 9.3.
Activity 9.2
1. Draw ∆ABC, right-angled at B on a stiff
paper, such that BC is longer than AB.
Construct the three squares on its sides, as
in Fig. 9.2.
Fig. 9.2
A
B
C
A
B
PYTHAGORAS’ THEOREM
9
87
2. Locate the centre P of the square on side BC.
Through P, construct a line perpendicular
to AC and another line parallel to AC, to
subdivide the square into four pieces as in
Fig. 9.3.
Fig. 9.3
3. Cut out the squares on AB and BC. Cut
up the square on BC into the four pieces
indicated. Arrange the pieces to cover
completely the square on AC (Fig. 9.4).
Note that the pieces can be moved into
their new positions without rotating any of
them or turning them over. (This is like a
jigsaw puzzle and is referred to as Perigal’s
dissection).
Fig. 9.4
What can you say about the areas of the
squares on the two shorter sides of the triangle?
Activity 9.3
1. Draw any right-angled triangle ABC, with
sides labelled as in Fig. 9.5
Fig. 9.5
2. Construct a square on each side of the
triangle.
3. Measure the length of the side of each square
and calculate the area of the square.
4. State the relationship between the areas of
the three squares.
From Activity 9.3, the Pythagorean
relationship or Pythagoras’ theorem, can be
stated as follows:
In a right-angled triangle, the area of the
square on the hypotenuse is equal to the
sum of the areas of the squares on the
other two sides, i.e.
c
2
= a
2
+ b
2
or a
2
+ b
2
= c
2
Point of interest
Pythagoras’ theorem is named after a great Greek
mathematician, Pythagoras of Samos, who lived
from about 580 BC to 500 BC. Pythagoras travelled
a great deal in Egypt before nally settling down
in a Greek colony in the south of Italy.
Pythagoras was both a philosopher and a
mathematician (a geometer). He was a leader of
a religious brotherhood whose members believed
in the transmigration of souls, i.e. that the spirit
of man or beast moved on after death to another
man or beast. They also believed in strict taboos
for self discipline (e.g. they would not eat beans)
and in the supreme importance of numbers in the
A
B
C
b
c
a
A
B
C
– – – – – – – – – – – –
2
3
4
1
5
2
3
4
1
5
– – – – – – – –
– – – – – – – – – – –
– – – – – – – – – – –
– – – – – – – –
– – – – – – –
– – – – – – – – – –
– – – – – – – – – – – –
A
B
C
– – – – – – – – – – – –
– – – – – – – – – – – –
2
3
4
1
5
P
88
creation of the universe. They sought to interpret
all things through numbers. Pythagoras learned
many facts about mathematics from the Egyptians.
One of the facts which certainly impressed him
was that a triangle of sides 3, 4 and 5 units is right
angled, a fact that was known long before his time.
However, the truth of the relationship a
2
+ b
2
= c
2
,
about any right-angled triangle, was not known.
It was probably in furthering religious ideas that
Pythagoras, or one of his disciples, discovered
the theorem.
Proof of Pythagoras’ theorem
Activity 9.4
Think of a triangle T whose sides are of lengths
a, b and c units, as in Fig. 9.6.
Fig. 9.6
1. Rotate triangle T through 90
0
in a clockwise
direction about the centre O of the square
on the longest side. Triangle T is mapped
onto triangle T′ as in Fig. 9.7.
Fig. 9.7
Note that CBC′ is straight and C′B′ is
at right angles to it.
2. Rotate triangle T
1
′ in a clockwise
direction about O twice through 90° in
each case, to positions T′′ and T′′′ as in
Fig. 9.8.
Note that the gure CC′C′′C′′′ that is
obtained after the three rotations, is a
square of side (a + b) units.
Fig. 9.8
3. Look at this square as follows:
(a) As the central square (of a side c) plus
four equal triangles of sides a and b
units. Thus, the area of the square is
c
2
+ 4 × × a × b
= (c
2
+ 2ab) square units ……… (i)
(b) As a square of side (a + b) units,
whose area is (a + b)
2
= (a
2
+ b
2
+ 2ab) square units ……(ii)
The two expressions, (i) and (ii), are for the
same area, and so,
c
2
+ 2ab = a
2
+ b
2
+ 2ab
i.e. c
2
= a
2
+ b
2
Using Pythagoras’ theorem
Pythagoras’ theorem deals with areas. Its
main use, however, is in calculating lengths. It
also provides us with a test for a right-angled
triangle.
A
B
C
a
c
b
T
1
2
–
A
B
C
a
b
B′
C′
a
a
b
T′′
T′
T′′′
a
b
T
b
C′′′
C′′
c
c
A
B
C
a
c
b
A′
B′
C′
a
c
b
•
O
T′
T
89
A triangle is right-angled whenever the
square of the length of the longest side equals
the sum of the squares of the lengths of the
other two sides.
Example 9.1
Calculate the length of the third side of the
triangle in Fig. 9.9.
Fig. 9.9
Solution
Using Pythagoras’ theorem,
25
2
= a
2
+ 7
2
i.e. 625 = a
2
+ 49
∴a
2
= 625 – 49
⇒a
2
= 576
⇒ a = √576 = 24 cm
i.e. the length of the third side is 24 cm.
Example 9.2
Find the length of AB in Fig 9.10.
Fig. 9.10
Solution
∆BCD is a right-angled triangle
∴ BD
2
= 8
2
+ 6
2
(by Pythagoras’ theorem)
= 100 cm
2
∆ABD is right-angled
∴26
2
= AB
2
+ BD
2
(Pythagoras’
theorem)
⇒ 26
2
= x
2
+ 100
∴x
2
= 676 – 100
⇒x
2
= 576
⇒ x = √576 = 24
∴AB = 24 cm.
Example 9.3
Find out whether a triangle with sides 11, 15
and 18 cm is right-angled.
Solution
The two shorter sides are 11 cm and 15 cm in
length. The sum of the squares of their lengths is
11
2
+ 15
2
= 121 +225
= 346
The square of the length of the hypotenuse is
18
2
= 324
Now 11
2
+ 15
2
≠ 18
2
∴ the triangle is not right-angled.
In solid geometry, it is often necessary to look
for right-angled triangles in different planes,
for example, Fig. 9.11 represents a rectangular
room.
Fig. 9.11
AB and BC are sides of the oor and CG is the
line in which two walls meet.
∆ABC is right-angled at B because AB and BC
are sides of a rectangle.
∆’s BCG and ACG are each right-angled at C
because CG is perpendicular to BC and AC.
7 cm
a cm
25 cm
8 cm
6 cm
26 cm
x cm
A
B
C
D
– – – – – – – – – – – – – – – – – – – – – – – – – – – –– – – – –
– – – – – – – –
– – – – – – –
– – – – – – – – – – – – – – – – – – – –– – – –– – – – – – –
F
D
E
G
C
B
A
H
– – – – – – – – – – – – – – – – – – – – – – –
Example 9.4
If Fig. 9.11 represents a rectangular hall
measuring 17 m long, 14 m wide and 9 m high,
nd the distance from a corner A of the oor to
the opposite corner G of the ceiling.
Solution
By Pythagoras theorem,
AC
2
= AB
2
+ BC
2
= 17
2
+ 14
2
90
(a) If b = 6 and c = 8, nd a.
(b) If b = 8 and c = 15, nd a.
(c) If b = 9 and a = 15, nd c.
(d) If a = 50 and c = 48, nd b.
3. In triangle ABC, AB = 3 cm, BC = 5 cm
and ∠ABC = 90°. Find AC.
4. In ∆LMN, LM = 4 cm, LN = 6 cm and
∠LMN = 90°. Find MN.
5. Which of the following measurements
would give a right-angled triangle?
(a) 6 cm by 8 cm by 10 cm
(b) 5 cm by 12 cm by 13 cm
(c) 4 cm by 16 cm by 17 cm
(d) 7 cm by 10 cm by 12 cm
(e) 9 cm by 30 cm by 35 cm
(f) 12 cm by 35 cm by 37 cm
(g) 12 m by 60 m by 61 m
(h) 21 m by 90 m by 101 m
(i) 20 m by 21 m by 28 m
(j) 28 m by 45 m by 53 m
(k) 27 m by 35 m by 50 m
(l) 33 m by 44 m by 55 m
(m) 4 m by 7 m by 8 m
(n) 14 m by 48 m by 50 m
(o) 2.7 m by 36.4 m by 36.5 m
(p) 2.9 m by 42 m by 42.1 m
6. Find the lengths marked by letters in
Fig. 9.14. All the measurements are in
centimetres.
!
b
B
c
a
A
C
C
B
A
1
2
–
1
2
–
1
2
–
1
2
–
1.8
8.2
c
1.6
3
a
9.9
2
b
From ∆ACG,
AG
2
= AC
2
+ CG
2
= 17
2
+ 14
2
+ 9
2
= 289 + 196 + 81
= 566
∴ AG = √566
= 23.79 (4 s.f. from tables)
i.e. AG is 23.8 m (3 s.f.)
Caution: It is not necessary to evaluate the
value of AC in this calculation. Such evaluation
may lead to inaccuracies due to approximating
too soon. Even if there was no inaccuracy
involved (as in the case of BD in Example 9.2),
it would be an unnecessary step.
Exercise 9.1
1. The triangle in Fig. 9.12 is right-angled
shown. A, B, and C are the areas of the
squares. Find the third area in each of the
following cases.
(a) A = 28 cm
2
, B = 17 cm
2
(b) B = 167 cm
2
, C = 225 cm
2
(c) A = 4.53 m
2
, C = 6.89 m
2
Fig. 9.12
2. Fig. 9.13 shows a right-angled triangle,
with all measurements in centimetres.
Fig. 9.13
91
that can be put into a rectangular box whose
internal measurements are 0.6 m by 0.5 m
by 0.4 m.
Pythagorean triples
Revisit your answers to Question 5 of Exercise
9.1. You notice that some of the values, like
(6 , 8 , 10), give right-angled triangles while
values such as (4 , 16 , 17) do not. Such sets
of values which give right-angled triangles are
known as Pythagorean triples or Pythagorean
numbers.
A Pythagorean triple (a, b, c,) is a group
of three numbers which give the respective
lengths of the sides and the hypotenuse of
a right-angled triangle and are related such
that a
2
+ b
2
= c
2
The group (3, 4, 5) is the most famous and most
commonly used Pythagorean triple. The triple
was known even before the time of Pythagoras.
It was, and is still, used for setting out the base
lines on tennis courts and other sports pitches.
There are many other Pythagorean triples.
Now complete the following patterns to
discover some more.
(a) (3 , 4 , 5) 3
2
= 4 + 5
(5 , 12 , 13) 5
2
= 12 + 13
(7 , 24 , 25) 7
2
= 24 + 25
(9 , 40 , 41) 9
2
= 40 + 41
(11 , … , …) 11
2
=
(13 , … , …) 13
2
=
(15 , … , …) 15
2
=
Fig. 9.14
7. In Fig. 9.15, all the measurements are in
metres. Find the lengths marked by letters.
Fig. 9.15
8. The sides of a rectangle are 7.8 and 6.4 cm
long. Find the length of the diagonal of the
rectangle.
9. The length of the diagonal of a rectangle is
23.7 cm and the length of one side is
18.8 cm. Find its perimeter.
10. The perimeter of a rhombus is 20 cm and
the length of its shorter diagonal is 6 cm.
Find the length of the longer diagonal.
11. A box is 15 cm long, 13 cm wide and 8
cm high. Find the distance from the corner
of the bottom face to the opposite corner of
the top face of the box.
12. Find the length of the longest straight rod,
measuring a whole number of centimetres,
b
5
9
13
12
23
7
8
c
– – – – – –
– – – – – – – – – – – – – – – – – –
F
D
E
G
C
B
A
H
d
6
4
12
– – – – –
– – – – – – – – – – – – – – – – – – – –
e
31.3
31.2
42.1
42
d
2.7
36.5
f
7
a
4
4
(b) (6 , 8 , 10) of 6
2
= 8 + 10
(8 , 15 , 17) of 8
2
= 15 + 17
(10 , 24 , 26) of 10
2
= 24 + 26
(12 , 35 , 37) of 12
2
= 35 + 37
(14 , … , …) of 14
2
=
(16 , … , …) of 16
2
=
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
92
Fig. 9.16
By Pythagoras’ theorem,
h
2
+ 1.2
2
= 3.9
2
⇒ h
2
= 3.9
2
– 1.2
2
∴ h = 3.9
2
– 1.2
2
h = 3.71 m (2 d.p.)
Thus the ladder reaches 3.71 m up the wall.
h
3.9 m
1.2 m
These two patterns provide ways of nding
Pythagorean triples. But they do not seem to
give all the possible sets of such triples. Is
there a general way of nding Pythagorean
triples?
Now consider the following.
(a) 2
2
– 1
2
, 2 × 2 × 1, 2
2
+ 1
2
(b) 3
2
– 1
2
, 2 × 3 × 1, 3
2
+ 1
2
(c) 4
2
– 2
2
, 2 × 4 × 2, 4
2
+ 2
2
(d) 5
2
– 3
2
, 2 × 5 × 3, 5
2
+ 3
2
Are these all Pythagorean triples?
It is a fact that:
Given any two positive integers m and
n, where m > n, we always obtain the
Pythagorean triple:
(m
2
– n
2
, 2mn, m
2
+ n
2
)
Exercise 9.2
1. Given (5, 12, 13) as a Pythagorean triple;
(a) write down four multiples of it.
(b) are all the four multiples in (a)
Pythagorean triples?
(c) Using a multiplier n and any
Pythagorean triple (a, b, c,) state the
general result for such multiples as
in (a).
2. Find out if the following are Pythagorean
triples.
(a) (7 , 24 , 25) (b) (8 , 15 , 17)
(c) (15 , 22 , 27) (d) (28 , 43 , 53)
(e) (11 , 60 , 61) (f) (20 , 21 , 29)
3. The following are the dimensions of two
triangles. Which one of them is a right-
angled triangle?
(a) 15 cm, 30 cm, 35 cm
(b) 33 cm, 56 cm, 65 cm.
4. Use the following numbers to generate
Pythagorean triples.
(a) 1 and 4 (b) 1 and 5 (c) 6 and 2
(d) 3 and 8
Application to real-life situations
Notice that, Pythagoras’ theorem connects the
areas of actual squares. Its main use, however,
is in calculating lengths without having to draw
any squares. The theorem also acts as a test for
right-angled triangles.
There are many real-life situations which
require the use of Pythagoras’ theorem.
Example 9.5
A ladder, 3.9 m long, leans against a wall. If
its foot is 1.2 m from the wall, how high up the
wall does it reach?
Solution
Fig. 9.16 is an illustration of the situation.
Note that the ground must be assumed to be
horizontal and level and hence at right angles
to the wall.
93
If the path is 48 m long, nd the distance
that the people save?
ILLUSTRATION
ome)
Fig. 9.19
7. A hall is 16 m long, 14 m wide and 9 m
high. Find
(a) The length of the diagonal of the oor.
(b) The distance from a corner of the oor
to the opposite corner of the ceiling.
8. Find the length of the diagonal of a cuboid
which measures 6 cm by 3 cm by 2 cm.
9. Fig. 9.20 represents a roof truss which is
symmetrical about QS. Beam PQ is 5 m
long, strut TS 2.4 m long and the distance
TQ is 1.8 m.
(a) Find the height QS.
(b) Hence, nd the span PR of the roof.
Fig. 9.20
16 m
A
B
C
12 m
P
S
T
Q
R
!
•
•
•
•
•
•
•
••• •
!
!
•
•
•
•
Fig. 9.18
Road
Road
path
45°
Exercise 9.3
1. Fig. 9.17 shows a television antenna. Find the
length of the wire AB holding the antenna.
Fig. 9.17
2. A ladder reaches the top of a wall of height
6 m when the end on the ground is 2.5 m
from the wall. What is the length of the
ladder?
3. The length of a diagonal of a rectangular
ower bed is 24.6 m and the length of one
side is 18.9 m. Find the perimeter and area
of the ower bed.
4. A piece of rope with 12 knots that are
equally spaced has been laid out and pinned
down on the ground as in Fig. 9.18.
(a) What can you say about the triangle
whose corners the stakes mark?
(b) Does it matter how great the distance
between the knots is?
5. The chalkboard in your classroom is
rectangular and it measures 2.2 m by 1.2 m.
What is the length of the longest straight
line that can be drawn on it?
6. Fig. 9.19 shows a road that turns through
a right angle to go round a rectangular
recreational garden in your town. To save
time, people on foot cut off the corner, thus
making a path that meets the road at 45°.
94
(e) (f)
Fig. 10.2
2. Name the adjacent side, the opposite side
and the hypotenuse relative to each of the
marked angles in Fig. 10.3.
Fig. 10.3
3. Using Fig. 10.4, in which all the lengths are
in centimetres, nd the length of the side
which is:
(a) opposite to ∠BAC
(b) adjacent to ∠ADC.
Introduction
The word ‘trigonometry’ is derived from two
Greek words: trigonon, meaning a triangle,
and metron, meaning measurement. Thus
trigonometry is the branch of Mathematics
concerned with the relationships between the
sides and angles of triangles.
In trigonometry, Greek letters are used to
indicate, in a general way, the sizes of various
angles. The most commonly used of these
letters are:
α
= alpha;
β
= beta; γ = gamma;
δ
= delta;
θ
= theta;
φ
= phi;
ω
= omega.
In Chapter 9, we studied the relationship
between the lengths of the sides of a right-angled
triangle, which is known as the Pythagorean
relationship. In this chapter, we shall study the
relationships between the acute angles and the
sides of a right-angled triangle. The sides of
such a triangle are named, with reference to the
angle marked
θ
, as shown in Fig. 10.1.
Fig. 10.1
If C is used as the reference angle AB becomes
the opposite side and BC the adjacent side.
These denitions of the sides apply to any right-
angled triangle in any position.
Exercise 10.1
1. Name the adjacent side, the opposite side
and the hypotenuse in each of the triangles
in Fig. 10.2, using the angle marked
θ
as the
angle of reference.
P
Q
S
R
γ
β
φ
θ
α
(a) (b)
(c) (d)
C
B
C
A
θ
A
B
θ
B
C
A
θ
B
C
A
θ
Adjacent side
Hypotenuse
Opposite side
C
B
θ
A
C
B
A
θ
B
A
C
θ
TRIGONOMETRY
10
95
Measure A
1
B
1
, A
2
B
2
, A
3
B
3
, A
4
B
4
and A
5
B
5
.
Copy and complete Table 10.1, where A
i
B
i
means A
1
B
1
, A
2
B
2
, etc.
i 1 2 3 4 5
A
i
B
i
OB
i
1 cm 2 cm 3 cm 4 cm 5 cm
A
i
B
i
OB
i
Table 10.1
What do you notice about the ratio
A
i
B
i
?
OB
i
Notice that the value of the ratio is roughly the
same, i.e. 0.7.
The ratio
A
i
B
i
is called the tangent of ∠AOB.
OB
i
The tangent of an angle,
θ
, is abbreviated as
tan
θ
.
Thus tan 35° = 0.7.
In general, given a right-angled ∆ABC
(Fig. 10.7),
Fig. 10.7
The ratio
AC
is called the tangent of ∠B.
BC
Similarly,
BC
is the tangent of ∠A.
AC
In short, we write
tan B =
AC
and tan A =
BC .
BC
AC
i.e. tan
θ
=
Opposite side
, where
θ
is one
Adjacent side
of the acute angles in a right-angled triangle.
7
B
D
C
A
5
3
C
A
B
Fig. 10.4
4. Using Fig. 10.5, in which all lengths are in
centimetres, name and give the lengths of
two sides which are:
(a) adjacent to angle
θ
,
(b) hypotenuse relative to angle
α
.
Fig. 10.5
Tangent of an acute angle
Activity 10.1
Draw ∠AOB = 35°.
On OB, mark points B
1
, B
2
, B
3
, B
4
, and B
5
such that OB
1
= B
1
B
2
= B
2
B
3
= B
3
B
4
= B
4
B
5
=
1 cm.
Construct perpendicular lines through B
1
, B
2
,
B
3
, B
4
and B
5
to cut OA at A
1
, A
2
, A
3
, A
4
, and
A
5
respectively (Fig. 10.6).
Fig. 10.6
A
5
A
B
1
A
4
A
3
A
2
A
1
B
2
B
3
B
4
B
5
B
35°
O
D
9
A
θ
α
B
12
20
C
96
Example 10.1
By drawing and measuring, nd the value of tan
30°.
Solution
Draw an angle AOB = 30°.
On OB, mark off OP = 5 cm.
At P draw a perpendicular line to cut OA at Q
(Fig. 10.8).
Measure PQ.
Tan 30° =
PQ
=
2.9
OP 5
=
2.9
5
= 0.58
Fig. 10.8
Example 10.2
By drawing and measuring, nd the angle
whose tangent is .
Solution
Let the angle be
θ
.
Tan
θ
=
Opposite
=
2 .
Adjacent 3
The lengths of the opposite and adjacent sides
are in the ratio 2 : 3.
So we can use 2 and 3 cm.
Draw AB = 3. At B, draw BC = 2 cm
perpendicular to AB. Join AC as in gure 10.9)
Measure ∠CAB.
By measurement, ∠CAB = 34°.
Fig 10.9
Degrees and minutes
Angles are usually measured to the nearest
degree. However, in calculations, it is possible
to have fractions of a degree.
One degree is divided into 60 equal parts
called minutes. Thus there are 60 minutes
in 1 degree. In symbols 1° = 60′.
Example 10.3
(a) Convert 30.5° to degrees and minutes.
(b) Express 15° 25′ in degrees only.
(c) Find the value of 15° 32′ + 18° 40′.
(d) Find the value of 48° 24′ – 25° 35′.
Solution
(a) 30.5° = 30° + 0.5°
= 30° + 0.5 × 60′
= 30° 30′
(b) 15° 25′ = 15° +
(
)
°
= 15° + 0.42°
= 15.42°
(c) 15° 32′ + 18° 40′
= 15° + 18° + 32′ + 40′
= 33° + 60′ + 12′
= 33° + 1° + 12′ (1° = 60′)
= 34° 12′ or 34.2°
(d) 48° 24′ – 25° 35′
= 48° – 25° + 24′ – 35′
= 23° + 24′ – 35′
= 22° + 60′ + 24′ – 35′
= 22° 49′ or 22.8°
Exercise 10.2
1. Find the tangents of the following angles by
drawing and measuring.
(a) 40° (b) 45° (c) 50°
(d) 25° (e) 33° (f) 60°
O
Q
P B
A
30°
2
3
–
25
60
––
A
B
C
θ
97
3. Convert the following into minutes.
(a) 1 ° (b) 2 ° (c) 6°
(d) 9° (e) 45° (f) 0.1°
(g) 0.3° (h) 0.4° (i) 0.6°
4. Convert the following into degrees.
(a) 75′ (b) 100′ (c) 300′
(d) 540′ (e) 6′ (f) 26′
(g) 41′ (h) 56′ (i) 557′
5. Express the following in degrees only.
(a) 28° 32′ (b) 36° 36′ (c) 40° 42′
(d) 67° 52′ (e) 80° 19′ (f) 75° 54′
6. Find the value of each of the following.
(a) 28° 32′ + 67° 62′ (b) 36° 36′ + 40° 42′
(c) 68° 42′ – 35° 34′ (d) 82° 16′ – 39° 57′
Table of tangents
Tables of natural tangents have been specially
prepared for use in nding tangents of angles
from 0
0
to 90
0
.
These tables are read in the same way as tables
of logarithms.
Example 10.4
Use tables to nd the tangents of the following.
(a) 67° (b) 60.55° (c) 38° 53′.
Solution
From the tables.
(a) Tan 67° = 2.355 9 (Row 67, main columns 0
(zero))
(b) To nd tan of 60.55°:
Row 60, main columns 5 give 1.767 5 +
Differences column 5 gives 3 6
1.771 1
∴ tan 60.55° = 1.771 1
(c) 38° 53′ = 38° +
(
)
° = 38° + 0.88°
= 38.88°
tan 38.8° = 0.804 0 (Row 38, main
columns 8)
Differences column 8 gives 23
∴ tan 38.88° = 0.804 0 + 0.002 3
= 0.806 3
Note: If your tables use degrees and minutes,
then get tan 38° 53′ as follows:
Row 38, main columns 48′ give 0.804 0 +
Differences column 5′ gives 2 3
0.806 3
∴ tan 38° 53′ = 0.806 3
Example 10.5
Use tables of tangents to nd the angle whose
tangent is:
(a) 0.500 8 (b) 2
Solution
(a) Let the angle be P.
Then, tan P = 0.500 8.
Locate .500 8 in the body of the table of
tangents.
.500 8 is in row 26 under the main column 0.6
∴ P = 26.6
0
(b) Let the angle be R.
Then, tan R = 2.
But 2 is between 1.997 0 and 2.005 7
(Read
from tables).
∴ R is between 63.4° and 63.5
0
.
0.003 must be added to 1.997 0 to get 2.
1
2
–
1
2
–
53
60
––
1
4
–
2
5
–
3
7
–
5
4
–
6
5
–
7
4
–
8
3
–
2. By drawing and measuring, nd the angles
whose tangents are as follows:
(a) (b) (c)
(d) (e) (f)
(g) (h) 1 (i) 3
98
5 cm
b
20°
a
18 cm
55°
From table of tangents
From table of reciprocals
b
10 cm
50°
a
6 cm
20°
Fig. 10.11.
4. Find the length of the side indicated in each
of the triangles in Fig 10.12. State your
answer in 3 s.f.
Sine and cosine of an acute angle
Activity 10.2
Construct ∆AOB such that ∠AOB = 30°,
∠ABO = 90° and OB = 6 cm.
On line OB mark points B
1
, B
2
, B
3
, B
4
and B
5
such that OB
1
= B
1
B
2
= B
2
B
4
= B
4
B
5
= B
5
B
= 1 cm. Through these points, draw lines
parallel to AB and meeting line OA at A
1
, A
2
,
A
3
, A
4
and A
5
respectively (See Fig. 10.11).
Fig. 10.13
Measure A
i
B
i
(i.e. A
1
B
1
, A
2
B
2
, …).
Measure OA
i
(i.e. OA
1
, OA
2
…).
Fig. 10.12.
So we look for 30 in the differences
columns. It is not there. 30 is nearer to 26
than to 35.
∴ we take 26 and add 0.03° to 63.4°
∴ R = 63.43
0
.
Example 10.6
Find the length indicated
as x in Fig. 10.10
Solution
The two indicated sides are the opposite and the
adjacent sides.
∴ Tan 35˚ =
⇒ x =
= = 4 × = 4 × 1.429
∴ x ≈ 5.7 cm (2 s.f)
Exercise 10.3
1. Use tables to nd the tangents of the
following angles:
(a) 7° (b) 13.6° (c) 50°
(d) 47.7° (e) 80.5° (f) 2.21°
(g) 18.46° (h) 55.68° (i) 66.99°
(j) 51° 45′ (k) 64° 15′ (l) 73° 30′
2. Use tables to nd angles whose tangents are
given as follows:
(a) 0.176 3 (b) 0.869 3 (c) 1.376 4
(d) 0.949 (e) 5.004 5 (f) 0.194
(g) 2.006 1 (h) 9 (i) 10
3. Find the length of the side indicated in each
of the triangles in Fig. 10.11. State your
answer to 3 s.f.
4 cm
35˚
x
Fig. 10.10.
A
5
A
B
1
A
4
A
3
A
2
A
1
B
2
B
3
B
4
B
5
B
O
30°
4
x
4
tan 35°
4
0.7002
1
0.7002
(a)
60°
17.3 cm
c
(c)
(b)
35°
10 cm
d
(d)
(a)
(b)
12 cm
c
75.5°
(c)
d
15.4 cm
38.8°
(d)
99
Exercise 10.4
1. Find, by drawing and measuring,
approximate values for the following.
(a) sin 20°, cos 20° (b) sin 42°, cos 42°
(c) sin 65°, cos 65° (d) sin 78°, cos 78°
2. Find , by drawing and measuring,
approximate sizes of the angles whose sines
and cosines are given below.
(a) sin A = (b) sin B =
(c) sin C = (d) cos D =
(e) cos E = (f) cos F =
(g) sin G = 0.42 (h) sin H = 0.84
(i) sin I = 0.65 (j) cos J = 0.23
(k) cos K = 0.34 (l) cos L = 0.56
Hint for parts (g) to (l): First convert the
decimals into fractions.
Sine and cosine tables.
Sine and cosine tables are read and used in the
same way as tables of tangents.
Note that as the angles increase from 0° to
90
0
:
(i) their sines increase from 0 to 1, and
(ii) their cosines decrease from 1 to 0.
This means that the values in the differences
columns in the sine tables have to be added
while those in the cosine tables have to be
subtracted.
Examples 10.7
Use tables to nd the value of:
(a) sin 53.4° (b) cos 71.2°
Solutions
Reading from tables,
(a) sin 53.4° = 0.802 8 (Row 53, main columns 4).
(b) cos 71.2° = 0.322 3 (Row 71, main columns 2).
2
3
–
3
4
–
3
5
–
1
3
–
3
7
–
5
7
–
θ
A
C
B
~
–
~
–
Copy and complete Table 10.2.
i 1 2 3 4 5
OA
i
(cm)
OB
i
(cm) 1 2 3 4 5
A
i
B
i
(cm)
A
i
B
i
OA
i
OB
i
OA
i
Table 10.2
In each case, what do you notice about the
ratios
A
i
B
i
and
OB
i
?
OA
i
OA
i
In each case
A
i
B
i
= 0.5 and
OB
i
= 0.87, where
OA
i
OA
i
OA
i
is the hypotenuse,
OB
i
is the adjacent side and
A
i
B
i
is the opposite side of the triangle.
The ratio
A
i
B
i
is called the sine of ∠AOB and
OA
i
the ratio
OB
i
is called the cosine of ∠AOB.
OA
i
The sine of angle
θ
is abbreviated as sin
θ
and the
cosine of angle
θ
is abbreviated as cos
θ
.
In general, given a right-angled ∆ABC as
shown in Fig. 10.14.
Sin
θ
=
Opposite side
=
AC , and
Hypotenuse AB
Cos
θ
=
Adjacent side
=
BC
Hypotenuse AB
Fig. 10.14
100
(a) (b)
(c) (d)
(e) (f)
(g) (h)
(a)
(b)
Example 10.8
Find the values of x and y in Fig. 10.15 to 2 s.f.
Solution
5. In Fig. 10.16, nd the lengths marked with
letters. The lengths are in centimetres.
Give your answer correct to 3 s.f.
Fig. 10.16
6. Find the lengths marked with letters
in Fig. 10.17. Measurements are in
centimetres. Give your answers to 3
signicant gures.
Fig. 10.17
30°
10
x
42°
y
28°
15
p
40°
q
33°
10
d
67°
8
c
37°
5
a
42
0
8
b
42.6°
11
h
20°
100
g
70°
20
f
42
0
100
e
y cm
8 cm
x cm
33.2°
Fig. 10.15
Sin 33.2° =
y
8
⇒ y = 8 sin 33.2°.
= 8 × 0.547 6 cm
= 4.380 8 cm
= 4.4 cm (2 s.f.)
Cos 33.2° =
x
8
⇒ x = 8 cos 33.2°
= 8 × 0.836 8 cm
= 6.694 4 cm
= 6.7 cm (2 s.f.)
Exercise 10.5
1. Use tables to nd the sine of:
(a) 3° (b) 13° (c) 70° (d) 63°
(e) 13.2° (f) 47.8° (g) 79.2° (h) 89.2°
2. Use tables to nd the cosine of:
(a) 18° (b) 27° (c) 49° (d) 70°
(e) 19.5° (f) 36.6° (g) 77.7° (h) 83.9°
3. Use tables to nd the angle whose sine is :
(a) 0.397 1 (b) 0.788 0 (c) 0.927 8
(d) 0.996 3 (e) 0.948 9 (f) 0.917 8
4. Use tables to nd the angle whose cosine
(a) 0.918 2 (b) 0.564 1 (c) 0.123 4
(d) 0.432 1 (e) 0.880 1 (f) 0.555 5
101
Sines and cosines of complementary
angles.
What are the values of the following?
(a) sin 25° and cos 65° (b) sin 38° and cos 52°
(c) sin 57° and cos 33°.
What do you notice ?
You should have noticed that:
sin 25° = cos 65°, sin 38° = cos 52° and
sin 57° = cos 33°
Recall that angles which add up to 90° are
called complementary angles and note that
in each of these cases, the two angles are
complementary .
Thus, for any two complementary angles,
a and b,
(i) sin a
= cos b, and
(ii) cos a = sin b
i.e. for any angle
θ
,
cos
θ
= sin(90° –
θ
) and
sin
θ
= cos (90° –
θ
)
Example 10.9
Solve the equations.
(a) sin x = cos 37° (b) cos y = sin 72°
(c) cos x = sin 5x
Solution
(a) sin x = cos 37°
∴ x + 37° = 90°
x = 90° – 37°
x = 53°
(b) cos y = sin 72°
∴ y + 72° = 90°
y = 90° – 72°
j = 18°
(c) cos x = sin 5x
x + 5x = 90°
6x = 90°
x = 15°
a
c
–
b
c
–
a
b
–
sin
θ
cos
θ
––––
a
c
–
b
c
–
—
a
c
–
c
b
–
a
b
–
sin
θ
cos
θ
–––––
θ
a
B
A
C
b
c
Exercise 10.6
Find the value of the unknown in each of the
following.
1. (a) sin w = cos 35° (b) sin 2x = cos 45°
2. (a) sin p = cos 2p (b) cos y = sin y
3. (a) cos 3y = sin 2y (b) cos 4y = sin 5y
4. (a) sin 3x = cos (x + 20°)
(b) sin (2x + 40°) = cos (3x + 20°)
5. (a) cos (3x – 30°) = sin (7x + 50°)
(b) sin y
2
= cos 26°
Relationship between sine, cosine and
tangent
Consider a right-angled ∆ABC with sides a, b
and c, as in Fig. 10.18.
Fig. 10.18
By denition,
sin
θ
= , cos
θ
= and tan
θ
= .
Now, = = × = = tan
θ
Therefore, for any angle θ, tan θ =
Thus given any two of the three trigonometric
ratios (sine, cosine or tangent), we can always
be able to nd the third one. It is also possible
to nd the other two ratios when given only
one.
102
2. Given the following trigonometric ratios,
nd the other two in each case.
(a) cos
θ
= (b) sin
θ
=
(c) sin
θ
= (d) tan
θ
=
(e) tan
θ
= (f) cos
θ
=
Trigonometrical ratios of special
angles (0
°
, 30
°
, 45
°
, 60
°
, 90
°
)
The trigonometric ratios of 30° and 60° can
be found using any equilateral triangle. Those
of 45
°
can be found using any isosceles right-
angled triangle.
Tangent, sine and cosine of 45
°
Fig. 10.20 shows ∆ABC, right-angled at B and
AB = BC = x cm.
Since ∆ABC is isosceles, then
∠ACB = ∠BAC = 45°.
Fig. 10.20
By Pythagoras’ theorem,
AC
2
= AB
2
+ BC
2
= x
2
+ x
2
= 2x
2
AC = 2x
2
= 2 x
2
= x 2.
tan 45° = = = 1,
sin 45° = = = and
cos 45° = = = .
Hence,
tan 45° = 1, sin 45° = , cos 45° = .
3
5
–
3
4
–
—
3
5
–
4
3
–
4
5
–
12
13
––
a
c
–
5
13
––
a
b
–
5
12
––
9
41
––
4
5
–
11
60
––
7
25
––
AB
BC
–––
BC
AC
–––
AB
AC
–––
x
x
–
x
x 2
–––
x
x 2
–––
1
2
––
1
2
––
A
B
C
x cm
x cm
45°
3
4
–
3
5
–
Example 10.10
Given that sin
θ
= and tan
θ
= , nd cos
θ
.
Solution
Tan
θ
=
sin
θ
tan
θ
–––––
Sin
θ
Ccos
θ
–––––
15
8
––
12
37
––
A
C
B
c = 13
a = 12
a
θ
1
2
–
5
12
––
3
5
–
4
5
–
1
2
–
1
2
–
8
15
––
12
13
––
√3
2
––
√3
2
––
√161
8
––
––
1
2
––
1
2
∴ Cos
θ
= = = × = .
Example 10.11
Find sin
θ
and tan
θ
given that cos
θ
= .
Solution
Fig. 10.19 is a sketch of the right-angled
triangle whose cosine is given.
Fig. 10.19
By Pythagoras theorem, c
2
= a
2
+ b
2
∴
a
2
= c
2
– b
2
and a = c
2
– b
2
Thus, a = 13
2
– 12
2
= 169 – 144 = 25
i.e. a = 5
Hence, sin
θ
= = and
tan
θ
= = .
Exercise 10.7
1. Given the following two trigonometric
ratios, nd the third one.
(a) cos
θ
= , sin
θ
=
(b) sin
θ
= , cos
θ
=
(c) tan
θ
= , cos
θ
=
(d) cos
θ
= , tan
θ
=
(e) sin
θ
= , cos
θ
=
(f) tan
θ
= √3, cos
θ
=
103
AD
BD
–––
BD
AB
–––
AD
AB
–––
3
1
––
1
2
–
BD
AD
–––
AD
AB
–––
BD
AB
–––
1
2
–
3
2
––
1
3
––
1
2
–
3
2
––
1
3
––
3
2
––
AD
BD
–––
10
x + 10
–––––
1
3
––
1
3
––
10
x + 10
–––––
C
A
A
1
B
A
2
A
3
Tangent, sine and cosine of 30
°
and 60
°
Fig. 10.21 shows an equilateral ∆ABC with
sides 2 units in length. AD is the perpendicular
bisector of BC.
∴ BD = DC = 1 unit
C
D
B
A
10 cm
45°
30°
x
C
D
B
A
2 units
2 units
2 units
3
2
––
1
2
–
Fig. 10.21
In ∆ABD,
AB
2
= AD
2
+ BD
2
(Pythagoras theorem)
2
2
= AD
2
+ 1
2
i.e. AD
2
= 3
∴ AD = 3
Since ∠B = 60°,
tan 60° = = =
3 ,
sin 60° = = and
cos 60° = = .
Since ∠BAD = 30°,
tan 30° = = ,
sin 30° = = and
cos 30° = = .
Hence,
tan 60° = 3, sin 60° = and
cos 60° = and
tan 30° = , sin 30° = and
cos 30° = .
Note that 30° and 60° are complementary
angles, and so sin 30° = cos 60° and
sin 60° = cos 30°.
