Vedanta Excel in Mathematics Book - 8 Final (2078)

Approved by the Government of Nepal, Ministry of Education, Science and Technology,
Curriculum Development Centre, Sanothimi, Bhaktapur as an Additional Learning Material

vedanta

Excel in

MATHEMATICS

8Book

Author
Hukum Pd. Dahal

Editor
Tara Bahadur Magar

vedanta
Vedanta Publication (P) Ltd.
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Vanasthali, Kathmandu, Nepal
+977-01-4982404, 01-4962082
[email protected]
www.vedantapublication.com.np

vedanta

Excel in

MATHEMATICS

8Book

All rights reserved. No part of this publication may
be reproduced, copied or transmitted in any way,
without the prior written permission of the publisher.

Second Edition: B. S. 2078 (2021 A. D.)

Published by:

Vedanta Publication (P) Ltd.
j]bfGt klAns];g kf| = ln=

Vanasthali, Kathmandu, Nepal
+977-01-4982404, 01-4962082
[email protected]
www.vedantapublication.com.np

Preface

The series of 'Excel in Mathematics' is completely based on the contemporary pedagogical teaching
learning activities and methodologies extracted from Teachers' training, workshops, seminars, and
symposia. It is an innovative and unique series in the sense that the contents of each textbooks of
the series are written and designed to fulfill the need of integrated teaching learning approaches.

Excel in Mathematics is an absolutely modified and revised edition of my three previous series:
'Elementary mathematics' (B.S. 2053), 'Maths in Action (B. S. 2059)', and 'Speedy Maths' (B. S. 2066).

Excel in Mathematics has incorporated applied constructivism. Every lesson of the whole series
is written and designed in such a manner, that makes the classes automatically constructive and
the learners actively participate in the learning process to construct knowledge themselves, rather
than just receiving ready made information from their instructors. Even the teachers will be able
to get enough opportunities to play the role of facilitators and guides shifting themselves from the
traditional methods of imposing instructions.

Each unit of Excel in Mathematics series is provided with many more worked out examples.
Worked out examples are arranged in the order of the learning objectives and they are reflective to
the corresponding exercises. Therefore, each textbook of the series itself plays the role of a ‘Text
Tutor’. There is a proper balance between the verities of problems and their numbers in each
exercise of the textbooks in the series.

Clear and effective visualization of diagrammatic illustrations in the contents of each and every
unit in grades 1 to 5, and most of the units in the higher grades as per need, will be able to integrate
mathematics lab and activities with the regular processes of teaching learning mathematics
connecting to real life situations.

The learner friendly instructions given in each and every learning content and activity during
regular learning processes will promote collaborative learning and help to develop learner-
centred classroom atmosphere.

In grades 6 to 10, the provision of ‘General section’, ‘Creative section - A’, and ‘Creative section -B’
fulfill the coverage of overall learning objectives. For example, the problems in ‘General section’
are based on the knowledge, understanding, and skill (as per the need of the respective unit)
whereas the ‘Creative sections’ include the Higher ability problems.

The provision of ‘Classwork’ from grades 1 to 5 promotes learners in constructing knowledge,
understanding and skill themselves with the help of the effective roles of teacher as a facilitator
and a guide. Besides, the teacher will have enough opportunities to judge the learning progress
and learning difficulties of the learners immediately inside the classroom. These classworks
prepare learners to achieve higher abilities in problem solving. Of course, the commencement of
every unit with 'Classwork-Exercise' plays a significant role as a 'Textual-Instructor'.

The 'project works' given at the end of each unit in grades 1 to 5 and most of the units in higher
grades provide some ideas to connect the learning of mathematics to the real life situations.

The provision of ‘Section A’ and ‘Section B’ in grades 4 and 5 provides significant opportunities
to integrate mental maths and manual maths simultaneously. Moreover, the problems in ‘Section
A’ judge the level of achievement of knowledge and understanding, and diagnose the learning
difficulties of the learners.

The provision of ‘Looking back’ at the beginning of each unit in grades 1 to 8 plays an important
role of ‘placement evaluation’ which is in fact used by a teacher to judge the level of prior
knowledge and understanding of every learner to select their teaching learning strategies.

The socially communicative approach by language and literature in every textbook, especially in
primary level of the series, plays a vital role as a ‘textual-parents’ to the young learners and helps
them overcome maths anxiety.

The Excel in Mathematics series is completely based on the latest curriculum of mathematics,
designed and developed by the Curriculum Development Centre (CDC), the Government of Nepal.

I do hope the students, teachers, and even the parents will be highly benefited from the ‘Excel in
Mathematics’ series.

Constructive comments and suggestions for the further improvements of the series from the
concerned are highly appreciated.

Acknowledgments

In making effective modification and revision in the Excel in Mathematics series from my previous
series, I’m highly grateful to the Principals, HODs, Mathematics teachers and experts, PABSON,
NPABSAN, PETSAN, ISAN, EMBOCS, NISAN, and independent clusters of many other Schools
of Nepal, for providing me with opportunities to participate in workshops, Seminars, Teachers’
training, Interaction programme, and symposia as the resource person. Such programmes helped
me a lot to investigate the teaching-learning problems and to research the possible remedies and
reflect to the series.

I’m proud of my wife Rita Rai Dahal who always encourages me to write the texts in a more
effective way so that the texts stand as useful and unique in all respects. I’m equally grateful to
my son Bishwant Dahal and my daughter Sunayana Dahal for their necessary supports during the
preparation of the series.

I’m extremely grateful to Dr. Ruth Green, a retired professor from Leeds University, England who
provided me with very valuable suggestions about the effective methods of teaching-learning
mathematics and many reference materials.

Thanks are due to Mr. Tara Bahadur Magar for his painstakingly editing of the series. I am thankful
to Dr. Komal Phuyal for editing the language of the series.

Moreover, I gratefully acknowledge all Mathematics Teachers throughout the country who
encouraged me and provided me with the necessary feedback during the workshops/interactions
and teachers’ training programmes in order to prepare the series in this shape.

I’m profoundly grateful to the Vedanta Publication (P) Ltd. for publishing this series. I would
like to thank Chairperson Mr. Suresh Kumar Regmi, Managing Director Mr. Jiwan Shrestha, and
Marketing Director Mr. Manoj Kumar Regmi for their invaluable suggestions and support during
the preparation of the series.

Also I’m heartily thankful to Mr. Pradeep Kandel, the Computer and Designing Senior Officer of
the publication house for his skill in designing the series in such an attractive form.

Hukum Pd. Dahal

Unit Topics Contents Page No.
Unit
Set 5-24
1
1.1 Set-Looking back, 1.2 Notation of set, 1.3, Methods of describing sets,
Unit 1.4 Cardinal number of sets, 1.5 Types of sets, 1.6 Subsets, proper subsets
and universal sets, 1.7 Number of subsets of a given set, 1.8 Venn-diagrams,
2 1.9 Set operations with Venn-diagrams, 1.10 Cardinality relations of sets

Unit Number System in different Bases 25-30

3 2.1 Denary, binary and quinary numbers - looking back, 2.2 Natural and
whole numbers, 2.3 Decimal numeration system, 2.4 Binary number system,
Unit 2.5 Formation of binary number system, 2.6 Conversion of decimal numbers
to binary numbers, 2.7 Conversion of binary numbers to decimal numbers,
4 2.8 Quinary number system, 2.9 Conversion of decimal numbers to quinary
numbers, 2.10 Conversion of quinary numbers to decimal numbers
Unit
Integers 31-38
5
3.1 Integers-Looking back, 3.2 Laws of addition of integers, 3.3 Sign rules of
Unit addition of integers, 3.4 Laws of multiplication of integers, 3.5 Sign rules of
multiplication and division of integers, 3.6 Simplification of integers
6
Real Numbers System 39-53
Unit
4.1 Rational numbers - Looking back, 4.2 Terminating and non terminating
7 recurring decimals, 4.3 Irrational numbers, 4.4 Real numbers, 4.5 Scientific
notation of numbers, 4.6 Introduction to surds, 4.7 Addition and subtraction
Unit of irrational numbers, 4.8 Multiplication and division of irrational numbers,
4.9 Rationalisation, 4.10 Conjugate
8
Ratio, Proportion and Unitary method 54-72
Unit
5.1 Ratio-Looking back, 5.2 Compounded ratio, 5.3 Proportion, 5.4 Types of
9 proportions, 5.5 Unitary method

Unit Percent and Simple Interest 73-84

10 6.1 Percent-Looking back, 6.2 Fundamental operations on percent,
6.3 Simple interest, 6.4 Rate of Interest, 6.5 Formula of simple interest
Unit
Profit and Loss 85-98
11
7.1 Profit and loss - Looking back, 7.2 Profit and loss percent, 7.3 To find
S.P. when C.P. and profit or loss percents are given, 7.4 To find C.P. when
S.P. and profit or loss percent are given, 7.5 Discount and discount percent,
7.6 Value Added Tax (VAT)

Algebraic Expressions 99-109

8.1 Algebraic terms and expressions - Looking back, 8.2 Polynomials,
and degree of polynomial, 8.3 Evaluation of algebraic expressions,
8.4 Some special products and formulae

Laws of Indices 110-116

9.1 Laws of indices - Review

Factorisation, H.C.F. and L.C.M. 117-133

10.1 Factors and factorisation - Looking back, 10.2 Highest Common Factor
(H.C.F.), 10.3 H.C.F. of polynomial expressions, 10.4 Lowest Common
Multiple (L.C.M.)

Rational Expressions 134-141

11.1 Rational expressions, 11.2 Reduction of rational expressions to their
lowest terms, 11.3 Multiplication of rational expressions, 11.4 Division of
rational expressions, 11.5 Addition and subtraction of rational expressions

S.N Chapter Page No.

Unit Equation, Inequality and Graph 142-165

12 12.1 Equation and inequality - Looking back 12.2 Properties of inequality,
12.3 Graphs of linear inequalities, 12.4 Solution of inequalities in one
Unit variable, 12.5 Linear equations with one variable, 12.6 Solution of equation,
12.7 Linear equation with two variables, 12.8 Graph of linear equation,
13 12.9 Simultaneous equations, 12.10 Methods of solving simultaneous
equations, 12.11 Application of simultaneous equations,12.12 Quadratic
Unit equation - Introduction, 12.13 Solution of quadratic equations,
12.14 Solving quadratic equations by factorisation method
14
Transformation 166-181
Unit
13.1 Transformation - Review, 13.2 Reflection, 13.3 Reflection of
15 geometrical figure using coordinates, 13.4 Rotation, 13.5 Rotation
of geometrical figures using coordinates, 13.6 Displacement,
Unit 13.7 Displacement of geometrical figures using coordinates

16 Geometry: Angle 182-193

Unit 14.1 Different pairs of angles - Looking back, 14.2 Experimental
verifications of pair of angles formed by two intersecting lines
17 14.3 Different pairs of angles made by a transversal with two straight lines,
14.4 Relation between pairs of angles made by a transversal with parallel
Unit lines

18 Geometry: Triangle 194-213

Unit 15.1 Triangle - Looking back, 15.2 Experimental verification of properties
of triangles, 15.3 Congruent triangles, 15.4 Experiment on the conditions of
19 congruency of triangles by construction, 15.5 Conditions of congruency of
triangles, 15.6 Similar triangles, 15.7 Conditions of similarity of triangles
Unit
Geometry: Quadrilateral and Regular Polygon 225-237
20
16.1 Quadrilaterals-Looking back, 16.2 Some special types of quadrilaterals,
Unit 16.3 Regular polygons, 16.4 Sum of the interior angles of a polygon,
16.5 Sum of the exterior angles of a polygon
21
Geometry: Construction 238-242
Unit
17.1 Construction of angle - Looking back, 17.2 Construction of rectangle,
22 17.3 Construction of regular polygons

Coordinates 232-242

18.1 Coordinates - Looking back, 18.2 Pythagoras Theorem,
18.3 Pythagorean Triples, 18.4 Distance between two points

Circle 243-248

19.1 Perimeter of circle - Looking back , 19.2 Area of circle

Area and Volume 249-274

20.1 Area of plane figures, 20.2 Area of triangle, 20.3 Area of quadrilateral,
20.4 Solids and their nets, 20.5 Area of solids, 20.6 Volume of solids

Bearing and Scale Drawing 273-280
21.1 Bearing - Looking back, 21.2 Scale drawing - Review

Statistics 281-301

22.1 Review, 22.2 Collection of data, 22.3 Frequency table, 22.4 Grouped
and continuous data, 22.5 Cumulative frequency table, 22.6 Graphical
representation of data, 22.7 Measures of central tendency, 22.8 Arithmetic
mean, 22.9 Median, 22.10 Quartiles, 22.11 Mode, 22.12 Range

Answers 302

Specification Grid 319

Model Questions 320

Unit Set

1

1.1 Set - Looking back
Classroom - Exercise

1. Let’s say and write the elements of these sets.

a) A = {first five multiples of 3 }, A = {....................................................}

b) B = { common factors of 10 and 20}, B = {....................................................}

c) C = { x : x ≤ 15, x ∈ prime numbers }, C = {....................................................}

2. Let’s rewrite these sets in set-builder forms.

a) M = {2, 4, 6, 8}, M = ........................................................................................

b) N = {3, 6, 9, 12}, N = .........................................................................................

