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Full text of "Mathematics (Std08 - English Medium)"See other formatsGovernment of Tamil Nadu
MATHEMATICS
VIII standard
Untouchability
Inhuman - Crime
Department of School Education
A publication under
Government of Tamllnadu
Distribution of Free Textbook Programme
(NOT FOR SALE)
/ ©
Government of Tamil Nadu
First Edition - 2011
(This Book published under Uniform System of School Education scheme)
CHAIRPERSON
Dr. K. Ravi
Associate Professor
Department of Mathematics
Sacred Heart College
Tinipattur (Vellore Dt.) 635 601 .
REVIEWERS
Dr. K. Vasudevan
Associate Professor
Department of Mathematics
Presidency College (Autonomous)
Chennai 600 005.
Dr. A. Singaravelu
Principal
Zigma Matric. Hr. Sec. School
Medavakkam
Chennai 600 100.
AUTHORS
S.P. Karthlkeyan
PG, Assistant
Gandhi Memorial Hr. Sec. School
Thinjvennatnallur, Villupuram Dt. 607 203.
N. ShanthI
Trained Graduate Teacher
Railway Mixed Hr Sec. School
Perambur, Chennai 600 011.
T.G. Narayanasami
B.T. Assistant
Govt. Boys Hr Sec. School
Pennadam, Cuddalore Dt. 606 105.
V. Mettlda Sornam Titus
B.T. Assistant
Doveton Corrie (G) Hr. Sec. School
Vepery, Chennai 600 007.
H. Lakshmi
Trained Graduate Teacher
Akshaya Matric. Hr. Sec. School
Velachery, Chennai 600 042.
D. Jane Clara
Post Graduate Teacher
Zion Matric. Hr. Sec. School
Tambaram East, Chennai 600 073.
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Preface
Mathematics is one of tiie most useful and fascinating subjects, finds many
applications in various sciences and it touches every major field of human
endeavour. The term mathematics is coined by the great mathematician
Pythagoras and it means "inclined to learn". The beauty of mathematics can be
realized through the most beautiful structures in the world. The nature behaves
on some mathematical order. Many objects in this world move and function on
the basis of some mathematical equations.
The subject mathematics is so wide, that it needs special care and
concentration to learn this subject. It will make man to think higher and higher
and it will force him to ask the questions "Why" and "How". It is a pleasant task to
write a book on mathematics for young students.
This book is prepared based on the principle of implementing the uniform
system of school education, in aU types of schools in Tamilnadu. This is a
common text book for aU the children studying in Government schools.
Matriculation schools, Anglo-Indian schools and Oriental schools. The content in
this book is so designed to meet the requirements of aU the above categories of
children studying in various schools.
A lot of steps and strain have been taken to give balanced curriculum at
this stage. The worked out examples, problems in the exercises, the proofs for the
theorems, the method of solving problems, simplicity and the style of presenting
the contents are carefully designed to attract various tj^e of children living in
various situations and places. More information regarding the subject matter and
{Hi)
the historical facts about the contents are given to motivate the students to learn
more and more about the various concepts available in different topics.
The pictures, figures and graphs are included in large number to make the
student understand well about the subject. The concepts like Activity, Try these,
Think it, Do you know are included in the text to motivate and create interest among
the students.
We, the authors feel that eighth standard stage of a child is a very important
stage in which the child tries to understand many things. So, we request every child
to go through each and every topic and understand well and make an attempt at all
the exercise problems. "The only way to learn mathematics is to do mathematics"
said Paul Halmos. So cultivate a practice of doing problems daily, repeatedly and
often remember the important points.
We hope teachers wiU give due importance to explain the concepts and
illustrate the worked out examples and build the ability of solving problems among
the students. Then we can achieve the dreams of great mathematicians that any one
who wants to learn mathematics has to first love it.
We welcome suggestions and constructive criticisms from learned teachers,
dear students and well wishers as there is always a room for further improvements.
On behalf of our team, I would like to place on record my sincere thanks to
the respective authorities for selecting and deputing us to do this noble work.
Dr. K. Ravi
Chairperson
(iv)
Contents
(^
No.
Name of the Chapter
Page No.
L
Real Number System
I
2.
Algebra
55
3.
Life Mathematics
94
4.
Measurements
139
5.
Geometry
160
6.
Practical Geometry
192
7.
Graphs
232
V-
8.
Data Handling
247
(v)
(vi)
Real Number System
1.1 Introduction
1.2 Revision : Representation of Rational
Numbers on the Number Line
1.3 Four Properties of Rational Numbers
1 .4 Simplification of Expressions Involving
Three Brackets
1.5 Powers: Expressing the Numbers in
Exponential Form with Integers as
Exponent
1 .6 Laws of Exponents with Integral Powers
1.7 Squares, Square roots,Cubes, Cube roots
1.8 Approximations of Numbers
1 .9 Playing with Numbers
1.1 Introduction
Number theory as a fundamental body of knowledge has played a
pivotal role in the development of Mathematics. The Greek Mathematician
Pythagoras and his disciples believed that "everything is number" and
that the central explanation of the universe lay in numbers.
The system of writing numerals was developed some 10,000 years
ago. India was the main centre for the development of the number
system which we use today. It took about 5000 years for the complete
development of the number system.
Integral numbers are fountain head of all Mathematics. The present
system of writing numerals is known as Hindu-Arabic numeral system.
In this system, we use the numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. It is
also called the decimal system with base 10. The word 'decimal' comes
from Latin word 'Decern' which means 'Ten'.
Mathematics is the 'Queen of Science' and
Number theory is the 'Queen of Mathematics'.
Paul Erdos
[26 March, 1913 -
20 September, 1996]
He was a
great prolific and
notable Hungarian
mathematician.
Erdos published
more papers
than any other
mathematician in
history, working
with hundreds
of collaborators
in many fields
including number
theory.
His fascination
with mathematics
developed early at
the age of three.
He could calculate
how many seconds
a person had
lived. His life was
documented in
the film "JV is a
Number: A Portrait |
of Paul Erdos",
while he was still
alive.
Erdos said, "I
know numbers
are beautiful.
If they aren't
beautiful, nothing
is."
In class VII, we have learnt about Natural numbers N = {1, 2, 3, ■• ■ }, Whole
numbers W = {0, 1, 2, • • • }, Integers Z = {• • • , - 2, - 1, 0, 1, 2 • •■ } and Rational numbers
Q and also the four fundamental operations on them.
1.2 Revision : Representation of Rational Numbers on the Number Line
Rational numbers
The numbers of the form — where p and q are integers and o ^ are known
q
as rational numbers. The collection of numbers of the form — , where q > is denoted
by Q. Rational numbers include natural numbers, whole numbers, integers and all
negative and positive fractions. Here we can visualize how the girl collected all the
rational numbers in a bag.
Rational numbers can also be represented on the number line and here we can
see a picture of a girl walking on the number line.
H — I— I — I I I I I I — h
-3 -2
1
iQii 1
3^4 2 ^
To express rational numbers appropriately on the number line, divide each unit
length into as many number of equal parts as the denominator of the rational number
and then mark the given number on the number line.
Illustration:
(i) Express ^ on the number line.
y lies between and 1.
< \ \ \ \
H \ h
_0.
7
J_
7
2_
1
(ii)
5 5
It lies between 3 and 4.
7
±
1
5_
1
7
J_
7
H h
j_
5
H h
4
2_
5
5
5
Real Number System
(iii)
_2_
3
It lies between - 1 and 0.
-1
H h
Every natural number
.^f is a rational number.
^ M / Is the converse true?
1.3 Four Properties of Rational Numbers
1.3.1 (a) Addition
(i) Closure property
The sum of any two rational numbers is always a rational number. This is called
'Closure property of addition' of rational numbers. Thus, Q is closed under addition.
If ^ and ^ are any two rational numbers, then ^ + ^ is also a rational number.
b d b d
Illustration: (i) ^ + ^ = ^ = ^ is a rational number.
(ii) 5 + ^ = Y + ^= ^^j" = ^ = 5^ is a rational number.
(ii) Commutative property
Addition of two rational numbers is commutative.
If J- and ^ are any two rational numbers, then j- + ^ = ^ + ^^
d
d d
1 2
Illustration: For two rational numbers y , ^ we have
1,2, ^ 2_ + 1
2 5 5 2
LHS =i + |-
^ 5 + 4 ^ 9
10 10
.-. LHS = RHS
.•. Commutative property is true for addition.
(iii) Associative property
Addition of rational numbers is associative.
RHS=f + l
_ 4 + 5 _
9
10
10
If -f , -^ and ^ are any three rational numbers, then ^ + (^ + -^) = (^ + ^) + -^^
b d f -^ b \d f I \b d I f
Illustration: For three rational numbers
2
3
1
' 2
and 2, we have
f-(^^) =
= (f
+
i)
+ 2
LHS =f + (^ + 2)
RHS
= (l4) + 2
=hihT)
=(M)-2
3 \2 11 3 2
=l-^=l-f
4 + 15 19 ol
6 6 6
7+12 19 ol
6 6 6
.-. LHS = RHS
.-. Associative property is true for addition.
(iv) Additive identity
The sum of any rational number and zero is the rational number itself.
If -^ is any rational number, then -^ + = -^ =
b ■^ b b
Zero is the additive identity for rational numbers.
Illustration: (i) ^ + = ^=0 + ^
<"> (Tf) + '' = Tf = » + (Tf
(v) Additive inverse
Do you know?
Zero is a special
rational number. It
can be written as
= ^ where q ^ 0.
-^) is the negative or additive inverse of ~.
b I b
If -^ is a rational number, then there exists a rational number (^^)
such that ^ + (^) = 0.
o o
Illustration: (i) Additive inverse of ^ is ^^
o o
(ii) Additive inverse of — ^^ is ^
(iii) Additive inverse of is itself.
Numbers
Addition
Closure
property
Commutative
property
Associative
property
Natural numbers
Whole numbers
Yes
Integers
Rational numbers
Yes
Real Number System
1.3.1 (b) Subtraction
(i) Closure Property
The difference between any two rational numbers is always a rational number.
Hence Q is closed under subtraction.
If -^ and 4- are any two rational numbers, then -^ - -^ is also a rational number.
b d ■^ b d
Illustration: (i)
2 i
111
is a rational number.
1 o 1 1
(ii) 1 - y = = y is a rational number.
(11) Commutative Property
Subtraction of two rational numbers is not commutative.
If ^ and ^ are any two rational numbers, then ^ — ^i^^ — ^^
d
A 9
Illustration: For two rational numbers ^ and ^, we have
d d
9
1^1
5 "^ 5
LHS
9 5
20- 18
45
_2^
45
9
RHS =
2__A
5 9
18-20
45
-2
45
(Do you know?
When two
rational numbers
are equal, then
commutative
property is true
for them.
.-. LHS ^ RHS
.•. Commutative property is not true for subtraction.
(Ill) Associative property
Subtraction of rational numbers is not associative.
If -y- , -^ and -^ are any three rational numbers, then ^ — (-^--^]^(^--^
b d J b \d J ) \ b a
Illustration: For three rational numbers ^, ^ and -j, we have
1_/1_1W/1_U 1
2 13 Ar\2 3
LHS
l_/i_n
2 13 a)
2 \ n I
^-(^
2 \\2
1
RHS
1_1
2 3
1
4
2 V12/ 12
.•. Associative property is not true for subtraction.
12
12
i 1^ {S^ede
Numbers
Subtraction
Closure
property
Commutative
property
Associative
property
Natural numbers
No
Whole numbers
Integers
Rational numbers
No
1.3.1 (c) Multiplication
(i) Closure property
The product of two rational numbers is always a rational number. Hence Q is
closed under multiplication.
If -^ and -^ are any two rational numbers, then -^ x -7- = -r^ is also a rational number.
b d -^ b d bd
Illustration:
1 7 1
(i) ^ X 7 = ^ = 2^ is a rational number.
(ii) 4^ X ^ = ^ is a rational number.
(ii) Commutative property
Multiplication of rational numbers is commutative.
If V- and -^ are any two rational numbers, then -y- x -^ = -^ X v-.
b d ■^ b d d b
T Q
Illustration: For two rational numbers ^ and — p, we have
^x
m - n
:8
1
3 ../-8
V 1
24
8
Xi^
LHS-fx(^)
55
RHS
Tfx(y)
-24
55
.-. LHS = RHS
.-. Commutative property is true for multiplication.
(iii) Associative property
Multiplication of rational numbers is associative.
If ^ , -^ and -J are any three rational numbers, then t^ X (^ X 7^) = (^ X 77) X 7^
Real Number System
Illustration: For three rational numbers ^, (— p) and ^, we have
-X
LHS = lx(-^)
1 ^^/^x^^
4 ^3)
l\ - -1
24
.LHS
(yx(^))4
RHS = (^)x
RHS
1 \ ., 1 _ ^
24
8 r^3
.-. Associative property is true for multiplication.
(iv) Multiplicative identity
The product of any rational number and 1 is the rational number itself. 'One' is
the multiplicative identity for rational numbers.
If -7^ is any rational number, then -^x 1 = -y- = 1 X -y-.
b -^ b b b
Illustration:
(i) yxl=y
^^y Is 1 the multiplicative
■ m /f identity for integers?
(v) Multiplication by
Every rational number multiplied with gives 0.
If -7^ is any rational number, then |'^xO = = Ox^.
b -^ b b
Illustration: (i) -5x0 =
(ii) (^)xO =
(vi) Multiplicative Inverse or Reciprocal
For every rational number -^, a^O, there exists a rational number -^ such that
b a
-T^x -^ = 1 . Then -^ is called the multiplicative inverse of -y-.
baa b
If -|- is a rational number, then — is the multiplicative inverse or reciprocal of it.
Illustration: (i) The reciprocal of 2 is
1
(ii) The multiplicative inverse of (— f^) is (— ^)-
Do you know?
i) has no reciprocal,
ii) 1 and - 1 are the only rational numbers
which are their own reciprocals.
#
'0k Is 0.3 the
W* reciprocal of
3^?
|t^ t^e^e-
Numbers
Multiplication
Closure
property
Commutative
property
Associative
property
Natural numbers
Whole numbers
Yes
Integers
Yes
Rational numbers
1.3.1 (d) Division
(i) Closure property
The collection of non-zero rational numbers is closed under division.
If -^ and ^ are two rational numbers, such that ^ 7^ 0, then ^h- ^ is
ha aba
Illustration:
always a rational number.
(i) Y^^ = YXj^ = y = 2isa rational number,
(ii) ^-^y = ^x^ = ^isa rational number.
(11) Commutative property
Division of rational numbers is not commutative.
If -^ and -^ are any two rational numbers, then -^ h- -^ 7^ -^ h- -^
b a b a a b
Illustration: For two rational numbers ^ and ^, we have
5 o
1^1 . l^A
5 8 ^ 8 5
LHS
5 3 15
R«S-fxf = l|
.-. LHS ^ RHS
.•. Commutative property is not true for division.
(ill) Associative property
Division of rational numbers is not associative.
If ^, ^ and -^ are any three rational numbers, then ^ h- (^ h- ^W (^ -h 4
b a J b \a J I \b a
O 1
Illustration: For three rational numbers ^, 5 and -^, we have
4 2
2^4
f
LHS
Real Number System
h'H
RHS = f.(5.i
-4)4
= f^(fxf)
20^1
= ^-10
4
3
10
- 3 y 1 - 3
4 10 40
.-. LHS ^ RHS
Associative property is not true for division.
\tu i^eae
Numbers
Division
Closure
property
Commutative
property
Associative
property
Natural numbers
No
W^hole numbers
Integers
Rational numbers
No
1.3.1 (e) Distributive Property
(i) Distributive property of multiplication over addition
Multiplication of rational numbers is distributive over addition.
a
If -^, -^ and -^ are any three rational numbers, then-f x(-^ + -^) = -^x4 + t^x
b a J b \d J I b a b
7 4 3
Illustration: For three rational numbers ^, ^ and ^, we have
3^19+5
LHS = fx(f + f
2 4 2 3
RHS = fxf + fxf
x
20 + 27
-^X
"3^45
45
47 94
135
•.LHS
_8_+ 2_
27 5
40 + 54
135
94
135
RHS
.•. Multiplication is distributive over addition.
(ii) Distributive property of multiplication over subtraction
Multiplication of rational numbers is distributive over subtraction.
If -^,-^and -^ are any three rational numbers, then -^xf-^--^') = -^x-S---T-x-^
b a J b \d J I b a b J
Illustration: For three rational numbers ^, ^ and ^, we have
fKf
LHS = lx(}-i
7
"7^10
5 2
8-5
10
_9_
70
.-. LHS
3^4
3^1
7^2
35 14
= 24-15 ^ _9_
70
70
RHS
.•. Multiplication is distributive over subtraction.
EXERCISE 1.1
1. Choose the correct answer:
i) The additive identity of rational numbers is
(A) (B) 1 (C)
ii) The additive inverse of —^ is
(A)
(B)
iii) The reciprocal of — ^ is
(A)
(B)
13
13 ' ' 5
iv) The multiplicative inverse of - 7 is
(C)
(C)
13
(D)2
(D)
(D)
13
(A) 7
(B)
V)
(A)0
has no reciprocal.