Example 10.12
In Fig. 10.22, nd x, leaving your answer with
sign.
Fig. 10.22
Solution
Since ∠ACD = 45° and ∠ADC = 90°, then
∠DAC = 45°.
Since ∆ACD is an isosceles triangle, then
CD = AD = 10 cm.
In ∆ABD, tan 30° = =
Since tan 30° = ,
Then =
⇒x + 10 = 10 3
∴ x = 10 3 – 10
= 10 ( 3 – 1) cm.
Tangent of 0° and 90°
Activity 10.3
Consider the right-angled ∆ABC in which
AC = BC = 1 unit (Fig. 10.23).
Let ∠ABC be
θ.
What is tan
θ
?
Fig. 10.23
Now consider points A
1
, A
2
, A
3
, … on AC such
that A
1
C > A
2
C > A
3
C >…
For each ∆A
i
BC, let
θ
1
= ∠A
i
BC and
θ
2
= ∠BA
i
C.
Note that
θ
1
and
θ
2
are complementary angles,
i.e.
θ
1
+
θ
2
= 90°.
104
0
1
1
0
1
0
a
0
A
1
A
2
A
3
A
4
– – – – – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – –
– – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – –
O
B
4
B
3
B
2
B
1
– – – – – – – – – – – – – – – – – – – – – – – – – –
B
A
Sine and cosine of 0
°
and 90
°
Activity 10.4
Consider Fig. 10.24 in which OA
1
= OA
2
= OA
3
= … = 1 unit, and ∆s A
1
OB
1
, A
2
OB
2
, … are
right-angled.
For each ∆A
i
OB
i
, let
θ
1
= ∠A
i
OB
i
, and
θ
2
= ∠OA
i
B
i
.
Fig. 10.24
Copy and complete Table 10.4.
OB
i
A
i
B
i
sin
θ
i
cos
θ
i
0.1 0.995 0.995 0.1
0.2 0.980 0.980 0.2
0.5 0.866 0.866 0.5
0.8 0.600
0.9 0.436
0.92 0.392
0.94
0.96
0.98
Table 10.4
As OA
i
moves closer and closer to OB, what
happens to ∠ A
i
OB
i
? Do you notice that it
approaches 0˚? The side A
i
B
i
becomes shorter
and shorter, while OB
i
gets longer and closer
to 1 unit.
What happens to:
(a) sin
θ
(b) cos
θ
?
What can you say are the values of:
(a) sin
θ
(b) cos
θ
?
Copy and complete Table 10.3, for the given
values.
A
i
C tan
θ
1
tan
θ
2
A
1
C = 0.8 0.8 1.25
A
2
C = 0.6 0.6 1.67
A
3
C = 0.4 0.4 2.5
A
4
C = 0.1 0.1
A
5
C = 0.01 0.01
A
6
C = 0.00 1
1 000
A
7
C = 0.000 1
A
8
C = 0.000 01
Table 10.3
As A
1
C becomes smaller, what happens to
(a)
θ
1
(b)
θ
2
(c) tan
θ
1
(d) tan
θ
2
?
If A
i
C = 0, what will be the value of
(a)
θ
1
(b)
θ
2
(c) tan
θ
1
(d) tan
θ
2
?
You should have noted that as A
i
C becomes
smaller,
(a)
θ
1
becomes smaller, getting nearer to 0° (zero).
(b)
θ
2
becomes larger, getting nearer to 90°.
(c) tan
θ
1
becomes smaller getting nearer to 0
(zero).
(d) tan
θ
2
becomes a very large number.
When A
i
C = 0, then:
(a)
θ
1
= 0°, (b)
θ
2
= 90°,
(c) tan
θ
1
= tan 0° = = 0, and
(d) tan
θ
2
= tan 90° = , which is undened.
Thus, tan 0° = 0, and tan 90° is undened.
Note that or , where a is any non-zero
number, is an innitely large number, i.e
an undened value, usually denoted by ∞
105
As
θ
, approaches 0˚,
θ
2
approaches 90˚, using
the observation above about sides A
2
B
2
and
OB
i
What can you say are the values of:
(a) sin 90˚? (b) cos 90˚?
Note that:
sin 0˚ = 0 and cos 0˚ = 1; and
sin 90˚ = 1 and cos 90˚ = 0
Thus, sin 0° = cos 90° = 0, and
sin 90° = cos 0° = 1
Exercise 10.8
In this exercise do not use mathematical tables.
Find the lengths marked x in Fig. 10.25 leaving
your answers with
signs. All dimensions
are in centimetres.
1. 2.
Logarithms of tangents, sines and
cosines
So far we have used logarithms in computation
of numbers which are not trigonometric ratios.
In this section we see how to use logarithms of
trigonometric ratios in computations.
Example 10.13
Evaluate: (a) 11.4 sin 32.8° (b)
Solution
(a) Method 1
11.4 sin 32. 8° = 11.4 × 0.541 7
(From tables, sin 32.8° = 0.541 7.)
Now either use long multiplication or use
logarithm tables.
No. Log.
11.4 1.056 9 +
0.541 7 1.733 8
6.175 (6) 0.790 7
∴ 11.4 sin 32.8° = 6.175
Method 2
The logarithm of sin 32.8° is read direct
from the table of logarithms of sines.
No. Log.
11.4 1.056 9 +
sin 32.8° 1.733 8
6.175 (6) 0.790 7
∴ 11.4 sin 32.8° = 6.175 or (6)
Note that Method 2 is less involving and is
shorter.
(b) Using Method 2, evaluate .
No. Log.
12.8 1.107 2 –
tan 42.7° 1.965 1
1.387 × 10
1
1.142 1
∴ = 13.87.
12.8
tan 42.7°
––––––––
(From table
of logarithms
of sines.)
(From table
of logarithms
of tangents)
12.8
tan 42.7°
––––––––
–
–
12.8
tan 42.7°
–––––––––
4
60˚
x
30˚
10.
x
45˚
12
30˚
5.
x
30˚
8
3.
x
45˚
10
4.
x
4
30˚
60˚
10
45˚
8.
x
x
5
60˚
x
60˚
45˚
6.
5
7.
45˚
60˚
4.5
x
60˚
15˚
x
9.
6
Fig. 10.25
106
Note:
1. Since sines and cosines of angles are less
than 1 (except for 0°), all their logarithms
have negative characteristics.
2. Since tangents of angles are either less than,
equal to or greater than 1, the characteristics
of the logarithms of tangents are either
positive, zero or negative depending on the
size of the angle.
Exercise 10.9
Use tables of logarithms of trigonometric ratios
to write down the following (Questions 1–3).
1. (a) log sin 14° (b) log sin 20.5°
(c) log sin 36.48° (d) log sin 54°
(e) log sin 69.3° (f) log sin 89.87°
2. (a) log cos 75° (b) log cos 18°
(c) log cos 34.8° (d) log cos 72.4°
(e) log cos 15.45° (f) log cos 69.83°
3. (a) log tan 11° (b) log tan 80°
(c) log tan 87° (d) log tan 24.6°
(e) log tan 50.56° (f) log tan 32.98°
4. In each of the following nd the value of
the unknown.
(a) log sin x = 1.512 6
(b) log sin x = 2.948 9
(c) log sin x = 1.926 1
(d) log sin x = 1.988 8
(e) log cos y = 1.988 5
(f) log cos y = 1.938 3
(g) log cos y = 1.687 8
(h) log cos y = 1.457 4
(i) log tan u = 1.886 2
(j) log tan u = 1.645 5
(k) log tan u = 0.037 9
(l) log tan u = 1.201 2
5. Evaluate each of the following.
(a) cos 30° sin 25° (b) sin 48° sin 42°
(c) cos 38° cos 52° (d) sin 72° cos 68°
(e) sin 24.5° (f) sin 71° cos 11.5°
cos 35.4° cos 24°
(g) 4.69 sin 52.5° (h) 6.5 cos 47°
5 sin 35° cos 45° tan 65°
(i) 4 cos 40° + 6 sin 40°
10 tan 50°
(j) 2 sin 50° cos 30°
tan 78° – 3 sin 0°
Application of trigonometry
There are may physical applications of
trigonometry, for example, if we wanted to nd
the distance across a river, we would proceed as
follows:
• Take some visible point B on the opposite
side of the river (banks parallel) and nd
a point A directly opposite it (say using
the surveyors’ instrument known as a
theodolite), as in Fig. 10.26.
Fig. 10.26
• Move to a point C on the bank and measure
the distance AC and ∠ACB =
θ
, by means
of instruments.
• Since ∆ABC is right angled, and
tan
θ
= , then
AB = AC tan
θ
.
How do we nd the height of a tower when the
foot is inaccessible?
This may be done as follows:
• Let A be the foot of the tower and D the top
AB
AC
–––
θ
C
B
A
– – – – – – – – – – – –
– – – – – – – – – – – – – – – –
–
–
–
–
–
–
–
–
–
–
107
Solution
Let the distance from the tree T to the chicken C
be x m.
To nd x, we use any of the following two
methods.
Fig. 10.28
Method 1
∠HCT = ∠BHC = 25° (alternate angles) and
∆CHT is right-angled at T.
∴ Using ∆HCT, tan 25° =
⇒ x tan 25° = 15
⇒ x = .
Using logs:
No Log
15 1.176 1 –
tan 25° 1.668 7
32.17 1.507 4
⇒ x = 32.17 m
Note that to do this calculation, you could also use a
calculator as follows:
Step 1: Enter 15
Step 2: Press ÷
Step 3: Press tan
Step 4: Enter 25
Step 5: Press = 32.167 603 807 6
Round off the value to 2 d.p.
Method 2
In ∆HCT, ∠CTH = 90°.
∴∠CHT = 180° – (90° + 25°) = 65°
Now, tan 65° = .
( )
15
tan 25°
––––––
x
15
––
A
x
a
C B
α
β
D
– – – – – – – – – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – –– – – – – – – – – – – – – – –
15
x
––
–
(Fig. 10.27). Let B and C be two points on
the same level as A such that C, B and A are
collinear. By means of instruments (e.g. a
clinometer or theodolite), measure
∠DBA =
α
and ∠DCA =
β
.
• Measure the distance BC = a.
Fig. 10.27
• Let x be the height of the tower.
Now tan
α
=
∴ AB =
Also, tan
β
=
∴ AC =
Thus, a = AC – AB = –
i.e. a = x –
Since angles
α
and
β
and the distance a
are known, x may then be found simply by
substitution into the formula.
The following examples illustrate some of the
calculations involved when faced with real-life
situations.
Example 10.14
A hawk is perched on a tree at a height of 15
m above the ground. It spots a chicken, sitting
on the ground, at an angle of depression of
25° (Fig. 10.28). How far from the tree is the
chicken?
1
tan
α
––––
1
tan
β
––––
x
tan
α
––––
x
tan
β
––––
x
AB
–––
x
tan
α
––––
x
AC
–––
x
tan
β
––––
C
15 m
x
B
25°
H
T
108
( )
Exercise 10.10
1. A ladder of length 4.5 m rests against a
vertical wall so that the angle between the
ladder and the ground is 68°. How far up
the wall does the ladder reach?
2. A ladder of length 5.5 m rests against a
vertical wall so that the angle between the
ladder and the ground is 60°. How far
from the wall is the foot of the ladder?
3. A boy is ying a kite using a string of
length 56 m. If the string is taut and it
makes an angle of 62° with the horizontal,
how high is the kite? (Ignore the height of
the boy).
4. Find the dimensions of the oor of a
rectangular hall given that the angle
between a diagonal and the longer side is
25° and that the length of the diagonal is
10 m.
5. A plane takes off from an airport and after a
while, an observer at the top of the control
tower sees it at an angle of elevation of 9°.
At that instant, the pilot reports that he has
attained an altitude of 2.4 km. If the height
of the control tower is 50 m, nd the
horizontal distance that the plane has own?
6. After walking 100 m up a sloping road, a
man nds that he has risen 30 m. What is
the angle of slope of the road?
7. The tops of two vertical poles of heights
20 m and 15 m are joined by a taut wire
12 m long. What is the angle of slope of
the wire?
8. A man walks 1 000 m on a bearing of 025°
and then 800 m on a bearing of 035°. How
far north is he from the starting point?
9. Two boats A and B left a holiday resort
at the coast. Boat A travelled 4 km on a
bearing of 030° and boat B travelled 6 km
on a bearing of 130°.
h
AC
–––
h
tan 20°
––––––
h
tan 30°
––––––
1
0.364 0
––––––
1
0.577 4
––––––
h
BC
–––
1
tan 20°
––––––
1
tan 30°
––––––
h
tan 20°
––––––
h
tan 30°
––––––
500
1.015
––––
( )
– – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – –
20°
30°
h
K
CB
A
500 m
⇒ x = 15 tan 65°
= 32.17 m (2 d.p
)
Note that 65° is the complement of 25°, and that
it is easier to multiply 15 by tan 65° (Method 2)
than to divide 15 by tan 25° (Method 1). Hence,
given a problem like the one above, it is better
to nd the complement of the given angle and
then use it to solve the problem. However, if
you are using a calculator, either method will do.
Example 10.15
Two observers, A and B, 500 m apart can see a
kite in the same vertical plane and from angles
of 20° and 30° respectively. If they are on the
same side, nd the height of the kite.(Hint:
Disregard the height of the observers.)
Solution
Fig. 10.29 shows the kite and the relative
positions of A and B.
Fig. 10.29
From Fig. 10.29,
tan 20° =
∴ AC =
Also tan 30° =
∴ BC =
Now, AC – BC = –
i.e. AC – BC = h –
⇒ 500 = h –
500 = h (2.747 – 1.732)
500 = 1.015 h
∴ h = ≈ 492.6 m.
109
(a) Which boat travelled further
eastwards and by how much.
(b) How far to the north is boat A from
boat B?
10. A bridge crosses a river at an angle of 60°.
If the length of the bridge is 170 m, what is
the width of the river?
11. A man sitting at a window with his eye
20 m above the ground just sees the sun
over the top of a roof 45 m high. If that
roof is 30 m away from him horizontally,
nd the angle of elevation of the sun.
12. The shaft of a mine descends for 100 m at
an angle of 13° to the horizontal and then
200 m at an angle of 7° to the horizontal.
How far below the starting point is the end
of the shaft?
13. Two girls, one east and the other west of
a tower, measure the angles of elevation
of the top of its spire as 28° and 37°. If
the top of the spire is 120 m high, how far
apart are the girls?
14. Atieno and Baabu stand on one side of a
tower and in a straight line with the tower.
They each use a clinometer and determine
the angle of elevation of the top of the
tower as 30° and 60° respectively. If their
distance apart is 100 m, nd the height of
the tower.
15. Fig. 10.30 shows the side view of an
ironing board. The legs are all 95 cm
long and make 60° with the oor when
completely stretched.
(a) How high is the ironing surface from
the oor, given that the board is
2.5 cm thick?
(b) How far apart are the legs at the oor?
Fig.10.30
60°
60°
30 cm
110
We would proceed as follows:
In ∆BCD, sin B =
i.e. sin 45.58° =
⇒ CD = 7 sin 45.58°
Hence, area of the triangle is × AB × CD
= × 8cm × 7 sin 45.58°
= 20 cm
2
(= 19.998 395 410 9 by
calculator)
Now consider any triangle ABC with sides a, b,
c and height h.
Fig. 11.3
Using ∆ACD,
h
= sin C.
b
∴ h = b sin C.
Thus, Area of ABC = ah
= ab sin C
This is the sine formula.
Similarly, it can be shown that:
Area of ∆ABC = bc sin A = ac sin B
Now consider the obtuse-angled triangle ABC
in Fig. 11.4.
Fig. 11.4
1
2
–
The sine formula
The area of a triangle is given by the formula,
Area = bh,
where b is the length of the base and h is the
height - see Example 11.1
Example 11.1
Calculate the area of the triangle shown in
Fig. 11.1.
Fig. 11.1
Solution
The height is 5 cm and the corresponding base
is 8 cm. For purposes of this question, we do
not need the 7 cm side.
Area of the triangle = bh
= × 8cm × 5 cm
= 20 cm
2
Suppose that, in Example 11.1, we are not given
the height CD but we are given
∠B = 45.58° (Fig. 11.2).
How would we nd the
area?
Fig. 11.2
1
2
–
1
2
–
– – – – – – – – – – – – – – – –
A
D
C
B
7 cm
8 cm
45.58°
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
A
C
b
– – – – – – – – – – – – –
B
D
c
a
h
– – – – – – – – – – – – – – – –
A
5 cm
D
C
B
7 cm
8 cm
1
2
–
CD
7
–––
CD
7
–––
A
B
c
– – – – – – – – – – – –
C
D
b
a
θ
h
– – – – – – – – – –
AREA OF A TRIANGLE
11
111
Using ∆ACD, h = b sin
θ
Area of ∆ABC = ah
= ab sin
θ
But
θ
= 180° – C, where C = ∠ACB.
Thus area of ∆ABC
= ab sin (180° – C ), where C is obtuse.
Note that these formulae apply when two sides
and the included angle of a triangle are given.
Example 11.2
Find the area of ∆ABC to the nearest cm
2
, given
that AB = 6 cm, BC = 9 cm and ∠B = 37°.
Solution
Let the height of ∆ABC be h cm (Fig. 11.5).
Fig. 11.5
In ∆ABN, = sin 37°
∴ h = 6 sin 37°
Area of ∆ABC = × BC × AN
= × 9cm × 6cm sin 37°
= × 9cm × 6cm × 0.601 8
= 16. 248 6 cm
2
= 16 cm
2
(to the nearest cm
2
)
Example 11.3
Find ∠C given that AC = 6 cm, BC = 7 cm
and area of ∆ABC = 16.1 cm
2
, and that ∠C is
an obtuse angle.
Solution
Fig. 11.6 is a sketch of ∆ABC .
Fig. 11.6
Area of ∆ABC = × 7cm × 6 sin C
× 7cm × 6 sin C = 16.1
⇒ sin C =
= 0.766 7
∴ ∠C = 50.1°
But ∠ACB is obtuse
∴ obtuse ∠C = 180° – 50.1°
= 129.9˚
Hero’s formula
The area of ∆ABC with sides a, b and c known,
can be found using the formula:
Area of ∆ABC = √s(s – a)(s – b )(s – c)
where s = (a + b + c)
i.e. s = half the perimeter of
the triangle
(i.e semi-perimeter)
Fig. 11.7
This is known as Hero’s formula, the proof of
which is beyond the scope of this book
Point of interest
Hero (or Heron) of Alexandria was a 1st Century (AD)
Greek mathematician and engineer who authored
a number of works on mensuration. In addition to
showing how to work out the volumes of cones,
prisms, pyramids, spherical segments, some regular
polyhedra, and other gures, Hero described a method of
approximating square roots and the formula for the
1
2
–
1
2
–
1
2
–
– – – – – – – – – – – –
A
C
6 cm
B
h
37°
N
9 cm
h
6
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
32.2
42
––––
A
B
C
b
a
c
A
B
C
6
h
7
– – – – – – – – – – – – –
– – – – – – – – – –
112
3. Find ∠B given that:
(a) AB = 3 cm, BC = 8 cm, area of
∆ABC = 7.05 cm
2
(b) AB = 10 cm, BC = 11cm, area of
∆ABC = 51cm
2
(c) AB = 12 cm, BC = 35 cm, area of
∆ABC = 210 cm
2
(d) AB = 7 cm, BC = 24 cm, area of
∆ABC = 84 cm
2
(e) AB = 6 cm, BC = 6 cm, area of
∆ABC = 10.3cm
2
(f) AB = 5 cm, BC = 9 cm, area of
∆ABC = 18.4cm
2
and ∠B is an obtuse
angle.
(g) AB = 8 cm, BC = 10 cm, area of ∆ABC
= 25.7 cm
2
and ∠B is an obtuse angle.
4. Find the areas of the triangles in Fig. 11.8.
All the measurements are in centimetres.
(a) (b)
(c) (d)
(e) (f)
(g) (h)
4
3
7
7
7
20°
20°
6
5
5
25°
5
6
140°
50°
8
3
4
58°
4.5
75°
25°
6
3
area of a triangle that bears his name. He also
invented many mechanical devices that were
steam driven. He is best remembered for the
formula of nding the area of a triangle using the
lengths of its three sides.
Example 11.4
Find the area of a triangle with sides 6 cm,
10 cm and 12 cm. State your answer correct to
4 s.f.
Solution
s = (6 + 10 + 12)cm = 14 cm.
Area of the triangle
= s(s – a)(s – b)(s – c)
= 14(14 – 6)(14 – 10)(14 – 12) cm
2
= 14 × 8 × 4 × 2
= 896 cm
2
= 2.993 3 × 10 cm
2
= 29.93 cm
2
(4 s.f.)
Exercise 11.1
1. Find the area of ∆ABC given that:
(a) AB = 10 cm, BC = 13 cm, ∠B = 48°
(b) AB = 6 cm, BC = 8.5 cm, ∠B = 37°
(c) AC = 7 cm, AB = 6 cm, ∠A = 57°
(d) AC = 4 cm, BC = 6 cm, ∠C = 130°
(e) BC = 9 cm, BA = 7 cm, ∠B = 145°
(f) AC = 10 cm, BA = 10 cm, ∠A = 165°
2. Find the areas of the following triangles
whose measurements are given in
centimetres.
(a) a = 9, b = 8, c = 5
(b) a =12, b = 6, c = 7
(c) a = 7, b = 6, c = 7
(d) a = 13, b = 12, c = 6
(e) a = 6.5, b = 5.5, c = 4
(f) a = 8.8, b = 10.6, c = 6.4
1
2
–
113
(i) (j)
(k) (l)
2
5
4
5
6
3
3
5
7
7
5
7
2
bc
–––
1
2
–
1
2
–
1
2
–
1
2
–
A
C
M
B
D
– – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
J K
M
N
h
L
P
A
G
C
E
F
B
D
X
Z
Fig. 11.8
5. By expressing the area of ∆ABC in two
ways, show that
sin A = √s(s – a)(s – b)(s – c).
Obtain similar expressions for sin B and
sin C.
Hence nd the angles of a triangle whose
sides are 7 cm, 8 cm and 9 cm.
Areas of triangles with equal bases
and between parallel lines
Consider triangles ABC and BCD in Fig. 11.9.
Fig. 11.9
The area of ∆ABC is × BC × h
1
and the
area of ∆BCD is × BC × h
2
Since WX ⁄⁄ YZ, h
1
= h
2
.
∴ area of ∆ABC = area of ∆BCD.
Fig. 11.10 shows two other triangles JKL and
MNP in which JK = MN = b.
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
D
A
W
Y
B
C
– – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – –
h
2
h
1
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
Fig. 11.10
Area of ∆JKL is bh and area of ∆MNP = bh,
i.e. Area of ∆JKL = area of ∆MNP.
The areas of triangles with equal bases and
between the same two parallel lines are
equal.
Exercise 11.2
1. In Fig. 11.11, AB ⁄⁄CD. Name two pairs of
triangles which have the same area.
Fig. 11.11
2. In Fig. 11.12, AB//CD and M is the
midpoint of AB. There are two, three and
four triangles which have the same areas.
Name them, ensuring that in each case, they
are different.
Fig. 11.12
114
Q
P
S
R
O
3. Fig. 11.13 PQRS, is a trapezium and O is
the point of intersection of its diagonals.
Show that triangles SOP and QOR have
equal areas.
Fig. 11.13
4. In Fig. 11.14, ∆PRS is isosceles. If PS = 13cm,
RS = 10 cm, and ∠QSR = 30°. What is the
length QS?
Fig. 11.14
5. In Fig. 11.15, AB⁄⁄CD, AB = 7 cm,
BC = 12 cm, AD = 9 cm, ∠BAD = 46° and
∠ACB = 15°. Find the length AC.
Fig. 11.15
Application of area of a triangle
In life, we often encounter situations requiring
the use of the knowledge of calculating the area
of a triangle, for example, we might want to
know the area of a triangular plot of land, most
of which is covered by a deep lake or a soggy
swamp. In such a case, measure the lengths of
the edges of the plot and possibly the angles,
but not the altitudes. Knowing two sides and an
included angle or all the three sides, can help
calculate the area.
At other times, we might know all the sides
and we wish to nd the angles of such a plot.
In this case, we would use the formulae given
below.
Sin A = √s(s – a)(s – b)(s – c)
Sin B = √s(s – a)(s – b)(s – c)
Sin C = √s(s – a)(s – b)(s – c)
where A, B and C, and a, b and c, are the
respective angles and sides of a triangle
ABC, and s = (a + b + c).
Example 11.5
The sides of a triangular plot of land are 170 m,
190 m and 210 m, but the altitudes of the plot as
well as the angles are not known. Find:
(a) the area of the plot in hectares.
(b) the angles of the plot.
Solution
(a) The three sides are known (Fig. 11.16).
Fig. 11.16
2
bc
––
2
ac
––
2
ab
––
1
2
–
46°
A
B
D
C
12 cm
15°
9 cm
7 cm
C
190 m
B
A
170 m
210 m
30°
R
S
P
Q
13 cm
10 cm
115
Using the available measurements, nd the
area of the plot.
2. Aiko nds that her farm has the shape of a
pentagon PQRST such that PQ = 60 m,
QT = 90 m, RT = 110 m, RS = 70 m,
∠PQT = 30°, ∠QTR = 43° and ∠TRS = 65°.
Make a sketch of the farm and hence
calculate its area in: (a) m
2
(b) hectares.
3. Fig. 11.18 shows a triangular eld. Find:
(a) the area of the eld, in m
2
.
(b) the length of a dividing fence from the
largest angle, meeting the opposite side
at right angles.
Fig. 11.18
4. A coastguard ship patrols the area dened
by the three points A, B and C, where B is
due north of A, AB = 14 km, BC = 12 km
and AC = 10 km. Find the bearings of C
from both A and B.
5. A spy plane is positioned to monitor the
area between points X, Y and Z in a certain
country. Y is 1 500 km from X on a bearing
of 158° and Z is 900 km from Y on a
bearing of 030°. Calculate the area under
the plane’s surveillance.
39 m
42 m
45 m
1
2
–
1
2
–
∴ Using Hero’s formula,
Area = √s(s – a)(s – b)(s – c), where
s = (a + b + c)
Now, s = (170 + 190 + 210)
= 285
Thus,
Area = √285(285 – 170)(285 – 190)(285 – 210)
= √285 × 115 × 95 × 75
= √233 521 875 ≈ 15 281 m
2
(using
calculator)
≈ 1.528 ha
(b) From (a), √s(s – a)(s – b)(s – c) = 15 281m
2
∴ Sin A = × 15 281
= × 15 281
≈ 0.766 0 (from calculator and
to 4 d.p.)
∴ ∠A ≈ 50.00° (2 d.p)
Sin B = × 15 281
≈ 0.8561
∴ ∠Β ≈ 58.88°
Sin C = × 15 281
≈ 0.946 2
∴ ∠C ≈ 71.12°
Alternatively,
∠C = 180° – (50° + 58.88° )
= 71.12°
Exercise 11.3
1. Figure 11.17 shows how Auma triangulated
her plot of land during a survey.
Fig. 11.17
2
190 × 210
––––––––
2
170 × 210
––––––––
2
bc
––
2
170 × 190
––––––––
20 m
68°
30°
24 m
30 m
116
Then ∠ABN = 180° –
θ
(angles on a straight
line).
h is the height of the parallelogram.
Introduction
In Form 1, we learnt that a polygon is a many-
sided plane gure. Hence a quadrilateral is a
polygon with four sides. Some of the areas of
quadrilaterals that we studied are as follows:
1. Area of a square = (length)
2
2. Area of a rectangle = length × breadth
3. Area of a parallelogram = hb, where b is the
length of any pair of parallel sides and h is the
perpendicular distance between these sides.
4. Area of a rhombus = hb (as in
parallelogram) or
Area of a rhombus = sum of areas of four
equivalent right-angled triangles (Fig. 12.1).
Fig. 12.1
Area of a rhombus
= area 1 + area 2 + area 3 + area 4
= 4 × area of any of the triangles.
Note: Lengths of diagonals must be known.
5. Area of a kite = sum of areas of four
triangles, which are two pairs of identical
right-angled triangles. Note that the
dimensions of the triangles must be known.
Area of a parallelogram
Fig. 12.2 shows parallelogram ABCD in which
AB = a, BC = b and ∠ABC =
θ
.
Fig. 12.2
Since diagonal AC bisects parallelogram
ABCD,
area of ∆ABC = area of ∆ADC = ab sin
θ
.
Area of parallelogram ABCD
=
ab sin
θ
+ ab sin
θ
= ab sin
θ
Alternatively:
Fig. 12.3 shows parallelogram ABCD in which
AB = a, BC = b, ∠ABC =
θ
and AN = h, i.e.
the height of the parallelogram.
Fig. 12.3
Area of parallelogram ABCD
= base × height = bh.
In ∆ABN, = sin
θ
∴ h = a sin
θ
Then, area of parallelogram ABCD = ab sin θ.
What if
θ
is obtuse?
Fig. 12.4 shows parallelogram ABCD with
∠ABC =
θ
, an obtuse angle.
Fig. 12.4
1
2
–
1
2
–
1
2
–
1 2
4
3
C
A
D
– – – – – – – – – –
B
θ
a
b
h
N
– – – – –
h
a
–
A
C
D
– – – – – – – – – –
B
θ
a
b
h
N
A
C
D
– – – – – – – – – – – – – – – – –
B
θ
a
b
AREAS OF QUADRILATERALS AND
OTHER POLYGONS
12
117
Fig. 12.5
Area of ∆ABN = × BN × h
Area of rectangle ANMD = ah
Area of ∆DMC = × MC × h
∴ Area of trapezium ABCD
= Area of ∆ABN + Area of ANMD +
Area of ∆DMC
= × BN × h + ah + × MC × h
= h (BN + a + a + MC)
= h (2a + BN + MC)
= h (2a + b – a) since BN + MC = b – a
= h (a + b)
Alternatively:
Diagonal BD divides trapezium ABCD into two
triangles (Fig. 12.6)
Fig. 12.6
Area of ∆BCD = bh
Area of ∆ABD = ah
∴ area of trapezium ABCD
= Area of ∆ABD + area of ∆BCD
= ah + bh
= h (a + b)
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
h
a
–
1
2
–
MN
A
h
h
C
D
B
b
a
– – – – – – – – – – – –
– – – – – – – – – – – –
⇒
A
h
C
D
B
– – – – – – – – – – – –
– – – – – – – – – – – –
– – – – – – – – – – –– – – – – – –
h
b
a
Hence, = sin (180° –
θ
)
h = a sin (180° –
θ
)
So, area of parallelogram ABCD
= ab sin (180° –
θ
).
In general;
If a and b are sides of a parallelogram and
θ
is the angle between them, then area of
the parallelogram
= ab sin
θ
, if
θ
is acute
= ab sin (180° –
θ
), if
θ
is obtuse.
Example 12.1
Find the area of parallelogram PQRS given that
PQ = 6.5 cm, QR = 11 cm and ∠PQR = 35°.
Give your answer correct to 4 s.f.
Solution
Area of parallelogram = ab sin
θ
i.e. Area of PQRS = 6.5cm × 11cm × sin 35°
= 6.5cm × 11cm × 0.573 6
= 41.01 cm
2
(4 s.f.)
Example 12.2
Find the area of parallelogram PQRS given
that PQ = 7.5 cm, QR = 12.5 cm and
∠PQR = 125°. Give your answer correct to
3 s.f.
Solution
Area of parallelogram
= ab sin (180° –
θ
), since 125° is obtuse.
Area of PQRS
= 7.5cm × 12.5cm × sin (180° – 125°)
= 7.5cm × 12.5cm × sin 55°
= 7.5cm × 12.5cm × 0.819 2
= 76.8 cm
2
(3 s.f.)
Area of a trapezium
Trapezium ABCD has sides AD⁄⁄BC, AD = a,
BC = b and AN = DM = h, the height
(Fig. 12.5).
Thus, area of a trapezium is given by
A = (sum of parallel sides) × height
= h (a + b)
1
2
–
1
2
–
118
4. PQRS is a parallelogram in which
PQ = 3 cm, QR = 12 cm and the
perpendicular from Q to PS is 2.5 cm. Find
(a) ∠PQR
(b) the area of the parallelogram.
5. Find the area of a rhombus whose diagonals
are 5 cm and 6 cm long. What are the sizes
of the angles of the rhombus?
Fig. 12.8
2. Find the areas of the trapezia in Fig. 12.9.
Fig. 12.9
1
2
–
1
2
–
(d) (e)
(f)
50°
3
8
40°
7
8
150°
5
5
50°
12
4
(a) (b)
5
9
4
65°
48°
6
10
5
(c) (d)
6
6
6
12
55°
5
10
5
3. ABCD is a parallelogram whose area is 18 cm
2
.
AB = 5 cm and BC = 4 cm. Find ∠ABC.
6
2
(c)
115°
4
7
Example 12.3
Find the area of trapezium ABCD in which
AD⁄⁄BC, AD = 8 cm, BC = 14 cm,∠BCD = 70°
and DC = 7 cm. Give your answer to 4 s.f.
Solution
Fig. 12.7 is a sketch of the trapezium.
Fig. 12.7
Draw a perpendicular from D to meet BC at N.
DN is the height of the trapezium.
In ∆DNC, DN = 7 sin 70° cm
∴ area of the trapezium
= (8 + 14cm) × 7cm sin 70°
= × 22 cm × 7cm × 0.9397
= 72.36 cm
2
Exercise 12.1
Attempt all questions and give your answers
correct to 4 s.f.
1. Find the areas of the parallelograms in
Fig. 12.8. All measurements are in
centimetres.
(a)
(b)
– – – – – – – – – – – – – – – – –
8 cm
N
B
C
D
A
70°
7 cm
14 cm
119
From this example we see that:
Given the distance, d, from the centre of a
regular n-sided polygon to a vertex, the area
of the polygon is given by:
Area = n × × d × d × sin
= nd
2
sin
Note that instead of being given the distance
from the centre of a regular polygon to a vertex,
we may be given the length of a side. In this
case, the area is found as in Example 12.5.
Example 12.5
Find the area of a regular nonagon of side 6 cm
stating your answer correct to 4 s.f.
Solution
Fig. 12.11 shows a sketch of the nonagon.
Consider ∆AOB.
Fig. 12.11
The bisector of ∠AOB also bisects AB at right
angles.
1
2
–
1
2
–
1
2
–
360°
n
––––
1
2
–
1
2
–
360°
n
––––
– – – – – – – – – – – – –
6 cm
B
A
O
– – – – – – – – – – – – – –
– – – – – – – – – – – – – –
h
C
6. PQRS is a trapezium in which PQ and SR
are parallel. If PQ = 7 cm, SR = 3 cm,
QR = 5 cm, the area of PQRS is 18 cm
2
,
nd ∠PQR.
7. The parallel sides AB and DC of a trapezium
ABCD are 4 cm apart. If AB = 8 cm and
DC = 5 cm, nd:
(a) the area of ABCD
(b) ∠ABC given that BC = 5 cm.
Area of a regular polygon
When the centre of a regular polygon is joined
to the vertices, isosceles triangles are formed.
The number of isosceles triangles formed is
equal to the number of sides of the polygon.
The area of the polygon can be found by nding
the sum of the areas of the triangles.
Example 12.4
Find the area of a regular hexagon if the
distance from the centre of the hexagon to any
vertex is 5 cm, giving your answer correct to
4 s.f.
Solution
Fig. 12.10 is a sketch
of a hexagon.
Fig. 12.10
Area of ∆OAB = × 5cm × 5cm sin 60° (since
there
are six equal angles at the
centre)
= × 5cm × 5cm × 0.866
Since the hexagon has 6 sides, there are 6
triangles.
∴ Total area of hexagon
= 6 × × 5cm × 5 cm× 0.866 cm
2
= 64.95 cm
2
(4 s.f.)
Since ∠AOB = 40° and OC is the angle
bisector, then
∠AOC = 20° ⇒ ∠OAC = 70°
∴ h = 3 tan 70°.
∴ Area of ∆OAB = × 6cm × h
= × 6 cm× 3 tan 70°
= (9 × 2.747 5) cm
2
= 24.727 5 cm
2
= 24.73 cm
2
1
2
–
1
2
–
5 cm
B
A
O
5 cm
60°
– – – – – – – – – – – –
– – – – – – – – – – –
120
a
O
h
C
D
– – – –
– – – – – –
– – – – – – – – – –
– – – – – – – – – – – – – – – – –
θ
θ
A
B
360°
n
––––
1
2
–
360°
n
––––
180°
n
––––
2h
a
––
a
2
–
1
2
–
a
2
–
1
2
–
1
4
–
simpler shapes it subdivides into. Usually such
a polygon subdivides into triangles whose areas
can then be found using any of the formulae for
nding area of a triangle.
Exercise 12.2
1. Find the area of a regular pentagon if the
distance from the centre of the pentagon to
any of the vertices is 8 cm.
2. Find the area of a regular octagon given that
distance from the centre of the octagon to
any of the vertices is 6 cm.
3. Calculate the area of a regular pentagon
given that the length of the side is 10 cm.
4. A regular polygon has an internal angle of
162° and a side of length 16 cm.
(a) How many sides does the polygon have?
(b) Find the area of the polygon.
5. Calculate the area of a regular duodecagon
of side 6 cm.
6. The area of a regular decagon is 294 cm
2
.
What is the distance from its centre to a vertex?
7. The area of a regular nonagon is 185.1 cm
2
.
What is the length of each side?
8. A piece of paved ground is in the shape of
an irregular hexagon with vertices A, B, C,
D, E and F. The vertices lie on a circle of
radius 8 m. Given that AB = 8 m,
BC = 10.8 m, CD = 7 m, DE = EF = 6 m
and AF = 10 m, nd the area of the piece
of ground.
9. An irregular hexagon PQRSTU is such that
PQ = 5 cm, ST = 4 cm, TU = 6 cm,
UP = 5 cm, PR = 8 cm, PS = PT = 10 cm
∠QPR = 30°and ∠PRS = 90°. Make a
sketch of the polygon and hence calculate
its area.
1
4
–
1
4
–
1
4
–
(
)
h
a
2
–––
180°
n
––––
(
)
Hence area of the nonagon
= (9 × 24.73) cm
2
= 222.57 cm
2
= 222.6 cm
2
(4 s.f.)
We see that if we are given the length of a
side of a regular polygon, as in Example 12.5,
its area may be found as follows: (Refer to
Fig. 12.12.)
Fig. 12.12
∆’s AOB, BOC, COD, etc. that are formed at
the centre are isosceles and congruent.