3. a) Let’s write all possible subsets of the set S = {1, 4, 9}

............................ ............................ ............................ ............................

............................ ............................ ............................ ............................

b) Among these subsets, let’s write an improper subset .............................................

4. Let’s say and write the members of the following set operations from the
Venn-diagram given alongside.

a) P ‰ Q = {.........................................................} PQ
b) P ˆ Q = {.........................................................}
c) P – Q = {.........................................................} 2
13
5 11
7
9 13

d) Q – P = {.........................................................}
5. If X = {1, 3, 5, 7, 9} and Y = {2, 3, 5, 7, 11}, say and write the elements of these

set operations.

a) X ‰ Y = {.....................................} b) X ˆ Y = {.....................................}

c) X – Y = {.....................................} d) Y – X = {.....................................}

Let’s take a collection of even numbers less than 10. The members of this collection
are definitely 2, 4, 6, 8. These members are distinct objects when considered
separately. However, when they are considered collectively, they form a single set
of size four, written {2, 4, 6, 8}. It is a set of even numbers less than 10.
Here, any even number less than 10 is definitely the member of the set.

Therefore, a set is a collection of ‘well-defined objects’.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 5 Vedanta Excel in Mathematics - Book 8

Set

1.2 Notation of set

We usually denote sets by capital letters. The members or elements of a set are
enclosed inside the braces { } and the members are separated with commas. The
table given below shows a summary of the symbols which are used in the notation
of sets.

Symbol Name Example Explanation
{} Set W = {0, 1, 2, 3, 4} The members of the sets
are enclosed inside braces
∈ Membership O = {1, 3, 5, 7, 9} { } and separated with
∉ Non- commas.
⊂ 1 ∈ W, 4 ∈ W, The symbol '∈' denotes the
⊆ membership membership of an element
⊃ Proper 5 ∈ O, 9 ∈ O of the given set.
subset 5 ∉ W, 6 ∉ W, The symbol '∉' denotes
the non-membership of an
Improper 2 ∉ 0, 4 ∉ 0 element to the given set.
subset {0, 3} ⊂ W, {1, 2, 4} ⊂ W A set which is contained in
{1} ⊂ O, {1, 5, 7, 9} ⊂ O another set.
Super set {0, 1, 2, 3, 4} ⊆ W, It A set which is contained in
means {0, 1, 2, 3, 4} ⊂ W or equal to another set.
and {0, 1, 2, 3, 4} = W
W ⊃ {0, 1, 2} Set W includes {0, 1, 2,}
and set O includes {5, 7, 9}
O ⊃ {5, 7, 9}

1.3 Methods of describing sets

We usually use three methods to describe a set. These methods are description,
listing (or roster), and set-builder (or rule) methods.

Method Example Explanation
Description
Listing M is a set of multiples of The common properties of elements
(or roster) 3 less than 15. of a set are described by words.
Set-builder
(or rule) M = {3, 6, 9, 12} The elements of a set are listed inside
braces (or curly brackets) { }.

M = {x : x ∈ multiples of A variable such as ‘x’ is used to
3, x < 15} describe the common properties of
the elements of set by using symbols.

1.4 Cardinal number of sets

Let’s consider a set A = {2, 3, 5, 7}. Here, the cardinal number of the set A
represented by n(A) is 4. Thus, the number of elements contained by a set is called
its cardinal number.
Similarly, if B = {0, 1, 2, 3, 4, 5}, then n(B) = 6, and so on.

Vedanta Excel in Mathematics - Book 8 6 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Set

1.5 Types of sets

On the basis of the number of elements of sets, there are four types of sets: null or
empty set, unit or singleton set, finite set, and infinite set.

Types of sets Examples Explanation
Null or empty The set of odd numbers It does not contain any element.
set between 5 and 7. It is denoted by empty braces { }
Unit or or by I (phi)
singleton set A = { } or A = I
Finite set The set of prime numbers It contains only one element.
between 5 and 9.
Infinite set P = {7} It contains finite numbers of
The set of natural numbers elements. It means counting of
less than 50. elements can be ended.
N = {1, 2, 3, 4, 5, … 49} It contains infinite number of
elements. It means, counting of
The set of natural numbers. elements is never ended.
N = {1, 2, 3, 4, 5, …}

On the basis of the types of elements contained by two or more sets, the types of
their relationship can be defined in the following ways.

Types of Examples Explanation
relationship
Equal sets A = {a, e, i, o, u} Equal sets have exactly the
Equivalent B = {o, i, u, a, e} same elements.
sets ?A=B
Equivalent sets have the
Overlapping P = {2, 3, 5, 7, 11} equal cardinal numbers,
sets Q = {2, 4, 6, 8, 10} i.e. they have the equal
n(P) = n(Q) = 5 number of elements.
Disjoint sets ?P~Q ‘~’ is the symbol used to
denote equivalent sets.

C = {1, 3, 5, 7, 9}

D = {1, 2, 3, 4, 5} elements is Overlapping sets have at
Set of common least one element common.
{1, 3, 5}

? C and D are overlapping sets

M = {s, v, u, 3, ª}

N = {c, cf, O, O}{ Disjoint sets do not have
any element common.
There are no element common.

? M and N are disjoint sets.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 7 Vedanta Excel in Mathematics - Book 8

Set

1.6 Subset, proper and improper subsets, and universal set

Subset

Let’s consider any two sets A = {1, 2, 3, ... 10} and B = {1, 3, 5, 7}. Here, every element
of B is contained by A. In other words, the set B is contained by the set A. So, set B is
called the subset of set A.
Thus, between two sets A and B, the set B is said to be a subset of A if every element of
B is contained by A. It is denoted as B ⊂ A. Here, A is said to be the super set of B and
it is denoted as A ⊃ B.

Proper subset

Let's consider any two sets A = {1, 2, 3, 4, 5, … 10} and B = {2, 4, 6, 8}, where B is a
subset of A. Here, B is called a proper subset of the set A.
Thus, between any two sets A and it's subset B, the set B is said to be a proper subset of
A, if it contains at least one element less than set A. It is denoted as B ⊂ A

Improper subset

Let B be a subset of A. Then, B is said to be the improper subset of A when B is equal to
A i.e. every element of A are contained by B. For example,

If A = {s, v, u, 3, ª} and B = {s, v, u, 3, ª}, B is the improper subset of A. It is denoted

as B ⊆ A.

Universal set

Let's consider a set of whole numbers less than 21, U = {0, 1, 2, 3, … 20}.
From this set, we can make many other subsets such as:
A = {1, 2, 3, 4, 5}, set of natural numbers less than 6.
B = {2, 4, 6, 8}, set of even numbers less than 10.
C = {1, 4, 9, 16}, set of square numbers less than 20, and so on.
Thus, in a given situation, the set of all the elements being considered from which many
other subsets can be formed is called a universal set.
A set of students of a school is also a universal set. From this universal set, the subsets
like the set of girls, the set of boys, the set of 8 class students, etc. can be formed.
A universal set is denoted by U or ξ (pxi).

1.7 Number of subsets of a given set

If n be the cardinal number of a given set, then, the number of subsets of a given set = 2n.
For example,
Let A = {x, y, z} be a given set, where n = 3.
Here, the number of possible subsets of A = 2n = 23 = 8
These subsets are {x}, {y}, {z}, (x, y), {x, z}, (y, z), (x, y, z) and { } or φ.
Remember that an empty (or null) set φ is always a subset of every given set.

Vedanta Excel in Mathematics - Book 8 8 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Set

EXERCISE 1.1
General Section – Classwork

1. If A = { 2, 4, 6, 8, ...} , let's say and write ‘true’ or ‘false’ as quickly as possible.

a) 6 ∈ A ......................... b) 12 ∈ A ......................... c) 5 ∈ A .........................

d) 10 ∉ A ......................... e) 15 ∉ A ......................... f) 18 ∉ A .........................

2. Let's say and write whether the following sets are empty, unit, finite or infinite.

a) A = { 1, 3, 5, 7, ...} ...............................

b) B = { 2, 4, 6, 8, ... 100} ...............................

c) C = { x : x ∈ natural numbers and x < 1 } ...............................

d) D = { composite numbers between 7 and 10} ...............................

3. Let's say and tick the correct answers as quickly as possible.

a) W = { 0, 1, 2, 3, 4} and A = { whole numbers less than 5 } sets W and A are

(i) Equal sets (ii) Equivalent sets

b) P = { 2, 3, 5, 7} and N = { x : x is a natural number, x < 5} sets P and N are

(i) Equal sets (ii) Equivalent sets

c) If A = { 3, 6, 9, 12} and B = { 5, 10, 15}, sets A and B are

(i) Overlapping sets (ii) Disjoint sets

d) If C = { 1, 2, 3, 4, 6, 12} and D = {1, 2, 5, 10}, sets C and D are

(i) Overlapping sets (ii) Disjoint sets

4. Let's say and write which one the universal set is and the other is its subset.

a) A = {teachers of a school}, B = {maths teachers of the school}

Universal set is ...................................... and its subset is ......................................

b) N = {natural numbers less than 100} , W = {whole numbers less than 100}

Universal set is ...................................... and its subset is ......................................

Creative section - A

5. a) A set of composite numbers less than 12. Express it in listing and set-builder
methods.

b) B = {the first five multiples of 2}. Express it in roster method and rewrite in
set-builder method.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 9 Vedanta Excel in Mathematics - Book 8

Set

c) Z = { x : x ∈ integers – 2 < x < 2}. List the elements of this set and also express
it in description method.

d) P = {2, 3, 5, 7}. Express it in description method and in rule method.
e) If A = {x : x is a letter in the word ‘SCHOOL’}, list the members and find n (A).
f) If B = {p : p is a factor of 12}, list the elements of this set and find n (B).

6. How many subsets are possible from the following sets? Also, write the subsets.

a) A = {7} b) B = {s, v} c) C = { a, e, i} d) D = {1, 3, 5, 7}

7. What can be the universal sets from which the following subsets can be formed?

a) The set of cricket players of class 8.
b) The set of cricket players of a school.
c) The set of odd numbers less than 10.

Creative section - B

8. Let's answer the following questions.

a) Is A = {1, 4, 9, 16} a subset of B = {1, 2, 3, … 15}? Why?

b) Is P = {a, e, i, o, u} a subset of Q = {a, b, c, d, e, ... z}? Why?

c) Is X = {1, 2, 4, 5, 10, 20} a proper or improper subset of
Y = {x : x ∈ factors of 20}? Why?

d) Are M = {x : x ∈ multiplies of 2, x < 10} and N = {y : y ∈ multiples of 3, y < 10}
overlapping sets? Why?

e) Are A = {p : p ∈ factors of 15, p > 1} and B = {q : q ∈ factors of 16, q > 1},
disjoint sets? Why?

It's your time - Project work!

9. a) Let's write a pair of sets of your own in each of the following cases.
(i) Equal sets (ii) Equivalent sets (iii) Overlapping sets (iv) Disjoint sets

b) Let's write a universal set of your own and write as many possible subsets of the
universal set. What is the improper subset of this universal set?

10. a) Make different sets from the object which you find in your classroom. Such as
students, furniture, stationeries, maths teacher of your class, etc. Then, classify
them by the number of elements they have.

b) Make a group of your friends and write a set with the names of members of the
group. Then make different subsets from the set. For example: A {students who
like football}, B = {students who like maths}, C = {girls students}, and so on.

1.8 Venn – diagrams

We can represent sets and set operations by using different types of diagrams.

We can use different types of diagrams to represent sets and operations of sets. Usually,
a rectangle is used to represent a universal set and circles (or oval shapes) are used to
represent the subsets of the universal set. The concept was first introduced by Swiss
Mathematician, Euler and it was further developed by the British mathematician John
Venn. So, the diagrams are famous as Venn Euler diagrams or simply Venn diagrams.

Vedanta Excel in Mathematics - Book 8 10 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Set

Let's study the following illustrations and learn to represent various set relations by
using Venn-diagrams.

U UU

A A BA B

A ⊂ U (A is a subset of U.) A and B are overlapping sets. A and B are disjoints.

U U U

A, B and C are overlapping A and B are overlapping, A and B are overlapping,
sets. A and C are overlapping, B and C are overlapping, but
but B and C are disjoint A and C are disjoint sets.
sets.