(B)l
(C)-7
(C)-l
(D)
(D)
Name the property under addition used in each of the following :
(iii)8 + i = i + 8
(iv)(=^) +
15
'H^)
(V)
+
i=f)-
=
3. Name the property under multiplication used in each of the following:
(i) fxA = Axf
(") {=r)
Xl
= IX
Real Number System
(m)(-^)x(^)=l (iv)ix(|xf) = (ix|)xf
Verify whether commutative property is satisfied for addition, subtraction,
multiplication and division of the following pairs of rational numbers.
(i) 4 and
(ii)
and
5 ^"' 4 7
Verify whether associative property is satisfied for addition, subtraction,
multiplication and division of the following pairs of rational numbers.
(i) |,f and-
(ii)
4 and 9
3' 5 10
Use distributive property of multiplication of rational numbers and simplify:
(i)
-5^/8 ,5
4 ^\9^ 1
(ii)fx(l
1.3.2 To find rational numbers between two rational numbers
Can you tell the natural numbers between 2 and 5?
11**111111 '
123456789 10
They are 3 and 4.
Can you tell the integers between - 2 and 4?
' I 1*****1 I '
-3-2-10 1 2 3 4 5
They are - 1, 0, 1, 2, 3.
Now, Can you find any integer between 1 and 2?
No.
But, between any two integers, we have rational numbers.For example,
1 o a
between and 1, we can find rational numbers j^, jtt, -ttt, ■ • • which can be written as
0.1,0.2, 0.3, ••-.
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
XAJ_AAA^A_2_ 1
10 10 10 10 10 10 10 10 10
113
Similarly, we know that the numbers -j, y, -j are lying between and 1. These
are rational numbers which can be written as 0.25, 0.5, 0.75 respectively.
0.25
0.5
0.75
J_
4
J_
2
_3_
4
1
9 4 9 4
Now, consider ^ and ^. Can you find any rational number between ^ and ^?
Yes. Tiiere is a rational number ^.
^-\ \ \ + \ \ >
1_
5
5
1 2 3
In the same manner, we know that the numbers ^, -^, ^ and ^ are lying between
and 1.
9 3
Can you find more rational numbers between ^ and ^?
9 90 3 30
Yes. We write ^ as ^ and ^ as ^, then we can find many rational numbers
between them.
2_20
21
22
23
24
25
26
27
28
29
30_3
5 50
50
50
50
50
50
50
50
50
50
50 5
We can find nine rational numbers
212223_2425_262128
50' 50' 50' 50' 50' 50' 50' 50
and
29
50'
22
99 93
If we want to find some more rational numbers between ^ and ^, we write
220
23 _ 230
''' 500 ^"^ 50 ^' 500
226 227 228 . 229
500' 500' 500 500'
Let us understand
this better with the help of
the number line shown in
the adjacent figure.
Observe the number
line between and 1 using
a magnifying lens.
. Then we get nine rational numbers
221 222 223 224 225
500' 500' 500' 500' 500
25 26 27 28 29 30^3
50 50 50 50 50 5
22 220 221_222223224225226227228229 230 23
50 500 500 500 500 500 500 500 500 500 500 500 50
Real Number System
Similar ily, we can observe many rational numbers in the intervals 1 to 2, 2 to
3 and so on.
If we proceed like this, we will continue to find more and more rational numbers
between any two rational numbers. This shows that there is high density of rational
numbers between any two rational numbers.
So, unlike natural numbers and integers, there are countless rational numbers
between any two given rational numbers.
To find rational numbers between two rational numbers
We can find rational numbers between any two rational numbers in two methods.
1. Formula method
Let 'a' and 'b' be any two given rational numbers. We can find many number of
rational numbers q^ q^, q^, ... in between a and b as follows:
q, = jia + b)
^2 = T ("^ + Ql)
2
_ 1
1 —
a
1
1
b
a
1
I2
-i — h-
b
1
(a + q ) and so on.
fli
The numbers q^, q^ lie to the left of q^. Similarly, q^, q^ are the rational numbers
between 'a' and V lie to the right of q^ as follows:
I * I 1
_ 1
^1
^ (q^ + b) and so on. a
^1
Average of two numbers always lie between that numbers.
2. Aliter
Let 'a' and 'b' be two rational numbers,
(i) Convert the denominator of both the fractions into the same denominator
by taking LCM. Now, if there is a number between numerators there is a
rational number between them,
(ii) If there is no number between their numerators, then multiply their
numerators and denominators by 10 to get rational numbers between them.
To get more rational numbers, multiply by 100, 1000 and so on.
u5»(Do you know?.
By following different methods one can get different
rational numbers between 'a' and 'b'.
Chapter 1
Example 1.1
Find a rational number between ^ and ^.
4 5
So/ution
Formula method:
Given: a=^,h = ^
4 5
Let Q, be the rational number between ^ and ^
^1 4 5
Qi = ^{a + b)
- 1/3 , 4\_ 1/15 + 16\
2U 5/ 21 20 1
^ _ 1 yl^\\_ 31
"^1 2>^l20i~40
The rational number is 47r-
40
Aliter:
Given: a = ^, t> = |-
We can write a and b as 4 X ^ = ^ and 4^ X ^ = ^
To find a rational number between ^ and ^, we have to
multiply the
numerator and denominator by 10.
15 ,, 10 150 16 ,, 10 160
20 '^ 10 200 ' 20 '^ 10 200
.-. The rational numbers between ]^r and ^^ are
151 152 153 154 155 156 157 158 ^ 159
200 ' 200 ' 200 ' 200 ' 200 ' 200 ' 200 ' 200 200 "
Example 1.2
Find two rational numbers between ^^ and ^ .
Solution
Given: a = =^,b = ^
Let q^ and q2be two rational numbers.
Qi = y(« + b)
^1 2^( 5 ' 2)" 2'^( 10 )~ 2^^(10) 20
q,= l(a + q,) = =lx(-/ +(-!))
_ 1 ../-12 + (-l)\_ 1 , /-12-Ix_ 1 ../-13\_
- 2^\ 20 j~ 2>H 20 J- 2>H 20 i~
-13
40
The two rational numbers are -^ and |, .
Note: The two rational numbers can be inserted as ^^ < ~}^ <
5 40
-1 <1
20 2
— ^m
Real Number System
EXERCISE 1.2
1 . Find one rational number between the following pairs of rational numbers.
(i) 4 and ^ (ii) ^ and J (iii) tt ^d
2. Find two rational numbers between
2
8
(iv) I and I
(i) ^ and 1
(ii)fandA
(iii) y and ^
(iv)
^andi
Find three rational numbers between
(in)
(i) |and 1
(ii) j^ and I
1 and 3
3 2
1.4 Simplification of Expressions Involving Three Brackets
Let us see some examples:
(i) 2 + 3 = 5 (ii) 5- 10 = -5
(iv)landX
2X^
(iii) fx^ = ^ (iv)
In examples, (i), (ii) and (iii), there is only one operation. But in example (iv) we
have two operations.
Do you know which operation has to be done first in problem (iv)?
In example (iv), if we do not follow some conventions, we will get different
solutions.
For example (i) (4
2)x^ = 2x^= 1
1
(ii) 4 — /2x^] = 4— 1 =3, we get different values.
So, to avoid confusion, certain conventions regarding the order of operations
are followed. The operations are performed sequentially from left to right in the order
of 'BODMAS'.
B - brackets, O - of, D - division, M - multiplication, A - addition, S - subtraction.
Now we will study more about brackets and operation - of
Brackets
Some grouping symbols are employed to indicate a preference in the order of
operations. Most commonly used grouping symbols are given below.
Grouping symbols
Names
Bar bracket or Vinculum
Parentheses or common brackets
{}
Braces or Curly brackets
[]
Brackets or Square brackets
Operation - 'Of"
We sometimes come across expressions like 'twice of 3', 'one - fourth of 20',
'half of 10' etc. In these expressions, 'of means 'multiplication with'.
For example,
(i) 'twice of 3' is written as 2 x 3,
(ii) 'one - fourth' of 20 is written as ^ X 20,
(iii) 'half of 10' is written as ^x 10.
If more than one grouping symbols are used, we first perform the operations
within the innermost symbol and remove it. Next we proceed to the operations within
the next innermost symbols and so on.
Example 1.3
Simplify: (l^
Solution
3.f)x^
15
H
3 ^ 3ri5
= (^ + ^)x 8
3
3/ 15
(^ X -j^) [ bracket is given preference ]
9 V-?- -16-1 1
^><15 15 ^5
Example 1.4
Simplify: 5^ + 1 of |
Solution
5^ + 4- of I-
2 +4>^9
1L^2A = ii . 2_
2 36 2 3
33 + 4 = 37 = gl
[ 'of is given preference ]
Example 1.5
Simplify: ( ^
Solution
1 ^5
-[h(\-\)
-{
3^xf)-[f-(^
) [Innermost bracket is given preference]!
x-) + [i^^
1
>[
x4 +^X4
3 4J L5
_ -25 + 144 _ 119 _
= -5 . 12
12 5
60
60
59
60'
Real Number System
Example 1.6
Simplify: f-{(l-f)-f}
Solution
2_f/l_2x_A-. = 2
7 lU ■ 3^ 6 J 7
-{(i
4)-f}
_ 2
7
f3
18
51_ 2 r9-2C
6 J 7 I 24
1
_ 2
r-
111 _ 2 , 11
7
l 24 / 7 ^ 24
_ 48 + 77
_ 125
1.
168
168
EXERCISE 1.3
Choose the correct answer:
(i)
2x4-
(qM
(D)f
3
(A) ^ (B) 2|
(ii)
2x4=
5 7
(A) 1* (B) 38^
(Of
(D,f
(iii)
(A) IO3 (B) ^^
(C)38
(D) ^3
(iv)
52'^
(A) ,2^ (B) 1
(C) '«
^"' 10
(V)
('-i)-(f-i)
(A) (B) 1
<c)i
(D)|
2.
Simplify:
« i^-d-f )
(ii)
(2^,'o)-('2 +
f)
("') fofd-D^if
(iv)
f^ofd-f)
w f-lyoflf-y]-'}
(vi)
(viii)
(l|x3l)-(4|-
-4)
(vii)(l+23ofl7)^l
1
6
(^)
-{l^(fx-:
5-i--i
2 4
Chapter 1
1.5 Powers: Expressing the Numbers in Exponential Form with Integers
as Exponent
In this section, we are going to study how to express the numbers in exponential
form.
We can express 2x2x2x2= 2"^, where 2 is the base and 4 is the index or
power.
In general, a" is the product of ' a ' with itself n times, where ' a ' is any real
number and 'n' is any positive integer .' a ' is called the base and 'n' is called the index
or power.
Definition
If 'n' is a positive integer, then x" means x.x.x x
n factors
i.e, x" = xxxxxx XX ( where 'n' is greater than 1)
n times
Note : x' = X.
How to read?
7^ is read as 7 raised to the power 3 (or) 7 cube. f
2
Here 7 is called the base, 3 is known as exponent
(or) power (or) index. '
To illustrate this more clearly, let us look at the following table
Index
Or Exponent
Or Power
w
ower or
Donent «
^^^^^Kd multipuc^^^^^^H
1
2X2X2X2
r
2
4
2
(-4)x(-4)x(-4)
(-4)3
-4
3
3
(|)x(|)x(f)x(f)x(f)x(f)
(ff
2
3
6
4
aXaXaX ... m times
a'"
a
m
Example 1.7
Write the following numbers in powers of 2.
(i) 2 (ii) 8 (iii) 32 (iv) 128
Solution:
(i) 2 = 2'
(ii) 8 = 2x2x2 = 2'
(v) 256
Real Number System
(iii) 32 = 2x2x2x2x2^ 2'
(iv) 128 - 2 X 2 X 2 X 2 X 2 X 2 X 2 - 2'
(v) 256 = 2X2X2X2X2X2X2X2 = 2'
1.6 Laws of Exponents with Integral Powers
With the above definition of positive integral power of a real number, we now
establish the following properties called "laws of indices" or "laws of exponents".
(i) Product Rule
Law 1
a'" X a" = a'"*", where 'a' is a real number and m, n are positive integers
Illustration
(f y ^ (f y " (f r' ^ (f y ^^'^"^ ^^^ ^'''^' ^ " ^ ^" ^ ^"' "' ^^^^^^ « = y, ^ = s, n = 4)
(ii) Quotient Rule
Law 2
^^ = fl'" ", where a / and m, n are positive integers with m> n
a"
Illustration
e
= e-' = e
m
(Using the law^ = a ", where a = 6,m=4,n=2)
(iii) Power Rule
Law 3
(«'")" = a"""', where m and n are positive integers
= 3'
Illustration
(32)4=3^X3^X3^X3^ = 3^^^
we can get the same result by multiplying the two powers ^:Mm*
i.e, (3^)' = 3^^^ = 3\
(iv) Number with zero exponent
Show that a'"'- x (f'''' X a'' = 1
For m i^ o,
m" ^m' = m^'^ = m° (using law 2);
Aliter:
m'-m'- '^
mXmXm
m mXmXm
= 1
Using these two methods, nv' ^ nv' = m° = 1.
From the above example, we come to the fourth law of exponent
Law 4
If 'a' is a rational number other than "zero", then a° = \
Illustration
(i)2«=l (ii)(ff=l (iii) 25° =1 (iv)(-|)°=l (v)(-100)°
Chapter 1
(v) Law of Reciprocal
The value of a number with negative exponent is calculated by converting into
multiplicative inverse of the same number with positive exponent.
Illustration
1 1
^^ 4' 4X4X4X4 256
Hi) 5 ' = ^= i = ^L
^ ^ 5' 5x5x5 125
(iii) 10^ - 1 - 1 - 1
10^
10x10 100
1
Reciprocal of 3 is equal to ^
oO
J — oo-i — o-i
3-\
1 fi°
Similarly, reciprocal of 6^ = ^ = ^ = 6° ' = 6
6 6
Further, reciprocal of (^] is equal to —^ ^ ,
iy)
From the above examples, we come to the fifth law of exponent.
Law 5
If 'a' is a real number and 'm' is an integer, then a " =
1
(vi) Multiplying numbers with same exponents
Consider the simplifications.
(i)
(ii)
4^'X7^^= (4X4X4)X(7X7X7)=(4X7)X(4X7)X(4X7)
= (4X7)'
5-'x4-
= ^X^X^ X 4-x4-x4-
5 5 5 4 4 4
= (ixi)x(ixi)x(ixi) = (i)
5 'M/ \5 4
= 20-3 = (5x4)-3
(iii)
&-&
- /3
(fxf)x(ixi) = (|xl)x(
1
5 '^ 2
5^2
= [W'
In general, for any two integers a and b we have
a'xb' = {axby = {aby
.•. We arrive at the power of a product rule as follows:
{aXaXaX ....m times) x{bxbxbx m times) = abxabxabx m times ={ab)"'
(i.e.,) a'" X b'" = {abf
Law 6
a'" X b'" = (ab)'", where a, b are real numbers and m is an integer.
Real Number System
Illustration
(i) 3'X4' = (3x4)' = 12^
(ii) rx2' =(7X2/ = 14^ = 196
(vii) Power of a quotient rule
Consider the simplifications,
(i)
(ii)
4^2
.3
(ir -
3 3 9
3'
and
1 = 1 = 5i = /lY
/1a2 /3^\ 3^ bi
a
3^3
3:1
5
5x5
5 - C-2 V 1 - C-2 V ■3-2-
3X3 3'=^^'<F = ^^'<3"=i^
X3"
2
Hence (^Y can be written as ^
(!_)'" = (f Xf X f X....m times) = /'\''\'' "^ ^™^'
V/?/ \b b b I bxbxb:
/a_Y = aZ_
\ b I b'"
< X ....m times
Law 7
^) ~ ^' ^h^'"^ b ^ 0, a and b are real numbers, m is an integer
Illustration
a V
a
b'
Example 1.8
(ii) f^f = 51 = 121
'^ ^ 1 3 i 3' 27
256
Simplify: (i)2'x2' (11)10' -10* (Hi) (:^°)'
(iv) {rr
(V)
(vi) (2')^
(vii) (2x3)'
(viii) If 2P ^ 32, find the value of p.
Solution
(1) 2^x2^ = 2^ + ^ = 2^
(11) 10'- 10*^ =10'-'= 10^
(ill) {xj = {iy= 1 [-.■ a° = l]
(iv) (2^^)" = 8° = 1 [•.• a° = 1]
Chapter 1
(V) f3f- 3^ = 243
^^^ \2) 2' 32
(vi) {2J = r^' = 2'" = 1024
(vii) (2x3y = 6*= 1296
(or) (2 X 3y = 2^X3' = 16x81 = 1296 2
2
(viii) Given : 2? = 32 2
32
16
8
2P = 2^ 2
2
Therefore p = 5 (Here the base on both sides are equal.)