In ∆AOB, ∠AOB = , where n is the
number of sides, and
θ
= 180° – = 90° –
Now, tan
θ
= =
∴h = tan
θ
Hence, area of ∆AOB = ah
= a × tan
θ
= a
2
tan
θ
Hence, area of an n-sided regular polygon of
side a units
= n × a
2
tan
θ
= a
2
n tan
θ
= a
2
n tan 90° –
Area of an irregular polygon
Whatever information is given about an irregular
polygon, you must rst sketch it and see what
121
Parts of a circle
An arc is a part of the circumference of a circle.
Thus in Fig. 13.1, AXB and AYB are arcs.
AXB is known as a minor arc and AYB as a
major arc.
Fig. 13.1
A sector of a circle is that area that is bounded
by two radii and an arc. Thus, in Fig. 13.1, the
shaded area AOB is a sector of the circle.
A chord is a line segment joining any two
points on the circumference of a circle, for
example in Fig. 13.2, AB is a chord.
A chord divides the circle into two parts known
as segments. Thus a segment is an area bound
by a chord and an arc. The larger segment is
known as the major segment while the smaller
one is the minor segment (Fig. 13.2).
Fig. 13.2
If a chord passes through the centre of the
circle, then it is known as a diameter.
The diameter divides a circle into two equal
segments called semicircles.
With the knowledge of these parts, we can now
study the area of part of a circle.
1
4
–
Area of a sector
The area of a sector is a fraction of the area of
the whole circle.
Thus, in Fig. 13.3 the area of sector AOB is of
the area of the whole circle. The fraction can
be obtained as
90°
, i.e.
sector angle
.
360°
whole revolution
What fraction of the area of the whole circle is
sector:
(a) COD (b) EOF (c) AOC (d) AOD?
Fig. 13.3
Generally, the area of a sector is proportional to
the sector angle.
In Fig. 13.4,
θ
is the angle subtended at the
centre and the area of the whole circle is
πr
2
. Therefore,
area of sector AOB = × πr
2
.
Fig. 13.4
1
4
–
B
major segment
minor segment
•
X
Y
•
A
Y
B
X
A
•
•
O
θ
D
B
A
F
45°
60°
90°
E
C
O
θ
360°
––––
B
A
θ
r
O
θ
AREA OF PART OF A CIRCLE
13
122
Example 13.1
A sector with an angle of 72° is cut off from a
circle of radius 14 cm. Find the remaining area
of the circle.
Solution
Required is the area of the shaded part in Fig. 13.5.
Fig. 13.5
Angle of remaining sector = 360° – 72° = 288°
Area of remaining sector = × × 14
2
cm
2
= × × 14cm × 14 cm
=
492.8 cm
2
Exercise 13.1
1. Each circle in Fig. 13.6 is of radius 8 cm.
Calculate the areas of the shaded sectors,
leaving π in your answers.
(a) (b) (c)
Fig. 13.6
2. Find the area of a sector with angle 54°
given that the radius of the circle is 10 cm.
(Use π = 3.142.)
3. An arc of a circle of radius 35 cm subtends
an angle of 144° at the centre. Use π =
to calculate the area of the sector.
4. The area of a sector of a circle of radius
14 m is 462 m
2
. Using π = , calculate the
angle of the sector.
288
360
–––
288
360
–––
22
7
––
22
7
––
45°
135°
60°
22
7
––
22
7
––
72°
21
21
21
21
80°
110°
50°
O
C
D
A
B
5. The area of a circle is 99 m
2
. What is the area
of a sector of the circle with an angle of 140°?
6. In Fig. 13.7, the dimensions are in
centimetres and the arcs are circular.
Calculate the shaded area in Fig. 13.7(a)
and hence nd the shaded area in Fig.
13.7(b).
(a) (b)
Fig. 13.7
7. Fig. 13.8 shows a circle which has been
subdivided into sectors. Find the ratio:
(a) area of sector AOB : area of sector BOC,
(b) area of sector COD : area of major
sector AOC.
Fig. 13.8
8. Given that the area of sector COD in
Fig. 13.8 is 45 cm
2
. Find the area of the
circle.
9. An arc of length 13.2 cm subtends an angle
of 108° at the centre of the circle. Find the:
(a) radius of the circle,
(b) area of the sector subtended by the arc.
Area of a segment
Consider the segment of
the circle shaded in
Fig. 13.9.
Fig. 13.9
81°
B
A
7 cm
7 cm
O
123
( )
( )
1
2
–
4
5
–
1
2
–
106.26
360
––––––
1
2
–
106.26
360
––––––
1
2
–
4
α
1
2
α
5
5
4
O
A
B
–
–
Fig. 13.13
C
1
2
α
M
O
P
B
A
N
5
β
5
7
7
8
•
•
81°
360°
––––
22
7
––
1
2
–
22
7
––
9
40
––
1
2
–
20
1
1
11
1
2
–
1
2
–
θ
360°
––––
θ
360°
––––
r
r
θ
We see that the segment together with ∆AOB make
up sector AOB. Thus, the area of the segment is the
area of the sector less that of ∆AOB.
Thus, area of segment
= area of sector AOB – Area of ∆AOB
= × × 7
2
– × 7cm × 7cm× sin 8
0
= × × 7 × 7 – × 49cm
2
× sin 81°
= (34.65 – 24.198) cm
2
= 10.452 cm
2
= 10.5 cm
2
(3 s.f)
For any segment, formed by an arc and chord
which subtends an angle
θ
at the centre of
the circle (Fig. 13.10), the area is obtained
as shown below:
Area of segment
= area of sector – area of triangle
= × πr
2
– r
2
sin
θ
= r
2
π – sin
θ
Fig. 13.10
Area of a common region between
two circles
When two circles intersect, they form an area
that is common between them, as shown in Fig.
13.11.
Fig. 13.11
To nd this area, calculate the sum of the
areas of the two segments which comprise the
common region.
( )
Example 13.2
Two circles of radii 5 cm and 7 cm intersect.
If they share a common chord of length 8 cm,
calculate the common area of intersection of the
circles. (Use π = 3.142.)
Solution
Fig. 13.12 shows the intersection of the circles,
with the common area shaded.
Fig. 13.12
Now, OA and OB are radii and so ∆AOB is
isosceles.
Thus, the perpendicular bisector of AB through
O divides angle
α
into two equal angles (Fig.
13.13).
∴ sin
α
= = 0.8
⇒
α
= 53.13° (from tables)
∴
α
= 106.26°
Note that OC = 5
2
– 4
2
= 3 cm
Area of segment AMB (Fig. 13.12)
= Area of sector OAMB – Area of ∆OAB
= × π × 5
2
– × AB × OC
= × 3.142 × 25cm – × 8cm × 3
= (23.19 – 12 )cm
2
= 11.19 cm
2
124
1
2
–
1
2
–
4
7
–
69.7
360
––––
1
2
–
22
7
––
– – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – –
•
4 m
4 m
3 m
60°
T
S
R
P
Q
O
– – – – – – – – – – – – – – – – – –
24.5 cm
24.5 cm
24.5 cm
24.5 cm
4. An arc of a circle subtends an angle of 81°
at the centre of the circle. If the area of the
sector is 138.6 cm
2
, nd the:
(a) radius of the circle,
(b) area of the segment bounded by the
same arc and the corresponding chord.
5. An arc of length 4.4 m subtends an angle of
72° at the centre of a circle. Calculate the:
(a) radius of the circle,
(b) area of the segment bounded by the arc
and the corresponding chord.
6. Fig. 13.16 shows a part of a circle that
remains when a segment is cut off.
Calculate its area.
Fig. 13.16
7. A doorway of a church is made up of a
circular arc PQR which subtends an angle
of 60° at the centre O, two upright bars PT
and RS each of length 4 m, and a horizontal
bar ST of length 3 m (Fig. 13.17).
Using π = 3.142, nd the area of the
doorway in m
2
.
Fig. 13.17
15 cm
54°
Likewise in ∆PAB,
sin
β
= = 0.571 4
∴
β
≈ 34.85°
⇒
β
= 69.7°
The height of ∆ABP = √7
2
– 4
2
≈ 5.74 cm
Hence, area of segment ANB
= Area of sector PANB – Area of ∆PAB
= × 3.142 × 7
2
– × 8cm × 5.74cm
= (29.81 – 22.96) cm
2
= 6.85 cm
2
Hence, common area of intersection
= area of segment AMB + area of
segment ANB
= 11.17 + 6.83
= 18 cm
2
Exercise 13.2
1. Calculate the area of the shaded part in
Fig. 13.14. (Use π = 3.142.)
Fig. 13.14
2. Calculate the area of the segment bounded
by an arc and a chord that subtend an angle
of 63° at the centre of a circle of radius
10 cm. (Use π = .)
3. Find the shaded area in Fig 13.15(a) and
hence nd the shaded area in Fig. 13.15(b).
(a) (b)
Fig. 13.15
– – – – – – – – –
81°
12 cm
– – – – – – – – –
125
60°
6 cm
60°
X Z
Y
10.5 cm
– – – – – – – – – – – – –
– – – – – – – – – – – – –
– – – – – – – – – – – – –
Fig. 13.19
10. Two circles of radii 7 cm and 9 cm intersect
such that they share a common chord of
length 10 cm. Find the common area of
intersection of the circles.
11. Two circles, each with radius 8 cm and
centres P and Q, intersect at A and B, with
∠APB = ∠AQB = 90°. Using π = 3.142,
calculate the common area of intersection of
the two circles.
8. Fig. 13.18 shows a closed shape comprising
of three arcs such that X Y and Z are
the centres of the circular arcs YZ, XZ
and XY respectively. If triangle XYZ is
an equilateral triangle of side 10.5 cm,
calculate the area enclosed by the arcs.
Fig. 13.18
9. Find the area of intersection of the circles
in Fig. 13.19. (Use π = 3.142.)
126
Example 14.1
A very thin sheet of metal is used to make a
cylinder of radius 5 cm and height 14 cm.
Using π = 3.142, nd the total area of the sheet
that is needed to make:
(a) a closed cylinder
(b) a cylinder that is open on one end.
Solution
(a) Radius of the circular face of the cylinder
= 5 cm
∴area of a circular face
= πr
2
= (3.142 × 5
2
) cm
2
∴area of the two circular end faces
= 2 × πr
2
= (2 × 3.142 × 5 × 5) cm
2
Recall that when a cylinder is opened
up to form its net, the curved surface
becomes a rectangle of length 2πr (i.e. the
circumference of the cylinder) and width h
(the height of the cylinder).
Thus, area of curved suface = 2πr × h
= 2 × 3.142 × 5 × 14 cm
2
Now, total area of sheet
= 2 × 3.142 × 5 × 5 + 2 × 3.142 × 5 ×
14 cm
2
= 2 × 3.142 × 5(5 + 14) cm
2
= 596.98 cm
2
(b) Surface area of open cylinder = πr
2
+ 2πrh
= 3.142 × 5 × 5 + 2 × 3.142 × 5 × 14
= 3.142 × 5(5 + 2 × 14)
= 3.142 × 5 × 33
= 518.43 cm
2
– – – – – – – – – – – – –
h
l
– – – – – – – – – – – – – – – –
– – – – – – – – –
– – – – – – –
b
Remember that in Form 1, you made models of
solids such as cubes, cuboids, cones, pyramids
and prisms. You learnt that such models could
be opened up and laid out at to form nets of
the respective solids.
Using nets, you found the surface areas of some
of the solids as the areas of their nets. This led
us to observe that the surface area of a solid
is obtained as the sum of the areas of all the
surfaces of the solid.
Surface areas of prisms
Recall that a prism is a solid which has
a uniform cross-section. Thus, we have
rectangular prisms (i.e. cubes and cuboids),
triangular prisms, circular prisms (i.e.
cylinders), and so on. Fig. 14.1 shows some
prisms.
Fig. 14.1
The surface area of a prism is found as
follows:
1. Find the area of the cross-section and
multiply it by 2.
2. Find the area of each rectangular side face
and add up these areas, or nd the area of
the curved surface in the case of a cylinder.
3. Add up the results to get the surface area
of the prism.
r
SURFACE AREA OF SOLIDS
14
Note: From the working in example 14.1,
we are reminded that:
Total surface area of closed cylinder
= 2πr
2
+ 2πrh = 2πr(r + h)
127
Area of each slanting face
= × 5 × √7
2
– 2.5
2
)
= × 5 × √49 – 6.25)
= × 5 × √42.75
= × 5cm × 6.538 cm
= 16.345 cm
2
∴ Total area of the slanting faces
= 4 × 16.345 cm
2
= 65.38 cm
2
1
2
–
1
2
–
1
2
–
1
2
–
Exercise 14.1
1. Calculate the total surface areas of the solids
in Fig. 14.2. (All measurements are in cm.)
8
3
4
7
8
16
9
7
10
15
4
14
End-face is a regular
hexagon of side 6 cm.
Fig. 14.2
2. Calculate the total surface area of a solid
cylinder whose radius and height are 9 cm and
12 cm, respectively, leaving π in your answer.
3. A paper label just covers the curved surface of
a cylindrical can of diameter 14 cm and height
10.5 cm. Calculate the area of the paper label.
4. The surface area of a cuboid is 586 cm
2
.
Given that its length and height are 12 cm
and 7 cm respectively, nd its breadth.
5. The solid in Fig. 14.3 has a total surface
area of 257 cm
2
. Find its length, l.
Fig. 14.3
(a) (b)
(e) (f)
(c) (d)
l
5 cm
6 cm
6.5 cm
4 cm
– – – – – – – – – – – –
– – – – – – – – – – – – – –
– – – – – – – – – – –
– – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – –
5 cm
5 cm
7 cm
– – – – – – – – – – – – – – – –
9
5
3
7
21
7 cm
2.5 cm
– – – – – – – – – – – – – – – – – – –
2.5 cm
h
Surface area of a pyramid
The surface area of a pyramid is obtained as
the sum of the areas of the slant faces and
the base.
Example 14.2
Find the surface area of the right pyramid
shown in Fig. 14.4.
Fig. 14.4
Solution
Area of the base = 5cm × 5cm = 25 cm
2
Each slanting face is an
isosceles triangle.
(Fig. 14.5).
Its height is √7
2
– (2.5)
2
cm
Since h
2
= 7
2
– 2.5
2
Fig. 14.5
128
Surface area of a cone
Activity 14.1
1. Draw a circle, radius l (say l = 10 cm ). At the
centre O of the circle, measure an angle AOB
(say of 150°) and form a sector as shown in
Fig. 14.7 (a) (shaded part). Cut out the sector.
2. Fold the sector so that OA and OB coincide.
This forms the curved surface of a cone as
shown in Fig. 14.7(b).
Such a cone is said to be a right-circular
cone since the base is a circle and the vertex
is vertically above the centre of the base. The
word ‘right’ here means ‘upright’.
(a) (b)
Fig. 14.7
O
AB
l
– – – – – – – – – – – – – – – –
r
– – – – – – –
h
– – – – – – – – – – – – –
– – – – – – – – –
– – – – – – –
– – – – – –
– – – – – – – – –
5 cm
– – – – – – – – –
12 cm
5 cm
5 cm
– – – – – – – – – –– – – – – – – –
– – – – – – – –
– – – – – – – –
– – – – – – – – – –
– – – – – – – – – – –
8 cm
8 cm
– – – – – – – – – – –
15 cm
8 cm
A
C
B
O
Total surface area of the pyramid
= (25 + 65.38) cm
2
= 90.4 cm
2
. (3 s.f)
Note: The area of a slant face could also be
found using Hero’s formula.
Exercise 14.2
In Questions 1–9, nd the total surface area of
the given right pyramid.
1. Height 4 cm; square base, side 6 cm.
2. Height 6 cm; square base, side 9 cm.
3. Height 5 cm; rectangular base, 6 cm by 4 cm.
4. Height 6 cm; rectangular base, 4 cm by 5 cm.
5. Height 16 cm; triangular base, sides 6 cm,
8 cm, 10 cm.
6. Slant edge 12 cm; rectangular base 6 cm
by 8 cm.
7. Height 10 cm; equilateral triangular base,
side 6 cm.
8. Slant edge 4cm; square base, side 4 cm.
9. Slant height 8 cm; square base, side 5.3 cm.
10. Fig. 14.6 shows solids which comprise of
cubes surmounted with pyramids.
Calculate the surface area of each solid.
(a)
(b)
The top is a right pyramid.
Fig. 14.6
3. What fraction of the circumference is arc
ACB in Fig. 14.7(a)? Calculate the length
of the arc.
4. What relationship is there between the
length of arc ACB and the circumference
of the base of the cone in Fig. 14.7(b)?
5. Using the relationsip in 4, calculate the
length of the radius of the base of the cone
in Fig. 14.7(b).
6. Using your result in 5, calculate the ratio .
7. Find, in terms of l the circumference of the
circle in Fig. 14.7(a).
Also, nd in terms of r the circumference
of the base of the cone in Fig. 14.7(b).
Hence, write down, in terms of l and r, the
ratio.
r
l
–
129
= 3.142 ×3
2
cm
2
= 28.278 cm
2
Hence, total surface area = πr
2
+ πrl
= (28.278+47.13) cm
2
= 75.408 cm
2
= 75.41 cm
2
(2 d.p)
Exercise 14.3
In this exercise, take π = 3.142.
In Questions 1 to 7, nd the surface area of the
given cone.
1. Slant height 8 cm; base radius 6 cm
2. Slant height 13 cm; height 5 cm
3. Height 8 cm; base diameter 12 cm
4. Height 8 cm; base radius 3 cm
5. Slant height 8.5 cm; height 6.5 cm
6. Slant height 9 cm; perimeter of base 12 cm
7. Height 4 cm; area of base 15 cm
2
8. A circle has a radius of 5 cm. The length of the
arc of a sector of the circle is 6 cm.
r
l
–
r
l
–
r
l
–
Find the:
(a) area of the sector
(b) surface area of the closed cone made
using this sector.
9. The height of a conical tent is 3 m and the
diameter of the base is 5 m. Find the area
of the canvas used for making the tent.
Frustum of a cone or pyramid
If a cone or pyramid is cut through a plane
parallel to its base, the top part will be a smaller
cone or pyramid. The bottom part is called a
frustum or frustrum of the cone or pyramid.
(See Fig. 14.8.)
Circumference of base of cone
Circumference of circle
8. What fraction of the area of the circle is
the sector shaded in Fig. 14.7(a)? What is
this fraction in terms of l and r?
9. Hence, what is the area of the curved
surface of the cone? Give your answer in
terms of l and r.
You should have found that:
(a) area of the curved surface of the cone =
area of the sector
(b) circumference of base of cone
circumference of circle
=
area of sector
=
area of circle
Hence, area of sector = of the area of the
circle = of πl
2
= πrl.
∴ area of curved surface of a cone = πrl.
Hence, total surface area of a closed cone
= πr
2
+ πrl.
Note, the values h
1
,r and l (Fig 14.7) are
connected by the relation l
2
= h
2
+ r
2
.
Example 14.3
Find the surface area of a cone whose height
and slant height are 4 cm and 5 cm respectively.
(Use π = 3.142.)
Solution
l = 5 cm, h = 4 cm
Since l
2
= h
2
+ r
2
, then r
2
= l
2
– h
2
i.e. r
2
= 5
2
– 4
2
∴ r = 3 cm
Area of curved surface = πrl.
= 3.142 × 3cm × 5 cm
= 47.13 cm
2
Area of circular base = πr
2
130
Example 14.4
Fig. 14.9 shows a lampshade in the form of a
frustum whose top and bottom diameters are
18 cm and 27 cm. Find the area of the material
used in making it, if the vertical height is 12 cm.
AB
BC
–––
x + 12
13.5
–––––
AD
DE
–––
x
9
––
27x
2
–––
27 cm
18 cm
12 cm
cut
– – – – – – –
– – – – – – – – – –
– – – – – – – – – – – –
– – –
– – –
– – – – – –
– – – – – –
– – – – –
– – – – –
– – – – – – –
– – –
– – –
– – – – – – –
– – – – – – – – – –
– – – – –
– – – – –
– – – – –
– – – – –
– – – – –
frustum
– – – – – –
– – – – – –
(a)
(b)
Fig. 14.8
To nd the surface area of a frustum:
1. Extend the slant height of a frustum of
a cone or the slant edges of the frustum
of a pyramid to obtain the solid from
which the frustum was cut.
2. Find the surface area of the complete
solid.
3. Find the curved surface area of the small
cone that was cut off or the total area of
the side faces of the small pyramid that
was cut off.
4. Subtract the area obtained in (3) from
that obtained in (2).
5. Find the area of the top face of the
frustum and add it to the result in (4).
Fig. 14.9
Solution
Fig. 14.10 shows the complete cone of which
the lampshade is a frustrum, which is open at
both ends.
BC = 9 cm, DE = 13.5 cm, BD = 12 cm.
Let AB = x cm.
Using similar triangles,
=
⇒ =
⇒ 13.5x = 9x + 9 × 12
⇒ = 9x + 108
⇒ 27x = 18x + 216
9x = 216
x = 2.
i.e. AB = 24 cm
Using Pythagoras’ theorem, AC
2
= AB
2
+ BC
2
= 24
2
+ 9
2
∴ AC = 25.63 cm
Similarly, AE
2
= 36
2
+ 13.5
2
∴ AE = 38.45 cm
Surface area of the lampshade (frustum)
= (π × 13.5 × 38.45 – π × 9 × 25.63) cm
2
= π(519.1 – 230.7)cm
2
= 3.142 × 288.4 cm
2
= 906.2 cm
2
Fig. 14.10
– – – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – –
– – – – – – – – –
– – – – – –
C
A
B
D
E
x cm
12 cm
9 cm
13.5 cm
Exercise 14.4
1. A frustum of a solid pyramid has a square
base of side 8 cm and a square top of side 6
cm. The height between the two ends is 2
cm. Calculate the surface area of the frustum.
131
Exercise 14.5
1. Calculate the surface area of a sphere whose
radius is:
(a) 3.2 cm (b) 1.2 cm (c) 4.2 cm
2. Find the radius of a sphere whose surface
area is:
(a) 78.5 cm
2
(b) 181 cm
2
3. Find the total surface area of a solid
hemisphere of diameter 10 cm.
4. A hollow sphere has an internal diameter of
18 cm and a thickness of 0.5 cm. Find the
external surface area of the sphere.
5. Figure. 14.14 shows a composite solid made
of a cylinder and a hemisphere. Find its
total surface area.
Fig. 14.14
14 cm
20 cm
The surface area of the sphere is given by:
Surface area = 4πr
2
square units.
Recall that half of a sphere is known as a
hemisphere.
Surface area of hemisphere
= half the area of the sphere + area of
at surface
= × 4πr
2
+ πr
2
= 3πr
2
Example 14.5
A solid hemisphere has a radius of 5.8 cm. Find
its surface area.
Solution
Surface area of hemisphere = 3πr
2
= (3 × 3.142 × 5.8 × 5.8) cm
2
= 317.1 cm
2
(4 s.f.)
– – – – – – – – –
– – – – – – – – –
r
r
1
2
–
8 cm
2 cm
6 cm
27 cm
18 cm
10 cm
2. A bucket is in the shape of a frustum of a
cone. Its diameters at the bottom and top
are 30 cm and 36 cm respectively. If its
depth is 20 cm, nd the area of the sheet of
material used in making the bucket.
3. Figure 14.11 shows a conical ask whose
external base diameter is 8 cm. The external
diameter of the mouth is 2 cm. Assuming that
the neck is cylindrical and ignoring the brim,
nd the external surface area of the ask.
Fig. 14.11
4. Fig. 14.12 shows a frustum of a solid cone.
Find the surface area of the frustum.
Fig. 14.12
5. A dustbin is in the shape of a frustum of
a right pyramid, with a square top of side
32 cm and square bottom of side 20 cm. If
the dustbin is 24 cm deep, nd the area of
the sheet of material used to make it.
Surface area of a sphere
Fig. 14.13 represents a solid sphere of radius r units.
Fig. 14.13
132
REVISION EXERCISES 2
8-14
Revision exercise 2.1
1. (a) Triangles PQR and XYZ are similar.
PQ = 24 cm, QR = 32 cm, RP = 36 cm
and YZ = 8 cm. What are the lengths of
the remaining sides of ∆XYZ?
(b) In Fig. R. 2.1, ∆s BDC and ADC are
similar.
(i) What is the ratio of the lengths
of corresponding sides?
(ii) What is the ratio of the areas?
Fig. R.2.1
2. Two cuboids are similar, with linear scale
factor 2. One has a surface area of 88 cm
2
and a volume of 48 cm
3
. What are the two
possible areas and volumes of the other?
3. In a right-angled triangle, the length of the
hypotenuse is x cm, and the lengths of the
other two sides are y and z.
(a) If z = 4.2. and y = 6.8, nd x.
(b) If y = 5.73 and z = 5.14, nd x.
(c) If x = 8.06 and y = 5.67, nd z.
(d) If x = 37.08 and z = 23.26, nd y.
4. (a) The sides of a rectangle are 6.75 cm
and 5.42 cm long. Find the length of a
diagonal of the rectangle.
(b) The length of a diagonal of a rectangular
courtyard is 32.7 m. The length of one
side is 18.5 m. Find the perimeter of
the yard.
51°
3 cm
a
51°
8 cm
d
63°
10 cm
c
5. (a) Find, by drawing, the value of:
(i) tan 20° (ii) sin 40°
(iii) tan 45° (iv) cos 60°
(b) Find the lengths of the sides marked by
letters in Fig. R. 2.2.
(i) (ii)
(iii) (iv)
Fig. R. 2.2
6. Use tables to write down:
(a) the angle whose:
(i) tangent is 0.425 5
(ii) sine is 0.275 4
(iii) cosine is 0.500 4
(iv) tangent is 5.217 5
(b) the value of:
(i) cos 66° 20′ (ii) tan 50° 18′
(iii) sin 81.55° (iv) tan 25.66°
7. Find the area of ∆PQR in which PQ = 6 cm,
QR = 7 cm, and ∠PQR = 34°.
8. In ∆ABC, AC = 5 cm, BC = 6 cm and
∠ACB = 118°. Find the area of the triangle.
36°
5 cm
b
– – – – – – – – – – – – – –
A
6 cm
B
C
D
8 cm
133
9. PQRS in Fig. R. 2.3, is a trapezium in
which PS||QR, PQ = 5 cm, QR = 15 cm
and PS = 9 cm. Find its area.
Fig. R.2.3
10. A pentagon ABCDE is such that AB = 6 cm,
BC = 9 cm, CD = 4.5 cm, AE = 12 cm,
∠BAE = ∠BCD = 90° and ∠DBE = 36°.
Find the area of the pentagon.
11. A chord of a circle, of radius 8 cm, subtends
an angle of 55° at the centre of the circle.
Calculate the area of the minor segment cut
off by the chord.
12. A test-tube has a hemispherical bottom of
radius 1cm and a height of 12 cm. Ignoring
the thickness of the material of which it is
made, nd its surface area.
Revision exercise 2.2
1. (a) An aeronautics engineer made a model
of a new plane to a scale of 1:20. Copy
and complete Table R. 2.1.
Model Plane
Length of body 11.2 m
Wing span 48 m
Number of engines 4
Surface area of wings 840 m
2
Number of seats 60
Table R. 2.1
(b) In Fig. R 2.4, ST//QR, PQ = QR = 14 cm,
ST = 4 cm and PT = 6 cm. Find the
lengths marked x and y.
Fig. R. 2.4
2. A cone of base radius 3 cm has a surface
area of 48 cm
2
and a mass of 0.75g.
Calculate the surface area and the mass of a
similar cone made of the same material and
with a base radius 9 cm.
3. (a) Complete the following Pythagorean
triples.
(i) (23, …, …) (ii) (29, …, …)
(iii) (26, …, …) (iv) (30, …, …)
(b) Use the following numbers to generate
Pythagorean numbers.
(i) 3 , 5 (ii) 7 , 2 (iii) 3 , 8 (iv) 5 , 6
4. Find by drawing, the angle whose:
(a) tangent is (b) sine is 0.6
(c) cosine is 0.3 (d) tangent is 2.3
5. From the top A of a tower AN, 75 m high,
the angle of depression of a point B is 49°.
Find the distance of B from N, giving your
answer correct to 1 d.p.
6. Given that the area of ∆XYZ is 20.4 cm
2
,
nd the length marked x. (Fig. R. 2.5)
Fig. R.2.5
1
4
–
T
P
R
Q
S
14 cm
14 cm
6 cm
x
y
4 cm
6 cm
x
Y
X
104°
Z
5 cm
R
S
P
Q
30°
15 cm
9 cm
134
7. Find the area of ∆ABC with sides 8 cm,
9 cm and 10 cm.
8. ABCD is a parallelogram such that AB
= 9 cm, AD = 7 cm and ∠ADC = 58°.
Calculate the area of ABCD.
9. A regular octagon has an area of 101.8 cm
2
.
Find to the nearest centimetre:
(a) the distance from its centre to a vertex.
(b) the length of a side.
10. A chord of a circle, radius 17.5 cm,
subtends an angle of 75° at the centre of
the circle. Calculate the area of the major
segment formed by the chord.
11. Find the surface area of the tetrahedron in
Fig. R.2.6, given that all measurements are
in cm.
Fig. R.2.6
12. A cone has a diameter of 14 cm and a
height 24 cm.
(a) Find the curved surface area of the cone.
(b) If the cone is made of paper, and the
paper is opened up and attened into a
sector of a circle, what is the sector
angle?
Revision exercise 2.3
1. X(2 , 2), Y(4 , 2) and Z(3 , 4) are the
vertices of a triangle. The triangle is
enlarged by a scale factor 2 and centre the
origin. What are the coordinates of the
vertices of the image ∆X′Y′Z′?
2. On squared paper, plot A(4 , 5), B(4 , 3)
and C(8 , 1) and draw ∆ABC. Plot also
A′(2 , 3) and B′(2 , 2), the images of A and
B under a certain enlargement.
By appropriate construction locate P, the
centre of enlargement. Locate also point C′,
a vertex of ∆A′B′C′. State the coordinates
of P and C′ and nd the scale factor of
enlargement.
3. Show that each of the triangles in Fig. R. 2.7
is right-angled. Hence write down the
tangent, sine and cosine of each of the
angles marked by the Greek letters.
(a) (b)
(c)
Fig. R. 2.7
4. A boat sails due north for 7.8 km and then
a further 12 km on a bearing of 090°.
Calculate the shortest distance of the boat
from the starting point.
5. (a) A man starts from a point A and walks
4.5 km due south to B and then 5.5 km
due west to C.
Find the bearing of C from A, stating
your answer in degrees and minutes.
5 cm
α
3 cm
4 cm
β
25 cm
β
α
γ
24 cm
7 cm
6 cm
5 cm
4 cm
– – – – – – – – – – – – – – – –
20 cm
β
α
γ
12 cm
9 cm
16 cm
15 cm
135
(b) The string of a kite is 80 m long,
and makes an angle of 72° with the
horizontal. How high is the kite?
6. The legs of a pair of dividers are each
9.5 cm long. The pair of dividers is opened
to an angle of 54°. What is the distance
between the points?
7. Find the area of the quadrilateral in
Fig. R. 2.8
Fig. R.2.8
8. Three ranger patrols A, B and C lay ambush
for poachers, in a certain national park, by
positioning themselves such that patrol B is
50 km on a bearing of 043° from patrol A,
and patrol C is on a bearing of 133° from A
and 160° from B.
(a) Make a sketch of a diagram showing
their relative positions.
(b) Calculate the triangular area within
their patrol.
B
40°
18 cm
C
D
43°
A
9. The area of a regular pentagonal region is
246.2 m
2
. Find the length of a side of the
region, giving your answer to the nearest
metre.
10. Two circles, centres A and B have radii
13.5 cm and 10.5 cm, respectively, and
intersect at points P and Q. If ∠PAQ = 67.5°
and ∠PBQ = 91°, calculate the common
area between the two circles.
11. Find the total surface area of each of the
solids shown in Fig. R.2.9.
(a) (b)
Pyramid Hemisphere
Fig. R.2.9
12. An open container is in the form of a
frustum of a right pyramid 4 m square at
the top, 2.5 m square at the bottom and 3 m.
What is the surface area of the material
used in making the container? (Ignore the
thickness of the material).
– – – – – – – –
– – – – – – – – – – – – – –
– – – – – – – – – – – –
6 cm
6 cm
4 cm
– – – – – – – – – – – –
6 cm
136
Volume of a prism
Note:
Volume of a cube
= l
3
, where l is the length of a side.
Volume of a cuboid
= lbh, where l is the length, b the breadth
and h the height of the cuboid.
Volume of a cylinder
= πr
2
h, where r is the radius and h is the
height of the cylinder.
A cuboid has a uniform cross-section (shaded in
Fig. 15.1) of area bh.
Fig. 15.1
Since the volume of a cuboid = lbh, then
volume of cuboid = l × area of cross-section.
We also learnt that the volumes of solids which
have uniform cross-sections which are not
rectangular are calculated in the same way as
that of a cuboid, i.e.
Volume = cross-section area × length.
Do you recall that solids which have a uniform
cross-section are known as prisms? The
following are examples of prisms. (Fig. 15.2).
Fig. 15.2
In general, for any prism:
Volume
= area of uniform cross-section × length
(or height) of the prism.
Example 15.1
The end-face of a beam is shaped as in Fig. 15.2,
with the measurements being in centimetres. If
the length of the beam is 5 m and all angles are
right angles, nd its volume.
Fig. 15.2
Solution
The end-face may be taken as comprising of two
rectangles.
∴ area of end-face = (14 × 4 + 8 × 4) cm
2
= 88 cm
2
length of beam = 5 m = 500 cm.
∴ volume of beam = 88 cm
2
× 500 cm
= 44 000 cm
3
Exercise 15.1
1. Calculate the volumes of the solids in Fig.
14.2 in Exercise 14.1 of Chapter 14.
2. A wooden beam has a rectangular cross-
section measuring 21 cm by 16 cm and is 4 m
long. Calculate the volume of the beam,
giving your answer in cm
3
and in m
3
.
8
5
5
14
4
b
h
l
VOLUMES OF SOLIDS
15
137
8. The volume of a prism with a regular
pentagonal base is 4 755 cm
3
. If the prism
is 1 m long, nd the distance in centimetres,
from the centre of the base to any of its
vertices.
Volume of a pyramid
By going through the following activity, let
us nd out how to determine the volume of a
pyramid.
Activity 15.1
Work in groups of three. Every member in the
group should construct a net of a square based
pyramid as shown in Fig. 15.4.
All measurements are in centimetres.
Fig. 15.4
Cut the net out and fold the triangles up to form
the pyramid.
Arrange the three pyramids to make a cube.
What is the volume of the cube?
What is the volume of each pyramid?
From Activity 15.1:
volume of the pyramid
= × volume of the cube
(Since the three pyramids are identical.)
Do you think this result is true for a cube of any
size?
9
1
1
10
8
8
3. A block of concrete is in the shape of a
wedge whose triangular end-face is such
that two of its sides are 16 cm and 19 cm
long and the angle between them is 50°. If
the block is 1 m long, nd its volume.
4. Figure 15.3 shows the shapes of cross-
sections of steel beams often used in
construction of buildings. Calculate the
volume of a 6 m length of each beam, given
that all dimensions are in centimetres and
that all angles are right angles.
(a) (b)
(c)
Fig. 15.3
5. A cylindrical container has a diameter of
14 cm and a height of 20 cm. Using π = ,
calculate how many litres of liquid it holds
when full.
6. If 8 800 litres of diesel are poured into a
cylindrical tank whose diameter is 4 m,
calculate the depth of diesel in the tank.
(Take π =
22
7
––
)
7. The volume of the prism in Fig. 15.4 is
1 170 cm
3
. Find its length.
Fig. 15.4
12
14
10
12
22
7
––
1
3
–
8
8
1
1
3
6
3
6
8.5
8.5
8.5
6
6
G
H
D
C
6
6
6
8.5
E
F
5 cm
8 cm
5 cm
B
138
Since the volume of a cube
= base area × height,
then the volume (V) of a pyramid is given by
V = Ah
where A = Area of the base, and
h = the vertical height of the
pyramid
Example 15.2
Fig. 15.5 shows a pyramid on a rectangular
base. Find its volume given that the dimensions
are in centimetres.
Fig. 15.5
Solution
Volume of a pyramid = (base area) × height
∴ V = (7cm × 5cm) × 6 cm
= 70 cm
3
Note: If, in Fig. 15.5, we are given the slant
edge, say VB, we need to use Pythagoras’
theorem to nd half the diagonal of the base
and hence the vertical height. Similarly, given
the slant height, say VM, we use Pythagoras’
theorem to nd the vertical height.
Exercise 15.2
In Questions 1 to 9, calculate the volume of the
given right pyramid.
1. Height 4 cm; square base, side 6 cm.
2. Height 6 cm; square base of side 9 cm
3. Height 5 cm; rectangular base, 6 cm by 4 cm.
4. Height 6 cm; rectangular base, 4 cm by
5 cm.
5. Height 16 cm; triangular base, sides 6 cm,
8 cm and 10 cm.
6. Slant edge 12 cm; rectangular base, 6 cm
by 8 cm.
7. Height 10 cm; equilateral triangle base, side
6 cm.
8. Slant edge 4 cm; square base, side 4 cm.
9. Slant height 8 cm; square base, side 5.3 cm.
10. A pyramid whose height is 8 cm has a
volume of 48 cm
3
. What is the area of its
base?
11. A pyramid has a square base of side 5 cm.
What is its height if its volume is 100 cm
3
?
12. A square-based pyramid has a height of
6 cm and a slant edge of 8 cm. What is its
volume?
13. Calculate the volume of each of the solids
in Fig. 14.6 in Exercise 14.2 of Chapter 14.
Volume of a cone
A cone may be regarded as a right pyramid with
a circular base.
Volume of cone
= × base area × height = πr
2
h.
Example 15.3
Find the volume of a cone whose height and
slant height are 4 cm and 5 cm, respectively.
(Take π = 3.142.)
1
3
–
1
3
–
– – – – – – – – – – – – –
– – – – – – – – – – – – – –
– – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
7
5
– – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – –
A
V
D
M
B
C
6
•
1
3
–
1
3
–
1
3
–
139
Solution
Fig 15.6 is a sketch of a cone
l = 5 cm and h = 4 cm
Hence, r
2
= l
2
– h
2
= 5
2
– 4
2
∴ r = 3 cm
Volume of cone = πr
2
h
= × 3.142 × 3cm × 3cm × 4cm
= 37.7 cm
3
Exercise 15.3
In Questions 1 to 7, nd the volume of the
given cone. Take π = 3.142.