1.9 Set operations with Venn-diagrams

Sets can be combined in a number of different ways to produce another set. It is known
as set operation. Here, we discuss about four basic set operations. These four basic
operations are:

(i) Union of sets (ii) Intersection of sets
(iii) Difference of sets (iv) Complement of a set

(i) Union of sets

Let's take any two sets, A = {1, 2, 3, 4, 5} and B = {2, 4, 6, 8, 10}

Then, the union of A and B, denoted by A ‰ B = {1, 2, 3, 4, 5, 6, 8, 10}.

When the elements of two or more sets are combined and listed A 2 6B
together in a single set, the operation is said to be the union of sets. 1

The shaded region of the diagram shows A ‰ B. 3 4 8
5 10

A‰B

Thus, the union of two sets A and B is the set which contains all the
elements that belong to A or B (or both). It is denoted by A ‰ B and read as
A union B. The symbol ‰ (cup) is used denote the union of sets.

In set builder form, the union of sets A and B can be defined as follows.
A ‰ B = {x : x ∈ A or x ∈ B}

Remember that in the case of overlapping sets, the common elements are mentioned
only once while making the union of the given sets.

11Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 8

Set

Let's study the following illustrations of the union of sets by using Venn-diagrams.

UU U

The shaded region contains The shaded region contains The shaded region contains

the elements of A ‰ B. Here, the elements of A ‰ B. Here, the elements of A ‰ B ‰ C.

A and B are disjoint sets. A and B are overlapping sets. A, B and C are disjoint sets.

U U U

The shaded region contains The shaded region contains The shaded region contains

the elements of A ‰ B ‰ C. the elements of A ‰ B ‰ C. the elements of A ‰ B ‰ C.

A, B and C are overlapping A and B, A and C are overlapping A and B, B and C are overlapping
sets. but B and C are disjoint sets. but A and C are disjoint sets.

(ii) Intersection of sets

Let's take any two sets A = {1, 2, 3, 6} and B = {1, 2, 4, 8}.

Then, the intersection of A and B denoted by A ˆ B = {1, 2} AB

When the common elements of two or more sets are taken and listed in 3 14
a separate set, the operation is said to be the intersection of sets. 6 28
The shaded region of the diagram shows A ˆ B.

Thus, the intersection of sets A and B is the set which contains all the elements that
belong to A and B. It is denoted by A ˆ B and read as A intersection B. The symbol
ˆ (cap) is used to denote the intersection of sets.

In set-builder form, the intersection of sets A and B can be defined as follows.

A ˆ B = {x : x ∈ A and x ∈ B}

Study the following illustrations of the intersection of sets by using Venn-diagrams.

U UU

AB

The shaded region contains Sets A and B are the disjoint C
sets. So, they do not have The shaded region contains
the elements of A ˆ B. any common elements.
the elements of A ˆ B ˆ C
U U
U
AB AB
AB

C C C

The shaded region contains The shaded region contains The shaded region contains the

the elements of only (A ˆ B). the elements of only (A ˆ C). elements of only (B ˆ C).

Vedanta Excel in Mathematics - Book 8 12 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Set

(iii) Difference of two sets

Let's take any two sets A = {2, 3, 5, 7, 11} and B = {1, 3, 5, 7, 9}.
Then, the difference of A and B denoted by A – B = {2, 11}

The difference of B and A denoted by B – A = {1, 9}

Between two sets, when the elements of one set which do not belong to another set, are

listed in a separate set, the operation is said to be the A BA B

difference of two sets. 2 3 1 2 3 1
11 57 9 11 75 9
The shaded regions of the diagrams show A – B and
B – A. A–B B–A

Thus, the difference of two sets A and B denoted by A – B is the set of all elements

contained only by A but not by B.

Similarly, the difference of two sets B and A denoted by B – A is the set of all elements
contained only by B but not by A.

In set-builder form, the difference of A and B or B and A can be defined as follows.

A – B = {x : x ∈ A, but x ∉ B} and B – A = {x : x ∈ B, but x ∉ A}

Study the following illustrations of the difference of sets by using Venn-diagrams.

UU UU

The shaded region The shaded region The shaded region The shaded region
contains the contains the contains the contains the
elements of A – B. elements of A – B. elements of B – A. elements of B – A.

(iv) Complement of a set U

Let's take a universal set U = {1, 2, 3, ... 10} and its subset A1
A = {1, 4, 9}.
Then, the complement of A denoted by 2 49 10
A = U – A = {2, 3, 5, 6, 7, 8, 10} 3 8
When the elements of a universal set, which do not belong to 567
its given subset are listed in a separate set, the operation is said
to be the complement of the given subset. The shaded region contains

the elements of A

Thus, if a set A is the subset of a universal set U, its complement denoted by A is the
set which is formed from the difference of U and A, i.e., U – A.

In set builder form, the complement of a subset A of the universal set U can be defined
in the following ways.

A= {x : x ∈ U, but x ∉ A}

13Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 8

Set

Worked-out examples

Example 1: Let A, B, and C are any three sets. Write the set operations represented
by the shaded regions. Define each operation by set-builder form.

a) b) C c) A B d)

A BB

C

Solution:

a) The shaded region represents A ˆ B.

A ˆ B = {x : x ∈ A and x ∈ B}

b) The shaded region represents B – C.

B – C = {x : x ∈ B, but x ∉ C}

c) The shaded region represents A ˆ B ˆ C.

A ˆ B ˆ C = {x : x ∈ A, x ∈ B and x ∈ C}

d) The shaded region represents A ‰ B.

A ‰ B = {x : A or x ∈ B}

Example 2: If A = {1, 2, 3, 4, 5}, B = {1, 3, 5, 7, 9}, and C = {2, 3, 5, 7, 11}, find
Solution: (i) A ‰ B (ii) A ‰ B ‰ C (iii) A ˆ B (iv) A ˆ B ˆ C (v) A – B
(vi) C – B. Illustrate the results in Venn-diagrams.

Here, A = {1, 2, 3, 4, 5}, B = {1, 3, 5, 7, 9}, and C = {2, 3, 5, 7, 11} 2 1

(i) A ‰ B = {1, 2, 3, 4, 5, 7, 9} 9

The shaded region contains the elements of A ‰ B. 9
2
(ii) A ‰ B ‰ C = {1, 2, 3, 4, 5, 7, 9, 11}
The shaded region contains the elements of A ‰ B ‰ C. 11

(iii) A ˆ B = {1, 3, 5} 21
The shaded region contains the elements of A ˆ B. 9

(iv) A ˆ B ˆ C = {3, 5}
The shaded region contains the elements of A ˆ B ˆ C.

11

Vedanta Excel in Mathematics - Book 8 14 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Set

(v) A – B = {2, 4} 9
The shaded region contains the elements of A – B.
12
(vi) C – B = {2, 11} 11
The shaded region contains the elements of C – B.

Example 3: A, B and C are the subsets of a universal set U. If U = {1, 2, 3, ... 20},
A ={1, 2, 3, 4, 5, 6, 7}, B ={2, 4, 6, 8, 10, 12}, and C ={3, 6, 9, 12, 15, 18},
perform the following set operations and illustrate them in Venn-
diagrams.

(i) (A ‰ B) ˆ C (ii) A ˆ (B ‰ C) (iii) A ‰ B (iv) A ‰ B ‰ C

(v) A ˆ C (vi) A ˆ B ˆ C (vii) (A – C) ˆ B
Solution:

Here, U = {1, 2, 3, ... 20}, A = {1, 2, 3, 4, 5, 6, 7}, B = {2, 4, 6, 8, 10, 12} and
C = {3, 6, 9, 12, 15, 18}
(i) A ‰ B = {1, 2, 3, 4, 5, 6, 8, 10, 12} U

?(A ‰ B) ˆ C = {3, 6, 12}

The shaded region contains the elements of 18 20
(A ‰ B) ˆ C. 16
(ii) B ‰ C = {2, 3, 4, 6, 8, 9, 10, 12, 15, 18} 19
17

U

?A ˆ (B ‰ C) = {2, 3, 4, 6}

The shaded region contains the elements of

A ˆ (B ‰ C). 18 19 20
(iii) A ‰ B = {1, 2, 3, 4, 5, 6, 7, 8, 10, 12} 16 17
U

? (A ‰ B) = U – (A ‰ B) 20
19
= {9, 11, 13, 14, 15, 16, 17, 18, 19, 20} 9 18
11 13 14 15 16 17
The shaded region contains the elements of (A ‰ B).
U
(iv) A ‰ B ‰ C = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 15, 18}
? A ‰ B ‰ C = U – (A ‰ B ‰ C) 14 16 18 17 19 20
= {11, 13, 14, 16, 17, 19, 20}

15Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 8

Set

(v) A ˆ C = {3, 6} U
? A ˆ C = U – (A ˆ C)
20
={1, 2, 4, 5, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20} 18 19
The shaded region contains the elements of A ∩ C.
(vi) A ˆ B ˆ C = {6} 16 17

U

? A ˆ B ˆ C = U – (A ˆ B ∩ ˆ)

={1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20} 16 18 20
The shaded region contains the elements of A ˆ B ˆ C. 17 19

U

(vii) (A – C) = {1, 2, 4, 5, 7}

? A – C ={3, 6, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}

? A – C ˆ B={6, 8, 10, 12} 16 18 17 19 20
The shaded region contains the elements of A – C ˆ B.

Example 4: From the given Venn-diagram, list the elements of the following sets. U
(i) A ‰ B (ii) B ‰ C (iii) A ‰ B ‰ C

(iv) A ˆ B (v) A ˆ C (vi) A ˆ B ˆ C

(vii) A ‰ B ‰ C (viii) A ˆ B ˆ C (ix) B 13

(x) A (xi) (A ‰ B) – C (xii) (A ˆ B) ˆ C 14 15
Solution:

Here, U = {1, 2, 3, … 15}, A = {2, 3, 5, 7, 11, 13}, B = {1, 3, 5, 7, 9}, C = {1, 2, 3, 4, 5, 6}

(i) A ‰ B = {1, 2, 3, 5, 7, 9, 11, 13}

(ii) B ‰ C = {1, 2, 3, 4, 5, 6, 7, 9}
(iii) A ‰ B ‰ C = {1, 2, 3, 4, 5, 6, 7, 9, 11, 13}
(iv) A ˆ B = {3, 5, 7}
(v) A ˆ C = {2, 3, 5}
(vi) A ˆ B ˆ C = {3, 5}
(vii) A ‰ B ‰ C = {8, 10, 12, 14, 15}

(viii)A ˆ B ˆ C = {1, 2, 4, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}

(ix) B = {2, 4, 6, 8, 10, 11, 12, 13, 14, 15} ∴ B = {1, 3, 5, 7, 9} It shows that B = B

(x) A = {1, 4, 6, 8, 9, 10, 12, 14, 15} ∴ A = {2, 3, 5, 7, 11, 13} It shows that A = A
(xi) (A ‰ B) – C = {7, 9, 11, 13}

(xii) (A ‰ B) ˆ C = {1, 2, 3, 5}

Vedanta Excel in Mathematics - Book 8 16 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Set

Example 5: If A = {a, e, i, o, u}, B = {a, b, c, d, e} and C = {e, f, g, h}, show that

Solution: A ‰ (B ˆ C) = (A ‰ B) ˆ (A ‰ C).
Here, A = {a, e, i, o, u}, B = {a, b, c, d, e}, C = {e, f, g, h}

Now, B ˆ C = {e} and A ˆ (B ˆ C) = {a, e, i, o, u}

Again, A ‰ B = {a, b, c, d, e, i, o, u}, A ‰ C = {a, e, f, g, h, i, o, u}

Then, (A ‰ B) ˆ (A ‰ C) = {a, e, i, o, u}

Thus, A ‰ (B ˆ C) = (A ‰ B) ˆ (A ‰ C) proved.

EXERCISE 1.2
General Section - Classwork

1. Let's say and write the answers as quickly as possible.
a) If A = {1, 3, 5, 7, 9} and B = { 2, 3, 5, 7, 11}, then
(i) A ‰ B = ..................................................................................................
(ii) A ˆ B = ..................................................................................................
(iii) A – B = ..................................................................................................
(iv) B – A = ..................................................................................................
b) A universal set U = { 1, 2, 3, ... 10}. Its subsets A = {2, 4, 6, 8} and
B = {1, 2, 3, 4, 5}. Let's tell and write

(i) A = ................................................ and A = ................................................