4
2
1
Example 1.9
Find the value of the following:
(i)3*x3- (ii)^ (iii)(|-)' (iv)10- (v)(^/
(vi)(^)\3(vii)f(f;f (viii)(f;x(f^(f)'
Solution
(i)
(ii)
3*X3-' = 34+(-3) = 34-3
1
3- = 3* = 81
(iv) 10
i6_
5' 25
1
1000
l\' - - 1
W (^) 32
(vi)
a/x3
= 1x3
= 3
[■■■({)"='
(vii)
(viii)
[(1)1" -
(l/x(i
(f)"-"
-)*^(l
, _ (1)
3'
+ 4
16
81
(1/ _ ,
' (I
)•
(f)' "
w (1)'"
' = (1
f=l
imple 1.10
Express 16
- as a power with base 4.
Solution
We know that 16 = 4'
.-.16^ =
{4T
Real Number System
= 4-
Example 1.11
Simplify
(i){2Tx{y)
Solution
(i)
(ii)
{3j
(2T'X(37 = 2'^^>'-^'x3
X-2)^ 0(2x2)
1
(ii)
Example 1.12
Solve
(i) 12^ = 144
Solution
= 2-^x3* = ^X3'
{2'f ^ ^^^2L =
3'^' 3'
2'
81
64
{3J
64
81
^mx&=if}
(i)
(ii)
Given 12^ = 144
12" = 122
.-. X =2 (•.• The base on both sides are equal)
(If ^(ir - (i[
( 8 ) ~ ( 8 ) ^ ■ ^^^ ^^^^ ^'^ ^°^^ sides are equal)
2x + x =
6
3x =
6
x =
1 = 2.
Example 1.13
c- yr {3')-'x{2')-'
^™P^^^y=(2rx3-x4-
So/ution
(3rx(2^)-^ _
3-^x2-'
(2*)-' X 3-^X4-'
2-** X 3-^x4-^
=
3-6+4^2-^+^x4^
=
3-' X 2^X4'
=
^1.X4X16=4X16
—
64 _7l
9 9'
Chapter 1
^^
^^^H
^^^H
1.
EXERCISE 1.4
Choose the correct answer for the foUowing:
(i)
a'" X a" is equal to
(A) a"' + a" (B) a™"
(C) «'"+" (D) a'""
(ii)
p" is equal to
(A) (B) 1
(C)-:
L (D)p
(iii)
In 10\ the exponent is
(A) 2 (B) 1
(C)10
(D) 100
(iv)
6" ^ is equal to
(A)6 (B)-l
(C)-
i (°)i
(V)
The multiplicative inverse of 2"^ is
(A) 2 (B) 4
(C)2^
(D)-4
(vi)
(-2)-'x(-2)Msequalto
(A)-2 (B)2
{O-i
S (D)6
(vii) (—2) ^ is equal to
(A) ^ (B) ^
(viii) (2° + 4 ') X 2' is equal to
(C)
(D)
(A) 2
(B)5
(C)4
(D)3
ix)
(^\ is equal to
(A) 3
(B)3^
(C)l
(D) 3-4
(X)
(- 1)^° is equal to
(A)-l
(B)50
(C) - 50
(D)l
2.
Simplify:
(i) (-iy^(-Af
(iv) (f)'x(|/x(i
(ii) 1 o3
2
5^^
(iii) (-3yx(|)
(V)(3-V3'°)X3- (vi) ^X3^^^,><3
(vii) /-" X /"' X /"" (viii) (4p)' x (2p)' Xp' (ix) 9^'^ -3x5°
1 x-l/2
81
(^) (1)
3. Find the value of
(i)(3'' + 4-')x2'
i^"' 3x82/3x40 + ^^^"^'"
1 V' , / 1 ^-'
(ii)(2-'x4-')^2-(iii) (^)" +(^)" +
(iv) (3- + 4-' + 5-')° (V) [[=f-y]
21 V3
(vi) 7-20-7-21.
4
Real Number System
4. Find the value of m for which
(i) 5'" - 5-' = 5' (ii) 4'" = 64 (iii) 8'"' = 1
(iv) {a'T = a' (v) (5'")' X (25/ X 125' = 1 (vi) 2m = (8)^ - i^Y'^
5. (a) If 2^ =16, find
(i) X (ii) 2t (iii) 2'=' (iv) 2'^' (v) /l^
(b)If 3' = 81, find
(i) X (ii) 3^^' (iii) y^' (iv) 3"' (v) 3''^
6. Prove that (i) ^x{^f = I, (ii) (^) .(^
1.7 Squares, Square roots, Cubes and Cube roots
1.7.1 Squares
m \m + n / „« \n + I j I \i + m
1
When a number is multiplied by itself we say that the number is squared. It is
denoted by a number raised to the power 2.
For example : (i) 3 x 3 = 3' = 9
(ii) 5x5 = 5^ = 25.
In example (ii) 5' is read as 5 to the power of 2 (or) 5 raised to the power 2 (or) 5
squared. 25 is known as the square of 5.
Similarly 49 and 81 are the squares of 7 and 9 respectively.
In this section, we are going to learn a few methods of squaring numbers.
Perfect Square
The numbers 1, 4, 9, 16, 25, • • • are called perfect squares or square numbers as
1 = 1\ 4 = 2\ 9 = 32, 16 = 4' and so on.
A number is called a perfect square if it is expressed as the square of a number.
Properties of Square Numbers
We observe the following properties through the patterns of square numbers.
1. In square numbers, the digits at the unit's place are always 0, 1, 4, 5, 6 or
9. The numbers having 2, 3, 7 or 8 at its units' place are not perfect square
numbers.
Chapter 1
2.
i)
Number Square
1
1
9
81
11
121
19
361
ii)
SjB^^
2
4
8
64
12
144
18
324
If a number has 1 or 9 in the unit's If a number has 2 or 8 in the unit's
place then its square ends in 1. place then its square ends in 4.
iii)
Number Square
3
9
7
49
13
169
17
289
iv)
Number Square
4
16
6
36
14
196
16
256
If a number has 3 or 7 in the unit's If a number has 4 or 6 in the unit's
place then its square ends in 9. place then its square ends in 6.
3. Consider the following square numbers:
10^ =
= 100
20' =
= 400
,30' =
= 900.
flOO' = 10000 1
200' = 40000
700' = 490000
Result
(i) When a number ends with '0' , its square ends with double zeros,
(ii) If a number ends with odd number of zeros then it is not a perfect square.
4. Consider the following:
(i)
100 =
_J
102
(Even number
of zeros)
(ii)
.-. 100 is a perfect square.
81,000 = 81 X 100 X 10
T
(Odd number
of zeros)
= 92 X 102 X 10 . 81,000 is not a perfect square.
Real Number System
5. Observe the following tables:
Square of even numbers
Square of odd numbers
Number ^|
2
4
4
16
6
36
8
64
10
100
^"
Number
I—
1
Square
^~
1
1
3
9
5
25
7
49
9
81
From the above table we infer that,
(i) Squares of even numbers are even,
(ii) Squares of odd numbers are odd.
Example 1.14
Find the perfect square numbers between
(i) 10 and 20 (ii) 50 and 60 (iii) 80 and 90.
Solution
(i) The perfect square number between 10 and 20 is 16.
(ii) There is no perfect square number between 50 and 60.
(iii) The perfect square number between 80 and 90 is 81.
Example 1.15
By observing the unit's digits, which of the numbers 3136, 867 and 4413 can
not be perfect squares?
Solution
Since 6 is in units place of 3136, there is a chance that it is a perfect square.
867 and 4413 are surely not perfect squares as 7 and 3 are the unit digit of these
numbers.
Example 1.16
Write down the unit digits of the squares of the following numbers:
(i) 24 (ii) 78 (iii) 35
Solution
(i) The square of 24 = 24 x 24. Here 4 is in the unit place.
Therefore, we have 4x4 = 16. .•. 6 is in the unit digit of square of 24.
Chapter 1
(ii) The square of 78 = 78 x 78. Here, 8 is in the unit place.
Therefore, we have 8x8 = 64. .-. 4 is in the unit digit of square of 78
(iii) The square of 35 = 35 x 35. Here, 5 is in the unit place.
Therefore, we have 5x5 = 25. .-. 5 is in the unit digit of square of 35.
Some interesting patterns of square numbers
Addition of consecutive odd numbers:
1 = 1 = 12
1 + 3 = 4 = 22
1 + 3 + 5 = 9 = 32
1 + 3 + 5 + 7 = 16 = 42
1 + 3 + 5 + 7 + 9 = 25 = 52
l + 3 + 5 + 7 + ---+n = n2 (sum of the first 'n' natura
The above figure illustrates this result.
To find the square of a rational number
odd numbers)
^ X — = — = Sq^^re of the numerator
b b b^ Square of the denominator
Illustration
'-3\.. /-3\_/-3V
J TpK^o you known-
_ (-3)x(-3) _ 9
(i) 452 = 2025 = (20+25)2
(ii) 552 = 3025 = (30 + 25)2
.". 45, 55 are Kaprekar
numbers
7X7
49
(ii)
5 = / 5_f = 25
gXg („
64'
EXERCISE 1.5
1. Just observe the unit digits and state which of the following are not perfect
squares.
(i)3136 (ii)3722 (iii) 9348 (iv) 2304 (v) 8343
2. Write down the unit digits of the following:
(i)782 (ii)272 (iii) 412 (i^)352 (^^ 432
3. Find the sum of the following numbers without actually adding the numbers,
(i) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15
(ii) 1 + 3 + 5 + 7
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13+15 + 17
Real Number System
4.
Express the following as a sum of consecutive odd numbers starting with 1
(i) 7' (ii) 92 (iii) 5' (iv) 11'
5.
Find the squares of the foUowing numbers
«8 ("'10 ("'>5 ('^'3
(V) 31
^^ 40
6.
Find the values of the following:
(i)(-3)^ (ii)(-7)^ (iii) (-0.3)^ (i^) ("f )'
(v)(-f)' (vi)(- 0.6)2
7.
Using the given pattern, find the missing numbers:
a)V + 2' + r- =3'' b)lP =121
2^ + 3^ + 6^ = 72 lOP = 10201
y + A' + XT = 132 10012 = 1002001
42 + 52+ =7V lOOOOP =1
2 1
5' + + 30' = 31'
6' + r + =
100000012 =
1.7.2 Square roots
Definition
When a number is multiplied by itself,
the product is called the square of that number.
The number itself is called the square root of
the product.
For example:
(i) 3 X 3 = 32 = 9
(ii) (-3)x(-3) = (-3)2=9
Here 3 and (- 3) are the square roots of 9.
The symbol used for square root is /~ .
.-. /9 = ±3 ( read as plus or minus 3 )
Considering only the positive root, we have /9 = 3
9 is the square of 3
3
3 is the square root of 9
Note: We write the square root of x as /x or x^ .
Hence, /4 = (4)2 and /lOO = (100)2
Chapter 1
In this unit, we shall take up only positive square root of a natural number.
Observe the following table:
Table 1
Single or double digit numeral has
single digit in its square root.
Perfect Square
Square Root
1
1
16
4
36
6
81
9
100
10
225
15
2025
45
7396
86
9801
99
10,000
100
14,641
121
2,97,025
545
9,98,001
999
10,00,000
1000
15,00,625
1225
7,89,96,544
8888
999,80,001
9999
3 or 4 digit numeral has 2 digits
in its square root.
5 or 6 digit numeral has 3 digits
in its square root.
7 or 8 digits numeral has 4 digits
in its square root.
From the table, we can also infer that
(i) If a perfect square has 'n' digits where n is even, its square root has y
digits.
(ii) If a perfect square has 'n' digits where n is odd, its square root has ^ 1"
digits.
To find a square root of a number, we have the following two methods.
(i) Factorization Method
(ii) Long Division Method
(i) Factorization Method
The square root of a perfect square number can be found by finding the prime
factors of the number and grouping them in pairs.
Example 1.17
Find the square root of 64
Solution
64 = 2X2X2X2X2X2 = 2' X 2- X 2'
Prime factorization
'64 = 72- X 2' X 2' = 2 X 2 X 2 = 8
^ = 8
2
2
2
2
2
2
64
32
16
8
4
2
Real Number System
Example 1.18
^^^^
Find the square root of 169
Prime factorization
Solution
13
169
169 =
13X13 = 13'
V '
13
13
yi69 =
yi3' = 13
1
Example 1.19
Find the square root of 12.25
Solution
Prime factorization
r 1 TTC
A2.25 =
/ 12.25x100
V 100
_ 5X'/
10
5
7
7
±Z.Z.lJ
225
49
yi225 _ ^5'
Aoo -J
xr
10'
7
i
yi2.25 =
35 -35
10 ''■^
Example 1.20
Find the square root of 5929
Solution
5929 = 7x7x11x11 =7'xll'
Prime factorization
5929
7
7
11
11
847
121
11
Prime factorization
2
2
2
5
5
200
100
50
25
5
75929 = y7'xll' = 7X11
.-.75929 = 77
Example 1.21
Find the least number by which 200 must be multiplied to
make it a perfect square.
Solution
200 = 2X2X2X5X5
'2' remains without a pair. 1
Hence, 200 must be multiplied by 2 to make it a perfect square.
Example 1.22 Prime factorization
Find the least number by which 384 must be divided to
make it a perfect square.
Solution
384 = 3x2x2x2x2x2x2x2
'3' and '2' remain without a pair.
Hence, 384 must be divided by 6 to make it a perfect
square.
3
2
2
2
2
2
2
2
384
128
64
32
16
8
4
2
Chapter 1
(ii) Long division method
In case of large numbers, factors can not be found easily. Hence we may use
another method, known as Long division method.
Using this method, we can also find square roots of decimal numbers. This
method is explained in the following worked examples.
Example 1.23
Find the square root of 529 using long division method.
Solution
Step 1 : We write 529 as 5 29 by grouping the numbers in pairs, starting
from the right end. (i.e. from the unit's place ).
Step 2 : Find the number whose square is less than (or equal to) 5.
Here it is 2.
5 29
Steps
Step 4
Steps
Step 6
Put '2' on the top, and also write 2 as a divisor as shown.
Step?
Multiply 2 on the top with the divisor 2 and write 4 under
5 and subtract. The remainder is 1.
Bring down the pair 29 by the side of the remainder 1, 2
yielding 129. 2
Double 2 and take the resulting number 4. Find that
number ' n ' such that Anxn is just less than or equal to
129.
For example : 42 x 2 = 84; and 43 x 3 = 129 and
so n = 3.
Write 43 as the next divisor and put 3 on the top along
with 2. Write the product 43x 3 = 129 under 129 and
subtract. Since the remainder is '0', the division is complete.
Hence /529 = 23 .
5 29
4
1
5 29
4 4-
1 29
2 3
2
5 29
4 si.
43
1 29
1 29
Example 1.24
Find 73969 by the long division method.
Solution
Step 1 : We write 3969 as 39 69 by grouping the digits into pairs, starting
from right end.
Real Number System
Step 2 : Find the number whose square is less than or equal to 39. It is 6.
6
Step 3 : Put 6 on the top and also write 6 as a divisor.
6 39 69
Step 4
Steps
Step 6
Step?
Multiply 6 with 6 and write the result 36 under 39 and n
subtract. The remainder is 3.
39 69
36
Bring down the pair 69 by the side of this remainder 3, ^
yielding 369.
39 69
36 nI/
3 69
Double 6, take the result 12 and find the number 'n'. Such
that \2nxn is just less than or equal to 369. 6 3
Since 122 X 2 = 244; 123 X 3 = 369, n = 3
Write 123 as the next divisor and put 3 on the top along
with 6. Write the product 123 x 3 = 369 under 369 and
subtract. Since the remainder is '0', the division is complete
6
39 69
36 vl^
123
3 69
3 69
Hence /3969 = 63.
1.7.2 (a) Square roots of Decimal Numbers
To apply the long division method, we write the given number by pairing off
the digits as usual in the integral part, and pairing off the digits in the decimal part
from left to right after the decimal part.
For example, we write the number 322.48 as
322 • 48
Integral Part Decimal Point Fractional Part
We should know how to mark the decimal point in the square root. For this we
note that for a number with 1 or 2 digits, the square root has 1 digit and so on. ( Refer
Table 1). The following worked examples illustrate this method:
Chapter 1
Example 1.25
Find the square root of 6.0516
Solution
We write the number as 6 . 05 16 . Since the number of digits in the integral part
is 1, the square root will have 1 digit in its integral part. We follow the same procedure
that we usually use to find the square root of 60516
2.4 6
2
44
486
6.05 16
4 ^
2 05
176
2916
2916
From the above working, we get 76.0516 = 2.46.
Example 1.26
Find the least number, which must be subtracted from 3250 to make it a perfect
square
Solution 5 7
5
32 50
25 ^^
107
7 50
7 49
1
This shows that 57^ is less than 3250 by 1. If we subtract the remainder from the
number, we get a perfect square. So the required least number is 1.
Example 1.27
Find the least number, which must be added to 1825 to make it a perfect square.