1. Height 4 cm; area of base 15 cm
2
2. Slant height 8 cm; base radius 6 cm
3. Slant height 13 cm; height 5 cm
4. Height 8 cm; base diameter 12 cm
5. Height 8 cm; base radius 3 cm
6. Slant height 8.5 cm; height 6.5 cm
7. Slant height 9 cm; perimeter of base 12 cm
8. Find the height of a cone whose base radius
is 3.72 cm and whose volume is 143 cm
3
.
9. The area of a sector of a circle of radius
4 cm is 20 cm
2
. What is the length of the
arc of the sector? Find the radius and
volume of the cone made using this sector.
10. The height of a conical tent is 3 m and
the diameter of the base is 5 m. Find the
volume of the tent.
11. The capacity of a conical tank is 66 m
3
.
The area of the circular base on which it
stands is 18 m
2
. Find the area of the surface
of the tank.
Volume of a frustum
The volume of a frustum is found as follows:
1. Extend the slant height of the frustum of
a cone or the slant edges of the frustum of
a pyramid to obtain the solid from which
the frustum was obtained.
2. Find the volume of the complete solid.
3. Find the volume of the small cone or
pyramid that was cut off.
4. Find the difference between the two
volumes to get the volume of the frustum.
Example 15.4
A bucket that is in the shape of a frustum of a
cone has a top radius of 12 cm and a bottom
radius of 8 cm. If it is 20 cm deep, nd its
capacity in litres.
Solution
Fig. 15.6 shows a sketch
of the bucket with the
cone from which the
frustum was obtained
shown in dotted lines.
Recall that in similar
triangles, the ratios of
corresponding sides
are equal.
Now, in Fig. 15.6, ∆s PRT
and QRS are similar.
∴ =
i.e. =
⇒12x = 8x + 160
4x = 160
Hence x = 40
QR
PR
–––
QS
PT
–––
x
x + 20
–––––
8
12
––
1
3
–
1
3
–
20 cm
x cm
Q
P
12 cm
– – – – – – –
– – – – –
– – – – – – – – – – – – – – – – – – – – – – –
R
Fig. 15.7
8 cm
– – – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
S
T
Fig. 15.6
4 cm
5 cm
140
5. Find the volume of the frustum of a solid
cone shown in Fig. 15.9.
Fig. 15.9
Volume of a sphere
Let A represent a small square area on the
surface of a sphere of radius r, centre O
(Fig. 15.10).
Fig. 15.10
If A is very small, we can look at it as almost at.
The solid formed by joining the vertices of A to
the centre O is a small ‘pyramid’.
Volume of the small ‘pyramid’ = Ar
Let there be such small ‘pyramids’ with base
areas A
1
, A
2
, A
3
, …
Their volumes are A
1
r, A
2
r, A
3
r, …
Total volume = r(A
1
+ A
2
+ A
3
+ …).
For the whole surface of the sphere, the sum of
all the base areas is 4πr
2
,
i.e., A
1
+ A
2
+ A
3
+ … = 4πr
2
.
Hence, total volume V of a sphere = r × 4πr
2
Volume of a sphere = πr
3
.
1
3
–
1
3
–
1
3
–
8 cm
4 cm
8 cm
(
)
20 cm
10 cm
12 cm
– – – – – – – –
– – – – – – –
– – – – – – – –
O
•
A
1
3
–
1
3
–
1
3
–
1
3
–
1
3
–
1
3
–
4
3
–
Volume of complete cone
=
1
3
–
× π × 12
2
× 60 cm
3
Volume of smaller cone
=
1
3
–
× π × 8
2
× 40 cm
3
∴ Volume of bucket (the frustum)
= × π × 12
2
× 60 – × π × 8
2
× 40 cm
3
= × π (8640 – 2560) cm
3
= 6 368 cm
3
(4 s.f.)
Hence, capacity of bucket = 6.368 litres.
Exercise 15.4
1. A frustum of a solid pyramid has a
rectangular base of sides 8 cm and 6 cm and
a rectangular top of side 4 cm and 3 cm.
Given that the vertical height of the frustum
is 5 cm, calculate the volume of the frustum.
2. A bucket, that is in the shape of a frustum,
has a diameter of 21 cm at the bottom and
28 cm at the top and its height is 40 cm.
Find the capacity of the bucket.
3. Find the capacity of the conical ask in
Question 3 of Exercise 14.4, Chapter 14.
4. Fig. 15.8 shows a conical salt-shaker which
contains some salt. Calculate the volume of
salt that is required to ll the shaker.
Fig. 15.8
141
Example 15.5
A solid hemisphere of radius 5.8 cm has density
10.5 g/cm
3
.
Calculate the: (a) volume,
(b) mass, in kg, of the solid
Solution
(a) Volume of hemisphere
= × volume of sphere
= × πr
3
= πr
3
= ( × 3.142 × 5.8
3
) cm
3
= 408.7 cm
3
(4 s.f.)
(b) Density =
Mass
Volume
∴Mass = Density × Volume
= 10.5 × 408.7 g
= 4 291.35 g
= 4.291 kg (4 s.f.)
Exercise 15.5
Take π = 3.142 or
22
7
––
.
1. Calculate the volume of a sphere whose
radius is:
(a) 3.2 cm (b) 1.2 cm (c) 4.2 cm
2. Find the radius of a sphere whose volume is:
(a) 73.58 cm
3
(b) 463 cm
3
3. Calculate the volume of a sphere whose
surface area is:
(a) 21.2 cm
2
(b) 972 cm
2
4. What is the volume of a solid hemisphere of
diameter 10 cm?
5. What is the mass of a solid gold hemisphere
of diameter 4 cm if the density of gold is
19.3 g/cm
3
?
6. A hollow sphere has an internal diameter of
18 cm and a thickness of 0.5 cm. Find the
volume of the material used in making the
sphere.
7. Ten plasticine balls of diameter 2.8 cm are
rolled together to form one large ball. What
is the radius of the large ball?
8. Ten marbles, each of radius 1.5 cm, are
placed in an empty beaker of capacity
150 ml. Water is then added to ll up the
beaker. What volume of water is added?
9. A solid cylinder has a radius of 18 cm and
height 15 cm. A conical hole of radius r
is drilled in the cylinder on one of the end
faces. The conical hole is 12 cm deep. If the
material removed from the hole is 9% of the
volume of the cylinder, nd:
(a) the surface area of the hole,
(b) the radius of a spherical ball made out
of the material.
2
3
–
4
3
–
2
3
–
1
2
–
1
2
–
142
Simple algebraic expressions
In Form 1, we learnt how to expand expressions
of the form a(x + y). In this case, each term
inside the bracket is multiplied by the number
outside the bracket.
Remember:
If the number outside the bracket is negative,
the sign of each term in the bracket changes
when the bracket is expanded or opened.
Example 16.1
Expand: (a) 3(4x – 5) (b)
–
4(2x + 7)
(c) (2x + 3)(
–
x)
Solution
(a) Note that 4x – 5 = 4x +
–
5, so that
3(4x – 5) = 3 × 4x + 3 ×
–
5 (Multiplication
is distributive over addition.)
= 12x +
–
15
= 12x – 15
(b)
–
4(2x + 7) =
–
4 × 2x +
–
4 × 7
=
–
8x +
–
28
=
–
8x – 28
(Note the sign change
from + to –)
(c) (2x – 3)(
–
x) =
–
x × 2x + (
–
x) × (
–
3)
=
–
2x
2
+ 3x
(Note the sign change
from +
to – and from – to +)
Exercise 16.1
1. Expand the following.
(a) 3(x + 4) (b) 4(7x – 3)
(c)
–
2(8a – 5) (d) 3(4 – 3a)
(e) 3(4a + 5b – c) (f)
–
5(x + 5)
(g) 5(6 – 2y) (h) b(b – 4)
(i) (4b – 1)(–b) (j)
–
x(2x – 3)
(k)
–
x(4x – 5) (l) (4x – 3)xy
(m)
–
2x(
–
3x – 5)
2. Expand and simplify:
(a) 2(x + 1) + 3(x + 2)
(b) 4(x – 5) + 3(2x + 1)
(c) 3(3y – 5) – 7(2y + 3)
(d) –x(x + 5) + 5(x – 5)
(e) 5(2x – 5) + 8(3x – 1)
(f) 3(4 – 5x) – 2(5 – 4x)
(g) y(y + 2) – 3y(y + 2)
(h) t(t – 5) – 5(t – 5)
Binomial expansions
An algebraic expression consisting of two terms
is called a binomial expression. Thus,
3x + 5, a + b, 2x – y are examples of binomial
expressions.
How do we nd the product of two binomial
expressions?
Recall: p(x + y) = px + py
Likewise, (a + b)(x + y) = (a + b)x + (a + b)y
[(a + b) taken as a single quantity]
= ax + bx + ay + by.
QUADRATIC EXPRESSIONS AND
EQUATIONS
16
143
Thus, each term in one bracket must be
multiplied by each term in the other bracket and
the results added.
i.e. (a + b)(x + y) = ax + ay + bx + by
Example 16.2
Multiply and simplify: (a) (x + 2)(x + 3)
(b) (3x – 2)(2x – 3)
Solution
(a) (x + 2)(x + 3) = x × (x + 3) + 2 × (x + 3)
= x × x + x × 3 + 2 × x + 2 × 3
(Each term in rst bracket multiplied
by each term in second bracket)
= x
2
+ 3x + 2x + 6
like terms
= x
2
+ 5x + 6 (Combine like terms)
(b) (3x – 2)(2x – 3) = 3x(2x – 3) – 2(2x – 3)
= 6x
2
– 9x – 4x + 6
like terms
= 6x
2
– 13x + 6
Exercise 16.2
Find the products of the following binomial
expressions and simplify where possible.
1. (a) (x + 2)(x + 5) (b) (x + 5)(x + 1)
(c) (x – 2)(x – 3) (d) (a – b)(a – b)
(e) (x + 4)(x – 4) (f) (x + 3)(x + 3)
(g) (x + y)(c + d) (h) (a + b)(a + b)
2. (a) (x + 3)(2x + 1) (b) (2x + 3)(3x + 2)
(c) (x + 7)(4x + 1) (d) (3a + 2)(a + 4)
(e) (y – 4)(5y – 3) (f) (4a – 3)(3a – 4)
(g) (7 + 4x)(x + 2) (h) (1 + x)(9 + 5x)
3. (a) (l + m)(l + n) (b) (p + q)(r + 4)
(c) (a + b)(c + z) (d) (m + 3)(m + n)
(e) (a + b)(c – d) (f) (3x – 2y)(x + 2)
(g) (p – q)(3p + 2q) (h) (x + y)(3x – 6y)
Binomial products (quadratic
identities)
Three special binomial products appear so often
in algebra that their expansions can be stated
with minimum computation.
Binomial squares
In arithmetic, we know that 2
2
means 2 × 2 = 4,
3
2
means 3 × 3 = 9, and so on.
In algebra, (a + b)
2
means (a + b) × (a + b).
Thus, (a + b)
2
= (a + b)(a + b)
= a(a +b) + b(a + b)
= a
2
+ ab + ba + b
2
= a
2
+ 2ab + b
2
(Since ab = ba.)
Also (a – b)
2
means (a – b) × (a – b).
Thus, (a – b)
2
= (a – b)(a – b)
= a(a – b) – b(a – b)
= a
2
– ab – ba + b
2
= a
2
– 2ab + b
2
(a + b)
2
and (a – b)
2
are called squares of
binomials or simply perfect squares.
The three terms of the product can be obtained
through the following procedure.
1. The rst term of the product is the square
of the rst term of the binomial, i.e.
(a)
2
= a
2
.
144
2. The second term of the product is two
times the product of the two terms of the
binomial, i.e. 2 × (a × b) = 2ab.
3. The third term of the product is equal
to the square of the second term of the
binomial, i.e. (b)
2
= b
2
.
Thus,
(a + b)
2
= a
2
+ 2ab + b
2
and not a
2
+ b
2
.
This is a common error which must be
avoided.
Similarly,
(a – b)
2
= a
2
– 2ab + b
2
and not a
2
– b
2
.
The square of a binomial always gives a
trinomial, (i.e. an expression having three
terms), also known as a quadratic expression.
A difference of two squares.
A third special product comes from multiplying
the sum and difference of the same two terms.
Consider the product (a + b)(a – b).
(a + b)(a – b) = a(a – b) + b(a – b)
= a
2
– ab + ab – b
2
= a
2
– b
2
(Since ab = ba, then
–
ab + ba =
–
ab + ab = 0.)
The above product may be obtained by:
1. Squaring the rst term of the factors.
2. Subtracting the square of the second
term of the factors.
The result (a + b)(a – b) = a
2
– b
2
is
called a difference of two squares.
The expansions
(a + b)
2
= a
2
+ 2ab + b
2
(a – b)
2
= a
2
– 2ab + b
2
, and
(a + b)(a – b) = a
2
– b
2
are known as quadratic identities.
Use of area to derive the quadratic
identities
In this section, we use the idea of area of a
rectangle to derive the three identities. It is
hoped that this section will help you appreciate
that when expanding the algebraic expressions,
we are looking for areas of some rectangles
(and squares).
Fig. 16.1(a) is a square ABCD with sides of
length (a + b).
Hence area of ABCD = (a + b)
2
………(1)
Fig. 16.1(b) is the same square ABCD
[Fig.16.1(a)]. In it is a small square AEFG of
lengths a.
The square ABCD [Fig. 16.1(b)] can be divided
as shown in Fig. 16.1(c).
(a) (b)
(c)
Fig. 16.1
Hence area of ABCD = area of AEFG + Area of
EBHF + Area of GHCD
Area of AEFG = a
2
Area of EBHF = ab
Area of GHCD = b(a + b) = ab + b
2
Thus area of ABCD = a
2
+ ab + ab + b
2
= a
2
+ 2ab + b
2
………(2)
B
C
A
D
a + b
a + b
a
E
F
G
C
A
D
a
a + b
B
a
E
F
G
B
C
A
D
a
a + b
b
H
– – – – – – – – – –
145
Since area of ABCD = (a + b)
2
(from (1))
Then
(a + b)
2
= a
2
+ 2ab + b
2
Fig. 16.2(b) shows the same square ABCD of
length a.
(a) (b)
Fig.16.2
PQRC is a square contained in ABCD such that
DP = SQ = AT = SA = RB = b
This means PQRC is a square of length (a – b).
Area of PQRC = (a – b)
2
………(1)
But area of PQRC = Area of ABCD – (Area of
DPQS + area of SQTA + area of QTBR)
Area of ABCD = a
2
Area of DPQS = b(a – b) = ab – b
2
Area of SQTA = b
2
Area of QTBR = b(a – b) = ab – b
2
∴ Area of PQRC = a
2
– [ab – b
2
+ b
2
+ ab – b
2
]
= a
2
– (2ab – b
2
)
= a
2
– 2ab + b
2
………(2)
Hence from (1) and (2)
(a – b)
2
= a
2
– 2ab + b
2
Fig. 16.3(a) is a rectangle ABCD with sides of
length (a + b) and (a – b) ………(1)
Hence area of ABCD = (a + b)(a – b)
Fig. 16.3(b) is the same rectangle ABCD in Fig.
16.3(a), with PB = b hence AP = a.
(a)
(b)
Fig.16.3
Area of APQD = a(a – b) = a
2
– ab
Area of PBCQ = b(a – b) = ab – b
2
Area of ABCD
= Area of APQD + area of
PBCQ
= a
2
– ab + ab – b
2
= a
2
– b
2
………(2)
Comparing (1) and (2) we get
(a + b)(a – b) = a
2
– b
2
.
We have seen that given squares of sides (a + b)
and (a – b) and rectangle of sides (a + b) and
(a – b), their areas are given by
(a + b)
2
= a
2
+ 2ab + b
2
(a – b)
2
= a
2
– 2ab + b
2
(a + b)(a – b) = a
2
– b
2
These are known as the three quadratic
identities.
Example 16.3
Find: (a) 12
2
(b) 18
2
(c) 102 × 98
Solution
(a) 12 = 10 + 2
C
B
D
A
a + b
a – b
B
C
A
D
a
a
T
P
S
B
C
A
D
Q
– – – – –
– – – – –
R
b
b
a
C
B
D
A
Q
P
a b
a – b
146
∴ 12
2
= (10 + 2)
2
= 10
2
+ 2 × 10 × 2 + 2
2
= 100 + 40 + 4
= 144
(b) 18 = 20 – 2
∴ 18
2
= (20 – 2)
2
= 20
2
– 2 × 20 × 2 + 2
2
= 400 – 80 + 4
= 324
(c) 102 × 98 = (100 + 2)(100 – 2)
= 100
2
– 2
2
= 10 000 – 4
= 9 996
Execise 16.3
1. Use the quadratic identities to calculate the
following.
(a) 11
2
(b) 29
2
(c) 67
2
(d) 97
2
(e) 21 × 19 (f) 202
2
(g) 501
2
(h) 999
2
2. Use quadratic identities to nd the areas of
the rectangles whose dimensions are:
(a) 33 m by 27 m (b) 104 m by 96 m
(c) 99 m by 101 m (d) 998 m by 1 002 m
The following examples illustrate how to
expand binomial products using quadratic
identities.
Example 16.4
Perform the indicated multiplication and
simplify.
(a) (3x + 2)
2
(b) (4a – 5b)
2
Solution
(a) (3x + 2)
2
= (3x)
2
+ 2(3x)(2) + (2)
2
= 9x
2
+ 12x + 4
(b) (4a – 5b)
2
= (4a)
2
+ 2(4a)(
–
5b) + (
–
5b)
2
= 16a
2
+ (
–
40ab) + 25b
2
= 16a
2
– 40ab + 25b
2
Example 16.5
Perform the indicated multiplications and
simplify.
(a) (x + 2y)(x – 2y) (b) (3a – b)(3a + b)
Solution
(a) In (x + 2y)(x – 2y), the two factors are
(x + 2y) and (x – 2y).
Square of rst term is (x)
2
= x
2
Square of second term is (2y)
2
or
(
–
2y)
2
= 4y
2
The difference is x
2
– 4y
2
(x + 2y)(x – 2y) = x
2
– 4y
2
(b) (3a – b)(3a + b) = (3a)
2
– (b)
2
= 9a
2
– b
2
Exercise 16.4
1. Expand the following using the method of
Example 16.4.
(a) (i) (a + 1)
2
(ii) (a + 6b)
2
(iii) (x + y)
2
(iv) (m + n)
2
(v) (2a + 3b)
2
(vi) (3x + 4)
2
(vii) (4x + 3y)
2
(b) (i) (b – 1)
2
(ii) (4x – 3)
2
(iii) (3x – 12)
2
(iv) (5x – 3)
2
(v) (4z – 3b)
2
(vi) (7x – 2y)
2
2. Expand the following using the method of
Example 16.5.
(a) (a + 3)(a – 3) (b) (a + 5)(a – 5)
(c) (x – 9)(x + 9) (d) (f + g)(f – g)
(e) (2p – 1)(2p + 1) (f) (4x – y)(4x + y)
(g) (7 + 2x)(7 – 2x) (h) (2a + 3b)(2a – 3b)
(i) (5y + 3)(5y – 3) (j) (4x – 1)(4x + 1)
147
(k) (3x + 4)(3x – 4) (l) (2x – 3y)(2x + 3y)
(m)(8 – 3x)(8 + 3x) (n) (3x + 7y)(3x – 7y)
Quadratic expressions
An algebraic expression of the type ax
2
+ bx +
c where a, b and c are constants, a ≠ 0 and x is
the variable, is called a quadratic expression.
Thus, x
2
+ 5x + 6, 3x
2
– 5x + 3, 3x
2
+ 5x,
2x
2
– 16 are examples of quadratic expressions.
In x
2
+ 5x + 6, the term in x
2
is called the
quadratic term or simply the rst term. The
term in x, i.e. 5x, is called the linear term
or second term or middle term, and 6, the
numerical term or the term independent of x, is
called the constant term or third term.
In 3x
2
+ 5x, the ‘missing’ constant term is
understood to be zero.
In 2x
2
– 16, the ‘missing’ linear term has zero
coefcient.
Factorising quadratic expressions
It is easy to see that 2x
2
– 16 = 2(x
2
– 8) and
3x
2
– 5x = x(3x – 5).
However, it is not easy to see what the
factors of x
2
+ 5x + 6 are. Our experience in
multiplying binomials is of great help here.
Now, consider the product (x + 3)(x + 2).
(x + 3) and (x + 2) are prime binomial
expressions, since the two terms in each
bracket have no common factor.
(x + 3)(x + 2) = x(x + 2) + 3(x + 2)
= x
2
+ 2x + 3x + 6
= x
2
+ 5x + 6 (Since 2x and 3x
are like terms.)
This means that (x + 3) and (x + 2) are factors
of x
2
+ 5x + 6.
⇒ x
2
+ 5x + 6 = (x + 3)(x + 2) (In factor
form.)
Note: In x
2
+ 5x + 6:
1. the coefcient of the quadratic term is 1,
2. the coefcient of the linear term is 5, the
sum of the constant terms in the binomial
factors, and
3. the constant term is 6, the product of the
constant terms in the binomial factors.
In a simple expression like ax
2
+ bx + c, where
a = 1, the factors are always of the form
(x + m)(x + n), where m and n are constants.
Such an expression is factorisable only if there
exists two integers m and n such that m × n = c
and m + n = b.
To factorise a quadratic expression of the
form ax
2
+ bx + c, where a = 1, follow the
steps below.
1. List all the possible pairs of integers
whose product equals the constant term.
2. Identify the only pair whose sum equals
the coefcient of the linear term.
3. Rewrite the given expression with the
linear term split as per the factors in 2
above.
4. Factorise your new expression by
grouping, i.e. taking two terms at a time.
5. Check that the factors are correct by
expanding and simplifying.
Example 15.6
Factorise x
2
+ 8x + 12.
Solution
In this example, a = 1, b = 8 and c = 12.
1. List all the pairs of integers whose product
is 12. These are:
1 × 12 3 × 4 2 × 6
–
1 ×
–
12
–
3 ×
–
4
–
2 ×
–
6
2. Identify the pair of numbers whose sum is 8.
The numbers are 6 and 2.
148
= y(y + 7) – 5(y + 7) (we factor out
–
5)
= (y + 7 )(y – 5).
2. The order in which we write mx and nx
in the split form of the expression does
not change the answer.
Exercise 16.5
1. Factorise the following by grouping.
(a) ax + ay + bx + by (b) x
2
+ 3x + 2x + 6
(c) 6x
2
– 9x – 4x + 6 (d) x
2
– 3x – 2x + 6
(e) cx + dx + cy + dy (f) ax + bx – ay – by
Factorise the following quadratic expressions:
2. (a) x
2
+ 4x + 3 (b) x
2
+ 12x + 32
(c) x
2
+ 20x + 100 (d) x
2
+ 11x + 18
(e) x
2
+ 3x + 2 (f) x
2
+ 6x + 9
3. (a) x
2
+ 5x – 24 (b) x
2
+ 2x – 63
(c) x
2
+ x – 12 (d) x
2
+ 2x – 15
(e) x
2
+ x – 6 (f) x
2
+ 5x – 6
4. (a) x
2
– 8x + 15 (b) x
2
– 9x + 14
(c) x
2
– 2x + 1 (d) x
2
– 4x + 4
(e) x
2
– 10 x + 24 (f) x
2
– 6x + 9
5. (a) x
2
– x – 12 (b) x
2
– 5x – 24
(c) x
2
– x – 30 (d) x
2
– 3x – 18
(e) x
2
– 3x – 10 (f) x
2
– x – 20
Further factorisation of quadratic
expressions
Consider the product (2x + 3)(2x + 7).
(2x + 3)(2x + 7) = 2x (2x + 7) + 3(2x + 7)
= 4x
2
+ 14x + 6x + 21
= 4x
2
+ (14 + 6)x + 21
= 4x
2
+ 20x + 21
In this example, (2x + 3) and (2x + 7) are the
factors of 4x
2
+ 20x + 21. In 4x
2
+ 20x + 21,
a = 4, b = 20, and c = 21.
{
{
{
{
3. Rewrite the expresion with the middle term
split.
x
2
+ 8x + 12 = x
2
+ 2x + 6x + 12
4. Factorise x
2
+ 2x + 6x + 12 by grouping.
x
2
+ 2x + 6x + 12 has 4 terms which we can
group in twos so that rst and second terms
make one group and third and fourth terms
make another group.
i.e x
2
+ 2x + 6x + 12
In each group, factor out the common
factor.
Thus,
x
2
+ 2x + 6x + 12 = x(x + 2) + 6(x + 2)
We now have two terms, i.e. x(x + 2) and
6(x + 2), whose common factor is (x+2)
∴ x
2
+ 8x + 12 = (x + 2)(x + 6) (Factor
out the common factor (x + 2.))
Check that (x + 2)(x + 6) = x
2
+ 8x + 12.
Note: Since all the terms in the example are
positive, the negative pairs of factors of 12 in 1
above could have been omitted altogether.
Example 16.7
Factorise y
2
+ 2y – 35.
Solution
The pairs of numbers whose product is
–
35 are
–
5, 7; 5,
–
7; 1,
–
35; and
–
1, 35.
The only pair of numbers whose sum is 2 is
–
5, 7.
∴ y
2
+ 2y – 35 = y
2
– 5y + 7y – 35
= y(y – 5) + 7(y – 5)
= (y – 5)(y + 7)
Note:
1. If the third term in the split form of the
expression is negative, we factor out the
negative common factor.
e.g. y
2
+ 2y – 35 = y
2
+ 7y – 5y – 35
(the third term is negative)
149
Note:
1. ac = 4 × 21 = 84
2. b = 20
3. There is a pair of integers m and n such that
m × n = ac and m + n = b.
The pair is 14 and 6.
An expression of the form ax
2
+ bx + c can
be factorised if there exists a pair of numbers
m and n whose product is ac and whose
sum is b.
Example 16.8
Factorise the quadratic expression
6x
2
+ 13x +6.
Solution
In this example, a = 6, b = 13 and c = 6.
Step 1: Determine if the expression can be
factorised.
ac = 36.
Possible pairs of m and n are
2 × 18, 3 × 12, 4 × 9 , 6 × 6, 1 × 36
Since all the terms are positive, we will
only consider positive values of m and n.
Step 2: 6x
2
+ 13x + 6 = 6x
2
+ 9x + 4x + 6
(Split middle term.)
Step 3: = 6x
2
+ 9x + 4x + 6 (Group in pairs)
= 3x(2x + 3) + 2(2x + 3) (Factorise
the groups.)
common factor
Step 4: = (2x + 3)(3x + 2) (Factor out the
common
factor.)
Step 5: Check if the factors are correct by
expanding (2x + 3)(3x + 2) and
simplifying.
Note: When you factorise the groups in Step 3,
the factors inside the brackets must be identical.
If not, then there is a mistake.
Example 16.9
Factorise 12x
2
– 4x – 5.
Solution
In this example, a = 12, b =
–
4, c =
–
5 and
ac =
–
60.
By inspection, m and n are
–
10 and 6.
∴ 12x
2
– 4x – 5 = 12x
2
– 10x + 6x – 5
= 2x(6x – 5) + 1(6x – 5)
common factor
= (6x – 5)(2x + 1)
Note:
1. If we cannot determine m and n by inspection,
then we use the procedure of Example 16.8.
2. If m and n do not exist, then the expression
has no factors.
Exercise 16.6
Factorise the following expressions:
1. x(x + 1) + 3(x + 1)
2. 3(2x + 1) – x(2x + 1)
3. 4a(2a – 3) – 3(2a – 3)
4. 4b(b + 6) – (b + 6)
5. 3y(4 – y) + 6(4 – y) 6. 2x
2
+ x – 6
7. 3a
2
+ 7a – 6 8. 2x
2
+ 3x + 1
9. 4x
2
– 2x – 6 10. 4y
2
– 4y – 3
11. 9b
2
– 21b – 8 12. 7x
2
– 3x + 6
13. 2x
2
+ 6x – 20 14. 6x
2
+ 5x – 6
15. 15a
2
+ 2a – 1 16. 9a
2
+ 21a – 8
17. 8b
2
– 18b + 9 18. 10a
2
+ 9a + 2
19. 7x
2
– 36x + 5 20. 6x
2
+ 23x + 15
150
Perfect squares
Just like we have square numbers in arithmetic,
we also have square trinomials in algebra.
Remember ( a + b)
2
= (a + b)(a + b)
= a
2
+ 2ab + b
2
In this case, a
2
+ 2ab + b
2
is a perfect square.
It has two identical factors.
If a trinomial is a perfect square:
1. The rst term must be a perfect square.
2. The last term must be a perfect square.
3. The middle term must be twice the
product of numbers that were squared
to give the rst and last terms.
Example 16.10
Show that the following expressions are perfect
squares and give the factor of each.
(a) 9x
2
+ 12x + 4 (b) 9x
2
– 30x + 25
Solution
(a) 9x
2
+ 12x + 4
Condition (1): rst term 9x
2
= (3x)
2
Condition (2): last term 4 = (2)
2
Condition (3): middle term 12x = 2(3x)(2)
∴ 9x
2
+ 12x + 4 = (3x)
2
+ 2(2)(3x) + 2
2
= (3x + 2)
2
(b) 9x
2
– 30x + 25
First term 9x
2
= (3x)
2
Last term 25 = (
–
5)
2
Middle term
–
30x = 2(
–
5)(3x)
∴ 9x
2
– 30x + 25 is a perfect square which
factorises to (3x – 5)
2
.
Note: In 9x
2
– 30x + 25, middle term of the
expression is negative, hence the constant term
in the binomial factor must be negative.
Exercise 16.7
Show that the following are perfect squares.
Hence state their factors.
1. x
2
+ 8x + 16 2. x
2
+ 12x + 36
3. x
2
– 14x + 49 4. y
2
– 6y + 9
5. 4x
2
+ 20x + 25 6. 9x
2
– 42x + 49
7. 9x
2
– 6x + 1 8. 16x
2
+ 24x + 9
9. 25x
2
– 40xy + 16y
2
10. 144x
2
– 120x + 25
11. 4x
2
+ 12x + 9 12. 36x
2
– 108x + 81
Difference of two squares
Remember: We have already seen that
(a – b)(a + b) = a
2
– b
2
.
(a – b)(a + b) is the product of the sum and
difference of the same two terms. The product
always gives a difference of the squares of the
two terms.
To factorise a difference of two squares, we
reverse the process, i.e. nd the factors, given
the expression.
In order to use this technique, we must be able
to recognise a difference of two perfect squares.
We proceed as in Exmple 16.11
To factorise a difference of two squares,
follow the following steps:
Step 1: Confirm that we have a perfect
square minus another perfect
square.
Step 2: Rewrite the expression in the form
a
2
– b
2
.
Step 3: Factorise the expression.
Example 16.11
Factorise (a) x
2
– 9 (b) 4x
2
– 25y
2
(c) 3x
2
– 27
Solution
(a) In x
2
– 9, x
2
and 9 are perfect squares.
∴x
2
– 9 = (x)
2
– (3)
2
= (x + 3)(x – 3)
151
(b) In 4x
2
– 25y
2
, 4x
2
and 25y
2
are perfect
squares.
∴ 4x
2
– 25y
2
= (2x)
2
– (5y)
2
= (2x + 5y)(2x – 5y)
(c) In 3x
2
– 27, 3x
2
and 27 are not perfect
squares but they have a common factor.
∴ 3x
2
– 27 = 3(x
2
– 9) (x
2
and 9 are
perfect squares)
= 3[(x)
2
– (3)
2
]
= 3[(x – 3)(x + 3)]
= 3(x – 3)(x + 3)
Note that in 3x
2
– 27, it was necessary for us
to factor out the common factor 3 in order to
discover the difference of two squares therein.
We must not forget to include 3 in our answer.
Also note that an expression of the form a
2
+ b
2
is called the sum of two squares, and it has no
factors.
Exercise 16.8
Factorise the following completely.
1. (a) x
2
– 16 (b) x
2
– 4
(c) x
2
– 25
2. (a) x
2
– 1 (b) 36 – a
2
(c) 81 – a
2
3. (a) 25 – y
2
(b) x
2
– y
2
(c) x
2
– 4y
2
4. (a) b
2
– 49 (b) 4a
2
– 25b
2
(c) 9x
2
– 49y
2
5. (a) 9y
2
– 25x
2
(b) 16p
2
– 9q
2
(c) 4x
2
– 9b
2
6. (a) 81x
2
– y
2
(b) p
2
– 25q
2
(c) a
2
– 16b
2
7. (a) 144x
2
– 121y
2
(b) 1 – c
2
(c) 2x
2
– 8y
2
8. (a) 3x
2
– 48y
2
(b) 18x
2
– 2
(c) 20 – 5b
2
9. (a) 8x
2
– 32y
2
(b) 50 – 2x
2
(c) r
4
– 9
10. (a) 49x
2
– 64y
4
(b) x
4
– 1
(c) a
4
b
4
– 16c
4
Quadratic equations
In Form 1, we studied linear equations and their
solutions. Now, we shall consider the solution
of an equation that contains the unknown to
the second power. Such an equation is called a
quadratic equation.
Quadratic equations can be written in the form
ax
2
+ bx + c = 0 where a, b and c are constants,
and a ≠ 0. This is the standard form.
The solution of a quadratic equation in the
standard form is the value of the variable that
makes the LHS equal to zero, hence, forming a
true statement as we did with linear equations.
Solving quadratic equations
Consider the equation x
2
+ 5x + 6 = 0.
In factor form, the equation becomes
(x + 2)(x + 3) = 0.
This equation states that the product of (x + 2)
and (x + 3) is zero.
We can solve this equation only if we know the
following fact.
If a and b are real numbers, and a × b = 0,
then either a = 0 or b = 0 or a = b = 0.
By this property, if (x + 2)(x + 3) = 0, then
either x + 2 = 0 or x + 3 = 0.
Since each of these factors is linear, we use the
method of solving linear equations.
If x + 3 = 0 and if x + 2 = 0
then x =
–
3 then x =
–
2
∴
–
3 and
–
2 are the solutions of the equation
(x + 3)(x + 2) = 0.
152
6. (a) (3x – 9)(2x + 3) = 0
(b) (2x – 1)(x + 2) = 0
7. (a) (2x + 1)(3x – 2) = 0
(b) (4x – 1)(3x + 1) = 0
8. (a) (4 – 3a)(8 – 5a) = 0
(b) (5 – y)(y – 7) = 0
9. (a) (2x + 7)(x – 3) = 0
(b) (5x + 3)(x – 10) = 0
10. (a) (x – 10)(4x – 3) = 0
(b) 3x (x – 12) = 0
11. (a) (x – 12)(3x + 1) = 0
(b) 3(2x + 11) = 0
12. (a) (8x – 1)(2x + 7) = 0
(b) (7 + 3y)(2 – 3y) = 0
Solving quadratic equations by factor
method
Note that if the LHS of an equation is a product
of linear factors and the RHS equals zero, there
will be one solution of the equation for each
linear factor. Thus, we can solve some standard
quadratic equations of the form ax
2
+ bx +
c = 0 by factorising the left hand side of the
expression into linear factors.
In general, to solve a quadratic equation
by factor method, we use the following
procedure.
1. Write the equation in standard
quadratic form.
2. Factorise the left hand side completely.
3. Set each of the factors containing the
variable equal to zero and solve the
resulting equations.
1
2
–
4
3
–
1
2
–
4
3
–
From the above discussion, we can see that to
solve any equation, in factored form, whose
product is 0, we:
1. set each factor equal to zero,
2. solve the resulting equations for the variable.
Example 16.12
Solve the equation:
(a) (3x + 4)(2x – 1) = 0
(b) x(x – 5) = 0
Solution
(a) (3x + 4)(2x – 1) = 0
(For the product to be equal to zero, one or
the other of the factors must be equal to 0.)
If 3x + 4 = 0 or if 2x – 1 = 0
then 3x =
–
4 then 2x = 1
∴ x =
–
∴ x =
The solution of the equation
(3x + 4)(2x – 1) = 0 is x =
–
or x = .
(b) x (x – 5) = 0. This equation has two
factors, x and x – 5.
Either x = 0 or x – 5 = 0
∴ x = 0 or 5 is the solution of the
equation x(x – 5) = 0.
Note: The solutions of a quadratic equation are
also known as its roots.
Exercise 16.9
Solve the following equations.
1. (a) x + 6 = 0 (b) x – 9 = 0
2. (a) 3x + 9 = 0 (b) 6 – 4x = 0
3. (a) 5x – 7 = 0 (b) (x + 5)(x – 5) = 0
4. (a) (x – 1)(x + 2) = 0 (b) x(x + 6) = 0
5. (a) 5p(p + 9) = 0 (b) x(x – 8) = 0
153
4. Check the solutions by substituting into
the original equation. (Substitute one
solution at a time).
Note that a quadratic equation must have
two roots.
Example 16.13
Solve the quadratic equation x
2
– 2x – 8 = 0.
Solution
x
2
– 2x – 8 = (x – 4)(x + 2)
Thus, x
2
– 2x – 8 = 0 becomes
(x – 4)(x + 2) = 0
⇒ Either x – 4 = 0 or x + 2 = 0
i.e. x = 4 or x =
–
2
The solution of x
2
– 2x – 8 = 0 is x = 4 or x =
–
2.
Example 16.14
Solve the equation 4x
2
= 20x – 25.
Solution
Write the equation in standard form.
4x
2
= 20x – 25 ⇒ 4x
2
– 20x + 25 = 0
(2x – 5)
2
= 0 (Factorising the LHS.)
2x – 5 = 0 (Setting the repeated factor
equal to zero once.)
2x = 5
∴ x = (Solving the linear equation.)
Check the solution by substituting for x in
the original equation.
Note: In this example, we have two identical
factors. Since the two factors are the same, we
need to solve the linear equations only once,
and we say the equation has only one distinct
solution. This is an example of a repeated
root. It should be written as
x = twice.
Exercise 16.10
Solve the following equations.