(ii) B = ................................................ and B = ...............................................
2. Let's say and list the elements as quickly as possible.

a) P ‰ Q = ............................................................. PQ
b) P ˆ Q = .............................................................
c) P – Q = ............................................................. r ecat m
d) Q – P = ............................................................. i
hs

Creative Section - A

3. Let A, B, and C are any three sets. Let's write the set operations represented by
the shaded regions. Define each operation by set-builder form.

a) A B b) B C c) A C

17Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 8

Set

d) A B e) A B A B

f)

C
C

4. P and Q are two overlapping subsets of a universal set U. Draw Venn-diagrams
and shade the regions that contain the element of the following set operations.

a) P ‰ Q b) P ˆ Q c) P – Q d) Q – P e) P f) Q

5. If A = {a, b, c, d, e}, B = {d, e, f, g, h, i} and C = {a, e, i, o, u}, find the elements
of the following set operations and illustrate them in Venn-diagrams.

a) A ‰ B b) B ‰ C c) A ˆ B d) A ‰ B ‰ C e) A ˆ B ˆ C

f) B – C g) C – A h) A ‰ (B ˆ C) i) (A ‰ B) ˆ C j) A – (B ˆ C)

6. A, B, C are the subsets of universal set U. If U = {1, 2, 3, ... 15},
A = {2, 4, 6, 8, 10, 12}, B = {3, 6, 9, 12, 15} and C = {1, 2, 3, 4, 5, 6, 7}, find the
elements of the following set operations and illustrate them in Venn-diagrams.

a) (A ‰ B) ˆ C b) (A ˆ B) ‰ C c) A ‰ B ‰ C d) (A ‰ C) ˆ B

e) A – (B ˆ C) f) (A ‰ B) – (B ˆ C) g) (A – B) ‰ (B – C) h) (A – C) ‰ B

7. From the adjoining Venn-diagram, list the elements of the following sets.

a) P ‰ Q b) Q ‰ R c) P ‰ R P U
d) P ˆ Q e) Q ˆ R f) P ‰ Q ‰ R 8
g) P ‰ Q ‰ R h) P ˆ Q ˆ R i) P ˆ Q ˆ R 12 Q
j) P ‰ (Q ˆ R) k) (Q ‰ R) ˆ P l) P – (Q ˆ R) 10 2 9
m) P n) Q o) R 4
63 R
1
5
7
11 13 14 15

Creative Section - B

8. a) If A = {1, 2, 3, 4, 5}, B = {1, 3, 5, 7} and C = {2, 3, 5}, show that

(i) A ‰ (B ‰ C) = (A ‰ B) ‰ C (ii) (A ˆ B) ˆ C = A ˆ (B ˆ C)

b) If A = {a, b, c, d, e, f}, B = {b, c, d, f, g} and C = (a, e, i, o, u), show that

i) A ‰ (B ˆ C) = (A ‰ B) ˆ (A ‰ C) (ii) A ˆ (B ‰ C) = (A ˆ B) ‰ (A ˆ C)

9. a) A and B are the subsets of a universal set U. If U = {1, 2, 3, ... 10},

A = {1, 3, 5, 7, 9} and B = {1, 2, 3, 6}, show that

(i) A ‰ B = A ˆ B (ii) A ˆ B = A ‰ B

b) P and Q are the subsets of a universal set U. If U = {x : x is a whole number, x≤9},
P = {y : y is a multiple of 2} and Q = {z : z is a factor of 12}, verify that:

(i) P ‰ Q = P ˆ Q (ii) P ˆ Q = P ‰ Q

Vedanta Excel in Mathematics - Book 8 18 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Set

It's your time - Project work

10. a) Make a group of your 10 friends and collect the data to complete the table given
below.

Name of Friends who like Friends who Friends who like Mathematics
friends Mathematics like Science and Science both

Now, draw Venn-diagram and shade each diagram to show each of the following
sets.
(i) Set of friends who like Mathematics or Science or both
(ii) Set of friends who like Mathematics and Science both
(iii) Set of friends who like only Mathematics
(iv) Set of friends who like none of these two subjects.

b) Let's write any three non-empty and overlapping sets A, B and C. Then verify the
following operations.

(i) A ‰ (B ˆ C) = (A ‰ B) ˆ (A ‰ C) (ii) A ˆ (B ‰ C) = (A ˆ B) ‰ (A ˆ C)

1.10 Cardinality relations of sets

The cardinal number of a set is called its cardinality. Let's study the following relations
which are generalised by taking the cardinalities of different sets.

(i) Cardinality relation of union of two disjoint sets

Let A = {a, b, c, d} and B = {e, f, g, h, i} are two disjoint subsets of a universal set
U = {a, b, c, d, e, f, g, h, i, j}

Here, n (A) = 4, n (B) = 5 and n (U) = 10
Now, A ‰ B = {a, b, c, d, e, f, g, h, i}
? n (A ‰ B) = 9 = 4 + 5 = n (A) + n (B)

Thus, n (A ‰ B) = n (A) + n (B), where A and B are two disjoint sets.

Also, n (A ‰ B) = n (U) – n (A ‰ B)

(ii) Cardinality relation of union of two overlapping sets

Let A = {a, b, c, d, e} and B = {d, e, f, g, h, i}, are two overlapping subsets of a

universal set U = {a, b, c, d, e, f, g, h, i, j}.

Here, n (A) = 5, n (B) = 6 and n (U) = 10. U

Now, A ‰ B = {a, b, c, d, e, f, g, h, i} a d f h
? n (A ‰ B) = 9 g
Also, A ˆ B = {d, e} b ei
? n (A ˆ B) = 2 c
j

Again, n (A ‰ B) = 9 = 5 + 6 – 2 = n (A) + n (B) – n (A ˆ B)

19Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 8

Set

It’s clear that in n (A) + n (B), the number of common elements [n (A ˆ B)] appears
two times. So, to correct it n (A ˆ B) is subtracted from n (A) + n (B).

Thus, n (A ‰ B) = n (A) + n (B) – n (A ˆ B) where A and B are overlapping sets.
Also, n (A ‰ B) = n (U) – n (A ‰ B)

Let’s learn a few more relations about the cardinality relation of the operations of
two overlapping sets from the following Venn-diagrams.

U UU U

A BA B A B A B

The shaded region The shaded region The shaded region It is n (only B) or no (B)
represents n (A ‰ B). represents n(A ˆ B) or n(B – A)
represents n (only A) ? no (B) = n (B) – n (A ˆ B)

?ornnoo(A(A))=ornn((AA) – B). ˆ B).
– n(A

Worked-out examples

Example 1: A and B are the subsets of a universal set U, where n (U) = 100, n (A) = 70,
Solution: n (B) = 50 and n (A ˆ B) = 30.

(i) Illustrate this information in Venn-diagram.
(ii) Find n (A ‰ B) and (A ‰ B).

Here, n (U) = 100, n (A) = 70, n (B) = 50 and n(A ˆ B) = 30

(i) Venn-diagram (ii) From the Venn-diagram
U n (A ‰ B) = 40 + 30 + 20 = 90

A ? n (A ‰ B) = n (U) – n (A ‰ B) = 100 – 90 = 10
B
Alternative process
40 30 20 n (A ‰ B) = n(A) + n (B) – n(A ˆ B)

10 = 70 + 50 – 30 = 90
? n (A ‰ B) = n (U) – n (A ‰ B) = 100 – 90 = 10.

Example 2: P and Q are the subsets of a universal set U. If n (U) = 85, n (P) = 45,
n (Q) = 55 and n (P ‰ Q) = 65, find

(i) n (P ˆ Q) (ii) no (P) or (P – Q) (iii) no (Q) or (Q – P)

Also, illustrate the information in Venn-diagram.

Solution:

Here, n (U) = 85, n (P) = 45, n (Q) = 55 and n (P ‰ Q) = 65 U

(i) Now, n (P ‰ Q) = n (P) + n (Q) – n (P ˆ Q) P
or, 65 = 45 + 55 – n(P ˆ Q) Q
or, n(P ˆ Q) = 100 – 65 = 35.
10 35 20

(ii) no (P) or (P – Q) = n (P) – n (P ˆ Q) = 45 – 35 = 10 20
(iii) no (Q) or (Q – P) = n (Q) – n (P ˆ Q) = 55 – 35 = 20

Vedanta Excel in Mathematics - Book 8 20 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Set

Example 3: There are 45 students in a class. 24 of them like cricket, 30 like football
and 14 like cricket as well as football.
(i) Illustrate this information in Venn-diagram
(ii) How many students like both games?
(iii) How many students do not like both games?
(iv) How many of them like only cricket?
(v) How many of them like only football?

Solution:

Let, C and F denote the sets of students who like cricket and football respectively.

Here, n (U) = 45, n (C) = 24, n (F) = 30, and n (C ˆ F) = 14

(i) Venn diagram: (ii) n (C ‰ F) = n (C) + n (F) – n (C ˆ F)
U
(iii) n (C ‰ F) = 24 + 30 – 14 = 40
CF (iv) n (only C) = n (U) – n (C ‰ F) = 45 – 40 = 5
10 14 16 (v) n (only F) = n (C) – n (C ˆ F) = 24 – 14 = 10
= n (F) – n (C ˆ F) = 30 – 14 = 16
5

Example 4: In a survey of 5,000 Chinese tourists who recently visited Nepal during
'Visit Nepal 2020', it was found that 3,500 tourists visited Pokhara, 4,000
visited Lumbini and 500 tourists did not visit both places.

(i) How many tourists visited Pokhara or Lumbini?

(ii) How many tourists visited Pokhara and Lumbini?

(iii) Represent the above information in a Venn-diagram.

Solution:

Let the set of Chinese tourist who visited Pokhara be P and Lumbini be L.

Here, n(U) = 5,000, n(P) = 3,500, n(L) = 4,000 and n (P ‰ L) = 500

(i) Now, n(P ‰ L) = n(U) – n (P ‰ L) = 5,000 – 500 = 4,500

? 4,500 tourists visited Pokhara or Lumbini. (iii) 3000 U
(ii) Also, n(P ‰ L) = n(P) + n(L) – n(P ˆ L)
P L
or, 4,500 = 3,500 + 4,000 – n(P ˆ L) 500 1000
or, n(P ˆ L) = 7,500 – 4,500 = 3,000
? 3,000 tourists visited Pokhara and Lumbini. 500

Example 5: In a group of 175 people, everybody can speak at least one of the two
languages, Nepali or Maithili. 125 people can speak Nepali and 105
people can speak Maithili.

(i) Illustrate the above information in a Venn-diagram.

(ii) How many people can speak both Nepali and Maithili languages?

(iii) Find the number of people who can speak only one language.

Solution:

Let N and M denote the sets of people who can speak Nepali and Maithili language
respectively.

Here, n(U) = n(N ‰ M) = 175, n(N) = 125, and n(M) = 105

21Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 8

Set

Let, the number of people who can speak both languages, n(N ˆ M) = x.

(i) Illustration in Venn-diagram N U
M

125–x x 105–x

(ii) From the Venn-diagram,

125 – x + x + 105 – x = 175 Alternative process

or, 230 – x = 175 n (N ‰ M) = n (N) + n (M) – n (N ˆ M)
or, 175 = 125 + 105 – n (N ˆ M)
or, x = 230 – 175 = 55 or, n (N ˆ M) = 230 – 175 = 55
? 55 people can speak both languages.

(iii) Again, n(only N) = 125 – x = 125 – 55 = 70 Alternative process

n(only M) = 105 – x = 105 – 55 = 50 n (only N) = n (N) – n (N ˆ M)
= 125 – 55 = 70
Number of people who can speak n (only M) = n (M) – n (N ˆ M)
only one language = 70 + 50 = 120
= 105 – 55 = 50

EXERCISE 1.3
General Section - Classwork

1. Let's say and write the correct answers as quickly as possible.

a) A and B are two disjoint subsets of a universal set U . If n (U) = 20, n (A) = 9
and (B) = 8, find,

(i) n (A ‰ B) = ......................... (ii) n (A ‰ B) = .........................

b) P and Q are two overlapping subsets of a universal set U. If n (U) = 50,
n (P) = 25, n (Q) = 30 and n (P ˆ Q) = 15, find:

(i) n (P ‰ Q) = .............. (ii) n (P ‰ Q) = ..............

2. In the given Venn-diagram, T and D denote the sets of students who want to
be teacher and doctor respectively. Let's say and write the correct answers as
quickly as possible.

a) n(T ‰ D) = ................ b) n (T ˆ D) = ................ U
c) n (T ‰ D) = ................ d) n (T – D ) = ................ TD

578
10

e) n (D – T) = ................ f) n(U) = ................

Vedanta Excel in Mathematics - Book 8 22 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Set

Creative Section-A

3. In the given Venn – diagram, A and B are the sets of students U
who like apple and banana respectively. If n (U) = 48 and AB
n (A ‰ B) = 36, find:
27–x x 18–x
b) n (A ‰ B) c) no (A)
a) n (A ˆ B) d) n (B – A)

4. a) If n (U) = 50, n (P) = 27, n (Q) = 25 and n (P ˆ Q) = 12, find:

(i) n (P ‰ Q) (ii) n (P ‰ Q) (iii) n (only P) (iv) n (only Q)

Also illustrate the above information in a Venn-diagram.

b) A and B are the subsets of a universal set U. If n (U) = 100, n (A) = 45,
n (B) = 60 and n (A ‰ B) = 95, find

(i) n (A ˆ B) (ii) n (A ‰ B) (iii) no (A) (iv) no (B)

Also illustrate the above information in a Venn-diagram.

c) A and B are the subsets of a universal set U. If n (U) = 75, n (A) = 40,
n (B) = 35 and n (A ‰ B) = 10, illustrate this information in a Venn-diagram
and find:

(i) n (A ‰ B) (ii) n (A ˆ B) (iii) n (A – B) (iv) n (only B)

d) X and Y are two subsets of a universal set U in which there are n (U) = 54,
n (X) = 30, n (Y) = 20 and n (X ˆ Y) = 9.