Solution
4 2
4
18 25
16 ^^
82
2 25
164
61
This shows that 42' < 1825.
Real Number System
Next perfect square is 43^ = 1849.
Hence, the number to be added is 43"
Example 1.28
Evaluate 70.182329
Solution
1825 = 1849 - 1825
= 24.
0.4 2 7
4
82
847
0.18 23 29
16 i
2 23
164
59 29
59 29
We write the number 0.182329
as 0.18 23 29. Since the number
has no integral part, the square root
also will have no integral part. We
then proceed as usual for finding the
square root of 182329.
Hence 70.182329 = 0.427
Note: Since the integral part of the radicand is '0', the square root also has '0' in
its integral part.
Example 1.29
Find the square root of 121.4404
Solution
11.02
1
21
2202
1 21. 44 04
ls^
021
21
44 04
44 04
7121.4404 = 11.02
Example 1.30
Find the square root of 0.005184
Solution
70.005184 = 0.072
0. 7 2
7
0.00 5184
49 ,.
142
2 84
2 84
Chapter 1
Note: Since the integral part of the radicand is 0, a zero is written before the
decimal point in the quotient. A '0' is written in the quotient after the
decimal point since the first left period following the decimal point is 00 in
the radicand.
1.7.2 (b) Square root of an Imperfect Square
An imperfect square is a number which is not a perfect square. For example 2,
3, 5, 7, 13,... are all imperfect squares. To find the square root of such numbers we use
the Long division method.
If the required square root is to be found correct up to 'n' decimal places, the
square root is calculated up to n+1 decimal places and rounded to 'n' decimal places.
Accordingly, zeros are included in the decimal part of the radicand.
Example 1.31
Find the square root of 3 correct to two places of decimal.
Solution 1. 7 3 2
1
27
343
3462
3.00
1 -l
00
)
2 00
1 89 s
/
1100
1029 ,
/
71 00
69 24
1 7
6
Since we need the answer correct to
two places of decimal, we shall first
find the square root up to three places
of decimal. For this purpose we must
add 6 ( that is three pairs of ) zeros to
the right of the decimal point.
3 = 1.732 up to three places of decimal.
3 = 1.73 correct to two places of decimal.
Example 1.32
Find the square root of 10^ correct to two places of decimal.
10.66 66 66
Solution
10 2 = 32 =
3 3
In order to find the square root correct to two places of
decimal, we have to find the square root up to three places.
Therefore we have to convert ^ ^^ a decimal correct to six
places.
T _ - — . • . . X 6525
3. 2 6 5
3
62
646
10^
3.265 (approximately)
= 3.27 (correct to two places of decimal )
10. 66 6
9 4^
56
7
1 66
1 24 s
/
42 66
38 76 s
/
3 90 67
3 26 25
6
44
2
Real Number System
EXERCISE 1.6
1. Find the square root of each expression given below :
(i) 3 X 3 X 4 X 4 (ii) 2x2x5x5
(iii) 3x3x3x3x3x3 (iv) 5 X 5 X 11 X 11 X7 X7
2. Find the square root of the following :
(i)^ (ii)^ (iii) 49 (iv)16
3. Find the square root of each of the following by Long division method :
(i)2304 (ii)4489 (iii) 3481 (iv) 529 (v) 3249
(vi)1369 (vii)5776 (viii) 7921 (ix) 576 (x) 3136
4. Find the square root of the following numbers by the prime factorisation method :
(i) 729 (ii) 400 (iii) 1764 (iv) 4096 (v) 7744
(vi)9604 (vii)5929 (viii) 9216 (ix) 529 (x) 8100
5. Find the square root of the following decimal numbers :
(i)2.56 (ii)7.29 (iii) 51.84 (iv) 42.25 (v) 31.36
(vi) 0.2916 (vii) 11.56 (viii) 0.001849
6. Find the least number which must be subtracted from each of the following
numbers so as to get a perfect square :
(i) 402
(ii) 1989
(iii) 3250 (iv) 825
(v) 4000
7. Find the least number which must be added to each of the following numbers so
as to get a perfect square :
(i) 525 (ii) 1750 (iii) 252 (iv) 1825 (v) 6412
8. Find the square root of the following correct to two places of decimals :
(i)2
(ii)5
(iii) 0.016 (iv)
1_
8
(v)l
J_
12
9. Find the length of the side of a square where area is 441 nr.
10. Find the square root of the following :
(i)
225
3136
(ii)
2116
3481
(iii)
529
1764
(iv)
7921
5776
Chapter 1
1.7.3 Cubes
Introduction
This is an incident about one of the greatest mathematical
geniuses S. Ramanujan. Once mathematician Prof. G.H. Hardy
came to visit him in a taxi whose taxi number was 1729. While
talking to Ramanujan, Hardy described that the number 1729
was a dull number. Ramanujan quickly pointed out that 1729
was indeed an interesting number. He said, it is the smallest
number that can be expressed as a sum of two cubes in
two different ways.
ie., 1729 = 1728 +1 = 12' + 1'
and 1729 = 1000 + 729 = 10' + 9'
1729 is known as the Ramanujan number.
There are many other interesting patterns of
cubes, cube roots and the facts related to them.
Cubes
We know that the word 'Cube' is used in geometry.
A cube is a solid figure which has all its sides are equal.
If the side of a cube in the adjoining figure is 'a'
Srinivasa Ramanujan
(1887 -1920)
Ramanujan, an Indian Mathe-
matician who was born in Erode
contributed the theory of num-
bers which brought him world-
wide acclamation. During his
short life time, he independent-
ly compiled nearly 3900 results.
1729 is the smallest Ramanu-
jan Number. There are an infi-
nitely many such numbers. Few
are 4104 (2, 16 ; 9, 15), 13832
(18, 20 ; 2, 24).
units
then its volume is given by a x a x a = a^ cubic units.
Here a' is called "a cubed" or "a raised to the power three"
or "a to the power 3".
Now, consider the number 1, 8, 27, 64, 125, • • •
These are called perfect cubes or cube numbers.
Each of them is obtained when a number is multiplied by itself three times.
Examples: 1x1x1 = 1', 2x2x2 = 2', 3 x 3 x 3 = 3', 5 x 5 x 5 = 5^
Example 1.33
Find the value of the following :
(i)15^ (ii)(-4)' (iii)(1.2)' (iv)(^y
Solution
(i)
(ii)
153 = 15 X 15 X 15 = 3375
(-4)3 = (-4)x(-4)x(-4)=-64
Real Number System
3 —
(iii) (1.2)
Cv) (^f =
1.2x1.2x1.2 = 1.728
(-3)x(-3)x(-3)__27
4x4x4
64
Observe the question (ii) Here (- 4)^ = - 64.
Note: When a negative number is multiplied by itself an even number of times, the
product is positive. But when it is multiplied by itself an odd number of times, the
product is also negative, ie, (- if = f~ ^ ^^ ^ ^^ ""^^
1+1
if n is even
The following are the cubes of numbers from 1 to 20.
Numbers Cube
ICTffff^^
11
1331
^.^12
1728
13
2197
— -14
2744
15
3375
^16
4096
^^\1
4913
18
5832
■ -19
6859
20
8000
Table 2
Properties of cubes
From the above table we observe the following properties of cubes:
1. For numbers with their unit's digit as 1, their cubes also will have the unit's
digit as 1.
For example: 1' = 1; iP = 1331; 21' = 9261; 31' = 29791.
2. The cubes of the numbers with 1, 4, 5, 6, 9 and as unit digits will have the
same unit digits.
For example: 14' = 2744; 15' = 3375; 16' = 4096; 20' = 8000.
3. The cube of numbers ending in unit digit 2 will have a unit digit 8 and the
cube of the numbers ending in unit digit 8 will have a unit digit 2.
For example: (12)' = 1728; (18)' = 5832.
4. The cube of the numbers with unit digits as 3 will have a unit digit 7 and
the cube of numbers with unit digit 7 will have a unit digit 3.
For example: (13)' = 2197; (27)' = 19683.
5. The cubes of even numbers are all even; and the cubes of odd numbers are
all odd.
Chapter 1
Adding consecutive odd numbers
Observe the following pattern of sums of odd numbers.
1 = 1 = 1^
Next 2 odd numbers, 3 + 5
Next 3 odd numbers, 7 + 9 + 11
Next 4 odd numbers, 13 + 15 + 17 + 19
Next 5 odd numbers, 21 + 23 + 25 + 27 + 29
Is it not interesting?
8 = 2^
27 = 3^
64 = 43
125 = 53
There are three 5's
in the product but
only two 2's.
Example 1.34
Is 64 a perfect cube?
Solution
64 = 2X2X2 X 2X2X2
= 2' X 2' = (2 X 2/ = 4'
.-. 64 is a perfect cube.
Example 1.35
Is 500 a perfect cube?
Solution
500 = 2X2X5X5X5-
So 500 is not a perfect cube.
Example 1.36
Is 243 a perfect cube? If not find the smallest number by
which 243 must be multiplied to get a perfect cube.
Solution
243 = 3X3X3X3X3
In the above factorisation, 3x3 remains after grouping
the 3'^ in triplets. .-. 243 is not a perfect cube.
To make it a perfect cube we multiply it by 3.
243x3 = 3x3x3 x 3x3x3
729 = 3'x3^^ = (3x3y
729 = 9' which is a perfect cube.
.-. 729 is a perfect cube.
Prime factorization
2
2
2
2
2
2
64
32
16
8
4
2
1
Prime factorization
2
2
5
5
5
500
250
125
25
5
Prime factorization
3
3
3
3
3
243
81
27
9
3
Prime factorization
3
3
729
243
3
81
3
3
3
27
9
3
Real Number System
1.7.4 Cube roots
If the volume of a cube is 125 cm\ what would be the length of its side? To get
the length of the side of the cube, we need to know a number whose cube is 125. To
find the cube root, we apply inverse operation in finding cube.
For example: Symbol
We know that 2' = 8, the cube root of 8 is 2.
We write it mathematically as
3/8 = (8)'/ '=(2^3 = 23/3 = 2
Some more examples:
(i) 'Jus =
(ii) '/64 =
(iii)
V denotes "cube - root"
5. = (53)1/3 ^ 53/3 ^3. ^ 3
43" = (43)1/3 _ ^3/3 _ ^1 _ 4
W = {Wf = W^' = 10' = 10
Viooo =
Cube root through prime factorization method
Method of finding the cube root of a number
Step 1 : Resolve the given number into prime factors.
Step 2 : Write these factors in triplets such that all three factors in each triplet
are equal.
Step 3 : From the product of all factors, take one from each triplet that gives
the cube root of a number.
Example 1.37
Find the cube root of 512.
Solution
Prime factorization
'/5T2 = (512)T
= ((2x2x2)x(2x2x2)x(2x2x2))i
= (2'x2'x2')3
= (2^=2^^
^/512 = 8.
2
2
2
2
2
2
2
2
512
256
128
64
32
16
8
4
2
2
Example 1.38
Find the cube root of 27 X 64
Solution
Resolving 27 and 64 into prime factors, we get
'/27 = (3 X 3 X 3)T = (3')3
Prime factorization
3
3
27
9
3
3
Chapter 1
V27
= 3
^/64
= (2x2x2x2x2x2)t
= (2^)3 = 2^ = 4
Prime factorization
2 64
2 32
^/64
= 4
2
2
2
16
V27 X 64
o
4
= 3x4
= 12
2
2
1
V27 X 64
Example 1.39
Is 250 a perfect cube? If not, then by which smallest natural number should 250
be divided so that the
quotient is a perfect cube?
Solution
250 =
2x5x5x5
Prime factorization
2
5
5
5
250
125
25
5
1
The prime factor 2 does not appear in triplet. Therefore
250 is not a perfect cube.
Since in the factorisation, 2 appears only one time. If we
divide the number 250 by 2, then the quotient will not contain 2.
Rest can be expressed in cubes.
.-. 250-2 = 125
= 5X5X5 = 5'.
.-. The smallest number by which 250 should be divided to make it a perfect
cube is 2. i
Cube root of a fraction
Cube root of a fraction = Cube root of its numerator
Cube root of its denominator
(i.e.)
(f)^
M
(b)^
Example 1.40
Find the cube root of i^.
216
Solution
Resolving 125 and 216 into prime factors, we get
125 = 5X5X5
Prime factorization
5
5
5
125
25
5
Real Number System
Example 1.41
Find the cube root of
Solution
-512 =
'/^5l2 =
1000 = 5x5x5x2x2x2
Viooo =
10
'/-
-512 _
1000
-8
10
V-
-512 _
1000
-4
5
Example 1.42
Find the cube root of 0.027
Solution
'r
-X
= '^{-x)x{-x)x{-x)
-X.
VO.027
= J 27
V 1000
_ . / 3x3x3
V 10x10x10
_ ^/F _ 3
'/W 10
= 0.3
VO.027
Example 1.43
Evaluate ^^729-^/2^
V512 +V343
Solution
Prime fa
3
ctorization
27
Prime fa
7
ctorization
343
V729 = V9^^
= 9 3
3
9
3
7
7
49
7
'/21^'/¥
= 3 1
1
Chapter 1
^/729 -
Prime factorization Prime factorization
9-3
8 + 7
" 15
3
729
3
243
3
81
3
27
3
9
3
3
2
5
2
2
2
2
2
2
2
2
2
512
256
128
64
32
16
8
4
2
EXERCISE 1.7
1
(D) 100
(D) 100
1. Choose the correct answer for the foUowing :
(i) Which of the following numbers is a perfect cube?
(A) 125 (B) 36 CC) 75
(ii) Which of the following numbers is not a perfect cube?
(A) 1331 (B) 512 (C) 343
(iii) The cube of an odd natural number is
(A) Even (B) Odd
(C) May be even, May be odd (D) Prime number
(iv) The number of zeros of the cube root of 1000 is
(A) 1 (B) 2 (C) 3
(v) The unit digit of the cube of the number 50 is
(A) 1 (B) (C) 5
(vi) The number of zeros at the end of the cube of 100 is
(A) 1 (B) 2 (C) 4
(vii) Find the smallest number by which the number 108 must be multiplied to obtain
a perfect cube
(A) 2 (B) 3 (C) 4 (D) 5
(viii) Find the smallest number by which the number 88 must be divided to obtain a
perfect cube
(A) 11 (B) 5 (C) 7 (D) 9
(ix) The volume of a cube is 64 cm^ . The side of the cube is
(A) 4 cm (B) 8 cm (C) 16 cm (D) 6 cm
(x) Which of the following is false?
(A) Cube of any odd number is odd.
(B) A perfect cube does not end with two zeros.
(D) 4
(D) 4
(D) 6
Real Number System
(C) The cube of a single digit number may be a single digit number.
(D) There is no perfect cube which ends with 8.
Check whether the following are perfect cubes?
(i) 400 (ii) 216 (iii) 729 (iv) 250
(v) 1000 (vi) 900
Which of the following numbers are not perfect cubes?
(i) 128 (ii) 100 (iii) 64 (iv) 125
(v) 72 (vi)625
Find the smallest number by which each of the following number must be
divided to obtain a perfect cube.
(i) 81 (ii) 128 (iii) 135 (iv) 192
(v) 704 (vi)625
Find the smallest number by which each of the following number must be
multiplied to obtain a perfect cube.
(i) 243 (ii) 256 (iii) 72 (iv) 675
(v) 100
Find the cube root of each of the following numbers by prime factorisation method:
(i) 729 (ii) 343 (iii) 512 (iv) 0.064
23
(v) 0.216
(vi)5
64
(vii) - 1.331
(viii) - 27000
7. The volume of a cubical box is 19.683 cu. cm. Find the length of each side of the box.
1.8 Approximation of Numbers
In our daily life we need to know approximate
values or measurements.
Benjamin bought a Lap Top for ? 59,876.
When he wants to convey this amount to others, he
simply says that he has bought it for ? 60,000. This is
the approximate value which is given in thousands
only.
Vasanth buys a pair of chappals for ? 599.95. This amount may be considered
approximately as ? 600 for convenience.
A photo frame has the dimensions of 35.23
cm long and 25.91 cm wide. If we want to check the
measurements with our ordinary scale, we cannot
measure accurately because our ordinary scale is marked
in tenths of centimetre only.
Chapter 1
In such cases, we can check the length of the photo frame 35.2 cm to the nearest
tenth or 35 cm to the nearest integer value.
In the above situations we have taken the approximate values for our convenience.
This type of considering the nearest value is called 'Rounding off the digits. Thus the
approximate value corrected to the required number of digits is known as 'Rounding
off the digits.
Sometimes it is possible only to give approximate value, because
(a) If we want to say the population of a city, we will be expressing only in
approximate value say 30 lakhs or 25 lakhs and so on.
(b) When we say the distance between two cities, we express in round number
350 km not 352.15 kilometres.
While rounding off the numbers we adopt the following principles.
(i) If the number next to the desired place of correction is less than 5, give the
answer up to the desired place as it is.
(ii) If the number next to the desired place of correction is 5 and greater than
5 add 1 to the number in the desired place of correction and give the answer.