1. (a) x
2
+ 3x + 2 = 0 (b) a
2
+ 14a + 49 = 0
2. (a) x
2
+ 6x – 16 = 0 (b) y
2
– 3y – 4 = 0
3. (a) x
2
=
–
11x – 10 (b) a
2
– 11a = 12
4. (a) x
2
– 14x = 15 (b) x
2
– 32 = 4x
5. (a) x
2
= 27 – 6x (b) a
2
– 3a = 28
6. (a) b
2
– 7b =
–
10 (b) y
2
– y = 6
7. (a) x
2
+ 4x = 0 (b) a
2
– 4a = 0
8. (a) 3y
2
– 5y = 0 (b) 4y
2
+ 7y = 0
9. (a) 2x
2
+ 6x = 0 (b) 3x
2
= 9x
10. (a) 2x
2
– 18 = 0 (b) 7x
2
– 7 = 0
11. (a) x
2
– 100 = 0 (b) 2x
2
– 7x – 9 = 0
12. (a) 6x
2
– 5x + 1 = 0 (b) 9x
2
+ 20 =
–
27x
13. (a) 6z
2
+ z =
–
10z – 3
(b) 3x
2
– 4x – 28 = x
14. (a) 3a
2
– 3 = 5a
2
+ 3a – 5
(b)
–
6x =
–
3x
2
– 3
15. (a) x(x + 3) =
–
2 (b) 2x(2x + 6) =
–
8
Forming quadratic equations from given
roots
The values of the variable which satisfy an
equation are called the roots of the equation.
Do not confuse the roots of an equation with
square or cube roots.
Consider the equation x
2
– 3x + 2 = 0.
In factor form, the equation becomes
(x – 2)(x – 1) = 0.
From this, we know that x – 2 = 0 or x – 1 = 0
i.e. x = 2 or x = 1.
1 and 2 are called the roots of the equation
x
2
– 3x + 2 = 0.
5
2
–
5
2
–
5
2
–
154
(b) If
–
is a root, (x + ) is a factor.
If is a root,
(
x –
)
is a factor.
∴
(
x +
)(
x –
)
= 0
x
2
– x + x – = 0
Multiplying each term by LCM of
denominators in order to write the equation
in standard form:
12x
2
– 8x + 9x – 6 = 0
12x
2
+ x – 6 = 0.
Note that the coefcient of the middle term of
the resulting equation is equal to the negative
sum of the roots, i.e. the roots are m and n and
the middle term is
–
(m + n)x.
Note also that the constant term is equal to the
product of the roots, i.e. the constant term
is mn.
These facts can be used to form a quadratic
equation from given roots, with minimum
computation.
Consider the equation x
2
– 3x + 2 = 0
⇒ (x – 2)(x – 1) = 0
∴ roots of x
2
– 3x + 2 are 1 and 2.
Sum of roots is 1 + 2.
Thus, coefcient of middle term is
–
(1 + 2) =
–
3.
This is true from the given equation.
Also the constant term is ‘product of the roots’,
i.e. product of roots = 1 × 2 = 2.
If the equation ax
2
+ bx + c = 0 and a ≠ 1,
and its roots are m and n, then coefcient of
middle term is
–
(m + n).
–
(m + n) = (Since ax
2
+ bx + c = 0
x
2
+ x + = 0
⇒ m + n =
–
.
The constant term is m × n = .
If an equation has the value 1 as a root, then
(x – 1) must be a factor of the LHS of the
equation.
Similarly, if x = 2 is a root of the same equation,
then (x – 2) must be another factor of the LHS
of the same equation. Since the product of the
factors on the LHS equals zero, then
(x – 1)(x – 2) must be equal to zero.
∴ (x – 1)(x – 2) = 0
x(x – 2) – 1(x – 2) = 0
x
2
– 2x – x + 2 = 0
x
2
– 3x + 2 = 0 (The equation we
started with.)
In general, if x = m is a root of a quadratic
equation, then (x – m) must be a factor of
the equation. If x = n is another root of the
same equation, then (x – n) must be another
factor of the same equation.
Since m and n are the only two roots of the
same equation, then the equation must be
quadratic.
This equation is
(x – m)(x – n) = 0
x
2
– mx – nx + mn = 0
x
2
– (m + n)x + mn = 0
Example 16.15
Find the equation whose roots are:
(a)
–
1 and 2 (b)
–
and .
Solution
(a) If
–
1 is a root, then (x + 1) is a factor of the
the quadratic equation.
If 2 is a root, then (x – 2) is a factor of the
the quadratic equation.
Then, the equation is (x + 1) (x – 2) = 0
x
2
– 2x + x – 2 = 0
x
2
– x – 2 = 0
3
4
–
2
3
–
3
4
–
2
3
–
2
3
–
3
4
–
6
12
––
2
3
–
2
3
–
3
4
–
b
a
–
–
c
a
–
3
4
–
c
a
–
b
a
–
⇐⇒
155
2
5
–
3
4
–
3
4
–
2
3
–
Exercise 16.11
1. Find the equations whose roots are given,
and write your answers in standard
quadratic form.
(a)
–
2 and 2 (b)
–
3 and 4 (c)
–
1 and
(d)
–
and
–
(e)
–
2 and
–
3 (f)
–
3 and
2. If x = 1 is a root of the equation
x
2
+ kx + 2 = 0, nd the value of k, and
the other root.
3. If x =
–
2 is a root of x
2
+ kx + 8 = 0, nd the
value of k and the other root.
4. If x = 1 is a root of 5x
2
+ kx – 3 = 0, nd the
other root.
5. Use the method of sum and product of roots
to form the equation whose roots are given:
(a) 4 ,
–
(b) ,
–
(c) ,
–
6. Find the sums and products of the roots of
the following equations.
(a) x
2
– 13x + 40 = 0 (b) x
2
+ 5x – 50 = 0
(c) 2x
2
– 6x + 5 = 0 (d) 2x
2
– 6x = 0
(e) 7x
2
– 11x – 8 = 0
(f) x
2
– (p + q)x + pq = 0
Equations leading to quadratic equations
Many equations involving fractions eventually
lead to quadratic equations.
Example 16.16
Solve the equation + = 2.
Solution
Step 1: Multiply each term by the LCM of the
denominators in order to remove the
fractions:
LCM of denominators = (2x + 1)(5x – 1)
∴ (2x + 1) ( 5x – 1) + =
2(2x + 1)(5x –1)
(5x – 1) × 3 + (2x + 1) × 4
= (2x + 1)(5x – 1) × 2
Step 2: Multiply out and simplify:
3(5x – 1) + 4(2x + 1)
= 2(2x + 1)(5x – 1)
15x – 3 + 8x + 4
= 2(10x
2
– 2x + 5x – 1)
23x + 1 = 20x
2
+ 6x – 2
Step 3: Rearrange the equation in the standard
quadratic form:
20x
2
+ 6x – 2 – 23x – 1 = 0
20x
2
– 17x – 3 = 0
Step 4: Factorise and solve:
20x
2
– 20x + 3x – 3 = 0
20x (x – 1) + 3(x – 1) = 0
(x – 1)(20x + 3) = 0
x – 1 = 0 or 20x + 3 = 0
i.e. x = 1 or x =
–
.
Exercise 16.12
Solve the following equations.
1. y + = 2 2. =
3. = 4. 2 – = x
5. =
x – 1
6. + 1 =
7. x – + 3 = 0 8. x + =
–
2
9. x – = 0 10. x – 1 =
11. (x + 5) + – 9 = 0 12. x – = 0
13. x – 9 = 14. =
Word problems leading to quadratic
equations.
Many word problems require the use of
quadratic equations for their solution.
1
2
–
1
2
–
1
4
–
3
4
–
1
3
–
3
2x + 1
–––––
4
5x – 1
–––––
3
20
––
1
y
–
x + 2
x – 2
––––
x + 3
x – 9
––––
1
x
–
2
x + 1
––––
2x
2
x + 1
––––
3x + 5
6x + 5
–––––
4
x
–
1
x
–
9
x
–
1
x – 1
––––
18
x + 5
––––
35
x + 2
––––
72
x – 8
––––
x + 10
x – 5
–––––
7x
x – 5
––––
1
2
–
3
2x + 1
–––––
4
5x – 1
–––––
1
9x
––
1 – x
2
––––
156
Example 16.17
The product of two consecutive even numbers is
168. Find the numbers.
Solution
Let x be the smaller even number and x + 2 be
the larger. (Consecutive even numbers differ by
2).
From ‘the product of two consecutive even
numbers’, we get x (x + 2).
∴ x
2
+ 2x = 168
x
2
+ 2x – 168 = 0 (Rearranging in
standard form.)
(x + 14)(x – 12) = 0 (Factorising the LHS)
∴ x + 14 = 0 or x – 12 = 0 (Setting each
factor equal to zero and solving)
i.e. x =
–
14 or x = 12
When x =
–
14, the next even number x + 2 =
–
12.
When x = 12, the next even number x + 2 = 14.
The consecutive even numbers are
–
14 and
–
12
or 12 and 14.
Check:
–
14 ×
–
12 = 168
and 12 × 14
= 168.
Both pairs give consecutive even numbers and
the conditions of the problem are met.
Example 16.18
The length of a rectangle is 2 m longer than the
breadth. If the area of the rectangle is 80 m
2
,
nd its dimensions.
Solution
Let w be the width and l the length of the
rectangle (See Fig. 16.4.)
Fig. 16.4
Now, Area (A) = l × w
so, w × l = 80
w(w + 2) = 80
w
2
+ 2w = 80
w
2
+ 2w – 80 = 0
(w + 10)(w – 8) = 0
∴w =
–
10 or 8.
Since the width of a rectangle cannot be
negative, w =
–
10 is not a solution of the
problem.
∴ w = 8 is the only practical solution of the
problem.
l = w + 2 = 8 + 2 = 10 m.
The rectangle is 8 m wide and 10 m long.
Exerise 16.13
1. One integer is 3 less than another. If their
product is 340, nd the integers.
2. The length of a rectangle is double its
breadth. If its area is 200 m
2
, nd its
length.
3. The product of two consecutive whole
numbers is 42. Find the numbers.
4. Find a number such that the sum of the
number and its reciprocal is 4 .
5. A rectangle is of length (x + 2) m and width
(x – 1) m. Write an expression for its area
in the simplest form.
6. The sum of the squares of two consecutive
integers is 113. Find these integers.
7. Three times the square of a certain number
is decreased by 9 times the number. The
result is 120. Find the number.
8. Think of a number, square it and add the
original number. If your result is 56, what
is the number?
l = w + 2
w
1
4
–
157
9. Think of a number, multiply it by 3,
subtract 1 and square the result. If this
result is the same as the square of the
original number, nd the number.
10. Find two consecutive odd numbers such
that the sum of their squares is 650.
11. In Fig. 16.5, all the dimensions are in
centimetres, and all the angles are right
angles. If the area of the gure is 90 cm
2
,
nd the value of x.
Fig. 16.5
1
2
–
12. An n–sided gure has n(n – 3) diagonals.
How many sides has the gure if it has 135
diagonals?
13. ABC is a right-angled triangle (Fig. 16.6)
and its dimensions are given in centimetres.
Fig. 16.6
Find the lengths of the sides of the triangle.
x + 1
x – 1
x
A
B
1
2
–
C
x
x
11
10
158
Inequalities
We are familiar with the equals sign, =.
Remember that an algebraic statement which
has an equals sign, e.g. 3a = 6, is called an
equation.
In this section we are concerned with
statements which involve the following
symbols:
> meaning ‘greater than’, e.g. 6 > 2 means
6 is greater than 2.
< meaning ‘less than’, e.g. 3 < 7 means 3
is less than 7.
≥ meaning ‘greater than or equal to’ (or ‘not
less than’).
≤ meaning ‘less than or equal to’, (or ‘not
greater than’).
Statements containing these symbols are
called inequalities or inequations. For
example, x < 2, x ≥ 3, y ≤
–
3, y > 10 are
inequalities. These are also called simple
inequality statements.
A statement such as x > 2 means ‘all numbers
that are greater than 2’, which is a range of
values. Just as we represent individual numbers
on a number line, we can also represent such a
range of numbers on a number line as shown in
the following examples.
Example 17.1
Illustrate each of the following on a number
line.
(a) x > 3 (b) x ≥ 3
(c) x <
–
2 (d) x ≤
–
2
Solution
(a) x > 3:
Fig. 17.1
In Fig. 17.1, the number 3 is not included in
the list of numbers to the right of 3. The heavy
arrow shows that the values of x go on without
end. The open dot
is used to indicate that 3 is
not included.
(b) x ≥ 3:
Fig. 17.2
In Fig. 17.2, the number 3 is included in the
list of the required numbers. The closed dot is
used to show that 3 is part of the list.
(c) x <
–
2:
Fig. 17.3
In Fig. 17.3, the number
–
2 is not included.
(d) x ≤
–
2:
Fig. 17.4
In Fig. 17.4, the number
–
2 is included.
Compound statements
Sometimes, two simple inequalities may be
combined into one compound statement such
0 1 2 3 4 5 6
x > 3
–
4
–
3
–
2
–
1
0 1 2
x ≤
–
2
–
4
–
3
–
2
–
1
0 1 2
x <
–
2
0 1 2 3 4 5 6
x ≥ 3
LINEAR INEQUALITIES
17
159
as a < x < b. This statement means that a < x
and x < b or x > a and x < b.
Example 17.2
Write the following pairs of simple inequality
statements as compound statements and
illustrate them on number lines.
(a) x ≤ 3, x >
–
3
(b) x >
–
1, x < 2
Solution
(a) x ≤ 3, x >
–
3 becomes
–
3 < x ≤ 3 (Fig. 17.5).
∴x lies between
–
3 and 3 and 3 is included.
Fig. 17.5
(b) x >
–
1, x < 2 becomes
–
1 < x < 2
(Fig. 17.6).
∴ x lies between
–
1 and 2.
Fig. 17.6
Forming inequalities from word
statements
Inequality symbols can be used to change word
statements into algebraic statements.
Example 17.3
Anyango has 25 books. Wanjiku has more books
than Anyango. Write an algebraic statements for this.
Solution
If Wanjiku has b books then b > 25.
Example 17.4
The distance from Nairobi to Kericho is 270 km
and that from Nairobi to Murang’a is 80 km.
If Nyeri to Nairobi is a shorter distance than
Nairobi to Kericho, and Nyeri to Nairobi is
a longer distance than Nairobi to Murang’a,
write an algebraic statement for this.
Solution
Let the distance from Nairobi to Nyeri be x km.
Then, x < 270 and x > 80
Hence, 80 < x < 270.
Example 17.5
The area of a square is greater than 36 cm
2
.
Write an inequality for: (a) the length (b) the
perimeter of the square.
Solution
We must rst dene our variables, just as we do
when forming equations.
Let the length of the square be x cm.
Area of the square = x
2
.
(a) x
2
> 36
⇒ x
2
> 36
⇒ x > 6
(b) Perimeter = 4x
Since x > 6,
then 4 × x > 4 × 6
i.e. 4x > 24
Exercise 17.1
1. Rewrite each of the following statements
using either < , ≤ , > or ≥ instead of the
words.
(a) 5 is less than 7
(b) 5 is greater than 2
(c)
–
1 is less than 0
(d)
–
2 is greater than
–
3
(e) x is greater than or equal to 4
(f) y is less than or equal to
–
5
(g) a is not less than 3
(h) b is not greater than 0
(i)
–
1 is not less than p
(j) 10 is not greater than q
2. Copy and complete each of the following
statements by inserting the correct
–
4
–
3
–
2
–
1
0 1 2 3 4
–
3
–
2
–
1 0 1 2 3
160
symbol, < or >.
(a) 7 + 3 11 (b) 6 – 3 5
(c) 13 + 2 16 (d) 12 – 4 3
(e) 3 × 4 8 (f)
–
4 × 3 12
3. Illustrate each of the following on a
number line.
(a) x >
–
5 (b) x < 0 (c) x ≥
–
3
(d) x ≤ 4 (e) x >
–
0.5 (f) x ≤ 2.5
4. Write each of the following pairs of simple
statements as compound statement and
illustrate the answer on a number line.
(a) x > 2, x < 4 (b) x < 3, x ≥ 0
(c) x ≤ 5, x >
–
2 (d) x ≥
–
1, x ≤ 1
(e) x < 1.5, x ≥
–
0.5 (f) x ≤ 2.2, x ≥ 1.8
(g) x < , x ≥ 0 (h) x ≥ , x < 2
(i) x < 3 , x > 2 (j) x > , x <
(k) x ≥
–
0.75, x ≤ 0.75 (l) x < 4 , x >
–
In Questions 5–11, write down the statements
as mathematical sentences using appropriate
symbols.
5. When two is added to a certain number, the
result is greater than ten.
6. When ve is added to a number, the result
is less than twice the number.
7. Multiplying a number by six, then adding
ve, gives a greater result than multiplying
the number by ve, then adding six.
8. The sum of three consecutive whole
numbers is more than 260.
9. The area of a square is less than its perimeter.
10. The square of a number is less than the
number cubed.
11. The radius of a circle is not more than
4 cm. What can you say about the
circumference?
Solution of linear inequalities in one
unknown
Solving an inequality means obtaining all the
possible values of the unknown which make the
statement true. This is done in much the same
way as solving an equation.
Example 17.6
Solve the following inequalities.
(a) x – 3 < 7 (b) x + 5 > 11
Solution
(a) x – 3 < 7
⇒ x – 3 + 3 < 7 + 3
⇒ x < 10
Thus, x < 10 is the solution of the inequality
x – 3 < 7.
(b) x + 5 > 11
⇒ x + 5 – 5 > 11 – 5 (Subtracting 5 from
both sides.)
⇒ x > 6.
Adding (or subtracting) the same number to
(or from) both sides of an inequality does
not change it.
Example 17.7
Solve the following inequalities.
(a) 3x – 4 ≥ 5 (b) x + 5 ≤ 14
Solution
(a) 3x – 4 ≥ 5
⇒ 3x – 4 + 4 ≥ 5 + 4
⇒ 3x ≥ 9
⇒ ≥ (Dividing both sides by 3.)
⇒ x ≥ 3
(b) x + 5 ≤ 14
⇒ x + 5 – 5 ≤ 14 – 5 ⇒ x ≤ 9
⇒ x × 4 ≤ 9 × 4 ⇒ x ≤ 36
1
4
–
1
2
–
1
4
–
3x
3
––
9
3
–
1
4
–
1
4
–
1
4
–
3
4
–
1
2
–
2
3
–
1
5
–
1
4
–
1
2
–
1
2
–
1
4
–
(Adding 3 to both
sides)
161
Multiplying or dividing both sides of an
inequality by the same positive number does
not change it.
Multiplication and division of
inequalities by negative numbers
We know that 8 < 10.
Consider multiplying both sides of this
inequality by any negative number, say
–
4.
LHS = 8 ×
–
4 =
–
32
RHS = 10 ×
–
4 =
–
40
We know that
–
32 is greater than
–
40,
i.e. 8 ×
–
4 > 10 ×
–
4
Thus, the inequality is reversed.
Similarly, 8 ÷
–
4 =
–
2, and
10 ÷
–
4 =
–
2.5
We know that
–
2 >
–
2.5
i.e. >
Thus, the inequality is reversed.
In general, multiplying or dividing both
sides of an inequality by a negative number
reverses the inequality sign.
Example 17.8
Solve the inequality 3 – 2x ≥ 15.
Solution
3 – 2x ≥ 15
⇒ 3 – 2x – 3 ≥ 15 – 3
⇒
–
2x ≥ 12
⇒ ≤
⇒ x ≤
–
6
Solving simultaneous inequalities
Inequalities that must be satised at the same
time are called simultaneous inequalities.
–
2x
–
2
–––
12
–
2
––
8
–
4
––
10
–
4
––
Example 17.9
Solve the following pair of simultaneous
inequalities.
3 – x < 5, 2x – 5 < 7
Solution
3 – x < 5
⇒
–
x < 2
⇒ x >
–
2 …………(i)
Also 2x – 5 < 7
⇒ 2x < 12
⇒ x < 6 …………(ii)
Combining (i) and (ii), we have
–
2 < x < 6.
Thus x lies between
–
2 and 6.
This is represented on the number line as in
Fig. 17.7.
Fig. 17.7
Example 17.10
Solve the inequality 3x – 2 < 10 + x < 2 + 5x
Solution
3x – 2 < 10 + x < 2 + 5x
Split the inequality into two simultaneous
inequalities as:
3x – 2 < 10 + x …………(i)
and 10 + x < 2 + 5x …………(ii)
Solve each separately.
3x – 2 < 10 + x
⇒ 3x – x < 10 + 2
⇒ 2x < 12
⇒ x < 6 ……………(iii)
10 + x < 2 + 5x
⇒ 10 – 2 < 5x – x
⇒ 8 < 4x
⇒ 2 < x ……………(iv)
Combine (iii) and (iv) to get 2 < x < 6
–
3
–
2
–
1 0 1 2 3 4 5 6
162
x
2
1
0
–
1
1 2 3 4 5 6
– – – – – – – – – – – – – – – – – – – – –
y
x < 3
x
y
2
1
0
–
1
1 2 3 4 5 6
x > 3
– – – – – – – – – – – – – – – – – – – – –
Exercise 17.2
Solve the following inequalities and represent
the solutions on number lines.
1. (a) x + 4 > 11 (b) x – 6 ≤ 5
2. (a) 2x – 8 ≤ 4 (b) 3x + 4 > 19
3. (a) 3 > 4x – 2 (b) 7 ≤ 5x + 12
4. (a) 3 – 2x < 5 (b) 4 – 5x ≥
–
11
5. (a) x – 3 > 4 (b) x + 2 < 1
6. (a)
–
4 > 2 – x (b)
–
x + 4 ≤
–
6
7. (a) 4m – 3 < 7m (b) 2m + 1 ≥ 5m – 10
8. (a) 2 – 2p > 13 – 3p (b) + p < 4p +
9. (a) q > 2 – 4q (b) q + 2 < 8 – q
10. (a)
–
r < 5 (b) 4 – r ≥ 3 – r
11. (a) 2(1 + x) + 3(x – 2) ≥ 25
(b) 3(4 – 3x) – (5x – 3) ≤ 2
Solve the following simultaneous inequalities
and represent each solution on a number line.
12. (a) 2x < 10, 5x ≥ 15
(b) 3x ≤ 9, 2x > 0
13. (a) x + 7 < 0, x – 2 >
–
10
(b) x ≥ 3, 2x – 1 ≤ 13
14. (a) 4x – 33 <
–
1,
–
2 < 3x + 1
(b) 2x – 5 < 22 ≤ 5x – 6
15. (a) 3x – 4 < 8 + x < 2 + 7x
(b) 6x + 2 < 3x + 8 < 27x – 1
Graphical representation of linear
inequalities
So far, we have represented inequalities on a
number line. Often we are required to represent
inequalities on a Cartesian plane. Remember
that (x , y) denotes any point on the Cartesian
plane.
1
3
–
1
7
–
1
5
–
1
3
–
1
2
–
1
4
–
2
3
–
2
3
–
1
9
–
3
4
–
1
3
–
1
4
–
1
2
–
x
y
3
2
1
0
–
1
1 2
3
4 5 6
x = 3
Fig. 17.8 shows the graph of x = 3.
Fig. 17.8
Note that the x-coordinate for every point to the
left of the line x = 3 is less than 3, i.e. x < 3.
For all the points to the right of line x = 3, the
x-coordinate is greater than 3, i.e. x > 3.
Fig. 17.9 (a) shows the region containing all
points (x , y) for which x < 3.
Fig. 17.9 (b) shows the region containing all
points (x , y) for which x > 3.
(a)
(b)
Fig. 17.9
163
Note that in both cases, the unwanted region
(i.e. the region in which the inequality is not
satised) is shaded.
Note also that points on line x = 3 are not
wanted, so the line is ‘dotted’.
Fig. 17.10 shows the region for which y ≥
–
1.
Fig. 17.10
Note that the line y =
–
1 is continuous. This
means that the points on the line are included in
the required region.
To represent an inequality on a graph, we
use the equation corresponding with that
inequality as the equation of the boundary
line;
e.g. y >
–
1: Boundary line is y =
–
1.
If points on the line are included in the
required region, the line is continuous
(solid). If not, the line is dotted (broken).
The normal convention is to shade the
unwanted region.
Exercise 17.3
Show each of the following regions on a
Cartesian graph.
1. (a) x > 1 (b) x < 5
2. (a) x <
–
2 (b) x ≥
–
1
3. (a) y ≤ 2 (b) y > 0
4. (a) y < (b) y ≥
–
1.5
5. (a) 3 < x < 4 (b) 1 < x ≤ 5
6. (a)
–
2 ≤ y < 2 (b)
–
1 ≤ y ≤ 1
7. (a) 2 – x ≤ x (b) 2x > 7 and 3x ≤ 18
8. (a) 4x – x
2
≤ x(1 – x) + 18
(b) y + 5 ≤ 4y < 2y + 14
Linear inequalities in two unknowns
We have dealt with inequalities of the form
x ≤ a, x ≥ b, y ≥ c, etc. where a, b and c
are constants.
This section, deals with inequalities of the form
ax + by ≤ c, ax + by ≥ d, etc. where a, b, c and d
are constants. Such an inequality is represented
graphically by a region containing all the
ordered pairs of values (x , y) which satisfy that
inequality.
Example 17.11
Draw a graph of the inequality x + 2y < 4.
Solution
Change the inequality into an equation:
x + 2y = 4. This gives us the boundary line.
Obtain any two points on the line and draw the
line on a Cartesian plane.
Where convenient, obtain the two points by
letting x = 0 and y = 0, and working out the
corresponding values of y and x.
Thus, when x = 0, y = 2 and when y = 0, x = 4.
∴ (0 , 2) and (4 , 0) are two points on the
boundary line.
Since the inequality sign is <, points on this line
are not included in the required region. Hence,
the line is dotted (Fig. 17.11).
Take a point on any side of the boundary line
and test if it satises the inequality.
For example take the point (4 , 1) and substitute
into the LHS of x + 2y < 4.
x + 2y = 4 + 2 × 1 = 6
1
2
–
1
3
–
2
3
–
x
y
2
1
0
–
1
–
2
–
1
1 2 3
y =
–
1
–
2
164
0 2 4 6 8 10
y
8
4
2
L
2
x
L
1
L
3
–
2
–
2
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
12
6
Fig. 17.11
Since 6 > 4, the point (4 , 1) does not satisfy the
inequality.
∴ The point is in the unwanted region. Shade
the unwanted region ( Fig. 17.11).
Note: If the boundary line does not pass
through the origin, it is more convenient
and faster to determine the required region
using the origin.
Graphical solution of simultaneous linear
inequalities
When solving linear equations, one is looking for
values of the two unknowns that make the two
equations true at the same time. Similarly, linear
inequalities in two unknowns are solved to nd
a range of values of the two unknowns which
make the inequalities true at the same time. The
solution is represented graphically by a region.
Example 17.12
Draw the region which satises the following
inequalities simultaneously:
x > 0, y > 0, x + 2y ≤ 6
State the integral values of x and y that satisfy
the inequalities.
Solution
In Fig. 17.12, x + 2y = 6 (solid line), x =
0 (broken line), y = 0 (broken line) are the
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
y
4
1
–
1
3
2
0
1 2 3 4 5
x
x
6
x + 2y = 4
y
– – – – – – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – – – – – –
0
–
1
2
1
3
4
x = 0
x + 2y = 4
1 2 3 4 5
x
6
y = 0
boundary lines. All the ordered pairs of values
(x , y) that satisfy the three inequalities lie
within the unshaded region.
Fig. 17.12
The integral pairs of x and y values that satisfy
the inequalities are (1 , 1), (1 , 2), (2 , 1), (2 , 2),
(3 , 1), (4 , 1).
Linear inequalities from inequality
graphs
Example 17.13 shows how one can nd linear
inequalities from inequality graphs.
Example 17.13
Write down the inequalities which are satised
by the unshaded region in Fig. 17.13.
Fig. 17.13
165
y
20
40
5
10
0
6
10
––
6
10
––
4. (a) x + y ≥ 3 (b) 4y – 3x > 0
(c) 4x + 3y > 12
5. x + y ≥ 0, x < 1, y > 1
6. 2x + 3y > 6, y > 0, x > 0
7. y – x < 0, x < 6, y ≥ 0
8. 3x + 5y > 15, 5x + 3y < 30, x > 0, y > 0
9. y ≥ 0, y < 4, 4x + 3y > 0, 5x + 2y < 15
10. Write down the inequalities satised by the
region in Fig. 17.14.
1
4
–
1
3
–
1
2
–
1
2
–
Solution
Line L
1
is the x-axis. i.e. y = 0 (solid line)
Points required are on the line or above it.
∴ the inequality is y ≥ 0
Line L
2
is a solid line.
Let L
2
be y = mx + c.
Since L
2
intersects y-axis at y = 2 then c = 2.
Point (2, 4) is on line L
2
. Substituting in the
equation of the line, we get
4 = 2m + 2
∴ m = 1
∴ Equation of L
2
is y = x + 2
By inspection, the inequality is y – x ≤ 2.
Equation of line L
3
: y = mx + c
By inspection c = 6
(10 , 0) is on line L
3
.
Hence 0 = 10m + 6
∴ m =
–
.
∴ Equation of L
3
is y =
–
x + 6
or 10y + 6x = 60
i.e. 5y + 3x = 30
By inspection the inequality is 10y + 6x < 60 or
5y + 3x <30.
Thus the inequalities are
y ≥ 0, y – x ≤ 2, 5y + 3x < 30.
Exercise 17.4
Show the region which contains the set of
points represented by each of the following
inequalities.
1. (a) x + y < 1 (b) y – x < 4
(c) 2x + 3y ≥ 6
2. (a) x – y ≤ 0 (b) 3x + 5y ≤ 9
(c) y – 3x >
–
6
3. (a) 3x < y + 6 (b) x – 3y >
–
2
(c) x – y ≤
–
1
1
3
–
Fig. 17.15
11. Write down the inequalities satised by the
region in Fig. 17.15.
0
y
4
8
10
20
2
6
Fig. 17.14
166
Linear motion
Linear (or rectilinear) motion is the
movement of an object from one point to
another in a straight line.
The hundred metres race is an example of linear
motion since the race is run between two points
in a straight line.
Displacement
Displacement is the distance covered by an
object moving in a particular direction. It is
measured from some initial position.
Speed
When an object moves a certain distance, the
total distance it moves divided by the time taken
gives the average speed, i.e.
Average speed =
Distance covered
Time taken
Since the movement in this case is not in any
specic direction, the total distance covered is
not a displacement.
If the distance is in kilometres and the time is in
hours, then the speed is given in kilometres per
hour (km/h or kmh
–
1
or kph).
If the distance is in metres and the time in
seconds, then speed is given in metres per
second (m/s or ms
–
1
).
Speed is therefore the rate of change of distance
per unit time. Average speed is then the speed
an object would move at if the motion was
constant.
Example 18.1
A man walked 10.8 km in 2 h. Find his average
speed in: (a) km/h (b) m/s
Solution
(a) Average speed =
Distance covered
Time taken
=
10.8 km
2 h
= 5.4 km/h
(b) Average speed = 5.4 km/h
=
5.4 × 1 000 m
1 × 60 × 60 s
= 1.5 m/s
Example 18.2
Kamau ran 100 m in 12.5 s. What was his
average speed in: (a) m/s (b) km/h?
Solution
(a) Average speed =
Distance
Time
=
100 m
12.5 s
= 8 m/s
(b) Average speed = 8 m/s
=
8 × 60 × 60
1 000
= 28.8 km/h
Example 18.3
A train leaves town A and travels towards town
B at 48 km/h. At the same time, another train
leaves town B and travels towards A at a speed
(Multiplying by 60 × 60
to get distance covered in
1 hour, divide by 1 000 to
change distance into km.)
(5.4 × 1 000 m covered
in 60
×
60 s)
LINEAR MOTION
18
167
of 52 km/h. If the two towns are 500 km apart,
nd:
(a) how far apart the trains are after travelling
for 45 minutes
(b) how far, from town A, they will be when they
meet
(c) how long they will take to meet.
Solution
(a) 45 min = h.
Since speed =
Distance ,
Time
Distance = Time × Speed
Train A will have travelled a distance
= × 48 km
= 36 km
and train B will have travelled a distance
= × 52 km = 39 km
∴ the distance apart = 500 – (36 + 39) km
= 425 km.
(b) The problem may be sketched as in Fig.
18.1, where C is the meeting point and x is
the distance travelled by train A by the time
they meet.
Fig. 18.1
Since speed =
Distance
, Time =
Distance
Time Speed
The trains start at the same time. So they
have travelled the same length of time by
the time they meet.
∴ time taken = =
∴ 48(500 – x) = 52x
⇒ 24 000 – 48x = 52x
⇒ 24 000 = 52x + 48x
⇒ 24 000 = 100x
∴ x = 240 km
i.e. the trains meet 240 km away from town A.
(c) Time taken to meet = h
= h
= 5 h
Exercise 18.1
1. Convert the following speeds to m/s.
(a) 60 km/h (b) 5.4 km/h (c) 108 km/h
2. Convert the following speeds to km/h.
(a) 3 m/s (b) 2.75 m/s (c) 10 m/s
3. A matatu takes 5 seconds to cross a bridge
km long. Calculate its speed in km/h.
4. A train 182 m long takes 14 seconds to
pass a signal post. Calculate its speed in
km/h.
5. A motorist drove 72 km on tarmac road for
10 minutes and a further 90 km on earth
road for 2 h. Find the average speed for the
whole journey.
6. A matatu travelling at 60 km/h took 2 h
25 min between two towns. Find the
distance between the towns.
7. Mr. Nyamwange took a total of 2 h to
walk from his home to his work place
and back. If his average speeds for the
to-and-fro journeys were 2 km/h and
3 km/h respectively, nd the distance
between his home and work place.
8. A lady travelled between two towns
360 km apart in 5 h. If on the return
journey the average speed is increased by
8 km/h, nd the time taken for the return
journey.
9. A cyclist travelling at 21 km/h from town A
to town B, a distance of 105 km, started
2 h earlier than a car travelling the same
journey but arrived 1 h 44 min after the
car. Calculate the average speed of the car.
3
4
–
3
4
–
3
4
–
500 km
48 km/h
52 km/h
C
500 – x kmx km
A
B
x
48
––
500 – x
52
–––––––
x
48
––
240
48
–––
3
4
–
1
10
––
1
4
–
168
10. Mwangi travelled between two towns
180 km apart at an average speed of
72 km/h. Opiyo took half an hour less
to travel the same distance. Find Opiyo’s
average speed.
11. One train leaves a station 2 h before a
second train which follows in the same
direction. If the rst train travels at
40 km/h and the second at 48 km/h, how
long will it take for the second train to
catch up with the rst?
12. Mwangi left town A at 5.30 a.m. and
travelled towards town B at an average
speed of 48 km/h. At the same time Kezia
left town B and travelled towards town
A at an average speed of 64 km/h. If the
distance between the two towns is 588 km:
(a) at what time did they meet?
(b) how far from town B was their
meeting point?
(c) how far apart were they after travelling
for 2 h?
13. How long will a train 70 m long travelling
at 45 km/h take to pass a stationary train
80 m long?
14. A man had walked two-third way across
a bridge when he sighted an approaching
train 60 m away. He ran back only to reach
the end of the bridge at the same time as
the train. If the train was moving at 25 m/s
and the man ran at 10 m/s, nd the length
of the bridge.
Velocity
When the distance moved by an object is in a
specied direction, the speed of the object is
then called velocity.
Average velocity
=
Displacement
Time taken
=
Distance moved in a particular direction
Time taken
Like speed, velocity is measured in km/h or m/s.
Note that velocity has both magnitude and
direction while speed has only magnitude
but no direction.
Quantities, such as velocity, which have
both magnitude and direction are known as
vector quantities.
Quantities, such as speed, which have only
magnitude are known as scalar quantities.
The velocity of an object changes when either
the magnitude or the direction changes. When
the magnitude and direction remain the same,
the object is said to have constant or uniform
velocity. A vehicle moving round a bend
(or a corner) may have constant speed but
not constant velocity since the direction is
changing.
Under ‘linear motion’, only motion in a straight
line is considered. It is then assumed that all
changes in velocity are changes in magnitude
only.
Acceleration
When the velocity of an object changes,
the object is said to have an acceleration.
Acceleration is dened as the rate of change of
velocity and is given by:
Average acceleration =
Change in velocity
Time taken
=
Final velocity – Initial velocity
Time taken
1
2
–
169
Acceleration is measured in m/s
2
when velocity
is in m/s or km/h
2
when velocity is in km/h.
Example 18.4
A motor bike starts from rest and reaches a
velocity of 20 m/s in 4 s. Find its acceleration.
Solution
Initial velocity = 0 m/s
Final velocity = 20 m/s
Time taken = 4 s
Acceleration =
Final velocity – Initial velocity
Time taken
=
(20 – 0) m/s
= 5 m/s/s = 5 m/s
2
4 s
Example 18.5
A vehicle slows down from 17 m/s to 11 m/s in
10 s. What is its acceleration?
Solution
Acceleration =
Final velocity – Initial velocity
Time taken
=
(11 – 17) m/s
=
–
0.6 m/s
2
10 s
Note that when acceleration has a negative
value, it is also called deceleration or
retardation. It means that the body is slowing
down.
Example 18.6
An object moving at 30 m/s accelerated at
8 m/s
2
for 6 s. What is its nal velocity?
Solution
Average acceleration
=
Final velocity – Initial velocity
Time taken
Let the nal velocity = v m/s
Then 8 m/s
2
=
(v – 30) m/s
6 s
8 m/s × 6 s = (v – 30) m/s
v = 48 + 30
= 78
∴ the nal velocity = 78 m/s
Exercise 18.2
1. An object travelled 1 800 m in 90 s. What
was its average velocity in:
(a) m/s (b) km/h?
2. An aeroplane travelled 180 km eastwards
in 50 minutes. What was its average
velocity in:
(a) km/h (b) m/s?
3. A supersonic jet travels at 300 m/s. What is
its average velocity in km/h?
4. A ball travelled 45 m towards the goal area
in 2.5 s. What was its average velocity in
km/h?
5. A motor cyclist travelled for 30 s at an
average velocity of 120 km/h. What
distance did he cover?
6. A ball moves at a velocity of 60 km/h.
How much time will it need to cover 30 m?
7. A vehicle starts from rest and reaches a
velocity of 20 m/s in 30 s. What is its
acceleration?
8. A particle increased its velocity from 30
km/h to 75 km/h in 10 s. What was its
acceleration?
9. A vehicle accelerated at 12 m/s
2
to reach
a velocity of 115 m/s in 6 s. What was its
initial velocity?
10. A motorist reduced the velocity of a vehicle
from 125 km/h to 75 km/h at a rate of
4 m/s
2
. How long did this take?
170
Distance-time graph
Two cars X and Y started moving from a
particular point. The distances covered and the
corresponding times taken are recorded in
Table 18.1 (a) and (b).
Car X
(a) Time t (min) 0 1 2 3
Distance s (km) 0 1 2 3
Car Y
(b) Time t (min) 0 1 2 3 4
Distance s (km) 0 0.5 1.8 3 3
Table 18.1
Fig. 18.2 shows the graphs of the relationships
between the distance (s) and time (t) for the two
cars.