(i) Draw a Venn-diagram to illustrate the above information.

(ii) Find the value of n (X ‰ Y) .

e) A and B are the subsets of a universal set U. If n (A) = 63, n (B) = 72,
n (A ‰ B) = 115, n (A ‰ B) = 20, find

(i) n (U) (ii) n (A ˆ B) (iii) no (A) (iv) no (B)

Also, illustrate the above information in Venn-diagram.

Creative Section-B

5. a) Among 72 students, 24 are selected to participate in music only, 26 are selected
to participate in dance only and none of them participated in both activities.
The rest of the students are selected in other activities.
(i) How many students are participating in music or in dance?
(ii) How many students are participating in other than music or dance?

(iii) Illustrate this information in a Venn-digram.

b) In a survey of 110 people, 50 use internet, 75 use cellular data, and 25 use
internet as well as cellular data.

23Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 8

Set

(i) Draw a Venn-diagram to illustrate the above information.
(ii) Find the number of people who use either internet or cellular data.
(iii) Find the number of people who use neither internet nor cellular data.
(iv) Find the number of people who use only one type of Wi-Fi connection.

c) In a survey of some students, 55% of the students like basketball, 65% like
cricket, and 35% like basketball as well as cricket.
(i) Show the above information in a Venn-diagram.

(ii) How many percent of students do not like basketball as well as cricket?

6. a) In a survey of 250 people, 180 like farming, 120 like civil service and every
people like at least one occupation. Draw a Venn-diagram and find the number
of people who like farming as well as civil service. Also, find how many people
like only civil service.

b) In a survey of 486 tourists who visited Nepal during 'Visit Nepal 2020', it was
found that 275 visited Rara, 300 visited Bardiya, and 56 tourists did not visit
both places.
(i) How many tourists were there who visited Rara as well as Bardiya?

(ii) Represent the above information in a Venn-diagram.

c) In an Annual Day at school, 60 students were involved in Science Exhibition,
75 were involved in Mathematics Exhibition, 20 were involved in both
Exhibitions and the rest of 285 students involved in many other activities but
not in the Exhibitions.
(i) Illustrate the above information in a Venn-diagram
(ii) How many students took part on the School Annual Day?
(iii) How many students were involved in only one exhibition?

It's your time - Project work

7. a) Let's conduct a survey inside your classroom and collect data about how many
of your friends like singing, dancing and singing as well as dancing. Then, find
the following numbers by using cardinality relation of the sets.
(i) Number of friends who like singing and dancing.
(ii) Number of friends who like singing only.
(iii) Number of friends who like dancing only.
(iv) Number of friends who do not like singing and dancing.
(v) Also, draw Venn-diagrams to show these numbers.

b) Let's conduct a survey among the teachers of your school and collect the data
about their favorite fruits between apple and orange. Then, perform as many set
operations as you can by using the cardinality properties of sets. Illustrate each
operation in Venn-diagrams.

Vedanta Excel in Mathematics - Book 8 24 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Unit Number System in Different Bases

2

2.1 Denary, binary, and quinary numbers - Looking back
Classroom - Exercise

1. Let's say and write the short forms of the given bases of numbers.

a) The decimal number of 4 × 103 + 6 × 101 + 8 × 10° is ...........................

b) The binary number of 1 × 24 + 1 × 22 + 1 × 21 + 1 × 2° is ...........................

c) The quinary number of 3 × 53 + 4 × 52 + 2 × 51 + 1 × 5° is ...........................

2. Let's say and write the answers as quickly as possible.

a) The digits in denary system are ......................................................................

e) The digits in binary system are ......................................................................

f) The digits in quinary system are .........................................................................

3. Let's say and tick the correct value.

a) Value of 3 in binary number is (i) 102 (ii) 112 (iii) 1012

b) Value of 9 in quinary number is (i) 145 (ii) 1045 (iii) 415

c) Value of 1012 in denary number is (i) 3 (ii) 6 (iii) 5

d) Value of 135 in denary number is (i) 7 (ii) 8 (iii) 9

2.2 Natural numbers and whole numbers - Looking back

N = {1, 2, 3, 4, 5, ...} is the set of natural numbers which are also called the counting
numbers. The operations of addition and multiplication of any two natural numbers
always give natural numbers. For example,

3 + 4 = 7 (7 is also a natural number)

3 × 4 = 12 (12 is also a natural number)

Similarly, in 4 – 3 = 1, 1 is a natural number. However, in 3 – 3 = 0, 0 is not a natural
number. It demands another set of numbers that should include 0 (zero) also. So, 0 is
included in the set of natural numbers to develop a new set of numbers that we call the
set of whole numbers. It is denoted by W.

Thus, W = {0, 1, 2, 3, 4, 5, ...} is the set of whole numbers.

25Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 8

Number System in different Bases

2.3 Decimal numeration system

Let’s take 18 blocks of cubes and regroup them into the group of 10 blocks.
18 = 10 + 8 = 1 × 101 + 8 × 10°

Let’s take 36 pencils and regroup them into the group of 10 blocks.

36 = 30 + 6 = 3 × 101 + 6 × 10°

Similarly, 342 = 300 + 40 + 2 = 3 × 102 + 4 × 101 + 2 × 100.
In this way, whole numbers can be regrouped into the base of 10 with some power of 10.
It is called the decimal numeration system or denary system.

Example 1: Express 472510 in the expanded form.
Solution:

472510 = 4 × 1000 + 7 × 100 + 2 × 10 + 5

= 4× 103 + 7 ×102 + 2 × 101 + 5 × 100.

2.4 Binary number system

Decimal or denary number system is base 10 system. Besides this number system, there
are four alternative base systems that are most useful in computer applications. These
are binary (base two), quinary (base five), octal (base eight), and hexadecimal (base
sixteen) systems.

Computers and handheld calculators use the binary system for their internal calculations.
The system consists of only two digits 0 and 1. So, all numbers can then be represented
by electronic switches of one kind or another, where “on” indicates 1 and “off”
indicates 0.

The octal system is used extensively by programmers who work with internal computer
codes. In a computer, the CPU often uses the hexadecimal system to communicate with
a printer or other output devices.

2.5 Formation of binary number system

Again, let’s take 15 blocks of cubes and regroup them into the group of 2 blocks.

7 pairs blocks of cube and 1 cube

Vedanta Excel in Mathematics - Book 8 26 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Number System in different Bases

Now, let’s arrange the groups of 2 blocks into the base of 2 with the maximum possible
powers.

8 4 21
23 22 21 20

So, 15 = 1 × 23 + 1 × 22 + 1 × 21 + 1 × 2° = 11112
In this way, the denary number 15 can be expressed in binary number as 11112.
The system of numeration in base 2 with some power of 2 is called the binary numeration
system. In this system, we use only two digits 0 and 1.

2.6 Conversion of decimal numbers to binary numbers

We convert a decimal number into binary number, dividing the given number successively
by 2 until the quotient becomes 0. The remainder obtained in each successive division
is listed in a separate column. The remainders arranging in the reverse order is the
required binary number. For example,
Example 2: Convert 185 into binary system.

Solution Arranging the remainders from
the bottom up: 111011012
2 185 Remainder 1×28 1×27 1×26 1×25 1×24 1×23 1×22 1×21 1×20
256 128 64 32 16 8 4 2 1
2 92 1
185 = 128 + 0 + 32 + 16 + 8 + 0 + 0 + 1
2 46 0 = 1×27 + 0 + 1×25 + 1×24 + 1×23 + 0 + 0 + 1×10
= 101110012
2 23 0

2 11 1

25 1

22 1

10

? 185 = 101110012

2.7 Conversion of binary numbers to decimal numbers

We convert a binary number into decimal number, just by expanding the given binary
number in the power of 2. Then, by simplifying the expanded form of the binary number,
we obtain the required decimal number. For example,

Example 3: Convert 1101102 into decimal system.
Solution:

1101102 = 1 u 25 + 1 u 24 + 0 u 23 + 1 u 22 + 1 u 21 0 u 20
= 32 + 16 + 4 + 2 + 0 = 54

27Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 8

Number System in different Bases

EXERCISE 2.1

General Section - Classwork

1. Let's say and tick the short form of these numbers as quickly as possible.

a) 3 × 102 + 7 × 101 + 5 × 100 (i) 3075 (ii) 375 (iii) 370

b) 2 × 104 + 8 × 102 + 4 × 101 + 9 × 100 (i) 20849 (ii) 28049 (iii) 2849

c) 1 × 24 + 1 ×23 + 1 × 22 + 1 × 21 (i) 110112 (ii) 111012 (iii) 111102
d) 1 × 25 + 1 × 23 + 1 × 21 + 1 × 20 (i) 1010112 (ii) 1100112 (iii) 1010102

2. Let's say and write the values of these decimal numbers in binary numbers as
quickly as possible.

a) 1= ............. b) 2 = ............. c) 3 = ............. d) 4 = .............

e) 5 = ............. f) 6 = ............. g) 7 = ............. h) 8 = .............

3. Let's say and write the values of these binary numbers in decimal numbers as
quickly as possible.

a) 12 = ............. b) 102 = ............. c) 112 = ............. d) 1002 = .............

e) 1012 = ............. f) 1102 = ............. g) 1112 = ............. h) 10002= .............

Creative Section

4. Expand the following decimal numbers in power of 10s.

a) 276 b) 3816 c) 54027 d) 809005 e) 7000409

5. Convert the following decimal numbers into binary numbers:

a) 9 b) 10 c) 11 d) 12 e) 18

f) 55 g) 124 h) 216 i) 361 j) 490

6. Convert the following binary numbers into decimal numbers:

a) 10012 b) 10102 c) 101112 d) 1010102 e) 1111012
f) 10101112 g) 11001112 h) 110010112 i) 111011102 j) 1101110112

It's your time - Project work

7. a) Let's take a few number of matchsticks and represent the following binary
numbers by pairing the required number of matchsticks.

(i) 102 (ii) 112 (iii) 1012 (iv) 1112 (v) 1102 (vi) 11102 (vii) 11112

b) Let's draw the required number of pairs of dots in the table to convert the given
denary numbers into binary numbers.

Denary numbers 24 23 22 21 20

5

13

18

Vedanta Excel in Mathematics - Book 8 28 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Profit and Loss

6. a) Let the marked price be Rs x, S.P. is Rs 180 and discount percent is 10%. Find
the marked price. [Hints: x – 10% of x = 180]

b) If a shopkeeper allows 25% discount and sells an umbrella for Rs 450, find the
marked price of the umbrella.

Creative Section - B

7. a) A shopkeeper marked the price of a calculator as Rs 1,600 and she/he sold it
allowing a discount of 10%.Then, she made a profit of Rs 140.
(i) Find the amount of discount.
(ii) Find the selling price after allowing discount.
(iii) Find the cost price of the calculator.

b) Mr. Muhammad fixed the marked price of a cycle as Rs 6,000. He sold it after
allowing a discount of 15% and made a profit Rs 500, find the cost price of the
cycle.

c) Kamala marked the price of a cosmetic item as Rs 400. She offered her customers
a discount of 20% and made a loss of Rs 30, what was the actual cost of the item
to her?

8. a) The marked price of a fan is Rs ,1500. The shopkeeper allows 20% discount on
it and then 13% VAT is charged.

(i) Find the amount of discount.

(ii) Find the selling price after giving discount.

(iii) Find the VAT amount.

(iv) Find the selling price with VAT.
b) The price of a rice cooker is marked as Rs 2,500. If 40% discount is allowed and

then 13% VAT is charged, find the selling price of the cooker with VAT.

c) The marked price of a jacket is Rs 5,000. What is the price of the jacket if 10%
VAT was levied after allowing 20% discount?

9. a) A retailer bought a watch for Rs 800 and fixed its price 25% above the cost price.
He then allows 10% discount.
(i) Find the marked price of the watch.
(ii) Find the discount amount.
(iii) Find the selling price of the watch.
(iv) How much should a customer pay for it with 13% VAT?

97Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 8

Profit and Loss

b) A shopkeeper bought an electric fan for Rs 2,000 and fixed its price 20% above
the cost price. If he then gives 15% discount, how much should a customer pay
for it with 15% VAT?

c) A dealer marks the price of his mobiles 40% above the cost price and allows
20% discount. If his purchase price of a mobile is Rs 7,500, how much should a
customer pay for it with 13% VAT?