The symbol for approximation is usually denoted by ~.
ilctrnty
Take a A4 sheet. Measure its length and breadth. How do you
express it in em's approximately. -
^
Let us consider some examples to find the approximate values of a given number.
Take the number 521.
Approximation nearest to TEN
Illustration
Consider multiples of 10 before and after 521. ( i.e. 520 and 530 )
We find that 521 is nearer to 520 than to 530.
X' ^
1
■^X
519 520 521 522 523 524 525 526 527 528 529 530
.-. The approximate value of 521 is 520 in this case.
Approximation nearest to HUNDRED
Illustration
(i) Consider multiples of 100 before and after 521. ( i.e. 500 and 600 )
Real Number System
X' ^
1
300
400
500
521
^X
600
700
We find that 521 is nearer to 500 than to 600. So, in this case, the approximate
value of 521 is 500.
(ii) Consider the number 625
Suppose we take the number line, unit by unit.
X' <-
I
■>x
623
624
625
626
627
628
629
In this case, we cannot say whether 625 is nearer to 624 or 626 because it is
exactly midway between 624 and 626. However, by convention we say that it is nearer
to 626 and hence its approximate value is taken to be 626.
Suppose we consider multiples of 100, then 625 wiU be approximated to 600 and not
700.
Some more examples
For the number 47,618
(a) Approximate value correct to the nearest tens = 47,620
(b) Approximate value correct to the nearest hundred = 47,600
(c) Approximate value correct to the nearest thousand = 48,000
(d) Approximate value correct to the nearest ten thousand = 50,000
Decimal Approximation
Illustration
Consider the decimal number 36.729
(a) It is 36.73 correct to two decimal places. ( Since the last digit 9>5, we add 1
to 2 and make it 3 ).
.-. 36.729 ~ 36.73 ( Correct to two decimal places )
(b) Look at the second decimal in 36.729, Here it is 2 which is less than 5, so we
leave 7 as it is. .-. 36.729 - 36.7 ( Correct to one decimal place )
Chapter 1
Illustration
Consider the decimal number 36.745
(a) It's approximation is 36.75 correct to two decimal places. Since the last digit
is 5, We add 1 to 4 and make it 5.
(b) It's approximation is 36.7 correct to one decimal place. Since the second
decimal is 4, which is less than 5, we leave 7 as it is.
.-. 36.745 ~ 36.7
Illustration
Consider the decimal number 2.14829
(i) Approximate value correct to one decimal place is 2.1
(ii) Approximate value correct to two decimal place is 2.15
(iii) Approximate value correct to three decimal place is 2.148
(iv) Approximate value correct to four decimal place is 2.1483
Example 1.44
Round off the following numbers to the nearest integer:
(a) 288.29 (b) 3998.37 (c) 4856.795 (d) 4999.96
Solution
(a) 288.29 ~ 288 (b) 3998.37 ~ 3998
(Here, the tenth place in the above numbers are less than 5. Therefore all the
integers are left as they are.)
(c) 4856.795 ~ 4857 (d) 4999.96 ~ 5000
[Here, the tenth place in the above numbers are greater than 5. Therefore the
integer values are increased by 1 in each case.]
EXERCISE 1.8
1. Express the following correct to two decimal places:
(i) 12.568 (ii) 25.416 kg (iii) 39.927 m
(iv) 56.596 m (v) 41.056 m (vi) 729.943 km
2. Express the following correct to three decimal places:
(i) 0.0518 m (ii) 3.5327 km
(iii) 58.2936/ (iv) 0.1327 gm
(v) 365.3006 (vi) 100.1234
Real Number System
3. Write the approximate value of the following numbers to the accuracy stated:
(i) 247 to the nearest ten. (ii) 152 to the nearest ten.
(iii) 6848 to the nearest hundred, (iv) 14276 to the nearest ten thousand,
(v) 3576274 to the nearest Lakhs, (vi) 104, 3567809 to the nearest crore
4. Round off the following numbers to the nearest integer:
(i) 22.266 (ii) 777.43 (iii) 402.06 I
(iv) 305.85 (v) 299.77 (vi) 9999.9567
1.9. Playing with Numbers
Mathematics is a subject with full of fun, magic and wonders. In this unit, we
are going to enjoy with some of this fun and wonder.
(a) Numbers in General form
Let us take the number 42 and write it as
42 = 40 + 2 = 10 X 4 + 2
Similarly, the number 27 can be written as
27 = 20 + 7 = 10 X 2 + 7
In general, any two digit number ab made of digits 'a' and 'b' can be written as
ab = 10 xa + b= 10 a + b
ba = 10 ^b + a = 10 b + a
Now let us consider the number 351.
This is a three digit number. It can also be written as
351 = 300 + 50 + 1 = 100 X 3 + 10 X 5 + 1 X 1
In general, a 3-digit number abc made up of digit a, b and c is written as
abc ^ 100 X a + \Oxb + IXc
= lOOa + lOb + Ic
In the same way, the three digit numbers cab and bca can be written as
cab^ 100c + lOa + b
bca= lOOZ? + 10c + a
(b) Games with Numbers
(i) Reversing the digits of a two digit number
Venu asks Manoj to think of a 2 digit number, and then to do whatever he
asks him to do, to that number. Their conversation is shown in the following figure.
Study the figure carefully before reading on.
Chapter 1
Conversation between Venu and Manoj:
Now let us see if we can explain Venn's "trick". Suppose, Manoj chooses the
number ab, which is a short form for the 2 -digit number \0a + b. On reversing the
digits, he gets the number ba = lOb + a. When he adds the two numbers he gets :
{lOa + b) + {lOb + a) = lla + lib
= ll{a + b)
So the sum is always a multiple of 11, just as Venu had claimed.
Dividing the answer by 11, we gei{a + b)
(i.e.) Simply adding the two digit number.
(c) Identify the pattern and find the next three terms
Study the pattern in the sequence.
(i) 3, 9, 15, 21, (Each term is 6 more than the term before it)
If this pattern continues, then the next terms are , and .
(ii) 100, 96, 92, 88, , , . (Each term is 4 less then the next
term before it)
(iii) 7, 14, 21, 28, , , . (Multiples of 7)
(iv) 1000, 500, 250, , , . (Each term is half of the previous term)
(v) 1, 4, 9, 16, , , . (Squares of the Natural numbers)
Real Number System
(d) Number patterns in Pascal's Triangle
The triangular shaped, pattern of numbers given below is called
Pascal's Triangle.
6
11
10
13
9
5
8
7
12
3x3 Magic Square
Look at the above table of numbers. This is called a 3 X 3
magic square. In a magic square, the sum of the numbers in each
row, each column, and along each diagonal is the same.
In this magic square, the magic sum is 27. Look at the middle
number. The magic sum is 3 times the middle number. Once 9 is
filled in the centre, there are eight boxes to be filled. Four of them will be below 9 and
four of them above it. They could be,
(a) 5, 6, 7, 8 and 10, 1 1, 12, 13 with a difference of 1 between each number.
(b) 1, 3, 5, 7 and 1 1, 13, 15, 17 with a difference of 2 between them or it can be
any set of numbers with equal differences such as — 11,— 6, — 1,4 and 14, 19, 24, 29
with a difference of 5.
Once we have decided on the set of numbers, say 1, 3, 5, 7 and 1 1, 13, 15, 17
draw four projections out side the square, as shown in below figure and enter the
numbers in order, as shown in a diagonal pattern.
9
3
13
11
17
9
1
7
5
15
Chapter 1
The number from each of the projected box is transferred to the empty box on
the opposite side.
3
1
2
9
5
7
6
5
9
1
7
8
2
4
7
2
6
3
5
9
7
2
4
2
8
1
9
3
3
9
8
2
5
7
4
5
6
3
1
1
7
3
5
8
9
4
8
3
4
2
7
5
Use all the digits 1, 2, ..., 9 to fill up each rows, columns
and squares of different colours inside without repetition.
The Revolving Number
14 2 8 5 7
First set out the digits in a circle.
Now multiply 142857 by the number from 1 to 6.
142857 142857 142857
X 1 X 2 X 3
142857
285714
428571
142857
X4
142857
X5
714285
142857
X6
571428
857142
We observe that the number starts revolving the same digits in different
combinations. These numbers are arrived at starting from a different point on the
circle.
Real Number System
EXERCISE 1.9
1. Complete the following patterns:
(i) 40,35,30,
(ii) 1,2,4,
3 J
J J
(iii) 84,77,70,
y y
(iv) 4.4,5.5,6.6,
(v) 1,3,6, 10, _
J J
(vi) 1, 1, 2, 3, 5, 8, 13, 21,
J J
(This sequence is called FIBONACCI SEQUENCE)
(vii) 1,8,27,64, , , .
2. A water tank has steps inside it. A monkey is sitting on the top most step. ( ie, the
first step ) The water level is at the ninth step.
(a) He jumps 3 steps down and then jumps back 2 steps up.
In how many jumps will he reach the water level ?
(b) After drinking water, he wants to go back. For this, he
jumps 4 steps up and then jumps back 2 steps down in
every move. In how many jumps will he reach back the
top step ?
3. A vendor arranged his apples as in the following pattern :
(a) If there are ten rows of apples, can you find the total
number of apples without actually counting?
(b) If there are twenty rows, how many apples will be
there in all?
Can you recognize a pattern for the total number of apples? Fill this chart and try!
Rows
1
2
3
4
5
6
7
8
9
Total apples
1
3
6
10
15
Chapter 1
Concept Summary
^ Rational numbers are closed under the operations of addition, subtraction
and multiplication.
'^ The collection of non-zero rational numbers is closed under division.
^ The operations addition and multiplication are commutative and associative
for rational numbers.
'^ is the additive identity for rational numbers.
'^ 1 is the multplicative identity for rational numbers.
^ Multiplication of rational numbers is distributive over addition and
subtraction.
'^ The additive inverse of ^ is ^^ and vice-versa.
b b
^ The reciprocal or multiplicative inverse of -7- is — .
^ Between two rational numbers, there are countless rational numbers.
'^ The seven laws of exponents are :
If a and b are real numbers and m, n are whole numbers then
(i) a'" X a" = a'"+"
(ii) a'" - a" = a'"-", where a^O
(iii) a^ - \ , where a 7^
(iv) or'" — — , where a j^
a'"
(v) {a'")" = a'""
(vi) a"'Xb'"^{abT
(v..) $
j^Y where b ^ o
▼ Estimated value of a number equidistant from the other numbers is always
greater than the given number and nearer to it.
Algebra
2.1 Introduction
2.2 Algebraic Expressions - Addition and Subtraction
2.3 Multiplication of Algebraic Expressions
2.4 Identities
2.5 Factorization
2.6 Division of Algebraic Expressions
2.7 Solving Linear Equations
2.1 Introduction
The mathematical term Algebra' was derived from the Arabic
word 'al-jabr'. *A1' means 'The' and *jabr' means 'the restoration of
broken parts'. It was coined from the title of the book 'Kitab al-jabr wa
1-muqabala'. That is 'The Book of Integration and Equation'. It literally
means 'reduction and comparison' written by the Arab Mathematician
Al - Khwarizmi.
In ancient India, Algebra was called as 'Bija - Ganitham'. ('Bija'
means 'the other' and 'Ganitham' means 'Mathematics') Indian
mathematicians like Aryabhatta, Bhrahmagupta, Mahavir, Bhaskara
II, Sridhara have contributed a lot to this branch of mathematics.
The Greek Mathematician, Diophantus of Alexandria had
developed this branch to a great extent. Hence he is called as 'The father
of Algebra'.
Diophantus
(About 3rd
century B.C.)
The Greek
Mathematician
lived in
Alexandria.
He is called
"The father of
Algebra". The
equation
x" + y" = z"
is known as
Diophantine
equation and for
n > 2, there are
no solutions with
positive integral
values for x, y
andz.
Al-Khwarizmi
(780 - 850 A.D.)
I The Arab
Mathematician
I wrote the book
■'Kitab al-jabr wa
1 1-muqabala". It was
I the synthesis of
Indian Algebra and
Greek Geometry
which had the most
profound effect on
the development of
mathematics.
Chapter 2
2.2 Algebraic Expressions - Addition and Subtraction
In class VII, we have learnt about variables, constants, coefficient of terms,
degree of expressions etc. Let us consider the following examples of expressions :
Illustration
(i) X + 5 (ii) 3y-2 (iii) 5m' (iv) 2xy + U
The expression x + 5 is formed with the variable x and the constant 5.
The expression 3}^ - 2 is formed with the variable y and the constants 3 and - 2.
The expression 5m' is formed with the variable m and the constant 5.
The expression 2xy + 1 1 is formed with the variables x and y and the constants
2 and 11.
2.2.1 Values of the Algebraic Expression
We know that the value of the expression changes with the values chosen for
the variables it contains. For example take the expression x + 5. The following table
shows what is happening when x takes different values:
Value for x
Value of the expression x + 5
Notice, The constant 5 remains
unchanged, as the values of
the expression go on changing
for various values of x.
1
1 + 5 = 6
2
2 + 5 = 7
3
3 + 5 = 8
4
4 + 5 = 9
-1
-1 + 5= 4
-2
-2+5=3
-3
-3+5=2
1
2
1 +5 = 11
2 ^ 2
1. Assign different values for the variables given in the remaining
example 2.1 and find the values of the expressions.
2. Have you noticed any change in the values of the constants?
2.2.2 Terms, Factors and Coefficients
Take the expression 2x + 3. This expression is made up of two terms 2x and 3.
Terms are added to form expressions. Terms themselves can be formed as the product
of factors. The term 2x is the product of its factors 2 and x. The term 3 is made up of
only one factor.
Algebra
Tenii 1
Term 2
Expression
^^^
N
^^
Ix
= 2x + 3
A
\
Factor 1 Factor 2
Consider the expression 3ab - 5a. It has two terms 3ab and —5a. The term
Sab is a product of factors 3, a and b. The term — 5a is a product of - 5 and a. The
coefficient of factor in a term is another factor whose product with the given factor is
the term itself. The coefficient of ab is 3 and the coefficient of a is - 5 .
Y-^2!^^_J!2^^(a) Look for the + (plus) or - (minus) sign in any expression, you
can easily find out the number of terms in it.
(b) Constant term is the seperate term with no variable accompanied to it.
/^^tivity
Identify the number of terms and coefficient of each term in the
expression and complete the following table x^y'^
5x'y + -^xy'^
11,
SI. No.
Term
Coefficient of the term
1
X2/
1
2
3
4
Chapter 2
2.2.3 Basic concepts of Polynomial : Monomial, Binomial, Trinomial and
Polynomial
Monomial : An Algebraic expression that contains only one term is called a
monomial.
5
Example :
Binomial
Example
2x , 3ab, - Ip, jvorb, - 8, Zlxyz, etc.
An Algebraic expression that contains only two terms is called a
binomial.
X + J, 4a - 3Z?, 2 - Ix'y, ? -Im, etc.
Trinomial : An Algebraic expression that contains only three terms is called a
trinomial.
Example \ x^y ^z, 2a- 3b + 4, x'y + y^z - z, etc.
Polynomial : An expression containing a finite number of terms with non-zero
coefficient is called a polynomial. In other words, it is an expression containing a
finite number of terms with the combination of variables, whole number exponents of
variables and constants.
Example :
a + b + c + d, Ixy, 3abc — 10, 2x + 3y — 5z, 3x^ + ix'* — 3x^ + 12x + 5 etc.
Degree of the Polynomial : The monomials in the polynomial are called the
terms. The highest power of the terms is the degree of the polynomial.
The coefficient of the highest power of x in a polynomial is called the leading
coefficient of the polynomial.
Example :
2x^ - X* + Ix^ - 6x- + 12x - 4 is a polynomial in x. Here we have six monomials
x\ lx\ — 6x^ \2x and —4 which are called the terms of the polynomial.
Degree of the polynomial is 5.
The leading coefficient of the polynomial is 2.
When the teacher asked to find the degree of the polynomial 13x* - 2x'y
Two of the students did this as given below.
Can you find who is doing it correctly?
2x
V Gautham
Po\ynom'\a\ :
Ux-lxY-A
The higheet power of
•the -terins is ■4-.
.'. The degree of the
polynomial is A-.
H
Ayisha
Polynomial :
1 3x —2xy - 4
= 1 Sx" - 2x^y - 4x V
The highest power of
the terms is 5.
.'. The degree of the
polynomial is 5.
a
If your answer is Ayisha, then you are right.
Algebra
If your answer is Gautham, then where is the mistake?
Let us analyse the given polynomial: ISx" - 2x^y^ - 4.
Term 1 : ISx" ^coefficient of x" is 13, variable x, power of x is 4. Hence the
power of the term is 4.
Term 2 : - 2x^y^ -^ coefficient of x^y^ is - 2 and the variables are x and y the
power of X is 2 and the power of y is S.Hence the power of the term x^y^ is 2+3 = 5
[Sum of the exponents of variables x andy ].
Term 3 : — 4 -* the constant term and it can be written as - 4x°y\ The power
of the variables x°y° is zero. Therefore the power of the term -4 is zero.