Car X
(a)
Car Y
(b)
Fig. 18.2 Distance-time graphs
y
Between points P and Q in Fig. 18.2(a),
distance changes from 1 km to 3 km while time
changes from 1 min to 3 min.
i.e. change in distance = (3 – 1) km, and
change in time = (3 – 1) min.
The ratio
Change in distance
=
3 – 1
= 1 is the
Change in time 3 – 1
gradient of the distance-time graph and it is
equivalent to the velocity of the moving object.
The graph for car X is a straight line since the
velocity of car X is constant.
Thus, the graph of motion under constant
velocity is a straight line.
How would the graph of motion look like for a
stationary object?
The graph of Car Y (Fig 18.2(b)) has three
sections:
• From O to A, the gradient is increasing. This
shows that the velocity was increasing for
the rst 2 minutes.
• From A to B the curve is becoming less
steep i.e. the gradient is decreasing. This
shows that between the times t = 2 and t = 3,
the velocity was decreasing.
• From B to C, the gradient is zero. This
shows that the velocity is zero, thus the car
had stopped 3 km from the starting point.
Thus, the graph of motion under acceleration is
a curve.
In a motion under acceleration, velocity is
continuously changing. To nd velocity at a
particular time, draw a line touching the curve
only at the given point and work out the
gradient of the line. Such a line is called a
tangent to the curve. (This will be dealt with in
more detail in Form 3).
2 min
y
O
Time t (min)
3
2
1
0
1 2 3 4
x
Distance s (km)
×
B
C
A
×
×
×
Time t (min)
s
3
2
1
0
1 2 3 4
x
Distance s (km)
P
Q
2 km
Gradient = 2 km
2 min
= 1 km/min
In a distance-time graph
Gradient
=
Velocity
=
Change in distance
Change in time
171
Example 18.7
Table 18.2 shows the distance covered and the
corresponding time taken for a cyclist moving
downhill.
Time t (s) 0 0.5 1 1.5 2 2.5 3 3.5 4
Distance s (m) 0 2 8 17 28 41 60 81 104
Table 18.2
Draw a distance-time graph and from it nd
(a) the distance travelled in the rst 2.5 seconds.
(b) the velocity after 1 s.
(c) the velocity after 3 s.
Solution
Fig. 18.3 shows the graph.
(a) When t = 2.5 s, distance = 41 m.
To nd the velocity, we require the gradient.
Here we draw the tangents at t = 1 and t = 3.
(b) When t = 1, the gradient is = 14
The velocity is 14 m/s.
(c) When t = 3, the gradient is = 38
The velocity is 38 m/s.
Fig. 18.3
When drawing a distance-time graph, always
ensure that time is plotted on the horizontal
axis, otherwise the gradient will not give the
velocity.
Exercise 18.3
1. A man travelled part of the journey to his
work place by bus, then got a lift on a motor
bike before walking the rest of the journey
(See Fig. 18.4.)
Fig. 18.4
(a) Find: (i) the speed of the bus,
(ii) the speed of the motor bike,
(iii) his walking speed.
(b) What is the distance between his home
and his working place and how long did
he take to travel the whole distance?
(c) Calculate the average speed for the
whole journey.
(d) Calculate the gradient of the dotted line
and state its units.
2. Kendi travelled a distance of 30 km by car
at a speed of 80 km/h. She then cycled a
distance of 7.5 km at 10 km/h and rested
for 10 minutes before taking the rest of the
journey on foot, a distance of 4 km in 1
hour. Draw a distance-time graph and nd
her average speed for the entire journey.
3. A marble was rolled down a sloping plank
and the values in Table 18.3 were recorded.
22 – 8
2 – 1
–––––
98 – 60
4 – 3
–––––––
y
50
x
Time (h)
Distance (km)
40
30
20
10
0
1 2 3 4 5
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
walk
motor bike
bus
y
120
x
100
80
60
40
20
0
1 2 3 4 5 6
Time (s)
Distance (m)
– – – – – –
– – – – –
– – – – – –
– – – – – – – – – – –
2 – 1
4 – 3
98 – 60
22 – 8
172
Time t (s) 2 4 6 8 10 12 14
Distance s (m) 1.58 2.24 2.74 3.16 3.54 3.87 4.18
Table 18.3
Draw the graph of s against t and estimate the
speed when t = 2 and when t = 3.5.
4. A cyclist rides at a constant speed of 10 m/s
from his home to a market a distance of
1 km away and then immediately returns
home at the same constant speed. Draw the
distance-time graph for this motion.
5. In a Physics experiment, a ball is rolled
down a slope. The values in Table 18.4 are
recorded.
Time t (s) 0 1 2 3 4 5 6
Distance s (m) 0 0.7 2.8 6.3 11.2 17.2 25.2
Table 18.4
Draw a graph of s against t and estimate the
speed when t = 3 and when t = 5.
Velocity-time graph
Fig. 18.5 is the velocity-time graph for a car. A
description of the motion for the various parts is
provided after Fig 18.5.
Fig. 18.5
The velocity recorded when time is 0 seconds is
called the initial velocity. This is 0 m/s at A.
AB indicates a constant increase in velocity
from 0 m/s at A to 45 m/s at B. The time taken
for the velocity to increase from 0 m/s to
45 m/s is 3 seconds.
Acceleration = Change in velocity
Change in time
= (45 – 0) m/s
(3 – 0) s
= 45 m/s
2
3
= 15 m/s
2
This is the gradient of the line AB.
Over BC, the car maintained a constant velocity
of 45 m/s.
Over CD, the velocity of the car decreased at a
constant rate from 45 m/s to 0 m/s in 6 seconds.
Acceleration over CD =
(0 – 45) m/s
(14 – 8) s
=
–
45 m/s
2
6
=
–
7.5 m/s
2
The negative acceleration indicates that the
car’s velocity is decreasing at 7.5 m/s each
second.
Since average speed =
Distance , then
Time
Distance = average speed × time
∴ the distance covered for the various sections
of the graph is worked out as follows:
Section AB:
Distance =
(
0 + 45
)
m/s × 3 s = 67.5 m.
2
Section BC:
Distance = 45 + 0 m/s × (8 – 3) s = 225 m
(velocity over this section is constant).
y
50
x
Time (s)
Velocity (v) (m/s)
40
30
20
10
0
2 4
8 10 12
6
14
A
B
C
D
173
Section CD:
Distance =
(
45 + 0
)
m/s × (14 – 8) s
2
= 135 m.
The total distance covered
= (67.5 + 225 + 135) m
= 427.5 m
Note: The average velocity is calculated as
‘nal velocity minus initial velocity’ only if the
acceleration is constant.
Calculate the area of trapezium ABCD in
Fig. 18.5.
What do you notice?
You should have noticed that:
The area under a velocity-time graph
represents the total distance covered and the
gradient, at any point on the graph, gives the
acceleration.
Exercise 18.4
1. A body accelerates uniformly from 4 m/s to
10 m/s in 8 s. Calculate its acceleration and
draw its velocity-time graph.
2. Express an acceleration of 30 m/s
2
in km/h
2
.
3. Express an acceleration of 6 km/h
2
in m/s
2
.
4. From Fig. 18.6, calculate;
(a) the acceleration;
(i) at P,
(ii) between 23
rd
and 45
th
seconds,
(iii) during the last 15 seconds.
(b) the distance covered between;
(i) the 10
th
and 23
rd
seconds,
(ii) the 23
rd
and 45
th
seconds.
Fig. 18.6
5. Copy Table 18.5 and ll in the missing
values.
Initial Final Time
velocity velocity taken Acceleration
(m/s) (m/s) (s) (m/s
2
)
10 30 4
40 5 8
40 0
–
0.5
2 5 1.5
Table 18.5
6. A car accelerated from rest to a velocity
of 10 m/s in 10 seconds. It travelled at this
velocity for 20 s and then came to a stop in
5 s. Find;
(a) the initial acceleration,
(b) the distance travelled,
(c) the average velocity.
7. A lift accelerated from rest for 3 s and
reached a velocity of 4 m/s. It then
immediately decelerated taking 5 s to come
to rest. Calculate;
(a) the acceleration
(b) the retardation
(c) the total distance travelled.
y
x
30
Time (s)
P
10 20 30 40 50 60
20
10
0
Velocity (v) (m/s)
174
8. The results in Table 18.6 were obtained for
the velocity (v) of a vehicle measured at
intervals of time (t) in the rst minute of its
motion.
t (s) 0 5 10 15 20 25 30 35 40 45 50 55 60
v (m/s) 0 10 15 18 20 20 20 20 20 15 12 12 12
Table 18.6
Draw a graph to show these gures and
estimate the acceleration when t = 15 and
when t = 45.
Interpretation of graphs of linear
motion
This section brings together all the
interpretations that can be done of graphs of
linear motion.
Distance-time graph
Fig. 18.6 is a distance-time graph. It shows
Omari’s distance from his house, where he
started his journey, over a period of time. What
can we deduce about his journey from the
graph?
Fig. 18.6
The following is the interpretation from the
graph:
Point O is the starting point, i.e. Omari’s house.
y
50
x
Distance (km)
40
30
20
10
0
O
61 2 3 4 5
Time (h)
D
C
B
A
OA — Omari travels at a constant speed so that
after 1 h, he is 30 km from his house.
AB — He travels at a constant speed and takes
2 h to travel ( 45 – 30) km = 15 km. His
speed is lower than in the rst stage OA.
At B, he is 45 km from his house,
having taken a total of 3 h.
BC — He takes 2 h and travels 5 km in this
time. His speed is much slower than
before. The speed is constant.
At C, he is 50 km away from his home.
CD — The distance is not changing. This
means that he is not moving during that
hour.
The gradient of each portion of the graph gives
the speed over the duration of that portion.
Calculate it in each case!
Velocity-time graph
Fig. 18.7
Fig. 18.7 is a velocity-time graph of a certain
vehicle. It can be interpreted as follows:
• P is the point when time is 0 seconds.
Velocity: The initial velocity in this case is
0 m/s.
• PQ indicates constant increase of velocity
from 0 m/s at P to 30 m/s at Q.
• QR shows that the vehicle maintains
constant velocity of 30 m/s.
y
x
Velocity (v) (m/s)
40
30
20
10
0
P
122 4 6 8 10
Time (s)
S
R
Q
175
• RS indicates that the velocity decreases from
30 m/s to 0 m/s in 7 s.
The gradient of each portion of the graph gives
the acceleration, and the area between the graph
and the time – axis gives the total distance
travelled. Calculate the acceleration over each
portion, and the total distance travelled.
Relative speed
Let vehicles A and B travel in the same
direction with speeds 100 km/h and 80 km/h
respectively. If you are in vehicle B and
vehicle A overtakes you, it would appear to
pass at quite a slow speed. The speed at which
vehicle A would appear to overtake vehicle B is
referred to as the apparent speed. This is also
called the relative speed. It is calculated by
subtracting the speed of one vehicle from that
of the other vehicle.
Thus the relative speed of A to B (the speed at
which A appears to move when observed by
someone in vehicle B )
= speed of A – speed of B.
= (100 – 80) km/h
= 20 km/h
Similarly, the relative speed of B to A
= speed of B – speed of A
= (80 – 100) km/h
=
–
20 km/h.
This means that if you were in vehicle A,
vehicle B would appear to be moving at 20
km/h backwards.
On the other hand if vehicle B and A were
moving in opposite directions, one would
appear to move very fast. In this case the
relative speed (apparent speed) of one vehicle
to the other is
speed of A + speed of B,
i.e. (100 + 80) km/h = 180 km/h
The relative speed of a body B to another
one A
= speed of B – speed of A, if moving
in same direction,
= speed of A + speed of B, if moving
in opposite directions.
Exercise 18.5
1. (a) Describe the motions in the graphs of
linear motion in Fig. 18.8.
(i)
(ii)
(iii)
Fig. 18.8
(b) Calculate the speed in each portion of
the graphs in part (a).
50
1 2 3 4 5 6
x
40
30
20
10
0
y
Distance (m)
Time (s)
50
1 2 3 4 5 6
x
40
30
20
10
0
y
Distance (km)
Time (h)
50
1 2 3 4 5 6
x
40
30
20
10
0
y
Distance (m)
Time (s)
176
50
1 2 3 4 5 6
x
40
30
20
10
0
y
Velocity (m/s)
Time (s)
2. (a) Describe the motions in the graphs
shown in Fig. 18.9.
(i)
(ii)
(iii)
(iv)
50
1 2 3 4 5 6
x
40
30
20
10
0
y
Velocity (m/s)
Time (s)
50
1 2 3 4 5 6
x
40
30
20
10
0
y
Velocity (m/s)
Time (s)
50
1 2 3 4 5 6
x
40
30
20
10
0
y
Velocity (m/s)
Time (s)
(v)
Fig. 18.9
(b) Calculate:
(i) the acceleration in each portion of
the graphs in part (a) and
(ii) the total distance travelled in each
case.
3. A cyclist riding at 25 km/h is overtaken by
a bus travelling at 120 km/h . What is the
relative speed of the bus to the cyclist?
4. A motor vehicle travelling at 95 km/h, met
a bicycle moving at 35 km/h. What was the
relative speed of the bicycle to the motor
vehicle?
5. A passenger in a bus whose speed was
110 km/h noticed a sign post appear to pass.
What was the relative speed of the signpost?
6. Nafula was running at 15 km/h when she
overtook Atieno who was moving at 8
km/h. At what speed did Nafula appear to
be running according to Atieno?
7. Kirui ran a 5 000 m race in 12 min. Kamau
ran the same race in 20 min. If the race was
around a eld and Kirui overtook Kamau
once, at what speed did Kamau appear to
Kirui to be running?
50
1 2 3 4 5 6
x
40
30
20
10
0
y
Velocity (m/s)
Time (s)
177
What is statistics?
For a group of people, such as pupils in a class,
a lot of information can be collected about
each one of them. Some of the information
that could be collected about the group may be
their names, where they live, what food they
like most, etc. They could also state their ages,
heights etc. Note that some of the information
uses numbers.
Information in which numbers are used is
called statistics. Each such number is also
called a statistic (plural statistics). Any other
number that may be derived from some
computation using the original numbers is
also a statistic.
Thus, statistics is the study of information
represented in numerical form.
For example, Table 19.1 shows marks scored by
40 pupils in a Biology test.
78 46 55 47 77 63 52 52 62 46
77 47 40 35 67 61 58 52 42 40
48 57 66 54 75 78 75 59 75 47
59 35 62 53 72 57 51 69 55 57
Table 19.1
(a) What is the highest score?
(b) What is the lowest score?
(c) What is the difference between the highest
and the lowest scores?
(d) What mark was scored by most pupils?
(e) How many pupils scored above 70?
(f) If the pass mark was 45, how many pupils
failed the test?
(g) How many pupils scored
(i) 30 to 39, (ii) 40 to 49,
(iii) 50 to 59, (iv) 60 to 69,
(v) 70 to 79 marks?
(h) How did the pupils perform in that test in
general?
To answer questions (d) – (g), we have to count
correctly, while question (h) requires some
general conclusions.
Note:
1. Information such as in Table 19.1 is called
raw data or simply data.
2. Statistics, in general, is concerned with
collecting data, organising it, answering
questions about the data and making sensible
conclusions and decisions based on the
analysis of the data.
Collecting and organising data
In Statistics, data must be collected and
recorded accurately. The data in Table 19.1
could have rst been recorded in a class list
showing marks scored by each pupil e.g.
Abungu J. 78
Akunda J. 46
Bari D. 55
The list could then have been used to make
Table 19.1.
Information could also be collected as shown in
Table 19.2. The table shows types of vehicles
which passed at a given point over a given time
interval.
STATISTICS (1)
19
178
Vehicle Tally marks Totals (Frequency)
Bicycle // 2
Motor bike /// 3
Car //// //// // 12
Minibus //// 5
Lorry // 2
Bus //// // 7
Table 19.2
When a vehicle passes, a stroke (/) is put in the
‘Tally’ column in the row corresponding to that
type of vehicle. Every fth stroke against a
given type of vehicle is made to cross the other
four so that instead of having ve strokes as /////
we have ////. The next stroke will be next to the
group of ve strokes, i.e. //// /, etc. These strokes
are recorded in the ‘Tally’ column until the time
for collecting the information is over. The total
number of strokes against each type of vehicle
is then recorded in the ‘Totals’ column.
Note:
1. The tally method of recording makes
counting easier and more accurate
especially when big numbers are involved.
2. Bundling strokes into ves (////) makes it
easier to count them.
Frequency distribution table
A table such as Table 19.2 is called a
frequency distribution table or simply
a frequency table. Frequency means the
number of times an item or value occurs.
In Table 19.2, the last column shows the total
number of each type of vehicle. This number
is the frequency for each type of vehicle. Thus
we usually write ‘Frequency’ in that column
instead of ‘Totals’.
Example 19.1
Table 19.3 shows the grades scored by a class
of 30 students in a Mathematics examination.
C B C A C B B D A A
B B C C B D B C B D
A B A C B A C C B C
Table 19.3
Make a frequency table for the information in
this table.
Solution
Table 19.4 is the required frequency table.
Grade Tally Frequency
A //// / 6
B //// //// / 11
C //// //// 10
D /// 3
30
Table 19.4
Table 19.4 is made by putting a stroke (/) in
the ‘Tally’ column against the grade, for each
grade in Table 19.3. This is systematically done
so that in Table 19.4, there is a stroke for each
grade in Table 19.3. All the strokes against
each grade are counted and the counts written
in the Frequency column.
Exercise 19.1
1. Work in groups of 5 or 6. Collect the
following data from each member of your
group: height, mass, size of shoes worn,
favourite subject, favourite sport. Each
group leader should record information
about his/her group as shown in Table 19.5.
Each leader should then collect the data
from other groups so that each group has
data for the whole class.
179
Name Height Mass Size of Favourite Favourite
(cm) (kg) shoes subject sport
Table 19.5
Make a frequency table for each statistic
except the name.
2. Record the temperature, (a) outside the
classroom, and (b) outside the classroom at
the following times: 7.30 a.m, 10.30 a.m,
1.30 p.m and 4.30 p.m. This should be
done everyday for a whole week (Different
groups could be assigned different days).
3. Choose a place near your school where
trafc passes. Count the number of different
types of vehicles that pass there in the
morning (7 – 8 a.m), at lunch time (1 – 2
p.m), and in the evening (4 – 5 p.m). Keep
different records for different directions.
Different groups should be assigned
different days of the week so that the data
is recorded for a whole week. Each group
should ll their data in a table such as Table
19.6(a). Table 19.6(b) could be used to
combine information from all groups.
(a)
Group no. ………
Date/Day ……………… Time ………
Direction of trafc ………………………
Vehicle Tally Frequency
Bicycle
Car
Motor cycle
Mini bus
Lorry
Bus
Others
(b)
Place …………………
Time …………………
Direction of trafc ………………………
Vehicle Mon Tue Wed Thur Fri
Table 19.6
4. Find out from the school library how many
books there are in each of the following
subject categories.
(a) English
(b) Other languages (Kiswahili, French,
German, etc.)
(c) Mathematics (d) Science
(e) Religion (f) Others
5. On this page that you are reading, count
all the words having one letter, two letters,
three letters, four letters, and ve letters.
Make a frequency table for this data.
6. Table 19.7 shows the amount of milk in
litres produced by 36 cows on Keli’s farm
in one day.
8 10 11 9 9 4 8 18 13
16 15 12 13 4 6 11 15 4
8 7 9 13 7 10 11 8 5
14 12 9 14 9 12 5 14 14
Table 19.7
Make a frequency distribution table for this
data.
180
all scores from 30 to 39, inclusive, are in one
group; all scores from 40 to 49, inclusive, are in
the next group; etc.
Each of these groups is called a class. The
values 30 and 39, 40 and 49, 50 and 59, etc. are
called class limits for the respective class.
Note that when the number of items in a data
distribution is small, it is easy to deal with
them; but when the number of items is large it
becomes necessary to group the data.
Example 19.2
Table 19.11 shows the masses (in grams) of 50
carrots taken from a plot of land on which the
effect of a new fertilizer was being investigated.
103 95 105 117 93 112 111 108 73 109
66 99 87 98 76 67 107 119 103 95
77 88 65 107 85 94 101 104 72 92
82 90 118 103 100 75 102 116 82 105
114 106 70 116 112 97 63 111 118 91
Table 19.11
Make a frequency table for this data.
Solution
The smallest mass is 63 g and the largest is
118 g. The difference between the largest value
and smallest value is called the range.
Thus, range = 118 g – 63 g = 55 g.
We need to group the data into a convenient
number of classes. Usually the convenient
number of classes varies from 4 to 12.
By dividing the range by class size we get the
number of classes. Thus a class size of 5 will
give us i.e. 11 classes.
A class size of 8 will give us i.e. 7 classes.
A class size of 10 will give us i.e. 6 classes.
Let us use a class size of 10.
Table 19.12 is the required frequency table.
55
5
––
55
10
––
55
8
––
7. Table 19.8 shows the sizes of shoes worn
by 40 pupils in a Form 2 class in Twaimba
High school.
(a) Make a frequency table for the data.
6 10 7 6 7 8 7 9 11 11
10 7 8 6 8 7 9 11 6 7
9 9 8 7 10 10 8 8 7 8
8 7 8 8 7 9 8 7 10 9
Table 19.8
(b) Whom do you think would be interested
in this information?
8. Table 19.9 shows the grades scored by 40
pupils in a Mathematics examination. Make
a frequency table for the data.
E A B D C C B C
B C C B D C B D
A C B C D A B B
C B D A C B B C
D C C B C B D C
Table 19.9
Grouped data
With reference to Table 19.1, we were required
to nd the number of pupils who scored from
30–39, 40–49, 50–59, 60–69 and 70–79.
You should have obtained the results shown in
Table 19.10.
Marks No. of students
30–39 2
40–49 9
50–59 14
60–69 7
70–79 8
Total 40
Table 19.10
When data is presented as in Table 19.10, it
is said to be grouped data. This means that
181
Mass in grams Tally Frequency
60–69 //// 4
70–79 //// / 6
80–89 //// 5
90–99 //// //// 10
100–109 //// //// //// 14
110– 119 //// //// / 11
Table 19.12
Note: If the range is small, it is more
convenient to use class sizes which are even.
If it is large, multiples of 5 or 10 are more
convenient. This is helpful especially if there is
need to represent the data graphically.
Exercise 19.2
1. A handspan is the distance (length) from
the end of the thumb to the end of the small
nger when the hand is fully open. Table
19.13 shows the handspans of some 21
children measured in centimetres.
18.4 17.4 20.7 14.3 20.0 19.0 18.5
21.7 17.5 18.1 19.3 16.9 19.8 15.9
21.2 18.7 19.2 16.6 14.8 17.8 16.0
Table 19.13
Make a frequency distribution table,
grouping the data into four classes starting
with 14.0 – 15.9.
2. The lengths of 36 pea pods were measured
to the nearest millimetre and recorded in
Table 19.14.
71 92 88 52 73 84 73 73 82
66 85 78 63 90 76 89 53 77
68 59 55 80 79 91 86 75 60
93 76 84 83 62 86 70 65 72
Table 19.14
Put the data into a grouped frequency table
by choosing a convenient number of classes.
3. Table 19.15 shows marks scored by 60
pupils in a physics examination.
69 70 72 40 52 60 22 31 78 53 56 55
28 67 63 54 57 48 47 56 55 62 72 78
75 38 37 44 62 64 58 39 45 48 56 59
65 50 58 46 47 57 35 34 58 64 48 50
62 37 41 42 36 54 52 48 53 57 44 45
Table 19.15
Make a grouped frequency table using
classes 20–29, 30–39, 40–49, etc.
4. A survey was conducted to nd out how
much money (in sh.) students in Form 2 had
spent in a particular week. The data in Table
19.16 were obtained.
60 61 97 22 61 98 99 46 50 47
53 70 53 30 87 40 41 45 49 64
92 85 44 16 41 11 83 12 79 44
61 40 74 57 49 24 47 64 18 30
54 52 63 47 59 38 83 34 50 49
32 53 87 96 80 57 77 54 52 81
69 66 43 81 85 52 65 30 62 78
Table 19.16
Make a grouped frequency table using
classes 10–19, 20–29, 30–39, etc.
5. A student measured the amount of ink in
biro pens used in his class by measuring the
length (in cm) of the ink column that could
be seen. Table 19.17 was obtained. Make a
grouped frequency table with 6 classes.
0.1 8.2 7.7 4.5 1.2 0.7 1.0 6.3 7.6 3.5
1.9 4.8 5.6 2.7 4.4 8.7 0.5 0.4 7.9 8.8
5.3 4.3 3.7 1.5 2.6 5.1 8.5 1.3 1.3 0.8
1.1 3.9 0.3 3.4 2.5 5.3 6.1 2.8 8.8 1.5
5.7 2.1 7.7 5.2 0.5 0.4 2.2 2.6 0.7 4.7
2.3 0.5 4.7 4.7 2.5 0.7 1.6 3.9 3.3 6.6
3.5 0.7 4.4 7.5 5.7 9.0 1.2 0.2 0.7 5.4
Table 19.17
182
Presentation of data
Once data have been collected, they may be
presented or displayed in various ways. Such
displays make it easier to interpret and compare
the data. The following are some of the ways.
Rank order list
A rank order list shows items that have
been arranged in order from the highest to
the lowest or from the lowest to the highest.
For example, ve pupils had the following
scores in a Mathematics test: 15, 12, 21, 13, 18.
The rank order is 21, 18, 15, 13, 12 or 12, 13,
15,18,21.
The rank order list helps us to nd the:
(a) highest value.
(b) lowest value.
(c) most common value.
(d) value which is in the middle.
(e) number of those above or below a given
value, etc.
Frequency distribution table
As we saw earlier, a frequency distribution
table is a table which shows data items and the
number of times (frequency) they occur. Such a
table helps us to see the:
(a) highest value
(b) lowest value
(c) most common value, etc.
Pictogram (or pictograph)
A pictogram (pictograph) is a diagram that
represents statistical data in a pictorial form.
Each picture or drawing represents a certain
number or value from the data.
The picture to be used is chosen to represent the
data subject as closely as possible, e.g. a shoe to
represent the number of pairs of shoes, a car to
represent the number of cars etc.
Example 19.3
Represent the data in Table 19.18 in a pictogram.
Size 6 7 8 9 10 11
No. of shoes 4 11 11 6 5 3
Table 19.18
Solution
Let ?? represent 2 pairs of shoes. Since 4
pupils wore shoes of size 6, then 4 will be
represented by ?? ,
etc. So the data in Table
19.8 would be represented as in Fig. 19.1.
Size 6
Size 7
Size 8
Size 9
Size 10
Size 11
Fig. 19.1
Note that a fraction of 2 is represented by a
fraction of the drawing.
Although the pictogram is used to display
information, it is not an accurate way of
representing data. For instance, suppose each
picture represented 10 pairs, how would you
represent 2, 3, 7 or 8 pairs?
Pie-chart
A pie-chart is a graph or diagram in
which different proportions of a given data
distribution are represented by sectors of a
circle.
Since the diagram is a circle, it is looked at as a
circular ‘pie’, hence the name pie chart.
Example 19.4
Table 19.19 shows grades scored by 15
candidates who sat for a certain test.
Grade A B C D E
No. of candidates 2 5 4 1 3
Table 19.19
183
Draw a pie chart for this data.
Solution
Work out the fractions of numbers of candidates
who scored each grade. For example, for grade
A we have .
Since the angle at the centre of a circle is 360°,
we calculate the angle to represent grade A as
of 360°
i.e. × 360° = 48°.
A is represented by an angle of 48°.
Table 2.20 shows all the angles.
Grade No. of Fraction Angle at centre
candidates of total of circle
A 2 × 360° = 48°
B 5 × 360° = 120°
C 4 × 360° = 96°
D 1 × 360° = 24°
E 3 × 360° = 72°
Table 19.20
Fig. 19.2 shows the required pie chart.
Fig. 19.2
Note:
1. Usually there are no numbers on a pie chart.
2. The sizes of the sectors give a comparison
between the quantities represented.
3. The order in which the sectors are presented
does not matter.
4. Sectors may be shaded with different
patterns (or colours) to give a better visual
impression.
Bar chart
A bar chart (or bar graph) is a graph
consisting of rectangular bars whose lengths
are proportional to frequencies in a data
distribution.
Example 19.5
Table 19.21 shows the sizes of sweaters worn by
30 Form 2 students in a certain school.
Represent the data on:
(a) a horizontal bar chart
(b) a vertical bar chart.
Size Small Medium Large Extra large
(S) (M) (L) (XL)
No. of
5 13 8 4
pupils
Table 19.21
Solution
(a) Horizontal bar chart Fig. 19.3(a)(i).
In a horizontal bar chart, frequency is
represented on the horizontal axis.
Bars can be drawn with spaces between
them (as in Fig, 19.3(a)(ii)) or without, and
they may be shaded or not.
(b) Vertical bar chart Fig. 19.3(b).
In a vertical bar graph, frequency is
represented on the vertical axis.
2
15
––
2
15
––
2
15
––
5
15
––
4
15
––
1
15
––
3
15
––
2
15
––
5
15
––
4
15
––
1
15
––
3
15
––
A
B
C
D
E
2
15
––
XL
L
M
S
0 2 4 6 8 10 12 14
Frequency
(a) (i)
184
(b)
Fig. 19.3
Note: In a bar chart:
1. The widths of the bars are the same.
2. The height of a bar is proportional to the
corresponding frequency.
Exercise 19.3
1. In a survey on soft-drinks, 180 people were
asked to state the brand they preferred.
35 chose brand A, 30 chose brand B, 100
brand C and 15 chose brand D. Draw a pie-
chart to display this information.
2. At the semi-nal stage of a football
competition, 72 neutral observers were asked
to predict which team they thought would
win. Table 19.20 shows their predictions.
Team No. of predictions
Team A 9
Team B 40
Team C 22
Team D 1
Table 19.22
Draw a pie-chart to display the predictions.
3. Mr. Kamau has a monthly income of
sh. 6 000. The pie-chart in Fig. 19.4 shows
how he spends the money. How much does
he spend on
(a) Food, (b) Rent, (c) Savings,
(d) Entertainment, (e) Travel?
Fig. 19.4
4. Of the animals on Kibor’s farm, 35% are
cows, 20% are goats, 15% are sheep, 2%
are donkeys and 28% are pigs. A pie chart
is to be drawn to illusrate this information.
Find the angle of the sector representing
each type of animal.
5. In Exercise 19.1 Question 1, you collected
information about your class, i.e.
(i) size of shoes worn
(ii) favourite subject
(iii) favourite sport of each pupil.
Display this information in a pie-chart.
6. Display the information you collected in
Exercise 19.1 Question 4, in a pie chart.
7. Represent the information in Exercise 19.1
S
14
M L XL
12
10
8
6
4
2
0
Frequency
TRAVEL
35°
80°
20°
75°
60°
90°
OTHERS
FOOD
SAVINGS
ENTERTAIN-
MENT
RENT
XL
L
M
S
0 2 4 6 8 10 12 14
Frequency
(ii)
185
Question 5 on
(a) a pie-chart,
(b) a bar chart.
8. Display the information in Table 19.9 of
Question 8 (Exercise 19.1) on a bar chart.
9. Represent the information in Table 19.15 of
Question 3 (Exercise 19.2) on a bar chart.
10. Display the information in Table 19.16 of
Question 4 (Exercise 19.2) on a bar chart.
11. Fig. 19.5 shows the heights of pupils in a
certain class.
(a) How many pupils are over 150 cm tall?
(b) How many pupils have a height
between 125 cm and 145 cm?
(c) How many pupils are in the class?
Fig. 19.5
12. Some children were picked at random and
their handspans measured in centimetres.
This data was recorded as in Table 19.23.
18.4 17.4 21.2 18.7 21.7 17.5 18.1
19.2 16.6 14.3 19.3 17.8 20.7 18.5
16.9 20.0 15.9 19.8 16.0 14.8 19.0
Table 19.23
8
Frequency
6
4
2
0
120 – 124
125 – 129
130 – 134
135 – 139
140 – 144
145 – 149
150 – 154
155 – 159
160 – 164
(a) Draw a bar-chart for this data using
class intervals 14.0–15.9, 16.0–17.9, …
(b) How many children had a handspan
greater than or equal to 18 cm?
Histogram
Consider the grouped distribution in Table
19.24.
Marks No. of students
30–39 2
40–49 9
50–59 14
60–69 7
70–79 8
40
Table 19.24
The class 30–39 includes all values equal to or
greater than 29.5 but less than 39.5.
Similarly, the class 40–49 includes all values
equal to or greater than 39.5 but less than 49.5.
The values 29.5, 39.5, 49.5, etc. are called class
boundaries.
Thus, the class 30–39 can be represented
as 29.5–39.5. The value 29.5 is the lower
class boundary and 39.5 is the upper class
boundary for this class.
Similarly, the class 40–49 can be extended to
39.5–49.5, etc.
The difference between the upper class
boundary and the lower class boundary is called
the class interval, class width, or class size,
i.e. class interval = upper class boundary –
lower class boundary.
A histogram (or a frequency histogram)
is a graph that consists of a series of
rectangles
(a) drawn on a continuous scale (i.e. no gaps
between the rectangles) and
186
(b) with areas being proportional to class
frequencies. The height of the
rectangles are obtained as
, f being
the frequency and w the class width. The
value is called frequency density.
Example 19.6
Represent the data on Table 19.25 on a
histogram.
Solution
Table 19.25 represents the same data as in
Table 19.24, but with the class limits having
been changed to class boundaries and the
columns for frequency densities also included.
Marks
No. of Frequency
students density
29.5–39.5 2 0.2
39.5–49.5 9 0.9
49.5–59.5 14 1.4
59.5–69.5 7 0.7
69.5–79.5 8 0.8
Table 19.25
Choose a suitable scale and use it to represent
marks on the horizontal axis, indicating all the
class boundaries along this axis.
Since the class interval is constant, the
rectangles must have the same width.
Fig. 19.6 shows the required histogram.
Note that:
(i) The class boundaries mark the boundaries
of the rectangular bars in the histogram.
(ii) In Fig. 19.6, the horizontal axis is
compressed. We put to show that
there is no information being displayed on
the lower part of the axis.
(iii) Where the widths of the rectangles of a
histogram are equal (as in Fig. 19.6), the
histogram is similar to a bar chart.
Fig. 19.6
Example 19.8
In a certain function, the ages of the people
present were recorded as shown in Table 19.26.
Age Frequency
0–2 15
3–5 25
6–12 78
13–20 124
21–34 166
35–60 252
660
Table 19.26
Draw a histogram for this data.
Solution
Remember that in a histogram, the area of
each rectangle represents the frequency, and
the width of a rectangle in the class interval.
So we get the height of the bar by dividing the
frequency by the class interval. This result is
known as frequency density.
1.4
1.2
1.0
0.8
0.6
0.4
0.2
0
29.5 39.5 49.5 59.5 69.5 79.5
Marks
Frequency density
f
w
f
w
187
Age f
Height of column
(frequency density)
0–2 i.e. 0 and less than 3 15 15 ÷ 3 = 5
3–5 i.e. 3 and less than 6 25 25 ÷ 3 = 8.3
6–12 i.e. 6 and less than 13 78 78 ÷ 7 = 11
13–20 i.e 13 and less than 21 124 124 ÷ 8 = 15.5
21–34 i.e 21 and less than 35 166 166 ÷ 14 = 11.5
35–60 i.e 35 and less than 61 252 252 ÷ 26 = 9.7
Table 19.27
The height of each column (Table 19.27)
represents the average number of people in
each age group. It is assumed that there is a
uniform distribution within the class intervals.
Fig. 19.7 shows the required histogram.
Fig. 19.7
Frequency polygon
Table 19.28 shows the same information that
was in Table 19.22 but this time the mid-points
or class marks of the classes are included. Each
class mark is obtained as half the sum of the
class limits (or boundaries).
Marks Mid-points No. of students Frequency
= frequency density
30–39 34.5 2 0.2
40–49 44.5 9 0.9
50–59 54.5 14 1.4
60–69 64.5 7 0.7
70–79 74.5 8 0.8
40
Table 19.28
Frequency density
16
5
14
12
10
8
6
4
2
0
10 15 20 25 30 35 40 45 50 55 60 65
Age
When the frequency densities are plotted
against the corresponding midpoints and the
points joined with straight line segments, we
obtain Fig 19.8.
Fig. 19.8 is called a frequency polygon.
Fig. 19.8
Thus a frequency polygon is a graph in
which frequency densities are plotted against
class mid-points and the points are joined
with straight line segments.
Note that in Fig. 19.8, the rst point plotted is
the mid-point of the rst class whose frequency
is 2.
Normally, it is taken that the previous class (i.e
20–29) has frequency 0 (zero). So the graph
is extended to this class. Similarly it is taken
that the next class, after the last, has frequency
0 (zero). The graph is also extended to this
class. Often, more than one frequency polygon
may be drawn on the same axes for purposes of
comparing frequency distributions.
Exercise 19.4
1. Table 19.29 shows masses of 100 students
at Zed College.
Frequency density
1.6
24.5
1.4
1.2
1.0
0.8
0.6
0.4
0.2
0
34.5 44.5 54.5 64.5 74.5 84.5
Marks
×
×
×
×
×
×
×
188
Mass (kg) No. of students
58–62 4
63–67 19
68–72 40
73–77 35
78–82 2
100
Table 19.29
Draw a histogram for this data.
2. Table 19.30 shows marks in a Mathematics
test for some 80 pupils at Waya primary
school.
Marks No. of pupils
50 – 53 1
54 – 59 2
60 – 62 9
63 – 68 11
69 – 74 13
75 – 80 22
81 – 85 8
86 – 90 7
91 – 99 7
Table 19.30
Draw a histogram for this data.
3. Table 19.31 shows masses, to the nearest
kg, of 100 pupils, who were picked at
random, in Mbombo secondary school.
Mass (kg) No. of pupils
25–34 2
35–44 9
45–54 10
55–59 12
60–64 16
65–69 20
70–79 13
80–89 4
90–104 2
Table 19.31
Draw a histogram to represent this
information.
4. Draw a frequency polygon using the data in
(a) Question 1 Table 19.29
(b) Question 2 Table 19.30
(c) Question 3 Table 19.31.
5. Draw a frequency polygon using the data in
Table 19.26.
Line graphs and trend
Table 19.32 shows the temperatures, in degrees
Celsius, observed at 2-hourly intervals, of a
patient who was admitted at a hospital.