10. a) A gift house allowed 20% discount on the marked price of a doll, 13% VAT was
levied on it. If the doll was sold at Rs 1,808, what was its marked price?

b) After allowing 16% discount on the marked price of a watch, 13% Value Added
Tax (VAT) was levied on it. If the watch was sold for Rs 4,746, calculate the
marked price of the watch.

It’s your time - Project work!
11. a) Let’s become a problem maker and problem solver.

Write the values of the variables of your own. Then, solve each problem to find
unknown variable.

M.P. = ................... M. P. = ........................

Discount % = .............. S. P. = ........................

Find Find
S.P. discount percent

S. P. = ........................ S. P. with VAT = .........

VAT% = ........................ VAT% = ................

Find S.P. Find S.P.
with VAT without VAT

b) Let’s search the rate of VAT of different countries in the available website.
Compare these VAT rates with the VAT rate of our country.

c) Let’s make group of 5 friends and visit your local market to find the selling price
of some electrical and electronic items. Calculate the S. P. of these items with
the current VAT rate system.

Vedanta Excel in Mathematics - Book 8 98 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Unit Algebraic Expressions

8

8.1 Algebraic terms and expressions - Looking back
Classroom - Exercise

Let’s say and write the answer as quickly as possible.
1. a) In 5x2, the coefficient is ................., base is ................., exponent is .....................

b) In 4a3, the coefficient is ................., base is ................., exponent is ......................
2. a) Are 3x and 3x2 like or unlike terms? .......................................................................

b) Are y3 and 7y3 like or unlike terms? .......................................................................
3. a) How many terms are there in the expression 2xy? ..................................................

b) How many terms are there in the expression 2x + y? .............................................
c) How many terms are there in the expression 2 – x + y? .........................................

2x, 3x2, 5y – 7a2b2, etc. are a few examples of algebraic terms. An algebraic term is the
product form of any number and variables.

There are three important parts in an algebraic term. These parts are coefficient, base,
and exponent.
3x2 Exponent 7y3 Exponent
Base Base
Coefficient Coefficient

The sum or difference of two or more algebraic terms form an algebraic expression. For
example 3x + 2 is a binomial expression, a2 – 4ab + b2 is a trinomial expression, and
so on. However, an algebraic term itself is a monomial expression. 2x, y2, –3p3, etc. are
monomial expressions.

8.2 Polynomials and degree of polynomials

x + y, 2x2 + 4xy – 3y2 , a3 + a2 – 2a + 5, 2 x4 + 2x2 – 7, etc. are the examples of
polynomials. In these polynomials, the powers of variables are whole numbers.

The algebraic expressions in which powers of variables are whole numbers are known
as polynomials.

On the other hand, if the power of variable of an expression is not a whole number, the

expression is not a polynomial. For example, in x +1x , the power of x in 1 is – 1 which
x
1
is not a whole number. So x + x is not a polynomial.

Similarly, 1 + 7, 3 – 4x + 5, etc. are not polynomials.
x x2

99Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 8

Algebraic Expressions

Let’s consider an algebraic term 3x2.
In 3x2, the variable x is multiplied 2 times. So, the degree of 3x2 is 2.
Similarly, in 2y3, the variable y is multiplied 3 times. So, the degree of 4y3 is 3.

Thus, in a polynomial with a monomial expression, the exponent of the variable is the
degree of the polynomial.

In the case of a term that contains two or more variables, the sum of the powers of
each variable is the degree of the term. For example,
In 3x3y2, the sum of the powers of x and y is (3 + 2) = 5. So, the degree of 3x3y2 is 5.
In 7abc, the sum of the powers of a, b, and c is (1 + 1 + 1) = 3. So, the degree of 7abc is 3.

In the case of a polynomial with multinomial expression, the highest power of its any
term is the degree of the polynomial. For example,

In 2x2 + 3x + 4, the highest power of x is 2. So, its degree is 2.
In 4y5 – 5y3 – y, the highest power of y is 5. So, its degree is 5.

In 3p2q2 – pq + 5, the highest power of pq is (2 + 2) = 4. So, its degree is 4.

8.3 Evaluation of algebraic expressions

Let’s take an algebraic expression 2(l + b) and evaluate it when l = 5 and b = 4.
Here, if l = 5 and b = 4, then 2(l + b) = 2(5 + 4) = 18.
Thus, evaluation is the process of finding the value of an algebraic expression by
replacing the variables with numbers.

Worked-out examples

Example 1: Which of the following expressions are polynomials? Write with reason.

(i) x2 + 1 (ii) y2 + 1 (iii) 3 x – 2 (iv) 2 x + 5
2 y2
Solution:

(i) x2 + 1 is a polynomial because the power of x is 2 which is a whole number.
2
1 1
(ii) y2 + y2 is not a polynomial because the power of y in y2 is –2 which is not a whole

number.

(iii) 3x – 2 is a polynomial because the power of x is 1 which is a whole number.

(iv) 2 x + 5 is not a polynomials because the power of x is 1 which is not a whole number.
2

Example 2: Write the degrees of the following polynomials.

(i) 3x2 (ii) 2a2bc (iii) 2x3 – 4x2 + 7x (iv) x3y2 + x2y2 – xy
Solution:

(i) The degree of 3x2 is 2.

(ii) The degree of 2a2bc is 2 + 1 + 1 = 4

(iii) The degree of 3y3 – 2y2 + 5x is 3. The highest power of the variable y is 3

Vedanta Excel in Mathematics - Book 8 100 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Algebraic Expressions

(iv) In x3y2 + x2y2 – xy, the sum of the powers of x3y2= 3 + 2 = 5 and it is the highest
sum of the powers. Therefore, it is 5th degree polynomial.

Example 3: If x = 3 and y = 2, evaluate x2 – 2xy + y2.
Solution:

Here, when x = 3 and y = 2, then x2 – 2xy + y2 = 32 – 2 × 3 × 2 + 22 = 9 – 12 + 4 = 1

EXERCISE 8.1

General Section - Classwork

1. Let’s say and write the polynomial expressions in the table.

a) 2x + 3 b) 3y2 – 2y + 4 Polynomials Expressions

c) a + 1 d) 5x + 1
a f) x + 2 – 3
x2
e) 2 – 3x + 7

2. Let’s say and write the degrees of these polynomials.

a) 3x ................. b) 2p2 ................. c) 4a2b .................

d) 7y – 1 ................. e) 2x3 – 3x2 + 6 ................. f) 9xy – 7x + 2y ................

3. Let’s tell and write the values of the expressions as quickly as possible.

a) If l = 4 and b= 3, then (i) l × b = …….. (ii) 2 (l + b) = ……..

b) If l = 5, then (i) l2 = …….. (ii) l3 = …….. (iii) 6l2 = ……..

Creative Section

4. Let’s answer the following questions.

a) Why is x2 a polynomial but 2 is not a polynomial?
2 x2

b) Why is 2x not a polynomial but 2 x is a polynomial?

c) Define the degree of a polynomial with examples.

5. Find the degrees of the following polynomials.

a) 2xyz b) 5a2bc c) 3y2 – 4y + 8

d) 7x5 – 2x4 + 5x3 – 7x2 – 3x e) x2 + y2 + z2 – 3xyz f) x3y3 – 2x2y2 + 7xy – 5

6. a) If l = 5 and b = 3, find the values of (i) l × b (ii) 2(l + b)

b) If l = 10, b = 8 and h = 5, find the values of

(i) l × b × h (ii) 2h (l + b) (iii) 2 (lb + bh + lh)

c) If r = 7, S = 22 , find the values of (i) Sr2 (ii) 2Sr
7

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 101 Vedanta Excel in Mathematics - Book 8

Algebraic Expressions

d) If a = 3 and b = 2, find the values of (ii) a2 – 2ab + b2 and (a –b)2
(i) a2 + 2ab + b2 and (a + b)2 (iv) a3 – 3a2b + 3ab2 – b3 and (a – b)3
(iii) a3 + 3a2b + 3ab2 + b3 and (a + b)3 (vi) (a + b)2 – 2ab
(v) (a + b) (a –b) and (a2 – b2)

7. a) Find the value of P ×T× R when P = 200, T = 2 and R = 5
100

b) Find the value of P 1 + R T
100
when P = 100, R = 10, T = 2

c) Find the value of (i) 2Srh (ii) 2Sr(r + h) (iii) Sr2h when S = 22 , r = 7 and h = 5
7

8.4 Special products and formulae
(i) The product of (a + b) and (a + b) : Square of binomials

Let’s multiply (a + b) by (a + b)

(a + b) u (a + b)= a (a + b) + b (a + b)

or, (a + b)2 = a2 + ab + ab + b2

or, = a2 + 2ab + b2 Da bC
a a2 ab
Thus, (a + b)2 = (a2 + 2ab + b2) b ab
a+b
Also, a2 + b2 = (a + b)2 – 2ab b2 b

When the length a of the smaller square is increased
by b, each side of the bigger square ABCD is (a + b)

Now, area of ABCD = a2 + ab + ab + b2 A bB
a+b
(a + b)2 = a2 + 2ab + b2

(ii) The product of (a – b) and (a – b)

Let’s multiply (a – b) by (a – b)

(a – b) u (a – b) = a (a – b) – b (a – b)

or, (a – b)2 = a2 – ab – ab + b2

Thus, = a2 – 2ab + b2 D a–b bC
(a – b)2 = a2 – 2ab + b2

Also, a2 + b2 = (a – b)2 + 2ab (a – b)2 b(a – b) a–b
a
When the length and breadth of the square ABCD is

decreased by b, b(a – b) b2 b
Area of ABCD = (a – b)2 + b (a – b) + b (a – b) + b2

or, a2 = (a – b)2 + ab – b2 + ab – b2 + b2 A a–b bB

or, a2 = (a – b)2 + 2ab – b2

or, (a – b2) = a2 – 2ab + b2

(iii) The product of (a + b), (a + b) and (a + b) (cube of binomials)

(a + b) (a + b) (a + b)= (a + b) (a + b)2

or, (a + b)3= (a + b) (a2 + 2ab + b2) a+b
b
= a(a2 + 2ab + b2) + b(a2 + 2ab + b2) a a a b b a+b
ab
= a3 + 2a2b + ab2 + a2b + 2ab2 + b3 a a b b a+b

= a3 + 3a2b + 3ab2 + b3 a3 + 3ab2 + 3ba2 + b3 = (a + b)3

Thus, (a + b)3 = a3 + 3a2b + 3ab2 + b3

Vedanta Excel in Mathematics - Book 8 102 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Algebraic Expressions

Furthermore, (a + b)3 = a3 + 3a2b + 3ab2 + b3
or, (a + b)3= a3 + b3 + 3ab (a + b)
or, a3 + b3= (a + b)3 – 3ab (a + b)

(iv) The product of (a – b), (a – b) and (a – b)

(a – b) (a – b) (a – b) = (a – b) (a – b)2

(a – b)3 = (a – b) (a2 – 2ab + b2)

= a (a2 – 2ab + b2) – b (a2 – 2ab + b2)

= a3 – 2a2b + ab2 – a2b + 2ab2 – b3

= a3 – 3a2b + 3ab2 – b3

Thus, (a – b)3 = a3 – 3a2b + 3ab2 – b3

Furthermore, (a – b)3 = a3 – 3a2b + 3ab2 – b3

or, (a – b)3 = a3 – b3 – 3ab (a – b)

or, a3 – b3 = (a – b)3 + 3ab (a – b) (a – b)3

(v) The product of (a + b) and (a – b)

(a + b) (a – b) D a CD C Da b
b a2 – b2 C
= a (a – b) + b (a – b)
= a2 – ab + ab – b2 a b EF
a–b (a + b) (a – b)
a–b

A a+b B

= a2 – b2 A B AB Area of ABCD
Area of ABCD Area of ABCDEF a2 – b2 = (a + b) (a – b)

= a2 = a2 – b2

Thus,(a + b) (a – b) = a2 – b2

(vi) The product of (a + b) (a2 – ab + b2)

Here,(a + b) (a2 – ab + b2) = a (a2 – ab + b2) + b (a2 – ab + b2)

= a3 – a2b + ab2 + a2b – ab2 + b3

= a3 + b3

Thus, a3 + b3 = (a + b) (a2 – ab + b2)

(vii) The product of (a – b) and (a2 + ab + b2)

Here,(a – b) (a2 + ab + b2) = a (a2 + ab + b2) – b (a2 + ab + b2)

= a3 + a2b + ab2 – a2b – ab2 – b3

= a3 – b3

Thus, a3 – b3 = (a – b) (a2 + ab + b2)

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Algebraic Expressions

Worked-out examples

Example 1: Find the squares of: a) 3x + 2y b) x– 1
Solution: x

a) Square of (3x + 2y)

= (3x + 2y)2 Let’s consider, 3x = a and 2y = b
= (3x)2 + 2.3x.2y + (2y)2 We know that (a + b)2 = a2 + 2ab + b2
= 9x2 + 12xy + 4y2 Then, (3x + 2y)2 = (3x)2 + 2.3x.2y + (2y)2

b) Square of x– 1
x
1
= x– x 2 Let’s consider, x = a and 1 =b
x

= (x)2 – 2.x. 1 + 12 We know that, (a – b)2 = a2 – 2ab + b2
x x 1 1 12
Then, x– x 2 = (x)2 – 2.x. x + x

= x2 – 2 + 1
x2
1
Example 2: Expand a) (y + 2z)3 b) 3a – a 3
Solution:

Let’s consider, y = a and 2z = b
a) (y + 2z)3 = y3 + 3y2.2z + 3y. (2z)2 + (2z)3 We know that

= y3 + 6y2z + 12yz2 + 8z3 (a + b)3 = a3 + 3a2b + 3ab2 + b3
Then,

(y + 2z)3 =y3 + 3.y2.2z + 3.y. (2z)2 + (2z)3

b) 3a – 1 3= (3a)3 – 3. (3a)2. 1 + 3.3a. 1 2– 1 3
a a a a
9 1
= 27a3 – 27a + a – a3

Example 3: Express 16x2 – 40xy + 25y2 as a perfect square.