Why Gautham is wrong?
Gautham thought, the power of the second term - 2x^y^ as, either two or three.
But the right way is explained above. This confusion led him to wrong conclusion.
Standard form of the polynomial
For a polynomial in x, if the terms of the polynomial are in decending powers
of X, we say that it is in the standard form.
Example :
Write the polynomial 2 + 9x - 9x^ + 2x'' - 6x^ in standard form.
Now we write the polynomial in the standard form as 2x'' - 6x^ - 9x^ + 9x + 2
Remember: When no exponent is shown, it is understood to be 1.
Example : 9x = 9x'
Like terms and Unlike terms
Like terms contain the same variables having same powers.
Consider the following expressions.
2x + 3x - 5
_ 2 - 2 „
2a + 3a + /a
Three terms
^11 are unlike terms;
Like terms
Variable -^ a
Power -^ 2
Look at the following expressions :
3x, 5x^, 2xy, — lOx, — 7, — 3^, — 3x^, — 20yx, 20, 4x,
f3.
Chapter 2
We can list out the like terms as :
(i) 3x, — lOx and 4x
5x^ and — 3x^
2xy and — 20yx
-7, 20 and
^ThyTit!^
(ii)
(iii)
(iv)
2,
7
The fohowing are not hke terms. Why?
(i) 3x and 3y and (ii) 5x^ and - 3y^
2.2.4 Addition and Subtraction of Algebraic Expressions - Revision
In class VII, we have learnt to add and subtract the algebraic expressions. Only
like or similar terms can be added or subtracted.
Let us have a quick revision now.
Example 2.1
Add ■.3x' + x--l and Ix' + 5x+5.
Solution
We first arrange these two as follows and then add:
It can also be written as
3x' + x' -2
(+) 2x"- + 5x + 5
3x^ + 3x^ + 5x + 3
Or
3x'+ x' + 0x-2
(+) Or' + 2x' + 5x + 5
3x^ + 3x^ + 5x + 3
Observe, we have written the term 2x' of the second polynomial below the
corresponding term x- of the first polynomial. Similarly, the constant term +5 is
placed below the constant term - 2. Since the term x in the first polynomial and the x^
in the second polynomial do not exist, their respective places have been left blank to
facilitate the process of addition. Or, for the non existing terms, we annexe the terms
with zero coefficients.
Example 2.2
Find out the sum of the polynomials 3x -y,2y — 2x and x + y.
Solution
Column method of addition
3x- y
-2x + 2y
x+ y
(Rearranging 2y — 2;c as — 2jc + 2y)
(+)
2x + 2y
Row method of addition
{3x — y) + {2y — 2x) + {x + y)
= {3x-2x + x) + {-y + 2y + y)
= (4x - 2x) + {3y - y)
= 2x + 2y.
Therefore, polynomials may be added in a
row by combining the like terms.
Algebra
Example 2.3
(i) Subtract 5xy from Sxy (ii) Subtract 3c + 7cf from 5c - d'
(iii) Subtract 2x' + 2y' - 6 from 3x' - Vy' + 9
So/ution
(i) Subtract Sxy from 8xy.
The first step is to place them as
Sxy
- Sxy (The two terms Sxy, — Sxy are like terms)
3xy
.'. Sxy — Sxy = 3xy
(ii) Subtract 3c + Id' from 5c — d'
Solution
5c- d'
■{3c + ld')
(or)
5c- d'
5c- d'
3c - Id'
Often, we do
this as -^
3c + Id'
Ic - Sd'
Ic - Sd'
Alternatively, this can also be done as :
{Sc-d')-{3c + ld') = Sc-d'-3c-ld'
= {Sc-3c) + {-d'-ld')
= 2c + (-Sd')
= 2c -Sd'
(iii) Subtract 2x' + 2y'-6 from 3x' -ly' + 9
Solution
3x' -ly' + 9
2x' + 2y' — 6 [Change of the sign]
- - +
X
9y' + 15
Alternative Method
{3x' - ly' + 9) - {2x' + 2y' - 6)
= 3x' -7y' + 9- 2x' - 2y' + 6
= {3x' - 2x') + (- ly' - 2y') + (9 + 6)
= x' + {- 9y') + 15
= X
9y' + 15
EXCERCISE 2.1
1. Choose the correct answer for the following:
(i) The coefficient of x* in - Sx'' + 4j-x* - 3x^ + 7x" - 1 is
3
(A)-5
(B)-3
(C)
7
(ii) The coefficient of xy' in Ix' — lAx'y + \Axy' — 5 is
(A) 7 (B) 14 (C) - 14
(D)7
(D)-5
Chapter 2
(iii) The power of the term x'y^z!' is
(A) 3 (B) 2
(C)12
(D)7
(iv) The degree of the polynomial x^ — Sx" + ^x' — 73x + 5 is .
(A) 7
(B)
3
(C)4
(C)5
(v) The degree of the polynomial x^
(A) - 576 (B) 4
(vi) jc' + y' - 2z' + 5x - 7 is a
(A) monomial (B) binomial (C) trinomial
(vii) The constant term of 0.4x^ — ISy' — 0.75 is
(A) 0.4 (B) 0.75 (C) - 0.75
(D) - 73
Sl^xy is
(D)7
(D) polynomial
(D)-75
Identify the terms and their coefficients for the following expressions:
(i) Zabc - 5ca (ii) 1 + x + y^ (m)3x^y^ - 3xyz + z^
(iv) - 7 + 2pq - ^qr + rp
(v)f
y_
2
Q.?>xy
3. Classify the following polynomials as monomials, binomials and trinomials:
3x\ 3x + 2, x^ — Ax + 2, x^ — 1 , x'' + 3xy + y^ ,
s^' + 3st — 2f', xy + yz + zx, a^b + b^c, 21 + 2m
4. Add the following algebraic expressions:
(i) 2x^ + 3x + 5, 3x' -Ax-1 (ii) x^ -2x -3,x^ + 3x + \
(iii) 2r^ + ^ - 4, 1 - 3^ - 5r^ (iv) xy - yz, yz -xz,zx - xy
(v) a^ + b\ b^ + c', c^ + a\ 2ab + 2bc + 2ca.
5. (i) Subtract 2 a-b from 3a — b
(ii) Subtract —3x + Sy from —Ix— \Qy
(iii) Subtract 2ab + 5bc - 3ca from lab - 2bc + IQca
(iv) Subtract x^ — 2x^ — 3x from x^ + 3x^+1
(v) Subtract 3x^y - 2xy + 2xy^ + 5x-ly -\Q from
\5 -2x + 5y - \ \xy + 2xy^ + ^x^y
6. Find out the degree of the polynomials and the leading coefficients of the
polynomials given below:
(i) x' - 2x' + 5x' - ^x' - 70x - 8
(iii)-77 + 7x'-x'
(v) x' - 2x'f + 3xy'' - lOx}' + 1 1
(ii)13x'-y^'-113
(iv) - 181 + 0.8}' - 8/ + 115/ + /
Algebra
2.3 Multiplication of Algebraic Expressions
2.3.1 Multiplying two Monomials
We shall start with x + x + x + x + x = 5x
Similarly, we can write, 5 x (2x) = {2x) + {2x) + {Ix) + (2jc) + (2jc) = \0x
Illustration
Multiplication is the shortened
form of Addition.
(1) xx5y = xX5Xy = 5XxXy = 5xy
(ii) 2xX3y = 2xxX3xy = 2x3xxXy = 6xy
(iii) 2xx{-3y) = 2xi-3)xxxy =-6xxxy =-6xy
(iv) 2xX3x- = 2xxx3xx^ = (2x3)x(xxx^) = 6x'
(v) 2xX{- 3xyz) = 2 X (- 3) X (x X xyz) = - 6x-yz.
Note: 1. Product of monomials are also monomials.
2. Coefficient of the product = Coefficient of the first monomial x
Coefficient of the second monomial.
3. Laws of exponents a'" X a" = a'"^" is useful, in finding the product of the
terms.
4. The products of a and b can be represented as: a x fa, ab, a.b, a (b), (a) b,
(a) (b), (ab).
(vi) {3x')(4x')
= (3 X X X x) (4 X X X X X x) (Or) ^^x") (Ax') = (3x4) (x' X x') = 12 (x-^^')
= (3x4)(xXxXxXxXx) = 12^' (using a'" X a" = a'"^")
= 12x^
Some more useful examples are as follows:
(vii) 2xX3yx5z = (2x X 3y) X 5z
= (6xy) X 5z
= 30 xyz
(or) 2xx3yx5z= (2x3x5)x(xxyxz) = 30xyz
(viii) Aab X 3a- b- X 2a' b' = (Aab X 3a- b') X 2a' b'
= (na'b') xla^b""
= 2Aa''b'
(ox)Aab X 3a- b' X 2a' b'^ 4x3x2x(abX a'b' X a'b')
= 2A(a'^'^'xb'^'^')
= 24a*fo'
2.3.2 Multiplying a Monomial by a Binomial
Let us learn to multiply a monomial by a binomial through the following
examples.
Example 2.4
Simplify: (2x) X (3jc + 5)
Solution We can write this as:
step/
(2x) X (3;c + 5) = {2x X 3;c) + {2x X 5)
Step2 = ^x' + lOx
[Using the distributive lawj
Example 2.5
Simplify: (- 2x) x (4 - 5y)
Solution step /
(- 2x) X (4 - 5y) = [(- 2x) X 4] + [(- 2x) X (- 5y)]
Step 2 = ( — 8x) + ( 1 Ox}') [using the distributive law]
= —Sx+ lOxy
Note: (i) The product of a monomial by a binomial is a binomial.
(ii) We use the commutative and distributive laws while doing the sums.
axb = bxa (Commutative Law)
a{b + c) = ab + ac and a{b - c) = ab - ac (Distributive laws)
2.3.3. Multiplying a Monomial by a Polynomial
A Polynomial with more than two terms is multiplied by a monomial as
follows:
Example 2.6
Simplify: (i) 3 {5y' -3y + 2)
(ii) 2x' X i3x' - 5x + 8)
Solution
(i) 3(5/-3>' + 2) = (3x5y) + (3x-3};) + (3x2)
= 15r-9>' + 6
[or]
5/-
X
-3y + 2
3
15/
-9y + 6
(ii) 2x- X (3x'- - 5x + 8)
= i2x' X 3x') + {2x- X ( - 5;c) ) + (2x' X 8)
= 6jc'-10x' + 16x'
[or] 3x' - 5x + 8
^ X 2x'
6x' -\Qx' + \6x'-
Algebra
2.3.4 Multiplying a Binomial by a Binomial
We shall now proceed to multiply a binomial by another binomial, using the
distributive and commutative laws. Let us consider the following example.
Example 2.7
Simplify : (2a + 3b) (5a + Ab)
Solution
Every term in one binomial multiplies every term in the other binomial.
{2a + l>b) {5a + Ab) = {2a X 5a) + {2a X 4b) + Ob X 5a) + i3b X 4b)
= \0a- + Sab + l5ba + l2b'
= I0a' + Sab + I5ab + I2b'
= \0a- + 23ab + \2b-
[Adding like terms Sab and 15ab]
{2a + 3b) {5a + 4b) = lOa' + 23ab + I2b'
[■.■ ab = ba]
Note : In the above example, while multiplying two binomials we get only 3 terms
instead of 2 x2 = 4 terms. Because we have combined the like terms 8ab and
15ab.
2.3.5 Multiplying a Binomial by a Trinomial
In this multiplication, we have to multiply each of the three terms of the trinomial
by each of the two terms in the binomial.
Example 2.8
Simplify: {x + 3) {x^ -5x + l)
Solution
(x + 3) (x -5x + l) =
X (x -5x + 7) + 3 (x -5x + 7)(Using the distributive law)
x' - 5x' + 7x + 3x' - 15;c + 21
X^ — 5x~ + 3x^ + 7x — 15x + 21 (Grouping the like terms)
= X
2x' - 8x + 2 1 (Combining the like terms) $SS)
Alternative Method
X {x^ — 5x + l)
3 {x' -5x + l)
{x + 3)
X
{x^-
5x + l)
x'
-5x'
+ Ix
3x'
-15JC + 21
= x'
-2x'
-8x + 21
^r,
In this example, while
multipltying, instead of
expecting 2x3 = 6 terms,
we are getting only 4 terms
in the product, could you
reason out this?
Chapter 2
Quit the confusion.
1. Is 2xx = 2x ?
No. 2xx can be written as 2 (x) (x) = 2x^.It is the product of the terms 2,
X and X. But 2x means x + x or 2 (x).
2. Is 7xxy = 7xy ?
No. 7xxy is the product of 7, x, x and y and not the product of 7, x and y.
Hence the correct answer for 7xxy = 7 (x) (x) (y) = 7x^y.
EXERCISE 2.2
1. Find the product of the following pairs of monomials:
(i)3, 7x (ii)-7x, 3y (iii) - 3a, Sab (iy) Sa\ - 4a (v)^x\ ^x'
{vi)xy^,x^y (vii) x^y\xy- (ym)abc,abc (ix) xyz.x'-yz (x) a^ b' c\ abc^
2. Complete the following table of products:
First monomial -^
Second Monomial i
2x
-3y
4x2
- 5xy
Ix^y
- Qxy
2x
4x2
....
-3y
4x2
- 5xy
25xy
Ix^y
- ^xY
18xy
3. Find out the product :
(i) la, 3a' , 5a'
(iv) m, 4m, 3m', — 6m'
(vii) - 2p, - 3q, - 5p'
4. Find the product :
(i) (a') X i2a') X (4a'^^)
(iv) (3x + 2) (4x - 3)
(ii) 2x, 4y, 9z
(v) xyz,y'z,yx'
(iii) ab, fac, ca
(vi) /m\ m/i^ //I'
(ii) (5 - 2x) (4 + x) (iii) (x + 3y) {3x - y)
(V) (fab) ( g
15
fl'Z?')
5. Find the product of the following :
(i) (a + b) {2a' - Sab + 3b') (ii) (2x + 3y) (x' - xy + y')
(iii) (x + y + z)(yX + y-z) (iv) (a + fo) (a' + 2ab + b') (v) (m - n) (m^ + mn + n')
6. (i) Add 2x{x — y — z) and 2)^(2 — ^^ — x)
(ii) Subtract 3a! {a-2b + 3c) from 4a (5a + 2/? - 3c)
Algebra
2.4 Identities
Let us consider the equality (x + 2){x + 3) = x^ + 5x + 6.
We evaluate both the sides of this equality for some value of x, say ;c = 5 .
For x = 5, LHS = (x + 2)(x + 3) = (5 + 2)(5 + 3) = 7x 8 = 56
RHS = x' + 5x + 6 = 5' + 5(5) + 6 = 25 + 25 + 6 = 56
Thus the values of the two sides of the equality are equal when ;c = 5 .
Let us now take, x =- 5.
LHS = (x + 2)(x + 3) = (-5 + 2)(-5 + 3) = (-3)(-2) = 6
RHS = x' + 5:^ + 6 = (-5)^ + 5(-5) + 6 = 25-25 + 6 =6
Thus the values of the two sides of the equality are equal when x =- 5,
If LHS = RHS is true for every value of the variable in it, then it is called as an
Identity.
Thus,(x + 2)(x + 3) = x^ + 5x + 6 is an identity.
Identity: An equation which is true for all possible values of the variable is called
an Identity.
../IctMty
Check whether the following are Identities:
(i) 5{x + 4) = 5jc + 20 and (ii) 6x + 10 = 4x + 20.
2.4.1 Algebraic Identities
We proceed now to study the three important Algebraic Identities which are
very useful in solving many problems. We obtain these Identities by multiplying a
binomial by another binomial.
Identity 1
Let us consider (a + bf.
{a + bf
Thus,
(a + b) (a + b)
a^ + ab + ba + b~
a^ + ab + ab + b^
a^ + lab + i>'
{a + bf = a' + 2ab + b'
[•.• ab = ba]
Chapter 2
Geometrical Proof oi {a + bf
In this diagram,
side of the square ABCD is (a + b).
Area of the square ABCD
= {a + b){a + b)
= Area of the square AEHG +
Area of the rectangle EBIH +
Area of the rectangle GHFD +
Area of the square HICF
-(a + b)
= {aXa) + {bXa) + {aXb) + {bXb)
= a~ + ba + ab + b^
{a + bf = a^ + lab + b'
{a + bf = a' + lab + b'
Identity 2
Let us consider {a — bf
(a-bf = (a-b)Q-b)
= a~ — ab — ba + b
= a' — ab — ab + b
= a' - lab + b^
Thus, (a - by
a^ — lab + b~
Geometrical Proof oi (a- bf
The Area of the square ABCD is a^ sq. units.
The Area of the square AHFE with side
{a - b)is (a - bf sq. units.
This is the area of the blue coloured square
portion.
The Area of the rectangles,
BCIH = fl X ^ sq. units.
EGCD = fl X ^ sq. units.
The area of the square FGCI = b- square units.