Time 6 a.m 8 a.m 10 a.m 12 noon 2 p.m 4 p.m
Temp °C 38.2 38.6 38.9 38.8 38.8 38.5
Time 6 p.m 8 p.m 10 p.m mid-night 2 a.m 4 a.m
Temp °C 38.2 37.8 37.1 37.0 36.8 36.8
Table 19.32
When a graph of time against temperature is
plotted using the data in Table 19.32 and the
points joined by dotted line segments, Fig. 19.9
is obtained.
Fig. 19.9
A graph, such as Fig. 19.9, which is formed
by broken line segments joining the points
representing given data is known as a line
graph.
Temperature °C
40
6
39
38
37
36
Time
×
×
×
×
×
×
8 10 12 2 4 6 8 10 12 2 4 6 8
×
×
×
×
×
×
– – – –
– – – –
– – – –
– – – – – –
– – – –
– – – – – – – –
– – – –
– – – –
– – – –
– – – –
189
(a) Why are points joined by dotted lines?
When could such points be joined by
continuous line segments?
(b) Estimate the patient’s temperature at
(i) 9. a.m (ii) 3 p.m
(iii) 11 p.m. (iv) 3 a.m
(c) What can you say about the patient’s
temperature
(i) during the day? (ii) during the night?
Note:
A line graph helps us appreciate the pattern or
trend of a given variable i.e. how the variable
changes with time.
Exercise 19.5
1. The average masses of pupils of different
ages in a certain school were obtained and
recorded as in Table 19.33.
Age (years) 11 12 13 14 15 16 17
Average
36.7 38.6 42.4 46.8 51.9 60.4 65.6
mass (kg)
Table 19.33
(a) Represent this data on a line graph.
(b) What is the estimate for the average
mass of pupils who are
(i) 11 years, (ii) 13 years,
(iii) 15 years old?
2. Table 19.34 shows the population of a
certain country, in thousands, in the given
years.
Year 1901 1911 1921 1931 1941 1951 1961 1966
Pop. 5 400 7 200 8 800 10 400 11 500 14 000 18 200 20 000
Table 19.34
(a) Draw a line graph for the data.
(b) Estimate the population in
(i) 1926 (ii) 1971.
3. Table 19.35 shows, in degrees, the
maximum and minimum temperatures
1
2
–
1
2
–
recorded for the rst 12 days of September
in a certain town.
Day 1 2 3 4 5 6 7 8 9 10 11 12
Max (°C) 26 21 17 23 26 21 20 21 19 19 19 17
Min (°C) 15 13 12 13 13 14 16 11 11 10 10 12
Table 19.35
(a) Using the same axes, draw line graphs
to represent these temperatures.
(b) Which of the two sets of temperatures
shows the greater variation?
(c) If the temperatures for 5
th
and 11
th
September had been omitted, could you
have estimated them?
(d) Using the trends shown by the graph,
make a forecast for 13
th
September.
4. Table 19.36 shows deaths, in thousands,
from two diseases during a period of 10
years.
Year 1947 1948 1949 1950 1951 1952 1953 1954 1955 1956
TB
23.9 22.9 17.5 14.1 12 9.3 7.9 7.1 5.8 4.9
deaths
Pneu-
monia 26.7 20.7 23.6 20.3 25.6 21 23 20.4 23.8 24.8
deaths
Table 19.36
(a) On the same axes, draw line graphs to
represent this data.
(b) What can you say about deaths from
each disease?
5. Table 19.37 shows the time taken by a
certain test car to accelerate from 0 to the
given speeds.
Acceleration Time taken (s)
0 to 50 kph 4.8
0 to 65 kph 6.4
0 to 80 kph 9.0
0 to 95 kph 12.3
0 to 110 kph 16.1
0 to 125 kph 21.4
Table 19.37
190
(a) Represent the data on a line graph with
speed on the horizontal axis.
(b) How long would the car take to
accelerate from
(i) 0 to 70 kph, (ii) 0 to 120 kph?
(c) What speed would the car gain from
rest in (i) 8 s, (ii) 14 s?
Averages
The average of a group of numbers is a single
number which can be used to represent the
entire group. For example:
1. If the average age of pupils in a class is
6 years, it means that most pupils in that
class would have their ages close to that
value. A child of age 11 or 3 years would
ordinarily not be expected to be in that
class.
2. If the average life span of a certain make of
light bulb is 600 hours, it is expected that a
light bulb of that make should last that long
when used, maybe a little more or a little
less.
The most common types of averages are the
mean, the mode and the median.
Averages are also called measures of central
tendency because they show how the values
in a data distribution tend to cluster around a
central value.
The mean
The arithmetic mean (or simply the mean) of
a group of values is the most common average,
and it is given by
Arithmetic mean
=
the sum of all the values in the group
the total number of values in the group
In other words, if the group has n values x
1
,
x
2
, x
3
, …, x
n
, then the mean of those values is
given by
Mean =
x
1
+ x
2
+ … + x
n
n
Example 19.8
Calculate the mean of sh. 5, sh. 8, sh. 11, sh. 15
and sh. 6, the pocket money of some 5 pupils.
Solution
Mean = sh
5 + 8 + 11 + 15 + 6
5
= sh
45
= sh. 9
5
Example 19.9
Table 19.38 shows the masses, in grams, of
some 40 mangoes sold in a shop. Calculate the
mean mass of the mangoes.
70 100 90 120 110 80 110 80 100 90
120 110 120 80 110 100 110 90 110 100
80 70 100 80 90 110 110 90 100 90
100 100 90 100 90 90 80 100 100 80
Table 19.38
Solution
Method 1
Mean mass =
Sum of masses of all mangoes
Total number of mangoes
Add all the values in Table 19.35. The total
mass is 3 850 g.
∴ mean mass =
3 850
g
= 96.25 g
40
Method 2
Make a frequency table as shown in Table 19.39.
Mass x (g) Tally Frequency, f fx
70 // 2 140
80 //// // 7 560
90 //// //// 9 810
100 //// //// / 11 1 100
110 //// /// 8 880
120 /// 3 360
∑f = 40 ∑fx = 3 850
Table 19.39
191
The column ‘fx’ represents the total mass of
mangoes each of mass x g. For example, there
are 2 mangoes each of mass 70 g. Their total
mass is (70 × 2) g = 140 g.
Similarly, there are 7 mangoes, each of mass
80 g, having a total mass of (80 × 7) g = 560 g,
etc.
The symbol ∑ (Greek letter ‘sigma’), stands for
‘sum of’.
Thus, ∑f means ‘sum of frequencies’ and ∑fx
means ‘sum of products of f and x’.
The mean, denoted by x, is given by
mean, x =
∑fx
∑f
In the example, ∑f = 40, fx = 3 850
∴ x =
∑fx
=
3 850
= 96.25 g
∑f
40
Exercise 19.6
1. Calculate the mean of the following.
(a) 1, 3, 5, 7 (b) 2, 4, 6, 8
(c) 1, 2, 3, 4, 5 (d) 6, 7, 8, 9, 10
(e) 2, 2, 3, 4, 4, 5 (f) 3, 4, 4, 7, 8, 9
(g) 10, 4, 11, 13, 15, 19, 21, 5
(h) 8, 0, 3, 3, 1, 7, 4, 1, 4
2. Calculate the mean of the following.
(a) 2.1, 1.4, 3.5, 2.7
(b) 4.8, 4.5, 3.2, 1.8, 2.2
(c) 0.7, 0.9, 0.3, 0.8, 0.7, 0.9, 0.8, 0.6, 0.5, 0.2
(d) 3 , 4 , 2 , 3
3. Eight ladies had masses of 51 kg, 44 kg,
57 kg, 63 kg, 48 kg, 49 kg, 45 kg, 53 kg.
Find the mean of their masses.
4. The mean mark scored by 5 pupils in a
Maths test was 19. Four pupils had the
following scores 15, 18, 17, and 16. What
score did the 5
th
pupil have?
5. Table 19.40 shows the lengths, in centimetres,
of nails found in a certain container.
2 7 4 3 2 6 5 5 4 6
5 7 3 5 4 7 3 4 3 2
4 6 4 7 5 3 5 5 6 3
6 5 7 6 6 4 6 6 5 7
5 4 2 7 4 3 6 5 5 7
Table 19.40
Construct a frequency table and use it to
calculate the mean length of the nails.
6. Table 19.41 shows the number of goals
scored in a series of football matches.
Number of goals 1 2 3
Number of matches 8 8 x
Table 19.41
If the mean number of goals is 2, what is x?
7. In this question, letter grades are assigned
the values shown in Table 19.42.
A A
–
B
+
B B
–
C
+
C C
–
D
+
D
12 11 10 9 8 7 6 5 4 3
Table 19.42
Use the values in Table 19.43 to determine
the mean grade for each subject as obtained
in an examination by students in Paani high
school.
A A
–
B
+
B B
–
C
+
C C
–
D
+
D
Biology 20 15 26 12 16 7 4 1 – –
Commerce 14 9 16 11 9 3 – – – –
German 1 2 6 7 10 1 3 3 – –
French 1 1 6 9 5 5 1 7 3 5
Art – – – 3 2 5 3 4 2 1
Physics – – – 5 4 13 22 35 6 10
Table 19.43
–
–
–
1
2
–
1
4
–
1
8
–
3
4
–
192
The mode
In a given data distribution, the value or item
that has the highest frequency is called the
mode (from the French ‘a la mode’meaning
‘fashionable’)
Example 19.10
What is the mode of 71, 71, 72, 75, 73, 75, 76,
76, 75, 72, 78, 79, 75, 78, 79, 75, 71, 73, 75, 76?
Solution
Table 19.44 is the frequency table for the data.
Number Frequency
71 3
72 2
73 2
75 6
76 3
78 2
79 2
20
Table 19.44
Table 19.44 shows that 75 is the value with the
highest frequency (i.e. 6). Thus, 75 is the mode
and 6 is the modal frequency.
The median
When data is arranged in order from the
smallest to the largest or the largest to the
smallest, the middle value is called the
median.
Example 19.11
Find the median of the following numbers
(a) 15, 12, 13, 13, 9, 10, 8, 11, 10, 8, 7, 9,
10, 10, 11
(b) 12, 8, 21, 11, 4, 12, 13, 18, 20, 19, 21, 11
Solution
Arrange the values in order from the smallest to
the largest.
(a) 7, 8, 8, 9, 9, 10, 10, 10, 10, 11, 11, 12, 13,
13, 15
This is the middle value
i.e. median = 10
(b) 4, 8, 11, 11, 12, 12, 13, 18, 19, 20, 21, 21
In this case, there is no one value that is in
the middle. In such a case the median is the
mean of the two middle values.
Thus, median =
12 + 13
= 12.5
2
Note:
When the number of values is odd, the
median is the middle value but when the
number of values is even, the median is the
mean of the two middle values.
Example 19.13
Table 19.45 shows the masses of some tomatoes
bought from a farmer.
Mass (in g) 58 59 60 61 62 63
Frequency 2 6 12 9 8 3
Table 19.45
Find (a) the mean, (b) the median, (c) the mode
of the masses of the tomatoes.
Solution
(a) Mean mass =
(58 × 2) + (59 × 6) + (60 × 12) + (61 × 9) + (62 × 8) + (63 × 3)
(2 + 6 + 12 + 9 + 8 + 3)
= 2 424 = 60.6 g.
40
(b) Since there are 40 tomatoes, the median
mass is the mass between those of the 20
th
and 21
st
tomatoes.
– – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – –
– – –
– – –
193
From Table 19.46, we see that there are 20
tomatoes with mass 60 g and less. The 21
st
tomato must, therefore, have a mass of 61 g.
Mass (g) 58 59 60 61 62 63
Frequency 2 6 12 9 8 3
Cumulative
2 8 20 29 37 40
frequency
Table 19.46
Cumulative frequency is a running total
of frequencies showing what the total
frequency is at the end of each class.
Mass of the 20
th
tomato = 60g, and mass of
the 21
st
tomato = 61 g.
∴ median =
60 + 61
= 60.5 g.
2
(c) The mode is 60 g.
Exercise 19.7
1. Find the mean, median and mode of each of
the following groups of numbers:
(a) 4, 5, 8, 6, 4, 12, 3
(b) 9, 9, 7, 7, 4, 13, 3, 17, 2, 21, 7
(c) 7, 9, 4, 9, 3, 9, 10, 1, 1, 12
(d) 5, 4, 5, 5, 4, 2, 3, 2, 1, 3, 1, 4, 0, 2, 2, 0, 1
2. Write down 5 numbers such that the mean
is 6, the median is 5 and the mode is 4.
3. Seven pieces of luggage have masses of
48 kg, 45 kg, 49 kg, 63 kg, 57 kg, 44 kg
and 51 kg.
(a) Find the mean masses of the seven
pieces.
(b) If the lightest and the heaviest piece
are taken away, what is the mean masses
of the remaining ones?
(c) What is the median masses if the
lightest piece is removed?
4. Table 19.47 shows lengths of some nails
picked at random.
Length of nail (cm) 2 3 4 5 6 7
Frequency 3 6 7 11 8 5
Table 19.47
Find: (a) the mean,
(b) the mode and
(c) the median length of the nails.
5. A student did a survey of the number of
people in cars passing at a certain point.
Table 19.48 shows the data he collected.
No. of occupants 1 2 3 4
No. of cars 7 11 7 x
Table 19.48
(a) Find x if the mean number of occupants
is 2 .
(b) What is the largest possible value of x if
the mode is 2?
(c) What is the largest possible value of x if
the median is 2?
Mean, mode and median of
grouped data
Earlier in this chapter, we saw that when the
number of data items is large, it is sometimes
necessary to group the items into classes.
Table 19.49 is an example of grouped data. It
is a frequency table also called a grouped
frequency table since it shows the frequencies
of classes rather than of individual values.
Marks Frequencies
5–9 2
10–14 13
15–19 31
20–24 23
25–29 14
30–34 6
35–39 1
90
Table 19.49
1
3
–
194
For grouped data, the class with the highest
frequency is called the modal class and its
frequency is referred as modal frequency.
Thus in Table 19.49, 15–19 is the modal class
and the modal frequency is 31.
Example 19.13 shows how to determine the
mean and median of grouped data.
Example 19.13
Table 19.49 shows marks scored by 90 pupils in
a Mathematics test. Find: (a) the mean score,
and (b) the median score for the group.
Solution
Table 19.50 shows how to set out the working.
Class Mid-mark x Frequency f fx
5–9 7 2 14
10–14 12 13 156
15–19 17 31 527
20–24 22 23 506
25–29 27 14 378
30–34 32 6 192
35–39 37 1 37
∑f = 90 ∑fx = 1810
Table 19.50
Note that the classes could be represented using
either class limits or class boundaries.
The mid-mark is the mid-point of a class e.g. 7
is the mid-point of the class 5–9.
fx is the product of the mid-mark value and the
corresponding frequency.
(a) Mean mark = ∑fx =
1 810
~ 20.11
90
(b) There are 90 pupils. So the median will be
the mean of marks obtained by the 45
th
and
46
th
pupils when ranked from the lowest to
the highest mark.
Since up to 19.5 there are 46 pupils, the
45
th
and 46
th
pupils will be in the class
15–19 (see Table 19.51). Thus 15–19 is the
median class.
Class Frequency Cumulative frequency
4.5–9.5 2 2
9.5–14.5 13 15
14.5–19.5 31 46
19.5–24.5 23 69
24.5–29.5 14 83
29.5–34.5 6 89
34.5–39.5 1 90
Table 19.51
Up to 14.5, there are 15 pupils. To get to the
45
th
pupil, we need 30 more pupils from the
next class, i.e. 30 out of 31 pupils.
The class interval 15–19 has 5 marks evenly
distributed among 31 pupils in that class.
The mark for the 45
th
pupil = 14.5 + × 5
= 19.34
Similarly:
The mark for the 46
th
pupil = 14.5 + × 5
= 19.5
Hence, median =
= 19.42 marks
Alternatively, the median mark can be
calculated as follows:
The median position is 45.5
th
i.e halfway
between 45 and 46. From 15
th
-45.5
th
, we
need 30.5
∴ The median mark
= 14.5 + × 5
= 14.5 + 4.92
= 19.42.
From Example 19.14, we see that given some
grouped data, we nd the mean as follows:
– – –
– – –
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
30
31
––
31
31
––
30.5
31
––––
19.34 + 19.5
2
––––––––––
195
(a) Calculate the mean number of letters
received.
(b) State the modal class of the distribution.
2. The heights of 50 athletes in a university
team were as shown in Table 19.53.
Height (cm) Frequency
150–159 2
160–169 9
170–179 12
180–189 16
190–199 7
200–209 4
Table 19.53
Calculate:
(a) the mean height
(b) the median height of the athletes.
3. Table 19.54 shows the masses of 100
students of agriculture from a certain
university.
1. Calculate the mid-interval mark (or simply
the mid-mark), x, of each class interval as:
upper class boundary + lower class boundary
x =
2
2. Use the class mid-marks to nd the mean,
mean, x =
∑fx
∑f
Example 19.13 shows that to nd the median
for grouped data.
Median = l.c.b +
How many more values
× c
No. of values in median class
Note:
1. In Example 19.13, the median is the average
of the two middle scores. Hence the above
computation is done twice.
2. When calculating both the mean and
median of grouped data, it is assumed
that the values in any one class are evenly
distributed within the class. Thus, the mid-
mark (average of the limits) is used in the
computation of the mean and a proportional
part of the median interval is used in the
computation of median.
Exercise 19.8
1. Table 19.52 shows the number of letters
received in a month by 30 pupils of a
certain class.
Number of letters Number of pupils
received (i.e. frequency)
0–2 10
3–5 9
6–8 6
9–11 4
12–14 1
Table 19.52
Mass (kg) Frequency
60–62 5
63–65 20
66–68 40
69–71 25
72–74 10
Table 19.54
(a) Calculate the mean mass.
(b) State the modal class.
4. Table 19.55 shows the lengths of 40 mango
tree leaves.
196
5. Table 19.56 shows the ages of 55 people
who were picked at random for a certain
survey.
Age (years) Frequency
10 and less than 15 2
15 and less than 20 8
20 and less than 25 16
25 and less than 30 10
30 and less than 35 9
35 and less than 40 6
40 and less than 45 4
Table 19.56
(a) Calculate: (i) the mean, and
(ii) the median age of the
group.
(b) State the modal age group.
Length (mm) Frequency
118–126 3
127–135 4
136–144 10
145–153 12
154–162 5
163–171 4
172–180 2
Table 19.55
(a) Find the mean and the median length of
the leaves.
(b) State the modal frequency of the
distribution.
197
Introduction
Angles in the same segment
We have already seen that a chord divides the
circumference into two arcs, a major and a
minor arc.
If AB is a minor arc, and P is any point on
the major arc (Fig. 20.1), then ∠APB is the
angle subtended by the minor arc AB at the
circumference of the circle.
Fig. 20.1
If Q is another point on the major arc AB (Fig.
20.1), ∠APB and ∠AQB are both subtended
at the circumference by the minor arc AB or
chord AB.
These are called angles in the same segment.
Activity 20.1
1. Draw a circle centre O, with any convenient
radius.
2. Mark off a minor arc AB.
3. On the major arc AB, mark distinct points P,
Q, R and S.
4. Join each of the points in 3 so as to form
angles APB, AQB, ARB and ASB.
5. Measure the angles.
6. What do you notice about the sizes of the
four angles?
Do your classmates have the same observation?
You should have observed that all the four
angles are equal.
If AB is the major arc, and C and D are two
points on the minor arc (Fig. 20.2), the angles
ACB and ADB are subtended by the major arc
AB at the circumference. These are also angles
in the same segment. Are they also equal?
Fig. 20.2
Activity 20.2
1. Draw a circle (Fig. 20.3) with any convenient
radius.
2. On the circumference, mark points A, B,
C and D such that chord AB has the same
length as chord CD.
3. On major arc AB mark point P and on major
arc CD mark point Q.
4. Draw angles APB and CQD.
5. Measure the angles.
6. What do you notice about the sizes of the
angles?
7. Using a piece of sewing thread, measure
the lengths of the minor arcs AB and CD.
A
B
P
Q
B
A
•
D
C
O
ANGLE PROPERTIES OF A CIRCLE
20
198
What do you notice about their lengths?
Fig. 20.3
In general:
1. An arc or a chord of a circle subtends
equal angles in the same segment.
2. Equal arcs of the same circle or of
equal circles subtend equal angles at the
circumference.
3. Equal chords of the same circle or of
equal circles cut off equal arcs.
Example 20.1
Fig. 20.4 (a) shows a circle centre O. Find the
values of the angles marked with letters.
(a) (b)
30°
A
B
x
z
w
20°
y
C
D
O
30°
A
B
x
z
w
20°
y
C
D
P
O
D
A
E
B
C
•
D
A
E
B
C
•
P
A
D
B
Q
C
Thus, ∆s AOD and BOD are isosceles.
∴ In ∆AOD, ∠DAO = ∠ADO = 30° (base
angles of an isosceles ∆)
In ∆BOD, ∠DBO = ∠BDO = 20° (base
angles of an isosceles ∆)
∴ ∠ADB = x = 30° + 20°
= 50°
In ∆ADO, ∠DAO + ∠ADO = ∠AOP (Exterior
angle of a ∆ is equal to the sum
of the opposite interior angles.)
∴ 30° + 30° = ∠AOP
i.e. ∠AOP = 60°
Similarly, ∠BOP
= ∠DBO + ∠BDO
= 20° + 20°
= 40°
But ∠AOP + ∠BOP = y
∴ 60° + 40° = y
i.e. y = 100°
w + y = 360° (Angles at a point add up
to 360°.)
∴ w = 360° – 100°
= 260°
∠ADB = ∠ACB (Angles subtended by
the same minor arc AB in the
same segment.)
∴ ACB = 50°
i.e. z = 50°
Example 20.2
ABCDE is a regular pentagon inscribed in a
circle (Fig. 20.5). Show that ∠ACD = 2∠ACB
(a) (b)
Fig. 20.5
Fig. 20.4
Solution
Join D to the centre O of the circle so that DO
produced meets the circumference at a point P
(Fig. 20.4(b)).
Since OA, OD and OB are radii of the same
circle, OA = OD = OB.
199
Solution
With reference to Figure 20.5 (b) ∠ACB is
subtended at the circumference of the circle by
the minor arc AB.
∠ECA is subtended at the circumference of the
circle by the minor arc AE.
∠DCE is subtended at the circumference of the
circle by the minor arc DE.
But arcs AB, AE and DE are equal. Therefore,
they subtend equal angles at the circumference.
∴ ∠ACB = ∠ECA = ∠DCE
But ∠ACD = ∠ACE + ∠ECD
∴ ∠ACD = 2∠ACB
Exercise 20.1
1. Find x and y, in Fig. 20.6.
Fig. 20.6
2. Find a and b, in Fig. 20.7.
Fig. 20.7
3. (a) Find the angles of ∆ABC, in Fig. 20.8.
(b) What does your result tell you about the
minor arcs AC and BC?
72°
x
24°
y
c
30°
60°
a
b
d
e
40°
30°
A
a
B C
80°
40°
d
D
E
b
c
A
B
•
O
C
Fig. 20.8
C
65°
B
D
50°
A
17°
b
a
128°
4. In Fig. 20.9, nd the angles marked with
letters a, b, c, d and e.
Fig. 20.9
5. In Fig. 20.10, nd the angles marked by the
letters a, b, c and d.
Fig. 20.10
Angle in a semicircle
Activity 20.3
1. Draw a circle centre O using any convenient
radius (Fig. 20.11).
2. On your circle draw a diameter AB.
3. Mark another point C on the circumference
and join A to C and B to C.
4. Measure angle ACB.
Fig. 20.11
Compare your result with those of other
members of your class. What do you notice?
200
In general, the angle in a semicircle is
equal to 90°, i.e. the angle subtended by the
diameter at any point on the circumference
is a right angle.
Angles at the centre and the
circumference of a circle
If AB is an arc, and O is the centre of a circle
(Fig. 20.12), angle AOB is called the central
angle of the circle, subtended by minor arc AB.
Fig. 20.12
In Fig. 20.11, angles AOB, BOC, COD and
AOD are examples of central angles of the
circle.
If two central angles in the same circle, or
in equal circles, are equal, then the arcs that
subtend them must be equal.
Activity 20.4
1. Draw a circle centre O with any convenient
radius. (Fig. 20.13)
Fig. 20.13
2. Mark two points A and B, on the
circumference, such that AB is a minor arc.
3. Mark another point P on the major arc AB.
4. Draw in angles AOB and APB.
O
D
A
C
B
48°
O
Q
P
R
1
2
–
O
x
y
P
A
B
48°
O
Q
P
R
– – – – – – – – – – – – – –
– – – – – –
5. Measure ∠AOB and ∠APB.
6. What is the relationship between the two
angles?
Compare your observations with those of other
members of your class. What do you notice?
In general, the angle which an arc of a circle
subtends at the centre is double that which it
subtends at any point on the remaining part
of the circumference.
Example 20.3
PQ is the diameter of a circle centre O
(Fig. 20.14(a)). R is a point on the
circumference such that ∠POR = 48°. Find:
(a) ∠PQR (b) ∠PRQ (c) ∠QPR.
(a) (b)
Fig. 20.14
Solution
(a) Join R to Q (Fig. 20.14 (b)).
∠PQR = ∠POR = 24° (Angle at the
centre is twice angle
at the circumference.)
(b) Join P to R (Fig. 20.14(b)).
∠PRQ = 90° (Angle in a semicircle.)
(c) ∠QPR = 180° – (90° + 24°)
= 66° (Angle sum of a triangle.)
Example 20.4
In Fig. 20.15, ∠ABD = 58° and CBD is a
straight line. Calculate;
(a) obtuse ∠AOC (b) ∠AEC
201
58°
E
C
O
A
B
D
A
C
B
a
b
x
O
b
y
x
a
O
c
p
d
O
Fig. 20.15
Solution
(a) ∠ABC = 180° – 58° = 122° (Angles on a
straight line.)
Now reex ∠AOC = 122° × 2 = 244° (Angle
subtended at the centre by major arc AC.)
Thus, obtuse ∠AOC = 360° – 244° = 116°
(Angles at a point.)
(b) ∠AEC is subtended at the circumference by
the minor arc ABC.
∴ 2 ∠AEC = obtuse ∠AOC
= 116°
∴ ∠AEC = 116°
2
= 58°
Exercise 20.2
1. In Fig. 20.16, O is the centre and AC is the
diameter.
Fig. 20.16
(a) If a = 24°, nd x. (b) If x = 62°, nd b.
(c) If a = 52°, nd b. (d) If b = 28°, nd a.
(e) Express x in terms of a.
(f) Express b in terms of a.
2. In Fig. 20.17, O is the centre of the circle.
Fig. 20.17
(a) If a = 102°, nd the value of x.
(b) If b = 144°, nd the value of y.
(c) If x = 73°, nd the value of a.
(d) If a + b = 220°, nd (x + y).
(e) If x + y = 106°, nd a + b.
(f) Form an equation connecting a, b, x
and y.
3. In Fig. 20.18, O is the centre of the circle.
If c = 47°, nd d and p.
Fig. 20.18
4. AB is the diameter of a circle and C is
any point on the circumference.
If ∠BAC = 2∠CBA, nd the size of angle
BAC.
5. ABC is a triangle such that points A, B and
C lie on the circumference of a circle centre
O. If AC = BC, and angle AOB = 72°, nd
angle BAC.
6. ABCDE is a regular pentagon such that its
vertices lie on the circumference of a circle.
Find the sizes of angles ACE, BDE and
DCA.
202
Cyclic quadrilateral
A quadrilateral whose vertices lie on the
circumference of a circle is called a cyclic
quadrilateral.
Activity 20.5
1. Draw a circle centre O using any convenient
radius.
2. On the circle, mark points A, B, C and
D in that order and join them to form a
quadrilateral.
3. Measure angles ABC and ADC and nd
their sum.
4. Measure angles BAD and BCD. Find their
sum.
5. What do you notice about the two sums in
3 and 4?
6. Are the pairs of angles in 3 and 4 adjacent
or opposite?
7. Do the other members of your class have
the same observations as you do?
8. Produce side AB of the quadrilateral, and
measure the exterior angle so formed. How
does the size of this angle compare with that
of interior ∠ADC?
Generally, in a cyclic quadrilateral;
1. opposite angles are supplementary i.e.
their sum is 180°.
2. an exterior angle is equal to the interior
opposite angle.
3. any side is a chord of the circle. It
therefore subtends equal angles at the
other two vertices. It follows that given
a line segment, say PQ, which subtends
equal angles at two points, say R and S,
on the same side of it, then the four points
P, Q, R and S lie on the circumference of
a circle.
41°
A
B
D
81°
C
72°
E
C
B
A
y
x
41°
D
Note: A cyclic quadrilateral can be drawn
without the circle so long as condition 1, 2 or 3
is satised.
Example 20.5
In quadrilateral ABCD, ∠ADC = 122°,
∠BDC = 41° and ∠ACB = 81°. Show that the
quadrilateral is cyclic.
Fig. 20.19
Solution
∠ADB = ∠ADC – ∠BDC
= 122° – 41°
= 81°
∴ ∠ADB = ∠ACB.
But these angles are subtended by the same
side AB at the points C and D, on the same side
of AB.
Therefore ABCD is a cyclic quadrilateral.
Example 20.6
In Fig. 20.20, nd the values of x and y.
Fig. 20.20
Solution
∠ADC = 180° – 72° = 108° (Angles on a
straight line.)
x + 108° = 180° (Opposite angles of a cyclic
quadrilateral.)
∴ x = 180° – 108° = 72°.
Note that ∠ABC is the interior angle opposite
to the exterior angle at D.
203
∠DCE = 180° – (72° + 41°) (Angle sum
of a triangle.)
= 180° – 113°
= 67°
i.e. y = ∠DCE = 67° (Exterior angle
equals interior
opposite.)
Exercise 20.3
1. A, B, C, D and E are ve points, in that
order, on the circumference of a circle (Fig.
20.21).
(a) Write down all angles in the gure
equal to ∠ACB.
(b) Write down all angles in the gure
supplementary to ∠BCD.
(c) If ∠ACB = 40° and ∠ACD = 75°, nd
the size of ∠DEB.
Fig. 20.21
2. In Fig. 20.22, nd (a) ∠BCD, (b) ∠CDA.
Fig. 20.22
3. Fig. 20.23 consists of two intersecting
circles. Use it to nd the angles marked by
letters.
Fig. 20.23
4. ABCD is a cyclic quadrilateral in a circle
centre O.
(a) If ∠ABD = 32°, nd ∠ACD.
(b) If AOC is a straight line, write down the
size of ∠ABC.
5. AB is a chord of a circle centre O.
If ∠AOB = 144°, calculate the angle
subtended by AB at a point on the minor
arc AB.
6. The angles of a quadrilateral are 80°,
75°, 100° and 105°, in that order. Is the
quadrilateral cyclic?
7. In the quadrilateral ABCD, AC = CB, angle
CAB = 70° and angle ADB = 40°. Is the
quadrilateral cyclic?
8. Two isosceles ∆s ACB and ADB are
drawn on opposite sides of a line AB.
If ∠ACB = 2∠BAD, show that
quadrilateral ABCD is cyclic.
E
75°
40°
D
B
C
A
40°
A
55°
B
D
C
70°
105°
a
98°
d
c
b
204
Vector and scalar quantities
A group of rangers is preparing to go on a
treasure hunt trip from their camping base.
Before they leave the camp, they must know;
1. the specic distance to the treasure, and
2. the direction in which they have to move.
If they do not know both the distance and the
direction, they are not likely to locate the exact
position of the treasure.
Quantities which have both magnitude (size)
and direction are called vector quantities or
simply vectors. Velocity and acceleration are
examples of vectors.
Quantities which have no direction for
example, a packet of biscuits, the cost of
a pen etc., are called scalar quantities or
simply scalars.
The direction of a vector, in a diagram, is
shown by means of an arrow.
Example 21.1
Represent the vector 30 km/h due west by means
of a diagram.
Solution
Fig. 21.1 shows the required representation.
Fig. 21.1
Speed in a specic direction is known as
velocity.
Exercise 21.1
1. State whether the following are vector or
scalar quantities.
(a) A speed of 70 km/h due west
(b) A distance of 50 km due east
(c) 30 cows
(d) 15 km
(e) A speed of 400 km/h
(f) A distance of 10 km south east of the
city centre
(g) 86 litres of milk
(h) A distance of 70 km
(i) A speed of 370 km/h on a bearing
of 050
0
2. Represent each of the vectors in Question 1
by an accurate well-labelled diagram.
3. Name two examples, other than those in
question 1, of:
(a) vector quantities,
(b) scalar quantities.
4. Write down the magnitude and direction of
each of the vectors in Fig. 21.2.
(a) (b)
(c) (d)
Fig. 21.2
N
30 km/h
N
6 m/s
100 km
45°
N
80 km/h
20 km
N
VECTORS (1)
21
205
Displacement vector and notation
We have seen that a directed distance is a
vector. To travel from town A to town B,
along the shortest distance, we must travel in
a specic direction and for a denite distance
(Fig. 21.3).
Fig. 21.3
The distance from A to B in the given direction
is called the displacement vector AB, denoted
as AB. Sometimes, vectors are denoted by
specied small letters e.g. vector a or a.
In print vector AB is shown in bold.
In our handwriting we use arrow or wavy line
notation since we cannot write in bold. For
example, vector AB is written as AB or AB;
vector a is written as a.
A vector whose initial and terminal points
coincide is a null vector denoted as 0. Its
magnitude is 0 (zero).
Equivalent vectors
If two vectors have the same magnitude and
the same direction, then they are equivalent
vectors.
(a) (b)
Fig. 21.4
(i) AB and CD (Fig. 21.4(a)) are parallel.
They have the same sense of direction,
the same magnitude or length, denoted as
|AB| and |CD| respectively. AB is therefore
equivalent to CD. We write, AB = CD
(sometimes written as AB
≡
CD).
(ii) PQ and XY (Fig. 21.4(b)) are parallel.
They also have the same magnitude, but
are in opposite directions. Therefore, PQ
and XY are not equivalent. We can write
PQ ≠ XY. We have to change the direction
of either PQ or XY so that their sense of
direction is the same. A negative sign is
used to reverse the direction. Thus,
PQ =
–
XY.
In general, two vectors are equivalent if they
have the same magnitude and the same
sense of direction.
Addition of vectors
Triangle ABC (Fig. 21.5) represents routes
joining three towns A, B and C.
Fig. 21.5
If you go from A to B, then from B to C, the
effect is the same as going from A to C directly.
The required effect is to reach town C from A.
Since the effect is the same, then
AB + BC = AC.
Vector AC is called the resultant vector of AB
and BC. Such a vector is usually represented
by a line segment with a double arrowhead, as
in Fig. 21.5.
Example 21.2
Using Fig. 21.6, write down the single vector
equivalent to
(a) AB + BC (b) AE + ED
(c) BC + CD + DE (d) ED + DC + CB
(e) AB + BA (f) CD + DC
15 km
30°
Town B (terminal point)
Town A (initial point)
N
~
~
Y
Q
P
X
A
C
B D
A
C
B
206
(g) AE + EB + BC (h) CD + DE + EB
(i) AB + BC + CD + DE
(j) DE + EA + AB + BC + CD
Fig. 21.6
Solution
(a) AB + BC = AC (Moving from A to B, then
from B to C is equivalent to
moving from A to C directly.)
(b) AE + ED = A to E then to D.
= A to D
= AD
(c) BC + CD + DE = BD + DE = BE
(d) ED + DC + CB = EC + CB = EB
(e) AB + BA = AB – AB = 0
(f) CD + DC = 0 (from A to B then back to A)
(g) AE + EB + BC = AB + BC = AC
(h) CD + DE + EB = CE + EB = CB
(i) AB + BC + CD + DE = AC + CD + DE
= AD + DE = AE
(j) DE + EA + AB + BC + CD
= DA + AB + BC + CD
= DB + BC + CD
= DC + CD
= 0 (Start from D and back to D)
Exercise 21.2
1. Which of the vectors in Fig. 21.7 are
equivalent?
E
D
BA
C
T
R
U
S
(a) (b) (c)
Fig. 21.7
2. STUR is a quadrilateral (Fig. 21.8). Use it
to write down the single vector equivalent to
(a) ST + TU (b) TS + SR
(c) RS + ST (d) UR + RS
(e) UT + TR (f) UR + RT
(g) TS + ST (h) UR + RU
(i) RS + ST + TU (j) UT + TS + SR
(k) ST + TU + UR + RS
(l) UT + TS + SR + RU
Fig. 21.8
3. Draw a triangle STR and put arrows on its
sides to show TS + SR = TR.
4. Draw a quadrilateral ABCD and on it show
BC, CD and DA. State a single vector
equivalent to BC + CD + DA.
5. A man walks 10 km in the NE direction,
and then 4 km due north. Using an
appropriate scale, draw a vector diagram
showing the man’s displacement from his
starting point. When he stops walking, how
far from the starting point will he have
walked?
6. Vectors a, b and c are such that a = b and
b = c. What can you say about a and c?
7. Mr. Komu’s family planned a sight seeing
trip which was to take them from Nairobi to
Limuru, then to Thika and back to Nairobi.
Draw a vector triangle to show their trip.
207
What vector does NL + LT + TN represent
if N stands for Nairobi, L for Limuru and T
for Thika?
8. PQRS (Fig. 21.9) represents a
parallelogram.
(a) Copy the gure. Mark with arrows and
name two pairs of equal vectors.
(b) Simplify:
(i) PQ + QR (ii) PS + SR
(iii) SP + PQ (iv) SR + RQ
Fig. 21.9
9. For each of the following equations, use
Fig. 21.10 to nd a directed line segment
which can replace PQ.
(a) AE + PQ = AB (b) DE + PQ = DB
(c) DB + PQ = 0 (d) EB + PQ = EC
(e) EB + PQ = ED (f) PQ + DA = CA
(g) AE + ED + PQ = AD
(h) AD + PQ + EC = AC
(i) DC + PQ + ED = 0
Fig. 21.10
10. Use Fig. 21.11 to simplify the following.
(a) a – w (b) u + a (c)
–
w + u
(d) u + v (e) u – b
S
P
Q
R
A
D
C
B
E
D
A
C
B
a
u
b
w
v
Fig. 21.11
Multiplication of a vector by a scalar
Consider Fig. 21.12 below.
Fig. 21.12
PT = PQ + QR + RS + ST
= a + a + a + a
= 4 × a
= 4a
This means that the length (magnitude) of PT is
four times that of a.
Note that 4a and a have the same direction.
Similarly, PR = 2a and
PS = 3a.