Solution: = (4x)2 – 2.4x.5y + (5y)2 Let’s consider 4x = a and 5y = b.
16x2 – 40xy + 25y2 = (4x – 5y)2 Then, (4x)2 – 2.4x.5y + (5y)2 is in
the form a2 – 2ab +b2 and
a2 – 2ab + b2 = (a – b)2.

Example 4: Express 8a3 + 36a2b + 54ab2 + 27b3 as a perfect cube.

Solution: 8a3 + 36a2b + 54ab2 + 27b3
8a3 + 36a2b + 54ab2 + 27b3 = (2a)3 + .................... + (3b)3
= (2a)3 + 3.(2a)2.3b + 3.2a.(3b)2 + (3b)3 Consider, 2a = a and 3b = b, then
= (2a + 3b)3 write the expression in the form
a3 + 3a2b + 3ab2 + b3 = (a + b)3

Example 5: Find the product of a) (x – 3y) (x + 3y) b) (4a2 + 3b2) (4a2 – 3b2)
Using (a + b) (a – b) = a2 – b2
Solution:

a) (x – 3y) (x + 3y)= x2 – (3y)2
= x2 – 9y2

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Algebraic Expressions

b) (4a2 + 3b2) (4a2 – 3b2) = (4a2)2 – (3b2)2
= 16a4 – 9b4

Example 6: Find the product of a) (3x + y) (9x2 – 3xy + y2)

b) (2a – 5b) (4a2 + 10ab + 25b2)
Solution:

a) (3x + y) (9x2 – 3xy + y2) Consider 3x = a and y = b, then
= (3x + y) [(3x)2 – 3x.y + y2] (3x + y) (9x2 – 3xy + y2) is in the form
= (3x)3 + y3 (a + b) (a2 – ab + b2) = a3 + b3
= 27x3 + y3

b) (2a – 5b) (4a2 + 10ab + 25b2)
= (2a – 5b) [(2a)2 + 2a.5b + (5b)2]
= (2a)3 – (5b)3 = 8a3 – 125b3\

Example 7: If x + y = 7 and xy = 5, find the value of x2 + y2.

Solution:

We have, x2 + 2xy + y2 = (x + y)2

or, x2 + y2 = (x + y)2 – 2xy

1 = 72 – 2 u 5 = 49 – 10 = 39 1 1
a a2 a
Example 8: If a + = 6, find the values of (i) a2 + (ii) a – 2.

Solution:

Here, a + 1 = 6
a

(i) We have, a2 + b2 = (a + b)2 – 2ab

? a2 + 1 = a + 1 2 – 2.a. 1 = 62 – 2
a2 a a

= 36 – 2 = 34

(ii) We have (a – b)2 = a2 – 2ab + b2
1 = a2 – 2.a.a1 + 1
? a – a 2 a2

= a2 + 1 –2
a2

= 34 – 2 = 32

Example 9: Simplify (x + 3y)2 – (x – 3y)2

Solution: Alternative process

(x + 3y)2 – (x – 3y)2 (x + 3y)2 – (x – 3y)2

= x2 + 2.x.3y + (3y)2 – (x2 – 2.x.3y + (3y)2) = (x + 3y + x – 3y) [x + 3y – (x – 3y)]

= x2 + 6xy + 9y2 – x2 + 6xy – 9y2 = 2x(x + 3y – x + 3y)

= 12xy 2.7 u 2.7 – 1.3 u 1.3 = 2x × 6y = 12xy
2.7 – 1.3
Example 10: Simplify
Solution:

2.7 u 2.7 – 1.3 u 1.3 = (2.7)2 – (1.3)2 = (2.7 + 1.3) (2.7 – 1.3) = 2.7 – 1.3 = 4
2.7 + 1.3 2.7 + 1.3 2.7 + 1.3

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Algebraic Expressions

EXERCISE 8.2
General Section - Classwork

1. Let’s say and write the expanded forms of these expressions.

Expressions Expanded forms Expressions Expanded forms
a) (x + y)2 e) (x – y)2
b) (a + 1)2 f) (m – 1)2
c) (p + 2)2 g) (a – 2)2
d) (y + 3)2 h) (p – 3)2

2. Let’s say and write the following products in the forms of difference of two squared
terms.

a) (x + y) (x – y) = .......................... b) (x + 1) (x – 1) = ..........................

c) (p + 2) (p – 2) = .......................... d) (m + 3) (m – 3) = ..........................

e) (a + a1) (a – 1 ) = .......................... f) (x + 2x ) (x – 2 ) = ..........................
a x

g) ( 3 + p) ( 3 – p) = .......................... h) ( 4 + m) ( 4 – m) = ..........................
p p m m

3. Let’s say and write the following expressions in the forms of cube of sum or
difference of two terms.

a) x3 + 3x2y + 3xy2 + y3 = ..................... b) x3 – 3x2y + 3xy2 – y3 = .................

c) x3 + 3.x2.1 + 3.x.12 + 13 = ..................... d) a3 – 3.a2.2 + 3.a.22 – 23 = .................

e) p3+3.p2. 1 + 3.p. 1 + 1 = ..................... f) x3 – 3.x2. 1 + 3.x.x12 – 1 = .................
p p2 p3 x x3

4. Let’s say and write the following expressions in the forms of cube of sum or
difference of two cubed terms.

a) (x + y) (x2 – xy + y2) = ……….. b) (x – y) (x2 + xy + y2) = ………….

c) (a + 1) (a2 – a + 1) = …………. d) (p – 2) (p2 + 2p + 4) = …………..
e) (2 + 12) (22 – 1 + 41) = …………. f) (31 – 3) (91 + 1 + 9) = …………..

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Algebraic Expressions

Creative Section - A

5. Let’s look at these diagrams and write the areas in algebraic expression forms as

shown in the example.
Dx 1C

x x2 1.x x+1 Area of ABCD = x2 + 1.x + 1.x + 12
(x + 1)2 = x2 + 2x + 1

1 1.x 12

A x+1 B

D x 2C S x R Mx yE
x2 2.x
a) x+2 b) c) x2 x.y x+y

x x x

2 2.x y x.y

A x+2 B PQ P x+y R

Area of ABCD Area of PQRS Area of PREM

d) D x-1 C 1 e) H G f) S x – y R y

x–1 (x-1)2 1.x x (x-2)2 x x – y (x – y)2 x
EF
A 1.(x-1) B 12 P Q
1 x y

x x

Area of ABCD Area of EFGH Area of PQRS

6. Let’s find the area of each of these rectangles in algebraic expression forms.

a) D x 1C b) P x 2O c) H x yG

x2 – 12 x–1 x2 – 22 x–2 x2 – y2 x–y
x x x

A x+1 B M x+2 N E x+y F
12 1 y2 y
2

1 2 y

Area of ABCD Area of MNOP Area of EFGH

7. a) Find the squares of (i) 2a + 1 (ii) x – 3y (iii) a – 1 (iv) x + 1
a x

b) Find the cubes of (i) p + 2 (ii) 2a – b (iii) y + 1 (iv) x– 1
y (iv) (3x –3x31x )3
c) Expand (i) (3x – 1)2 (ii) (2y + 1 )2 (iii) (2p + q)3
2y
8. a) Express (i) x2 +2x + 1 (ii) a2 – 4ab + 4b2 (iii) 9p2 + 12pq + 4q2 as perfect

squares.

b) Express (i) x3 + 6x2 + 12x + 8 (ii) 8a3 – 36a2 + 54a – 27

(iii) p3 – 9p2q + 27pq2 – 27q3 as perfect cubes.

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Algebraic Expressions

9. Let’s find the expressions to be inserted to make these expressions perfect squares.

a) x2 + ……… + 16y2 (Hint. x2 + 2.x.4y + (4y)2, so the required terms is 8xy)

b) x2 + …....................…… + 4y2 c) x2 – …....................…… + 9y2

d) 4x2 + …....................…… + 25y2 e) 9a2 – …....................…… + 49b2

10. Let’s find the expressions for the blank space to make these expressions perfect cubes.

a) x3 + ……. + ……..+ 27 [Hint. x3 + 3. x2. 3 + 3. x. 32 + (3)3, so, the required

expression is 9x2 + 27x]

b) a3 + ….......….. + ….......….. + 1 c) x3 + ….......….. + ….......….. + 8

d) a3 – ….......….. + ….......….. – 27 e) x3 – ….......….. + ….......….. – 64y3

11. Let’s apply the appropriate formula to find the products.

a) (a + 3) (a – 3) b) (2x + 1) (2x – 1) c) (4 + 3p) (4 – 3p)

d) (2x – 3y) (2x + 3y) e) (x2 – y2) (x2 + y2) f) (2x2 + 5y2) (2x2 – 5y2)

g) (a + b + c) (a + b – c) h) (x – y + z) (x + y + z) i) (p – q – r) (p + q – r )

j) (x + y) (x – y) (x2 + y2) k) (x + 2) (x – 2) (x2 + 4) l) (2a + y) (2a – y) (4a2 + y2)

12. Let’s find the products using the formula (a + b) (a – b) = a2 – b2.

Hint. 99 × 101 = (100 – 1 ) × (100 + 1) = 1002 – 12 = 10000 – 1 = 9999

a) 19 × 21 b) 49 × 51 c) 78 × 82 d) 102 × 98

13. Let’s find the products by using formula a3 + b3 = (a + b) (a2 – ab + b2) or

a3 – b3 = (a – b) (a2 + ab + b2)

a) (x + 2) (x2 – 2x + 4) b) (x + 3) (x2 – 3x + 9) c) (x – 1) (x2 + x + 1)

d) (y – 4) (y2 + 4y + 16) e) (2x + 3y) (4x2 – 6xy + 9y2) f) (3a – 5b) (9a2 + 15ab + 25b2)

14. a) If (x + y) = 5 and xy = 3, find the value of x2 + y2.
b)
c) If (a – b) = 4 and ab = 2, find the value of a2 + b2.

d) If x + 1 = 3, find the value of x2 + 1 .
x x2
15. a)
If p – 1 = 7, find the value of p2 + 1 .
b) p p2

c) If a + 1 = 3, find the values of (i) a2 + 1 (ii) a – 1 2
a a2 a
d)
16. a) If x + 1 = 5, find the values of (i) x2 + 1 (ii) x – 1 2
x x2 x
b)
c) If p – 1 = 4, find the values of (i) p2 + 1 (ii) p + 1 2
p p2 p

If m2 – 1 = 6, find the values of (i) m2 + 1 (ii) m + 1 2
m m2 m

If (x + y) = 6 and xy = 2, find the value of x3 + y3.

If (x – y) = 5 and xy = 4, find the value of x3 – y3.

If a + 1 = 4, find the value of a3 + 1 .
a a3

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Algebraic Expressions

d) If m – 1 = 3, find the value of m3 – 1 .
m m3

17. a) If a + b = 2, find the value of a3 + b3 + 6ab.

b) If x + y = 5, find the value of x3 + y3 + 15xy

c) If x – y = 3, find the value of x3 – y3 – 9xy

d) If a – b = 7, find the value of a3 – b3 – 21xy
18. Simplify.

a) (a + b)2 + (a – b)2 b) (x + y)2 – (x – y)2 c) (2p – 3)2 + (2p + 3)2

d) (3x + y)2 – (3x – y)2 e) (x + 1 )2+ ( x – 1 )2 f) (2 – 1 )2 – (2 + 1 )2
x x a a

19. Simplify.

a) 2.1 u 2.1 – 0.9 u 0.9 b) 3.6 u 3.6 – 1.4 u 1.4
2.1 – 0.9 3.6 + 1.4

c) 2.5 u 2.5 u 2.5 – 1.4 u 1.4 u 1.4 d) 2.7 u 2.7 u 2.7 + 1.8 u 1.8 u 1.8
(2.5)2 + 2.5 u 1.4 + (1.4)2 (2.7)2 – 2.7 u 1.8 + (1.8)2

It’s your time - Project work

Paper folding:

20. a) Let’s take a few rectangular sheet of papers (photocopy paper). Then, fold them
to get square sheet of papers as shown in the diagrams.

b) Again, let’s fold each square sheet of paper as shown in the diagram and complete

(i) the sums. x1

x x2 x.y 1. = (x + 1)2 = x2 + 1.x + 1.x + 12

1 1.x 12 = x2 + 2x + 1

(ii) xy

x = ................ = ..........................
y x.y = ..........................