A
(a-b) E b D
H
b
B
(fl - bf
(a-b) G h
I
-I
b
C
Algebra
General Identity
Let us consider (x + a) (x + b)
ix + a)(x + b) = x' + bx + ax + ab
= x^ + ax + bx + ab
Thus,
{x + a){x + b) = x^ + {a + b)x + ab
Geometrical Proof of {x + a){x + b)
The area of the rectangle
ABCD = {x + a){x + b)
= Area of the square DHIG +
Area of the rectangle AEIH +
Area of the rectangle IFCG +
Area of the rectangle EBFI
= x^ + ax + bx + ab
= x^' + {a + b)x + ab
{x + a){x + b) = x^ + {a + b)x + ab
D
G
H
a
tsTsTsH
i?i?i?i
■" tiMtS
i?imi
■iTlTif
c ^
+
Algebraic
Identities
• (a + bf = a' + lab + b'-
• (a - bf = a'-
2ab + b'-
• {a + b){a-b)
= a'-b'
• {x + a){x + b)
= x^ + (a +
b)x
+ ab
ll
(Usually, in identities the symbol ' = ' is used. Here we use '=' for simplicity)
2.4.2 Applying the Identities
Example 2.9
Expand (i) {x + 5f (ii) {x + lyf (iii) {2x + Syf (iv) 105'.
Solution
(x + Sy = x' + 2{x){5) + 5'
= x'+ lOx + 25
(x + Sy = {x + 5){x + 5)
= x{x + 5) + 5{x + 5)
= x' + 5x + 5x + 25
= x'+ lOx + 25
(i)
Aliter:
Using the identity: V
(a + bf = a^ + 2ab + b^
Here, a = x, b.p^,.
Algebra
(ii)
{x + 2yf
=
x' + 2{x){2y) + {2yy
Using the identity:
=
x' + Axy + Ay^
(a + bf = a^ + 2ab + b^
Here, a - x, b - 2y. ^m
Aliter:
{x + 2yf
=
{x + 2y){x + 2y)
=
x{x + 2y) + 2y{x + 2y)
=
x' + 2xy + 2yx + 4y^ [•.• xy = yx]
=
x' + Axy + Ay^
(iii)
{2x + Syf
=
{2xf + 2{2x){3y) + {3yf
4x'+ 12x>' + 9/
Using the identity:
(a + bf = a^ + 2ab + b^
Here, a-2x,b- 3y.^B
Aliter:
{2x + 3yf
=
(2x + 3y){2x + 3y)
=
2x(2x + 3y) + 3y{2x + 3y)
=
(2x)(2x) + i2x){3y) + {3y){2x) + {3y){3y)
=
4x' + 6xy + 6yx + 9y' [•.• xy = yx]
{2x + 3yf
=
4x'+ 12x>' + 9/
(iv)
105'
=
(100 + 5)'
100' + 2(100)(5) + 5'
W Using the identity:
(a + by = a^ + 2ab + b^
Here, a = 100, b = 5. ^
=
(100x100)+ 1000 + 25
=
10000 + 1000 + 25
105'
=
11025
Example 2.10
Find the values of (i) (x -
-yf {ii){3p-2qj (iii) 97' (iv) (4.9)'
Solution
(i)
(x - yf
=
x'-2{x){y) + f
x' — 2xy + y^
Using the identity:
(a - bf = a^ - 2ab + b^
Here, a - x, b - y.
(ii)
{3p - 2qf
^
i3py-2{3p){2q) + {2qf
9p- - \2pq + 4^'
Using the identity:
(a - bf = a^ - 2ab + b^
Here, a - 3p, b - 2q.
(iii)
97'
=
(100-3)'
(100)'-2(100)(3)+3'
Using the identity:
(a - bf = a^ - 2ab + b^
Here, a = 100, b = 3.
=
10000 -600+9
=
9400 + 9
~-
9409
Chapter 2
(iv)
(4.9)' - (5.0 - 0.1 y Using the identity:
= (5.0f-2(5.0)(0.1) + (0.iy (a-bf^a^-Iab + b^
Here, a = 5.0, b = 0.1.
= 25.00- 1.00 + 0.01
= 24.01
Example 2.11
Evaluate the following using the identity (a + b){a — b) = a^ — b^
(i)(x + 3)(x-3) {\\){5a + 3b){5a - 3b) (iii)52x48 (iv)997'-3'.
Solution
(i) (x + 3)(x-3) = %'-?>-
= x'-9
Using the identity:
{a + b)ia-b)^a^- b^
Here, a - x, b - 3.
(ii) {5a + 3b){5a - 3b) = {5a)'-{3bf
= 25a' -W
Using the identity:
(a + b)(a-b)= V-b^
Here, a-Sa,b- 3b.
(iii)
(iv)
52 X 48 = (50 + 2)(50 - 2) Using the identity:
(a + b)(a-b)^a^- b^
■^^ ^ Here, a = 50, b = 2.
= 2500-4
= 2496
9972 _ 32 = (997 _|_ 3)(997 _ 3) Using the identity:
a^-b^^ia + b)ia-b)
= (1000)(994) Here, a = 997, b = 3.
= 994000
Example 2.12
Using the identity (x + a){x + b) = x' + {a + b)x + ab, find the values of the
following: (i) (m + 3){m + 5) (ii) {p - 2){p - 3) (iii) {2x + 3y){2x - Ay)
(iv) 55x56 (v) 95x103 (vi) 501x505
Solution
(i) (m + 3)(m + 5) = m' + (3 + 5)m + (3)(5) ^ J''T''"!'"!''''l
ah
- m' + 8m + 15
Here, x = m, a = 3, fa = 5.
(ii)
(p-2)(p-3)
= p^ + (-2-3)p + (-2)(-3)
= p^ + (-5)p + 6
= /?" - 5/? + 6
Using
the identity:
(x + a) (x + b)
= X- + (a + b) X
+ ab
Here
,x=p,a
= -2,b = -3.
(iii) {2x + 3y){2x - Ay) =
Algebra
{2xf + {3y-4y){2x) + {3y){-4y)
4x' + (-y){2x) - \2f ^^'"9 "'^ '■''''"">'•■
Ax^ — 2xy — 1 2y^ Here : X, a,b are 2x, 3y, -Ay.
(iv)
55x56 = (50 + 5)(50 + 6)
= 50- + (5 + 6)50 + (5)(6)
= (50x50) + (ll)50 + 30
Using the identity:
{x + d)(x + b)=x^ + {a + b)x + ah
2500 + 550 + 30 Here,x = 50,a = 5,b = 6.
(V)
= 3080
95x103 = (100-5)(100 + 3)
= (100)' + (-5 + 3)(100) + (-5)(3)
= (100xl00) + (-2)(100)- 15
10000 -200
9800 - 15
9785
-| c Using the identity:
{x + a){x + b)=x- + {a + b)x + ab
Here, x = 100, a = - 5, b = 3.
(vi)
501x505 = (500 + l)(500 + 5)
= (500y + (l + 5)(500) + (l)(5)
= (500x500) + (6)(500) + (l)(5)
= (500x500) + (6)(500) + 5
= 250000 + 3000 + 5
= 253005
2.4.3 Deducing some useful Identities
Let us consider,
(i) {a + by + {a-by = {a- + 2ab + b') + {a- - 2ab + b')
= a' + :yab + b' + a'- '/ab + b^
= 2a' + 2b'
(a + bf + ia-bf = 2{a' + b')
^[{a + bf + {a - bf] = a' + b'
(ii) {a + bf-{a-bf = {a' + 2ab + b') - (a' - 2ab + b')
= / + 2ab+^'-/ + 2ab-/>'
{a + b)- - {a - b)- = Aab
^[{a + bf - {a - bf] = ab
Using the identity:
(x + a) (x + b) = X- + (a + b) X + ab
Here, x = 500, a = 1, b = 5.
Chapter 2
(iii)
(iv)
(V)
(vi)
a + bf-2ab = a' + b' + ^ab - Yab
= a' + b'
a + bf - lab = a' + b'
a + bf - Aab = a" + lab + b^ - Aab
= a' - lab + b^
= {a -by
a + b)--4ab = {a - bf
a-bf + lab = a' - Ijdb + b' + l^b
= a' + b'
a-bf + lab = a- + b^
a- bf + Aab = a^ - lab + b^ + Aab
= a' + lab + b^
Deduced
[dentities
• ^[{a + bY + {a
- bf] = a' + b'
. L^^a + bf-{a
- bf] = ab
• (a + bf - lab =
--a' + b'
• (a + bf - Aab -
-- {a - bf
' {a - bf + lab --
--a' + b'
• {a - bf + Aab --
-- {a + bf
= {a + bf
(a-bf + Aab = {a + bf
Example 2.13
If the values of a + b and a- b are 7 and 4 respectively, find the values of
a^ + b^ and ab.
Solution
(i)
a' + b' = ^[{a + bf + {a-bf]
- 1
[T + A^] [Substituting the values of a + b = 7, a - b = 4]
a' + b'
(ii)
= ^(49+16)
= i(65)
^ 61
2
= 61
2
1
ab = ^[{a + bf - {a - bf]
= 4-(7^ — 4^) [Substituting the values of a + b = 7, a - b = 4]
^ 1
4
= 1
4
ab = ^
A
(49- 16)
(33)
Algebra
(i)
(ii)
Example 2.14
If {a + b)= 10 and ab = 20, find a' + b' and (a - bf.
Solution
a^ + b^' = {a + bf - lab [Substituting a + b = 10, ab = 20]
a' + b' = (10/ -2(20)
= 100-40 =60
a' + b' = 60
{a- bf = {a + bf - Aab [Substituting a + b = 10, ab = 20]
= (10^-4(20)
= 100 -80
(a-bf = 20
Example 2.15
If {x + l){x + m) = x^ + 4x + 2 find /' + m' and (/ - mf
Solution
By product formula, we know
(x + l){x + m) = x^ + {l + m)x + lm
So, by comparing RHS with x^ + Ax + 2,we have,
I + m =4 and Im = 2
Now, f + m' = (1 + mf - 2lm
= 4' -2(2) = 16 -4
1.
(i)
(ii)
(iii)
I' + m'
(1 - mf
(1 - mf
= 12
= {l + mf-4lm
= 4'- 4(2) = 16
= 8
- 8
(D) (a -
(D) (a -
(D)a^
EXERCISE 2.3
Choose the correct answer
(a + bY = {a + b)X
for the following:
+ fa)
+ 2ab + b'
(A) ah (B) 2ah (C) (a
ia-bV- = (a-b)X
-b)
(A) (a + b) (B) - 2ab (C) ab
(a' - b') = (a-b)X
-b)
(A) (a - fa) (B) (a + fa) (C) a'
- 2ab + b^
Chapter 2
(iv)
9.6^ =
(A) 9216 (B)93.6 (C) 9.216 (D) 92.16
(v)
{a + by - (a - by =
(A) 4ab (B) 2ab (C) a' + lab + b' (D) 2 (a' + b')
(vi)
m^ + (c + d)m + cd =
(A)(m + cy (B) (m + c) (m + d) (C) {m + dy (D) (m + c) (m - d)
2.
Using a suitable identity, find each of the following products:
(i) {x + 3){x + 3) (ii) (2m + 3)(2m + 3)
(iii) (2x-5)(2x-5) (iv) («-^)(«-^)
(v) (3x + 2)(3x-2) (vi) {5a - 3b){5a - 3b)
(vii) (2/ - 3m)(2/ + 3m) (viii) (| - x)(^ + x)
(ix) ('- + -V---) (x) (100 + 3)(100-3)
^ ' \x yl\x yl
3.
Using the identity {x + a){x + b) = x^ + {a + b)x + ab, find out the foUowing
products:
(i) (x + 4)(x + 7) (ii) (5x + 3)(5x + 4)
(iii) (7x + 3);)(7jc - 3}^) (iv) (8;c - 5)(8;c - 2)
(v) (2m + 3n)(2m + 4n) (vi) {xy - 3){xy - 1)
(vii) [a + l)(a + 1) (viii) (2 + x)(2 - y)
4.
Find out the following squares by using the identities:
(i) {p-qf (ii) {a-Sf (iii) (3x + 5)^
(iv) (5;c-4)' (v) {lx + 3yf (vi) (10m - 9?i)'
(vii) (0.4a - Q.Sb)' (viii) (x - ^)' (ix) (| - ^J
(x) 0.54x0.54-0.46x0.46
5.
Evaluate the following by using the identities:
(i) 103' (ii) 48' (iii) 54'
(iv) 92' (v) 998' (vi) 53x47
(vii) 96 X 104 (viii) 28 x 32 (ix) 81 x 79
(x) 2.8' (xi) 12.1' -7.9' (xii) 9.7x9.8
6.
Show that
(i) {3x + 7)' - 84;c = {3x - 7)'
(ii) {a - b){a + b) + {b- c){b + c) + {c- a){c + a) = Q
Algebra
7. li a + b = 5 and a- b = 4, find a^ + b^ and ab.
8. (i) If the values oi a + b and ab are 12 and 32 respectively, find the values of
a' + ^' and (a- bf.
(ii) If the values oi (a — b) and ab are 6 and 40 respectively, find the values of
a' + Z?' and {a + bf.
9. If (x + a){x + b) = x^ - 5x - 300, find the values of a^ + b'.
10. Deduce the Algebraic identity for {x + a){x + b){x + c) by using the product
formula. [Hint: (x + a){x + b){x + c) = (x + a)[{x + b){x + c)]]
2.5 Factorization
Let us take the natural number 20.
We can write it as the
product of
following numbers.
20 = 1 X 20
20 = 2 X 10
20 = 4 X 5
@^ ^)
&
X (20) X
0x0
1
We can arrange the twenty tea cups
in different ways as given below:
1x20
^^►©i5©©SSS>©
2x10
4x5
The number 20 has 6 factors : 1, 2, 4, 5, 10 and 20.
Among these factors 2 and 5 are the prime factors of 20.
The prime factor form of 20 = 2 x 2 x 5.
20
10
KsiK^o you know?
1. A whole number greater than 1 for which the only factors are 1 and itself,
is called a prime number. Example: 2, 3, 5, 7 etc.
2. A whole number greater than 1 which has more than two factors is called
a composite number. Example: 4, 6, 8, 9, 10 etc.
3. While writing a number as a product of factors, we do not normally write
1 as a factor, since 1 is a factor of any number.
4. Every natural number is either prime or composite.
5. 1 is neither prime nor composite.
Chapter 2
2.5.1. What is factorization?
We can also write any algebraic expression as the product of its factors.
Factorization: The process of expressing any polynomial as a product of its factors
is called factorization.
We can express the following algebraic expressions as the product of their
factors:
(i) 6x' = {2x){3x')
(ii) 3a'b + 3ab' = {3ab){a + b)
(iii) 2x' + X - 6 = (2x - 3) (x + 2)
We can also write the above examples as follows:
Algebraic
Expression
Factor 1
Factor 2
Can we factorize further?
Factor 1
Factor 2
6x'
2x
3x2
Yes. 2x = 2 X X
Yes. 3x2 = 3 X X XX
3a^ b + 3ab^
(Sab)
(a+b)
Yes. Sab = 3 x a x b
No. (a+b) can't be
factorized further
2x2 + X - 6
(2x-3)
(x + 2)
No. (2x - 3) cannot
be factorized further
No. (x + 2) cannot be
factorized further
Note: A factor that cannot be factorized further is known as irreducible factor.
In the above examples (a + b), {2x — 3), and {x + 2) are irreducible.
2.5.2. Factorization by taking out the common factor
In this method, we rewrite the expression with the common factors outside
brackets. Remember that common factors of two or more terms are factors that appear
in all the terms.
Example 2.16
Factorize the following expressions:
(i) 2x + 6 (ii) Ax' + 20xy (iii) 3x' - I2xy (iv) a'b- ab'
(v) 3x' - 5x' + 6x (vi) 11' m'--21lm'n + 2S Im
Solution
(i) 2x + 6 = 2x + (2x3)
.". 2x + 6 = 2 (x + 3) (Note that '2' is common to both terms. )
Note: (i) Here, the factors of {2x + 6) are 2 and {x + 3).
(ii) The factors 2 and (;c + 3) cannot be reduced further.
Therefore 2 and {x + 3) are irreducible factors.
Algebra
(ii)
(iii)
(iv)
(V)
4x^ + 20xy = i4xxXx) + (4x5XxXy)
Ax {x + 5y)
[Taking out the common term 4a:]
3x^-12xy = (3xxxx)-i3x4:XxXy)
= 3x ( X — 4y) [Taking out the common term 3x]
a'b-ab'- = {aXaXb)-{aXbXb)
— ab {a — b) [Taking out the common term ab]
3x'-5x^ + 6x = {3xxXxXx)-{5xxXx) + i6xx)
= X {3x — 5x + 6) [Taking out the common term x]
(vi) ll'm'-2llm'n + 2Slm
= {IXlXlXl XmXm)- {1x3x1 XmXmXn) + {IxAxlXm)
= llm (fm — 3mn + 4) [Taking out the common term 7/m]
2.5.3. Factorization by Grouping the terms
In this method, the terms in the given expression can be arranged in groups of
two or three so as to get a common factor.