If X is a point halfway between P and Q, PX
would be half of a.
i.e., PX = a.
In vectors PT, PS, PR and PX, the values 4, 3,
2 and are scalars.
A scalar multiplier can take any value, positive
or negative, whole number or fraction, or even
zero (Fig. 21.13).
(a) (b) (c) (d) (e)
Fig. 21.13
a
1
2
–
a
1
2
–
–
a
2a
–
a
P
a
Q
a
R
a
S
a
T
208
In Fig. 21.13 parts (b) to (e), vector a has been
multiplied by 2,
–
1, and
–
respectively.
If m is a scalar, and a is a vector, then ma is a
vector parallel to a, and m times its length.
If m > 0, then ma has the same sense of
direction as a, and its magnitude is m times
that of a.
If m = 0, then ma is a null vector whose
magnitude is zero.
If m < 0, then ma is a vector whose
direction is opposite that of a and whose
length is |m| times that of a , where |m|,
read as the modulus of m, means the value
of m irrespective of the sign.
Multiplying a vector by a negative number
reverses the direction of the vector.
Example 21.3
Towns A, B and C are along a straight road
which runs due north from A. From A to B is
6 km and from B to C is 12 km.
Express in terms of AB:
(a) BC (b) AC (c) CB
Solution
(a) AB is 6 km due north.
BC is 12 km due north.
BC = twice AB
= 2AB (see Fig. 21.14)
(b) AC = AB + BC due north
= 6 + 12 km due north
= 18 km due north
∴AC = 3AB
(c) CB is 12 km due south. CB has same
magnitude as BC but is in the opposite
direction.
∴CB =
–
BC
=
–
2AB
Exercise 21.3
1. Represent each of the following vectors by
a diagram.
(a) A velocity of 400 km/h due west.
(b) A speed of 60 km/h on a straight road
due east increased in the ratio 5 : 4.
(c) Half the reverse of a speed of 80 km/h
due north on a straight road.
2. Four railway stations L, M, E and R are on
a straight railway line (Fig. 21.15).
L M E
R
Fig. 21.15
(a) Express the following in terms of ME.
(i) LM (ii) LE (iii) ER
(iv) 2ER (v) MR (vi) 3LM
(b) Express the following in terms of LM.
(i) MR (ii) LE (iii) ML
(iv) ME (v) ME (vi) LR
(vii)RM (viii) ER
(c) Express the following in terms of RE.
(i) EM (ii) ME (iii) LE
(iv) RL (v) ML (vi) LM
(vii) EL (viii) LR
(d) Write the following in terms of ME.
(i) LM + EM (ii) LM + RE
(iii) ER + ML (iv) ME + RE
(v) ME + ML (vi) ER +
–
3ME
(vii) LM – ER
(viii) LM + ME + ER
Vectors in the Cartesian plane
In Fig. 21.16, AB represents a vector, namely
a translation of the points on the plane, which
moves A to B, H to K, P to Q, etc. This
translation transfers each and every point on the
plane 1 unit to the right and 3 units up. We can
indicate this vector by an ordered pair of
C
12 km
A
B
6 km
N
1
2
–
1
3
–
Fig. 21.14
10 km
5 km
15 km
1
2
–
1
2
–
209
Example 21.4
Using Fig. 21.16, nd the column vectors which
represent:
(a) (i) AB (ii) CD (iii) BA (iv) DC
(v) HL (vi) PR (vii) LH
(b) How can we obtain the column vectors for
HL and PR from those of:
(i) HK and KL
(ii) PQ and QR espectively?
Solution
(a) (i) Point A moves 1 unit to the right and 3
units up to map onto B.
∴ AB =
(ii) Point C moves 4 units to the right and
2 units down to map onto D.
∴ CD =
(iv) D moves 4 units to the left and 2 units
up to C.
∴DC =
(v) H moves 5 units to the right and 1 unit
up to L.
∴HL =
(vi) P moves 5 units to the right and 1 unit
up to R.
∴ PR =
(vii) L moves 5 units to the left and 1 unit
down to H.
∴ LH =
(b) (i) Since HL = HK + KL,
then HL = + =
–
5
–
1
5
1
5
1
4
–
2
1
3
5
1
1
3
x
0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
y
8
7
6
5
4
3
2
1
R
Q
P
D
C
B
A
K
L
H
Fig. 21.16
(
)
(
)
( )
( )( )
1
3
( )
( )
4
–
2
( )
–
1
–
3
( )
–
4
2
(
)
numbers which we write in column form as
(
)
.
This is called a column vector for the said
translation. 1 and 3 are called components of
the vector.
(iii) B moves 1 unit to the left and 3 units
down to A.
∴BA =
210
Note that given two vectors u and v such that
u = and v = where a, b, c and d
are scalars, u + v = + = .
Also, u – v = – = .
Exercise 21.4
Use Fig. 21.17 to answer Questions 1 to 3.
1. (a) Name all the vectors that are equal to
AB and state their column vectors.
(b) Name the vector which is equal to EF.
(c) Is PQ equal to KL? Give a reason for
your answer.
(d) Is KL equal to QP? Why?
(e) Simplify EF + FG + GH and give your
answer as a column vector.
(f) Name a resultant vector which is equal
to NM.
(g) Name three vectors which are equal to
2GL.
(h) Name a vector which is parallel to GH.
2. Write all the vectors in Fig. 21.17 as
column vectors.
3. Simplify FG + GH giving your answer in
column vector form. What is the meaning
of the rst component in your answer?
4. Draw diagrams on squared paper to show:
(a) + = (b) + =
(c) + =
5. Simplify:
(a) + (b) +
(c) + (d) +
a
b
c
d
a
b
c
d
a – c
b – d
6
4
0
2
–
2
1
3
3
3
1
3
2
3
4
0
3
(We add corresponding components
in the column vectors HK and KL)
(ii) PR = PQ + QR
= + =
( )
( )
c
d
a
b
a + c
b + d
( )
( )
( )
3
2
4
3
–
2
4
–
3
–
8
–
10
–
4
–
6
8
( )
( )
( )
( )
( )
( )
( )
( )
( ) ( )
( )
( )
( )
( )
( )
( ) ( ) ( )
( )( )
0
4
6
1
2
2
Fig. 21.17
x
0
2 4 6 8 10 12 14
y
8
6
4
2
H
B
S
R
M
P
G
E
F
A
Q
N
D
C
L
K
T
U
4
–
2
1
3
5
1
( )( )( )
211
6. Given that 2u means u followed by u, 2
means followed by which means a
total of 2 units to the right and 4 units up i.e.
2 = =
Simplify:
(a) 3 (b) 2 (c) 3
(d) (e)
–
(f)
–
3
(g) 4 (h) 4
7. If a = , b = and c = , evaluate:
(a) a + b (b) 2a (c) 3b
(d) b + c (e) 4a + 3b (f)
–
a + 2c
(g) 3(a + b) (h) a – b (i) a + c
(j) a + b + c (k) 2b + 5c (l) a + b
(m)
–
2 (a + c) (n) a – c
Position vector
On a Cartesian plane, the position of a point
is given with reference to the origin, O, the
intersection of the x- and y- axes. Thus, we can
use vectors to describe the position of a point
(Fig. 21.18).
From the origin, A is
+
2 units in the x direction
and
+
4 units in the y direction.
Thus, A has coordinates (2 , 4) and OA has
column vector .
Similarly, B is
–
3 units in the x direction and
+
1
units in the y direction.
Thus, B has coordinates (
–
3 , 1) and OB has
column vector .
C is
–
2 units in the x direction and
–
3 units in
the y direction.
Thus, C is (
–
2 ,
–
3) and OC = .
D is
+
3 units in the x direction and
–
6 units in
the y direction.
Thus, D is (3 ,
–
6) and OD = .
OA, OB, OC and OD are known as position
vectors of A, B, C, and D respectively.
All position vectors have O as their initial
point.
Example 21.5
P has coordinates (2 , 3) and Q (7 , 5).
(a) Find the position vector of (i) P, (ii) Q.
(b) State the column vector for PO.
(c) Find the column vector for PQ.
Solution
(a) (i) P is (2 , 3) (ii) Q is (7 , 5)
∴ OP = ∴OQ =
(b) OP =
PO =
–
OP
=
–
=
1
2
3
1
–
1
5
2 × 1
1 × 2
1
2
1
2
1
2
5
2
2
4
5
6
–
9
0
2
–
1
–
6
4
–
2
3
1
2
–
1
2
–
2
4
( )
( ) ( )
1
3
–
1
2
–
( )
( ) ( )
( ) ( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
4
1
–
1
2
–
3
2
( )
( )
–
2
–
3
–
3
1
3
–
6
( )
2
3
2
3
2
3
–
2
–
3
( )
( )
( )
( )
( )
7
5
Fig. 21.18
–
3
–
2
–
1
1 2 3
4
3
2
1
–
1
–
2
–
3
–
4
–
5
–
6
y
0
A
B
C
D
x
212
(c) PQ = PO + OQ
=
+ =
= .
Note that
PO + OQ = OQ + PO = PQ
∴PQ = OQ – OP (since PO =
–
OP)
= position vector of Q – position
vector of P
= –
=
Example 21.6
If OA = and OB = , nd
(a) the coordinates of A,
(b) the coordinates of B,
(c) the column vector for AB. .
Solution
(a) OA =
∴A is (2 ,
–
5)
(b) OB =
∴ B is (
–
4 ,
–
3)
(c) AB = AO + OB
=
–
OA + OB (since AO =
–
OA)
= OB – OA
= – = = .
In general, if P is (a , b) then OP = .
Similarly, if Q is (c , d) then OQ = .
PQ = position vector of Q – position
vector of P
= OQ – OP
= – = .
7
5
–
2
–
3
–
2 + 7
–
3 + 5
5
2
7
5
2
3
5
2
2
–
5
–
4
–
3
2
–
5
–
4
–
3
–
6
2
a
b
c
d
c – a
d – b
5
x
y
4
3
2
1
0
–
1
–
2
–
3
–
4
–
3
1
–
2
2
–
1
3 4
×
×
×
×
×
×
×
×
×
×
×
K
H
B
L
F
G
A
I
J
E
D
×
C
( ) ( )
( )
–
4
–
3
( )
2
–
5
( )
( )
( )
( )
( )
( ) ( )
( )
( )
–
4 – 2
–
3 –
–
5
( )
( )
( )
( )
( )
( )
( )
c
d
a
b
Exercise 21.5
1. Use Fig. 21.19 to write down the position
vectors of the marked points.
Fig. 21.19
2. On a squared paper, mark the points whose
position vectors are given below.
(a) (b) (c)
(d) (e) (f)
(g) (h)
3. (a) If OP = and OQ = , nd the
column vector for PQ.
(b) If OP = and OQ = , nd the
column vector for (i) PQ (ii) QP.
(c) If OF = and OG = , nd the
column vectors for FG and GF.
(d) If OM = , ON = and
OP = , nd the column vector:
(i) MN (ii) MP (iii) NM
4
–
1
1
1
2
3
1
–
1
5
–
2
–
4
–
1
–
1
2
–
2
3
5
0
4
1
2
5
–
2
4
7
4
–
1
4
1
3
–
2
–
2
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
3
8
213
(iv) NP (v) PN (vi) PM.
4. If a = and b = , evaluate the
following and illustrate them on a Cartesian
plane given that a and b are position
vectors.
(a) a + b (b) a – b (c) 2a + 3b
(d) 2b – a (e)
–
2a + 3b (f)
–
4a – b
Midpoints
In Fig. 21.20, O is the origin and M is the
midpoint of AB.
Fig. 21.20
Since O is the origin, OA, OB and OM are
position vectors.
AB = AO + OB
=
–
OA + OB (since AO =
–
OA)
= OB – OA
= b – a
Since M is the midpoint of AB, then
AM = MB = AB
= (b – a)
∴OM = OA + AM
= a + (b – a)
= a + b – a
= a + b = (a + b)
If A is (7 ,
–
3) and B is (
–
5 , 5),
then OA = a = and OB = b = .
Thus, OM =
+
= =
Since OM is a position vector, M is (1 , 1).
In general, if A is (a , b) and B is (c , d), the
midpoint M of AB has coordinates
M , .
Exercise 21.6
1. Calculate the coordinates of the midpoints
of the line segment joining the following
pairs of points.
(a) A (2 , 1), B (5 , 3)
(b) A (0 , 3), B (2, 7)
(c) A (4 ,
–
1), B (4 , 3)
(d) A (
–
2 , 3), B ( 2 , 1)
2. In each of the following cases;
(i) Find the column vector of PQ.
(ii) Hence or otherwise, nd the coordinates
of the midpoint.
(a) P (3 , 0), Q (4 , 3)
(b) P (
–
3 , 1), Q (5 , 1)
(c) P (
–
2 ,
–
1), Q (
–
12 ,
–
8)
(d) P (
–
9 , 1) Q (12 , 0)
(e) P (
–
8 , 7), Q (
–
7 , 8)
(f) P (
–
3 , 2), Q (3 ,
–
2)
3. P is (1 , 0), Q is (4 , 2) and R is (5 , 4). Use
vector method to nd the coordinates of
S if PQRS is a parallelogram. Find the
coordinates of the midpoints of the sides of
the parallelogram.
Length of a vector
We have already seen that a vector can be
represented by a directed line segment using
a horizontal and a vertical component, for
example, AB = means that, starting
from A, the horizontal distance covered is 6
units while the corresponding vertical distance
is 8 units (Fig. 21.21).
A
a
B
O
M
b
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
1
2
–
a + c
2
b + d
2
1
2
2
3
( )
( )
–
5
5
( )
7
–
3
( )
2
2
( )
( )
1
1
( )
( )
6
8
–
5
5
( )
(
)
7
–
3
( )
214
( )
Fig. 21.21
AB represents the hypotenuse of a right-angled
triangle whose two shorter sides are 6 and 8
units long.
Therefore, the length or magnitude of AB,
written as |AB|, is found by using Pythagoras’
Theorem.
Thus, |AB|
2
= 6
2
+ 8
2
= 36 + 64
= 100
|AB| = √100
= 10 units
In general, if P is (x , y) and Q is (a , b), then
PQ = and
|PQ| = √(a – x)
2
+ (b – y)
2
.
Example 21.7
Given that P is the point (7 ,
–
9) and Q is the
point (10 ,
–
5), nd:
(a) |OP| (b) |OQ| (c) |PQ|
(d) |OP| + |OQ| (e) |OP + OQ|
Solution
(a) OP =
∴|OP| = √7
2
+ (
–
9)
2
= √49 + 81 = √130
= √130 = 11.4 units
a – x
b – y
10
–
5
7
–
9
(b) OQ =
∴ |OQ| = √100 + 25
= √125
= 11.18 units
(c) PQ = – =
∴ |PQ| = √3
2
+ 4
2
= √25 = 5 units
(d) |OP| + |OQ| = 11.4 + 11.18 = 22.58 units
(e) OP + OQ = + =
∴|OP + OQ| = √17
2
+ (
–
14)
2
= √289 + 196
= √485 = 22.02 units
You should have noticed that
|OP| + |OQ| ≠ |OP + OQ|.
In general, |a + b| ≠ |a| + |b|.
Exercise 21.7
1. Calculate the length of each of the following
vectors.
(a) (b) (c) (d)
2. Calculate the distances between the
following pairs of points.
(a) A (5 , 0), B (10 , 4)
(b) C (7 , 4), D (1 , 12)
(c) E (
–
1 ,
–
1), F(
–
5 ,
–
6)
(d) P (4 ,
–
1), Q(
–
3 ,
–
4)
(e) H (b , 4b), K(
–
2b , 8b)
(f) M (
–
2m , 5m), N (
–
4m ,
–
2m)
3. Which of the following expressions
represent the distance between the points
A (a , b) and B (c , d)
(a) √(b – d)
2
+ (a – c)
2
(b) √(a – b)
2
+ (c – d)
2
3
4
7
–
9
10
–
5
7
–
9
10
–
5
17
–
14
6 units
B
C
A
8 units
( )
( )
( )
( )
( )
( )
( )
( )
–
3
–
4
6
–
8
12
5
1
–
8
( )
( ) ( ) ( )
215
( )
(c) √(c – a)
2
+ (d – b)
2
(d) √(a – c)
2
+ (b – d)
2
4. Which of the following statements are true
and which are false?
(a) Given A (3 , 1), B (6 , 5), P (
–
1 ,
–
2)
and Q (3 ,
–
5), |AB| = |PQ|.
(b) For the points given in (a) above,
AB = PQ.
(c) If A is (3 , 1), B is (
–
2 , 4), C is (0 , 8)
and D is (
–
5 , 11), then AB = CD.
(d) If OP and OQ represent u and v
respectively, then OR represents
(u – v), where R is the midpoint of PQ.
5. (a) Calculate the length of:
(i) (ii) (iii)
(b) If OP = , OQ = and
|OP| = 2|OQ|, nd two possible
positions of OP.
Translation and displacement vectors
Translation as a translation
We are going to consider a displacement
process under which each point of a plane
gure is moved onto another point, a certain
distance away, in a certain direction (Fig. 21.22).
(b)
(a)
Fig. 21.22
Activity 21.1
1. Copy triangle ABC (Fig. 21.22(a)) on a
tracing paper.
2. Draw the line segment joining B to B′ as
shown in Fig. 21.22.
3. Slide the tracing using line BB′ as a guide
line, to ensure that B moves onto B′ in a
straight line.
4. When B coincides with B′ stop the slide.
What do you notice about the positions of
A and C?
5. What do you notice about the new position
of ∆ABC? What can you say about the two
triangles?
6. Using a ruler, measure the lengths of AA′,
BB' and CC'. What do you notice about the
lengths? Are the lines parallel?
We notice that each point on triangle ABC has
moved the same distance in the same direction.
The process that moves triangle ABC onto
triangle A′B′C′ is called a translation.
In general, under a translation:
1. All the points on the object move the
same distance.
2. All the points move in the same direction.
3. The object and the image are identical
and they face the same direction. Hence,
they are directly congruent.
4. A translation is fully dened by stating
the distance and direction that each
point moves.
Translation in the Cartesian plane
When using the Cartesian plane, a translation is
fully dened by stating the distances moved in
the x and y directions. The column vector that
1
2
–
x – 1
2
3
1
C
A
B
B′
A′
C′
4
–
1
2
3
4
0
( )( )
( )
( )
216
A B C D
E
L
K
H
G
F
– – – – – – – – – – –
– – – – – –
– – – – –
A
– – – – – – – – – – –
– – – – – –
– – – – –
B
( )
2
5
( )
2
5
( )
2
5
Exercise 21.8
1. A packaging case is pushed (without
turning) a distance of 8 m, in a straight line
(Fig. 21.24). The corner of the box which
was originally at A ends up at B.
Fig. 21.24
How far will each of the following have
moved?
(a) The upper front edge
(b) Each vertex
(c) The centre of each face
(d) The centre of the box
2. Fig. 21.25 shows a tiling composed of
congruent parallelograms of sides 10 cm by
5 cm. Suppose the lines in the diagram are
used as guide lines along which tiles may be
slid.
Use the gure to answer Questions 2 to 5.
17 18
19 20
13 14 15 16
9 10 11 12
5
X
6 7 8
1 2 3 4
Fig. 21.25
(a) If tile 6 moves onto tile 7, in what
direction has each vertex of the tile
moved? How far has the point X
moved? Make a statement that is true
for every point on tile 6.
(b) Tile 6 moves along the guideline
parallel to line BC in th direction BC.
Observe and state the direction of
motion of
(i) each vertex of tile 6.
denes the translation is also called the
displacement vector of the translation.
Example 21.8
Triangle ABC has vertices A (0 , 0), B (5 ,
1) and C ( 1 , 3). Find the coordinates of
the points A', B' and C', the images of A, B
and C respectively, under a translation with
displacement vector
.
Solution
A displacement vector means that each
point on ABC moves 2 units in the positive
direction of the x-axis followed by 5 units in the
positive direction of the y-axis (Fig. 21.23).
Thus A(0 , 0)→ A′(0 + 2 , 0 + 5), i.e. A′(2 , 5)
B (5 , 1) → B′(5 + 2 , 1 + 5), i.e. B′(7 , 6)
C (1 , 3) → C′(1 + 2 , 3 + 5), i.e. C′(3 , 8)
∴ ∆A′B′C′ is the image of ∆ABC under
translation, displacement vector .
Fig. 21.23
0
2 4
6
y
8
6
4
2
7
5
3
1
1 3 5 7
x
A
C
A′
C′
B′
– – – – – – – – – – – – – – – – – – – – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – –
B
– – – – – – – – – – – – – – – – – – – – – – – – – –
217
(ii) the point X on tile 6.
(iii) every point on tile 6.
(c) Answer Question (b) for the case where
tile 6 is slid to position 14.
3. Name the tiles onto which tiles, 1, 11, 15
and 18 will be translated by a translation
equivalent to that of Question 2 (a).
4. (a) Name the tiles onto which tiles, 2, 8, 11
and 5 will be translated by a translation
equivalent to that of Question 2 (c).
(b) What will be the images of letters E, F,
G and H under the same translation?
5. Write down all the possible translations that
are equal to the translation:
(a) FH.
(b) AC.
6. State which of the following statements are
true, and which are false.
When a gure or object has a translation
applied to it;
(a) all points move in the same direction.
(b) not all points of the gure move in the
same direction.
(c) all lengths in the object remain
unchanged
4
2
–
3
2
(d) usually, at least one point on the gure
remains unchanged.
(e) a translation can be described by many
directed line segments, provided each
has the same length and same direction.
7. Find the image of ∆ABC, where A is
(
–
3 ,
–
2), B is (
–
1 , 1) and C is (2 ,
–
1),
under translation vector .
8. The image of A (6 , 4) under a translation is
A' (3 , 4). Find the translation vector.
9. ∆ABC with A (0 , 1), B (2 , 0) and C (3 , 4)
is given a translation equivalent to
followed by .
Find the coordinates of the nal image
10. Quadrilateral P′Q′R′S′ is the image of
PQRS under a certain translation. P is
(2 , 0), Q is (0 , 3), P′ is (5 , 4), R′ is
(6 , 9) and S′ is (9 , 4). Find;
(a) the translation vector
(b) the coordinates of R, S and Q'.
11. Given that a is a vector, solve the equation:
(a) 2a + = – a
(b) 4a + + =
0
5
9
3
1
4
–
1
–
2
4
( )
( )
( )
2
4
( )
( )
( )( )
218
REVISION EXERCISES 3
Revision exercise 3.1
1. A test tube has a hemispherical bottom of
radius 1 cm. It contains a liquid up to a
height of 10 cm.
(a) What is the volume of the liquid in the
test tube?
(b) If the density of the liquid is
0.8 g/cm
3
, what is the mass of the
liquid in the test-tube?
2. A solid metal cone has a diameter of 14 cm
and a height of 24 cm.
(a) What is the volume of the cone?
(b) If the cone is melted and recast into a
cylinder of the same diameter, what is
the height of the cylinder?
3. (a) Expand and simplify;
(i) (2n – 7)(3n – 1)
(ii) (2x + 1)(2x – 1) + 1
(iii) (3x – 2y)(2x + 2y) – 5xy
(iv) (2x + 3)(x – 5)
(b) Factorise and simplify where
necessary.
(i) c(y – z) – y + z
(ii) 9 – 9z
2
(iii) 18 – 8(r + s)
2
(iv)
x
2
– 9
5x
2
– 13x – 6
4. Solve the following quadratic equations.
(a) (i) (2x – 5)(x + 3) = 0
(ii) (x – 2)(3x + 4) = 0
(iii) ( 2x – 5)
2
= 0
(iv) (3x – 7)(x – 1) = 0
(b) (i) 2x
2
– 3x = 0
(ii) 2m
2
– 5m + 3 = 0
(iii) 16x
2
+ 9 = 24x
(iv) 3x
3
– 12x = 0
5. (a) Represent each of the following
inequalities on a number line.
(i) x > 3 (ii) x < 4
(iii) x ≥ 2 (iv) x ≤
–
4
(b) Write down the inequalities illustrated
in Fig. R.3.1.
(i)
(ii)
(iii)
(iv)
Fig. R.3.1
6. (a) Which of the following inequalities are
true?
(i)
–
3 <
–
2 (ii) 3(
–
1) > 2(
–
1)
(iii) 2 <
–
3 (iv) 4(2) < 5(2)
(v) (
–
2)(
–
1) > 3(
–
1)
(vi) 4(
–
2) < 5(
–
2)
Correct those that are false.
(b) The following pairs of inequalities are
equivalent. True or false?
(i) 3x > 5; x >
(ii) 2 – x > 4; x >
–
2
(iii) x < 6;
–
2x >
–
24
(iv) x – 3 ≥ 5; x > 8
–
1
0 1 2 3 4
•
°
–
4
0 1 2 3 4
–
3
–
2
–
1
°
50 1 2 3 4
–
3
–
2
–
1
4 5 6 7 8 9
•
1
2
–
5
3
–
15-21
219
7. An athlete runs 100 m in 9.6 seconds. Find
his average speed in km/h.
8. Fig. R.3.2 shows velocity-time graphs. Use
the graphs to nd;
(i) the distance travelled in each case.
(ii) the acceleration or deceleration
(retardation) in each case, for the
various sections of the graphs.
(iii) the average velocity in each case.
Fig. R.3.2
9. (a) The times in seconds, to the nearest 0.1
seconds, taken by 45 pupils in running
100 m were recorded as in Table R.3.1
below.
13.0 12.1 11.9 12.2 14.0 13.2 14.1 11.6 13.7
12.8 11.9 13.2 11.9 12.8 14.1 13.6 14.3 13.8
12.8 12.7 13.1 13.5 12.1 13.8 12.8 12.3 14.2
12.0 12.7 11.7 13.4 13.8 13.6 14.1 13.2 14.8
11.9 12.3 13.7 13.5 13.2 13.5 12.5 12.6 12.1
Table R.3.1
Make a frequency table with the times
grouped as 11.5–11.9, 12.0–12.4,
12.5–12.9, …
(b) Using the frequency table you made in
part (a), make a bar chart and state the
modal class.
0
1 2 3 4
5
20
15
10
5
Time (s)
Velocity (m/sec)
x
y
0
2 4 6 8
10
40
30
20
10
Time (s)
Velocity (m/sec)
x
y
0
1 2 3 4
5
10
5
Time (s)
Velocity (m/sec)
x
y
6
Velocity (m/sec)
x
y
6
5
4
3
2
1
0
1 2 3 4
Time (s)
(a)
(b)
(c)
(d)
220
O
Y
X
Z
42°
A
O
C
B
D
27°
10. (a) Calculate the arithmetic mean of
(i) 1, 4, 4, 2, 5, 2, 7, 7, 7, 6
(ii) 91, 94, 94, 92, 95, 92, 97, 97,
97, 96
Hence write down the arithmetic mean
of 551, 554, 554, 552, 555, 552, 557,
557, 557, 556
11. (a) In Fig. R.3.3(a), X, Y and Z are points
on the circumference of a circle centre O.
Given that ∠OXY = 42°, nd ∠XZY.
(a) (b)
Fig. R.3.3
(c) In Fig. R.3.2(b), O is the centre of
the circle. If ∠ABC = 27°, nd ∠AOC
and ∠ADC.
(d) A, B, C and D, in that order, are points
on the circumference of a circle centre
O. If ∠ACB = 53°, and AC is
diameter, nd ∠CDB.
12. (a) Given that a = , b = and
c = , nd;
(i) a + b (ii) b – c
(iii) c – a (iv) a + b + c
(v) |a + b + c| (vi) |a + |b| + |c|.
(b) If (i) + 3p = , nd p.
(ii) + SR = , nd SR.
(iii) a = , b = , c =
and ma + nb = c, form two
simultaneous equations and
solve them for m and n.
3
4
–
2
3
–
1
2
–
1
2
–
–
5
4
( )
( )
2
–
3
0
–
2
( )
– – – – – – – – – – – –
– – – – – – – –
– – – – – – –
– – – – – – – – –
•
8 cm
6 cm
8 cm
11 cm
–
8
2
( )
0
5
( )
4
–
4
( )
–
2
3
( )
–
5
21
( )
4
–
6
( )
1
3
( )
3
m
––
9
m
––
Revision exercise 3.2
1. Find the volume of each of the solids in
Fig. R.3.4.
(a) (b)
Fig. R.3.4
2. (a) Find the quadratic equation whose
roots are; (i)
–
1, 2 (ii)
–
, .
Give your answer to (ii) in the form
ax
2
+ bx + c = 0
(b) If x = 1 is a root of the equation
x
2
+ kx + 2 = 0, nd the value of k.
(c) If x = 2 is a root of the equation
3x
2
+ kx – 6 = 0, nd the other root.
3. (a) Given that – 4m = 2 – , nd the
value of m.
(b) Solve the equation + =
1.
(c) Factorise completely x
4
– 1 and hence
solve the equation x
4
– 1 = 0 as far as
possible.
4. (a) Solve the following inequalities.
(i)
–
2 + 3 ≤
–
5x
(ii)
–
4 + 7 <
–
2x – 5
(iii) 2x – 7 < 32 – x
(iv) (5x – 4) ≤ x + 5
(b) Find the range of values of x which
satisfy the inequality
2x + 1 < 11 < 6x – 1.
Illustrate your answer on a number line.
2
1 – x
––––
x
1 + x
––––
221
350 km apart. Use the graph to answer the
following questions.
(a) At what time did the two matatus start
from their respective towns?
(b) At what time did each reach its
destination?
(c) What is the average speed of the
Peugeot for the whole journey?
(d) What do you think could have
happened to the Peugeot for the driver
to stop at the time he did?
(e) At what time and how far from A did
the two matatus meet?
7. (a) Table R.3.2 shows the results of a
voting exercise by a class of 200
students to select their leader.
Make a pie chart to display these
results.
y
x
350
300
250
200
150
100
50
0
8.00 a.m 9.00 a.m 10.00 a.m 11.00 a.m 12 noon
1.00 p.m
Nissan
Peugeot
Distance (km)
Time
4
3
2
1
–
1
–
2
3 4
–
1
y
x
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
5
5
1
0
2
Fig. R.3.5
The unshaded region in Fig. R.3.5 is bounded
by the lines y = 4, x = 2 and y = x. State the
three inequalities that dene the region.
6. Fig. R.3.6 shows the motion of two express
matatus plying between towns A and B,
Fig. R.3.6
x = 2
y =
x
y = 4
B
A
5.
222
Weeks of term gone 0 1 2 3 4 5 6
Amount left (sh) 500 480 400 380 320 300 420
Weeks of term gone 7 8 9 10 11 12
Amount left (sh) 400 340 240 220 200 200
Table R.3.3
Draw a line graph to represent the
information.
8. (a) Table R.3.4 shows speeds of vehicles
measured to the nearest 10 kph as they
passed a certain point.
Speed (kph) 30 40 50 60 70 80 90 100 110
Frequency 1 4 9 14 38 47 51 32 4
Table R.3.4
(i) What is the mean speed of the
vehicles?
(ii) State the modal speed.
(b) In a mixed class of 45 pupils, the mean
score in a mathematics test was 45 for
the 24 boys and 58.8 for the girls.
(i) What was the mean score for the
whole class?
(ii) If the pupil with the highest mark
of 87 was found to have cheated,
what was the mean score for the
rest of the class?
9. (a) In Fig. R.3.7(a), RT is the diameter of
the circle centre O. Given that ∠PRT =
30°, nd ∠PQR.
(a) (b)
Fig. R.3.7
(b) In Fig. R.3.7(b), A, B, C and D are
points on the circumference of a circle
centre O. BA is produced to E.
If ∠DAE = 88° and ∠ADO = 54°, nd
∠BCD and ∠ABO.
10. In Fig. R.3.8, PS = 3 cm, PQ = 4 cm and
QS is a diameter of the circle. Find ∠PRQ.
Fig. R.3.8
11. Given that the position vectors of A, B, C
and D are , , and
respectively,
(a) state the column vectors of:
(i) AB (ii) BC (iii) CD (iv) BA
(b) evaluate: (i) AB + BC (ii) AB – BC
(iii) AB – CD
(c) nd the total distance from A to C
through B.
(d) nd, by calculation, the coordinates of
the midpoint of CD.
(e) nd the position vector of the midpoint
of AC.
30°
S
R
P
Q
T
O
•
O
C
B
E
88°
D
54°
A
•
4 cm
O
•
3 cm
R
P
Q
S
1
3
–
1
3
–
Candidate No. of votes
Gamba 65
Njogu 52
Korir 24
Mutie 42
Spoilt votes 10
Abstained 7
200
Table R.3.2
(b) Table R.3.3 shows the amount of
pocket money a pupil had left at the
end of each week in a particular term.
3
–
2
( )
–
6
4
( )
( )
1
2
( )
–
9
–
3
223
A
B
C
D
x
12
9
( )
–
1
2
12. (a) ABCD is a rectangle whose vertices
are A(2 , 4), B(6 , 4), C(6 , 7) and
D(2 , 7). After a translation equivalent
to , A, B, C and D are mapped
onto A′, B′, C′ and D′.
State the coordinates of A′, B′, C′ and
D′.
(b) After a certain translation, point
M(
–
3 ,
–
2) is mapped onto point
M′(5 ,
–
4).
(i) Find the translation vector.
(ii) Find the coordinates of point A,
whose image is A′(3 ,
–
1).
A
y m
C
B
D
wall
1 200 m
2
Fig. R.3.11
(a) State the four inequalities that satisfy
the region.
(b) List all the points in the region which
have integral coordinates.
(c) Calculate the area of the region.
y
5
4
3
2
1
0
–
5
–
2
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
1 2 3 4
x
5
Revision exercise 3.3
1. A container is in the form of a frustum of a
right pyramid 4 m square at the top, 2.5 m
square at the bottom and 3 m deep. What is
the capacity of the container?
2. (a)
Fig. R.3.9
Fig. R.3.9 represents rectangle ABCD
which measures 12 cm by 9 cm. If the
shaded area is 68 cm
2
, nd the value
of x.
(b) A man wishes to fence off a rectangular
plot. He has fencing wire 100 metres
long. AB = y m long, and the area to
be enclosed is 1 200 m
2
(Fig. R.3.10).
Side BC has already been catered for
by a wall. Write AD in terms of y.
Form a quadratic equation in y and
solve it to nd two possible dimensions
of the plot.
Fig. R.3.10
3. (a) I think of number x, square it and
subtract three times the original
number. My answer is
–
2. Find the
number x.
(b) Two consecutive odd numbers have a
product 195. Find the numbers.
4. Fig. R.3.11 shows a region bounded by
four lines.
224
Marks scored Frequency
0 – 9 9
10 – 19 18
20 – 29 12
30 – 39 10
40 – 49 8
50 – 59 3
60
Table R.3.6
5. (a) ABCD is a parallelogram with vertices
at A(5 , 0), B(0 , 5), C(
–
5 , 0) and
D(x , y).
(i) Locate point D and state its
coordinates.
(ii) Write down the inequalities which
determine the parallelogram.
(b) A parallelogram has two of its vertices
at (
–
2 , 0), and (1 , 0). If y – 2x ≤ 4 and
x + y ≤ 1 are two of the inequalities
that dene the parallelogram in
question, nd the other two inequalities
and state the coordinates of the other
two vertices.
6. A body accelerates uniformly from rest for
10 seconds during which time it travels
a distance of 150 m. It then decelerates
uniformly and comes to rest in 9 seconds.
(a) Find the maximum velocity attained
by the body.
(b) Represent this motion graphically.
(c) How far did the body travel in the rst
8 seconds?
(d) Find the average velocity for this
motion.
7. (a) A cyclist travels the rst x km of his
journey at 20 km/h, and the remaining
70 km of the journey at 15 km/h. The
whole journey takes him 6
hours.
Find the value of x.
(b) A man took 3 hours to travel from
Town A to Town B driving at an
average speed of 140 km/h. Without
stopping, he continued to town C,
driving at an average speed of 110
km/h and took 2 hours. Calculate his
average speed for the whole journey.
8. (a) P and Q are points whose coordinates
are (
–
2 , 4) and ( x , y) respectively.
B is another point (2 , 0) such that
PQ = 3QB. Find x and y.
(b) OABC is a parallelogram. O is the
origin, A is (6 , 4) and B is (4 , 8).
(i) Express OC and AB as column
vectors.
(ii) Given that M and N are the
midpoints of OC and AB
respectively, nd the position
vectors for M and N.
(c) If A is (4 , 3) and B is (8 , 13), nd
without plotting
(i) the coordinates of the midpoint
of AB,
(ii) the magnitude of AB.
9. (a) The mean mass of 5 boys is 54.4 kg.
Three boys of masses 52 kg, 55.1 kg
and 57.5 kg join the ve. What is the
mean mass of all the eight boys?
(b) Table R.3.5 shows the distribution
of ages of workers in a certain
organisation.
Age (years) Number of workers
20 – 24 5
25 – 29 10
30 – 34 18
35 – 39 15
40 – 69 12
Table R.3.5
Calculate;
(i) the mean age,
(ii) the median age of the workers.
(c) Table R.3.6 shows marks scored
by 60 pupils in a physics test.
2
3
225
R
P
Q
T
S
Fig. R.3.14
12. (a) In a pie chart representing the number
of animals that a farmer has, the angle
representing cattle is 216°. Find the
total number of animals that the farmer
has, given that he has 324 cattle. If the
farmer has 90 goats, nd the angle of
the sector representing goats.
If the rest of the animals are sheep,
how many are they?
(b) Table R.3.7 shows the number of
vehicles that were observed passing
a certain point, and the number of
passengers in the vehicles.
Number of Number of
passengers vehicles
0 – 4 50
5 – 9 25
10 – 18 18
19 – 25 14
26 – 50 12
51 – 70 10
Table R.3.7
Draw a histogram from this data.
(c) Using the data in Table R.3.5, draw a
frequency polygon.
30°
65°
Y
W
Z
X
2x
x + 2y
3x + y
2
x + 4 y
8 3
Calculate: (i) the mean score,
(ii) the median score.
State the modal class.
10. Answer the following using Fig. R.3.12.
(a) Name an angle(s) which is equal to:
(i) ∠PQT, (ii) ∠TQS (iii) ∠STR
(iv) ∠RTQ (v) ∠PTQ
(b) Name four cyclic quadrilaterals.
(c) If QS is a diameter, which angles are
right angles?
Fig. R.3.12
11. (a) Fig. R.3.13 shows a cyclic quadrilateral
WXYZ in which ∠XWY = 30° and
∠YXZ = 65°. If ∠WZY = 70°, nd
the remaining three angles of the
quadrilateral.
Fig. R.3.13
(b) Find the values of x and y in
Fig. R.3.14.