(iii) x xx x1
x2
x x2 x x2–12 x–1 x2–12 = ....x.2..–...1..2.... = ..........................
x

12 1 x+1
1
(iv) x x x xy

x x2 x2 x x2–y2 x–1 x2–y2 = ................ = ..........................
x x+y

y2 y
y

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Factorisation, H.C.F. and L.C.M.

(vi) Factorisation of trinomial expressions of the form a2 + 2ab + b2:
Let’s find the product of (a + b) and (a + b) ab

(a + b) (a + b) = a (a + b) + b (a + b) a a2 ab
= a2 + ab + ab + b2

= a2 + 2ab + b2 b ab b2 b

b

Thus, (a + b)2 are the factors of a2 + 2ab + b2. Similarly, (a – b)2 are the factors of
a2 – 2ab + b2. So, the expression of the form a2 + 2ab + b2 is factorised by making it

a perfect square.

(vii) Factorisation of trinomial expressions of the form px2 + qx + r:

In the trinomial expressions of the form px2 + qx + r, p and q are the numerical
coefficients of x2 and x respectively and r is any constant term.

To factorise such expressions, we need to find the two factors a and b of the product
of p and r such that a + b = q. Then, the expression is expanded to four terms and
factorisation is performed by grouping.

Worked-out examples

Example 1: Factorise a2 + 6a + 9

Solution: a2 + 6a + 9 is expressed in the form a2 + 2ab + b2
a2 + 6a + 9= a2 + 2.a.3 + 32

= (a + 3)2

Example 2: Resolve into factors 4x2 – 20x + 25.

Solution:
4x2 – 20x + 25 = (2x)2 – 2.2x.5+52 4x2 – 20x + 25 is expressed in the form a2 – 2ab + b2

= (2x – 5)2
Example 3: Resolve into factors 2x2 + 11x + 5.

Solution: 2 × 5 = 10
2x2 + 11x + 5 = 2x2 + (10 + 1)x + 5
In 2x2 + 11x + 5
= 2x2 + 10x + x + 5
= 2x (x + 5) + 1 (x + 5) 10 + 1
= (x + 5) (2x + 1) p × r = 2 × 5 = 10
q = 11 = ( 10 + 1)

Example 4: Factorise 6x2 + x – 2 6 × 2 = 12
Solution:
6x2 + x – 2 = 6x2 + (4 – 3)x – 2 In 6x2 + x – 2

= 6x2 + 4x – 3x – 2 4–3
= 2x (3x + 2) – 1 (3x + 2) p × r = 6 × 2 = 12
= (3x + 2) (2x – 1) q = 1 = ( 4 – 3)

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Unit Geometry: Angle

14

14.1 Different pairs of angles – Looking back

Classroom - Exercise

1. Let's say and write the pairs of adjacent angles, liner pairs, vertically opposite
angles, complementary angles, and supplementary angles from the following
figures.

a

db

x mn cp
y q

Adjacent Linear pairs Vertically Complementary Supplemen-
angles opposite angle
........and........ ........and........ angles tary angles
........and........ ........and........
........and........ ........and........ ........and........ ........and........
........and........ ........and........
........and........ ........and........
........and........
........and........
........and........
........and........
........and........
........and........
........and........

........and........

2. Let's say and write the pairs of alternate angles, corresponding angles and
co-interior angles.

Alternate angles Corresponding angles Co-interior angles ab
dc
........and........ ........and........ ........and........
ef
........and........ ........and........ ........and........ hg

........and........

........and........

(i) Adjacent angles C C
B
‘AOB and ‘BOC are a pair of adjacent B
angles. They have a common vertex O and a O AO A
common arm OB.

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Geometry: Angle

Linear pair C B
OA
If the sum of a pair of adjacent angles is 180q, they are
said to be linear pair.
In the figure alongside, ‘AOB + ‘BOC = 180q.

So, ‘AOB and‘BOC are the linear pair.

(ii) Vertically opposite angles A O C
D B
In the adjoining figure, ‘AOC and ‘BOD are vertically C B
opposite angles formed by intersected line segments. They 90° A
have a common vertex and they are lying to the opposite
side of the common vertex. ‘AOD and ‘BOC are another O A
pair of vertically opposite angles. B

Vertically opposite angles are always equal. O

‘AOC = ‘BOD and ‘AOD = ‘BOC.

(iii) Complementary angles

A pair of angles are said to be complementary if their sum is a
right angle (90q). In the figure, ‘AOB and ‘BOC are a pair of
complementary angles.

? ‘AOB + ‘BOC = 90q

Also, complement of ‘AOB = 90q – ‘BOC.
Complement of ‘BOC = 90q – ‘AOB.

(iv) Supplementary angles

A pair of angles are said to be supplementary if their
sum is two right angles (180q). In the figure, ‘AOB and
‘BOC are a pair of supplementary angles.

? ‘AOB + ‘BOC = 180q. C
Also, the supplement of ‘AOB = 180q – ‘BOC
The supplement of ‘BOC = 180q – ‘AOB.

14.2 Experimental verifications of pair of angles formed by two
intersecting lines

Experiment 1: The sum of a pair of adjacent angles formed by two intersecting lines
is 180°.

Step 1: Draw three pairs of intersecting lines AB and CD intersecting at O.

A A D A D
CO O O
D

B CB C B
fig (i) fig (ii) fig (iii)

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Geometry: Angle

Step 2: Measure ‘AOC and ‘AOD (or ‘AOC and ‘BOC or ‘BOC and ‘BOD or ‘AOD

and ‘BOD) with the help of protractor and write the measurements in the table.

Fig. No. ‘AOC ‘AOD ‘AOC + ‘AOD Result
(i) ‘AOC + ‘AOD = 180°
(ii)
(iii)

Conclusion: The sum of pair of adjacent angles formed by two intersecting lines
is 180°.

[Note: The pair of adjacent angles whose sum is 180° are said to be the linear pair.]

Experiment 2: Each pair of vertically opposite angles formed by two intersecting lines
are equal.

Step 1: Draw three pairs of intersecting lines AB and CD intersecting at O.

A A D A D
CO O O
D

B CB C B
fig (i) fig (ii) fig (iii)

Step 2: Measure ‘AOC and ‘BOD, ‘AOD and ‘BOC with the help of protractor. Write

the measurements in the table.

Fig. No. ‘AOC ‘BOD ‘AOD ‘BOC Result
(i) ‘AOC = ‘BOD
(ii)
(iii) ‘AOD = ‘BOC

Conclusion: Each pair of vertically opposite angles formed by two intersecting lines
are equal.

Worked-out examples

Example 1: If x° and (x + 6)° are a pair of complementary angles, find them.

Solution:

Here, xq + (x + 6)q = 90q [The sum of a pair of complementary angles]

or, 2xq = 90q – 6q
8=24°42=q
or, = 42q and xq = 42q = 48q.
? xq (x + 6)q + 6q

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Geometry: Angle

Example 2: If a pair of supplementary angles are in the ratio 4:5, find them.
Solution:
Let, the required supplementary angles be 4xq and 5xq.

Now, 4xq + 5xq = 180q [The sum of a pair of supplementary angles]

or, 9xq = 180q
or, xq = 1890°= 20q
? 4xq = 4 u 20q = 80q and 5xq = 5 u 20q = 100q.

Example 3 : Find the unknown sizes of angles.

a) B b) c)
D C

3x 3x x+80°
2x 5x°

Ox 4x+50°

A x+30° x B
AO

Solution:

a) x + 2x + 3x = ‘AOB b) x + 3x + (x + 30°) = ‘BOA

or, 6x = 90° or, 5x + 30° = 180°

or, x = 15° or, x = 30°

? x = 15°, ? x = 30°

2x = 2 × 15° = 30° 3x = 3 × 30° = 90°

3x = 3 × 15° = 45° x + 30° = 30° + 30° = 60°

c) (x + 80°) + 5x° + (4x + 50°) = 360°

or, 10x + 130° = 360°

or, x = 23°

? x + 80° = 23° + 80° = 103°

5x° = 5 × 23 = 115°

4x + 50° = 4 × 23° + 50° = 142°

EXERCISE 14.1

General Section- Classwork

1. Let's say and write the answers as quickly as possible. = .....................
a) If x° and 100° are a linear pair, then x° = .....................
b) If y° and 50° are vertically opposite angles, then y° = .....................
c) If a° and 30° are a pair of complementary angles, then a° = .....................
d) If p° and 120° are a pair of supplementary angles, then p°

Creative Section - A

2. a) If xq and 50q form a linear pair, find xq.

b) If 2xq and 3xq are adjacent angles in linear pair, find them.

c) If yq and 48q are a pair of complementary angles, find yq.

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Geometry: Angle

d) If 2pq and (p + 15)q are a pair of complementary angles, find them.
x°
e) If xq and 4 are a pair of supplementary angles find them.

f) If pair of complementary angles are in the ratio 4:11, find them.

g) If a pair of supplementary angles are in the ratio 7:5, find them.

3. Find the value of x in the each of figures given below.

a) b) c) d)

x° 58° x° 4x° 100° 3x° 2x°
125°

e) f) g) h)
5x° 4x° 3x°+50° 2x°
x° ) x )° ) x4 )°
2
x°

4. Find the unknown sizes of angles. c) d) 92°
4x° x° z° y°
a) b) x°
3x° 2x°
(x+20)° 2x° x° 2x° 3x°
x°

e) f) g) h) q° x°

x° 3x° 2x° x+80° 3q° x+25°
z° y° 2x–30° 5x° p°

4x+50°
x+35°
2x+30°

5. a) In the given figure, calculate the values of angles p and q, qp
where ‘p = 21‘q.

b) In the figure, given alongside, if x = 3y, find the sizes of x
angles represented by a and b. ay

b

Creative Section -B x c
ab
6. a) In the adjoining figure, if ‘a + ‘b + ‘c = 180q,
prove that ‘x = ‘b + ‘c. az b
xy
b) In the figure given alongside, ‘a = ‘x and ‘b = ‘y.
Show that ‘x + ‘y + ‘z = 180q.

Vedanta Excel in Mathematics - Book 8 186 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Geometry: Angle

7. a) Draw a line segment AB. Mark a point O on AB and draw an angle BOC. Measure
‘BOC and ‘AOC. Verify that ‘BOC + ‘AOC = 180°.

b) Draw two intersecting line segment AB and CD intersecting at O. Measure the
size of each pair of vertically opposite angles. Verify that each pair of vertically
opposite angles are equal.

14.3 Different pairs of angles made by a transversal with two
straight lines

In the adjoining figure, straight line segment EF intersects two E B
line segments AB and CD at the point G and H respectively. A 1G 2 D
Here, EF is called a transversal. 43

‘1, ‘2, ‘3, ‘4, ‘5, ‘6, ‘7 and ‘8 are the angles made by 5H 6
87
the transversal with two line segments. C F

Here, ‘1, ‘2, ‘7 and ‘8 are the exterior angles.

‘3, ‘4, ‘5 and ‘6 are the interior angles.

Also, ‘4 and ‘6, ‘3 and ‘5 are two pairs of alternate angles.

‘1 and ‘5, ‘2 and ‘6, ‘4 and ‘8, ‘3 and ‘7 are four pairs of corresponding angles.
‘4 and ‘5, ‘3 and ‘6 are two pairs of co-interior angles.

14.4 Relation between pairs of angles made by a transversal

with parallel lines

Two line segments are said to be parallel if they do not P RD
intersect each other when they are extended to either C
directions.

In the figure, AB and CD are two parallel lines. It is written A Q SB
as AB//CD.

The perpendicular distance between two parallel lines is always equal.

? PQ = RS.

Let's investigate the relation between the pair of alternate angles, pair of corresponding
angles and the pair of co-interior angles formed due to the intersection of two parallel
lines by a transversal.

Experiment 3: Each pair of alternate angles formed by a transversal with two parallel
lines are equal.

Step 1: Draw three pairs of parallel lines AB and CD intersected by a transversal EF at
G and H.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 187 Vedanta Excel in Mathematics - Book 8