Example 2.17
Factorize the following:
(i) x^ — 3x^ + X — 3 (ii) 2xy — 3ab + 2bx — 3ay
(iii) 2m' - 1 Omn - 2m + 1 On (iv) ab{x^ + \) + x (a' + b^)
Solution
(i) x'-3r + X-3 = x'{x-3) + \{x-3) [By grouping the first two and
''■^v— ^ '•-%—' the last two and taking out the
= {x' + \){x — 3) common terms]
(ii) 2xy — 3ab + 2bx — 3ay = 2xy + 2bx — 3ab — 3ay [Rearranging the terms]
= 2x {y + b) — 3a (y + b) [Taking out the common terms]
= {2x-3a){y + b)
(iii)2m' — lOmn — 2m + lOn = 2m (m — 5n) — 2{m — 5n)
= (2m — 2) (m — 5n) [Taking out the common terms]
{\v)ab{x' + \) + x{a^ + b^) = abx^ + ab + xa^ + xb^
= abx^ + a' X + b' X + ab [Rearranging the terms]
= ax (bx + a) + b {bx + a) [Taking out the common terms]
= {ax + b) {bx + a)
2.5.4. Factorization by using Identities
Recall:
(i)
{a
+ by^
a' + lab + b^
(ii)
{a
-by =
a- - lab + b^
(iii)
(a
+ b){a
-b) = a'-b'
Sometimes, the given polynomial or expression can be written in the form of
above mentioned Identities. The expressions
on the LHS are the factors of the expressions
of RHS.
In this method, we consider the
following illustrations and learn how to use
the identities for factorization.
Example 2.18
Expression
Factors
a^ + lab + b^
(a + b) and (a + b)
a^ - lab + b^
(a - b) and (a - b)
a'-b^
(a + b) and (a - b)
(iii) 49m' -56m + 16
(vi) m'-n'
Factorize the following using the Identities:
(i) x' + 6jc + 9 (ii) x'-lOx + lS
(iv) X--6A (v) 9x'y-Af
Solution
(i) x' + 6x + 9
Comparing x' + 6;c + 9 with a^ + lab + b^,we see that a = x,b = 3.
Now, x' + 6x + 9 = x' + l{x){3) + 3'
Using, a' + lab + b^ = (a + by, a = x and b = 3,
we get X' + 6x + 9 = {x + 3y .
.-. The factors of x' + 6x + 9 are (x + 3) and (x + 3).
(ii) x'-lOx + 15
Comparing x^ - \0x + 15 with a^ — lab + b\we see that a = x,b = S.
Now, x'-lOx + 15 = X- - l{x){5) + 5'
Using, a' - lab + b^ = {a — by, a = x and b = 5,
we get x^-lOx + 15 = (x-Sf.
.-.The factors of x^ - \0x + 25 are {x - 5)and (x - 5)
(iii) 49m- - 56m + 16
In this expression, we can express 49m' = (7m) 'and 16 = 4'
Algebra
Using the Identity a^ - lab + b^ = (a - ii)' with a = 7m and b = 4,
49m' - 56m + 16 = (7m)' - 2 (7m) (4) + 4'
= (7m - 4)'
.-.The factors of 49m' - 56m + 16 are (7m - 4) and (7m - 4)
(iv) Comparing x^ - 64 with a' - b^, we see that a = x and b = 8
Using, a- - b^ = {a + b){a-b),
x'-64 = x2-82
= (x + 8) (x - 8)
.-.The factors of x' - 64 are (x + 8) and (x— 8)
(v) 9x'y-Ay' = y [Bx^ - 4/]
= y[{3xr-{2yr]
Comparing (3x)' - (lyY with a' - b\ we see that a = 3x and b = 2y
Using a' - fo' = (a + fo) {a- b), where, a = 3x,b = 2y,
9x'y-4y' = y[{3x + 2y) i3x - 2y)]
(vi) m'-n' = (my - (ny
= (m* + n"^) (m* — n*) [Using the Identity a - b = {a + b) {a - b)]
2 2
[•.• m — n = (m + n) (m — n)]
= im' + n')[imy-iny]
= im' + n') Vim' + «') (m' - n')]
= (m* + n") (m' + n') [(m + n)(m — n)\
m^ — rf = (m* + n'^){m' + n') (m + n) (m — n)
2.5.5 Factorization by using the Identity {x + a){x + b) = x' + {a + b)x + ab
Let us now discuss how we can use the form {x + a){x + b) = x' + {a + b)x + ab
to factorize the expressions.
Example 2.19
Factorize x' + 5x + 6
Solution
Comparing x' + 5x + 6 with x' + (a + b)x + ab
We have, afa = 6, a + b = 5 and x = x.
If ab = 6, it means a and b are factors of 6.
Let us try with a = 2 and b = 3. These values satisfy ab = 6 and a + b = 5.
Therefore the pair of values a = 2 and b = 3 is the right choice.
For this ab = 2 X 3 = 6
and a + b = 2 + 3 = 5.
Using, X + (a + b)x + ab = (x + a)(x + b)
:.x' + 5x + 6 = ;c' + (2 + 3)x + (2x3)
= {x + 2){x + 3)
.-. {x + 2) and {x + 3) are the factors of x" + 5x + 6.
Example 2.20
Factorize: x^ + x- 6
Solution
Comparing x^ + x - 6 with x^ + (a + b)x + ab = {x + a){x + b) ,
we get ab = - 6 and a + b = 1
To find the two numbers a and b such that ab = - 6 and a + b = 1. The values of
a and b may be tabulated as follows:
a
b
ah
a + h
Choice
1
6
6
7
X
1
-6
-6
-5
X
2
3
6
5
X
2
-3
-6
-1
X
-2
3
-6
1
-/
Here, we have to select the pair of factors a = - 2 and b = 3, because they alone
satisfy the conditions ab = - 6 and a + b = 1
Using {x + a){x -\-b) = X + {a + b)x + ab, we get
x" + X-6 = {x-2){x + ?>). I
Example 2.21
Factorize: x^ + 6x + 8
Solution
Comparing x~ + 6x + S with x'- + (a + b)x + ab = {x + a){x + b),
we get ab = 8 and a + b = 6.
.-. x' + 6x + 8 = x' + (2 + 4)jc + (2 X 4)
= (x + 2)(x + 4)
The factors of X^ + 6x + 8 are {x + 2) and {x + A). Hence the correct
factors are 2, 4
Factors
of 8
Sum of
factors
1,8
9
2,4
6
Algebra
1. Choose the correct answer for the following :
(i) The factors of 3a + 2 lab are
(A) a , (3 + 21b) (B) 3, (a + 7b) (C) 3a, (1 + 7b)
(ii) The factors of jc^ — x — 12 are
(A)(x + 4),(x-3) (B)(x-4),(x-3) (C) (x + 2), (x - 6) (D) (x + 3), (x - 4)
(ill) The factors of 6x^ — x — 15 are (2x + 3) and
(A) (3x - 5) (B) (3x + 5) (C) (5x - 3)
(iv) The factors of 169Z' - 441m' are
(A) (13/ - 21 m), (13/ - 21 m)
(C) (13/ -21m), (13/ + 21m)
(v) The product of (x - 1) (2x - 3) is
(A) 2x' - 5x - 3 (B) 2x' - 5x + 3 (C) 2x' + 5x - 3
(B) (13/ + 21m), (13/ + 21m)
(D) 13 (/ + 21m), 13(/-21m)
(D)2x' + 5:^ + 3
(iv)- I2y + 20}^'
(viii)17/' + S5m'
Factorize the following expressions:
(i) 3x - 45 (n)lx - Uy (m)5a- + 35a
(v) I5a'b + 35ab iy\)pq — pqr (vii)18m^ - A5mn^
(ix) 6x'y - Ux'y + \5x' {x)2a'b' - Ua'b' + Aa'b
Factorize:
(i)2ab + 2b + 3a (ii)6xy - Ay + 6 - 9x (iii)2x + 3xy + 2y + 3y'
(iv) 15^1 — 3bx' — 5b + x^ (v) a^x^ + axy + abx + by
(vi) a^x + abx + ac + aby + b'y + be (vii) ax^ + bx^ + ax + by
(viii) mx — my — nx + ny (ix) 2m + 3m — 2m —3 (x) a' + lib + I lab + a
Factorize:
(i) a' + Ua + 49 (ii) x' - \2x + 36
(iv) 25x' - 2<^xy + Ay (v) 169m' - 625n'
(vii) 121a' + \5Aab^A9b^
(ix) 36 - 49x' (x) 1 - 6:x + 9x'
Factorize :
(i)x' + 7x + 12 (ii)p' - 6p + 8 (iii)m'
(iii) Ap - 25q
(vi) x~ + ^x + -
(viii) 3x —15x
Am -21
(iv) X - \Ax + 45 (v);c - 24;c + 108 (vi)a + 13a + 12
(vii) x' - 5x + 6 (viii)x' — 14x>' + 2Ay^
(ix)m'-21m-72 (x) x' - 28x + 132
2.6 Division of Algebraic Expressions
2.6.1 Division of a Monomial by another Monomial
Consider 10 ^ 2, we may write this as ^ = y = 5
Similarly, (i) lOx ^ 2 may be written as ^ = ^x^x^ = 5^
(ii) 10x=^2x = 1^ = 5xl>or = SxMxx = 5^
(iii) lOx^ - 2x = ^ = 5X2^X/XXXX = 5^2
^ ^ Ix ^X/
Civ) lOx' - 2x' = ^^•^^ = 5 x:^x/x/xxXxXx _ ^ 3
^ ^ ■ 2x' ^X^X/
Instead, we can also use the law of exponent ^^ = a'" ";
a"
Thus in (iv), we can write
lOx' _ 10
2x'
-X
5x'
(v) Sa'b'c' - I5abc = ^^^ = ^X^XaX^X^x/x c ^ ahcL = Labc
^^ \5abc 0X3X^XJ^X/ 3 3
(or) 5a^ b^ c'- ^ I5abc
\5abc
- 5
1
, ^ a' 'Z?' 'c' ' = ^a&c [using ^^ = a'" "]
1 J J a"
2.6.2 Division of a Polynomial by a Monomial and Binomial
Let us consider the following example.
Example 2.22
Solve: (i) {Ix' -5x)^x (ii) {x' - 3x' + 2x') -^ 3x' (iii) (8x' - 5x- + 6x) - 2x
Solution
(i) (Ix' — 5x)^x =
- Ix' _ 5x_
- 1 XxXx
\
Alternative Method:
/ J\i iJiAi —
5x.x:
X
= Ix —
(ii) {x'-3x' + 2x')^3x'
3?
- 7x2-1-5x1-1
= 7x1 _ 5;^o = 7^ _ 5 ^
[•.• a° = 1]
= 7x - 5 !
3x^
- 1
X — X' -\-
Alternative Method:
"^^
we can
find the
common term x^ and simplify
it as
x^
- 3x^ + 2x2
_ x^Cx''
-3x^ + 2)
3x2
_ x^
3
3x2
- 3x2 + 2) 1
1
Algebra
(ill) {Sx' - 5x' + 6x) ^ 2x
^^^^^^^P Alternative Method ^^^H
For 8x^ — 5x' + 6x, separating 2x from each term
- 8r'-5x' + 6x
2x ^^ 8^^ '
_ Sx^ 5x^ ^ 6x ^^' - 5^' + 6x = 2x(4x^) - 2x(|-x) + 2x(3)
^' ^' ^' =2x(4x^-|x + 3)
= 4x' - |-x + 3
8x^ - 5x^ + (
2x
2x(4x^ - ^x + 3)
3X _ 2
2x ^^^
Example 2.23
Solve: {5x'+ 10x)-(x + 2)
^^^^^^^^^^^^^g
So/ution
(5x^+1 Ox) : (x + 2) - 5x'+10x
X + 2
Let us factorize the numerator (5x" + lOx).
5x'+10x = (5xxXx) + (5x2xx)
— 5x (x + 2) [Taking out the common term 5x]
Now, (5x' + lOx) : (x + 2) - ^^^ + ^^^^
1.
Simplify:
_ 5x(jM^^ _
5x. [By cancelling (x + 2)]
EXERCISE 2.5
(i)16x*-32x
(ii) - 42/ - If (iii) Z^a'h'c' - 45a^c
(iv) {Im} - 6m) -^ 2m
(v) 25x>' - 15x'>' (vi) (- lll'm'rf) - (- 8/'m'n')
2.
Work out the following divisions:
ii)5y-Af + 3y^y
(ii)(9x'-15x'-21x')-(3x')
(in) (5x' - 4x' + 3x) - (2x)
(iv) Ax^y - 28xy + 4xy2 ^ (4xy)
(v) (Sx'yz - Axfz + 3x'>'z')
-(2x3;z)
3.
Simplify the follwing expre
>ssions:
(i)(x' + 7x+10)-(x + 2)
(ii)(a' + 24a+ 144) -(fl+ 12)
(iii) (m' + 5m - 14) ^ (x +
7) (iv) (25m' - An") - (5m + In)
(v) (4a'-4aZ?- 15^') -(2(
2 - 5b) (vi) (a' - b') -r{a-b)
Chapter 2
2.7 Solving Linear Equations
In class VII, we have learnt about algebraic expressions and linear equations in
one variable. Let us recall them now.
Look at the following examples:
(i) 2;c = 8 (ii) 3x' = 50
{w)^x + ^y = 4: (vi)3x' = 81
These are all equations.
(iii) 5x' - 2 = 102 (iv) 2x - 3 = 5
(vii) 2(5:x + 1) - (2x + 1) = 6x + 2
A statement in which two expressions are connected by ' = ' sign, is called as an
equation. In other words, an equation is a statement of equality which contains one
or more variables.
In the above equations (i), (iv), (v) and (vii), we see that power of each variable
is one. Such equations are called linear equations.
An equation which involves one or more variables whose power is 1 is called a
linear equation.
Therefore, the equations (ii), (iii) and (vi) are not linear equations. (Since the
highest power of the variable > 1)
Understanding the equations
Consider the equation 2x - 3 = 5.
(i) An algebraic equation is an equality involving [variab le | | Equali ty |
variables and constants. 2x - 3 = 5
(ii) Every equation has an equality sign. The expression
on the left of the equality sign is the Left Hand Side
(LHS). The expression on the right of the equality
sign is the Right Hand Side (RHS).
(iii) In an equation the values of
the expression on the LHS and
RHS are equal. It is true only for
certain values of the variables.
These values are the solutions
or roots of the equation.
Equation
2x-3 = LHS
5 = RHS
x =
4 is the solution of the
equation.
2x
-3 = 5 when x = 4,
LHS = 2 (4) - 3
= 8-3 = 5 = RHS
X =
5 is not a solution of this equation.
If A
= 5, LHS = 2 (5) - 3
= 10 - 3 =
7 ^ RHS.
Algebra
Rules for Solving an Equation
We may use any one or two or all the three of the following rules in solving
an equation.
1. We can add or subtract the same quantity to both sides of an equation
without changing the equality.
2. Both sides of an equation can be multiplied or divided by the same non-
zero number without changing the equality.
3. Transposition Method: For solving an equation, we need to collect
all the terms containing the variables on one side of the equation and
constant terms on the other side of the equation. This can be done by
transferring some terms from one side to the other. Any term of an
equation may be shifted from one side to the other by changing its sign.
This process is called as Method of Transposition.
2.7.1 Linear Equation in one variable
We have learnt in class VII to solve linear equations in one variable. Consider
the linear equation of the type ax + b = where a^ 0.
Example 2.24
Find the solution of 5x - 13 = 42.
Solution
Step 1 : Add 13 to both sides, 5x - 13 + 13 = 42 + 13
5x = 55
A Linear equation in one
variable has a unique solution.
Step 2 : Divide both sides
bys, f = 5|
Verification:
LHS = 5 X 11 - 13
X = 11
= 55-13
Transposition method:
= 42
5X-13
= 42
= RHS
5x
= 42 + 13 [Transposing - 13 to RHS]
5x
= 55
5x
5
= ^ [Dividing both sides by 5]
X
= 11
Example 2.25
Solve: Sy + 9 = 24
Solution
5y + 9 = lA
we get, 5y + 9-9 = 1A-
9 (Subtracting 9 on both the sides)
Chapter 2
5y = 15
5y _ 15
-^ — -^ [Dividing both the sides by 5]
y = 3
Verification:
LHS = 5(3) + 9 = 24 = RHS
Example 2.26
Solve: 2x + 5 = 23 - X
Solution
2x + 5 = 23 — X
2x + 5 — 5 = 23— X — 5
[Adding - 5 both sides]
2x = 18 —X
2x + X = 18— x + x
Alternative Method
5y + 9 = 24
5y =
24-9 [Transposing 9 to RHS]
5y =
15 and y = — . Hence y = 3
Alternative Method
2x + 5
= 23
— X
2x + x
= 23
— 5 [By transposition]
3x
= 18
X
18
3
[Dividing both sides
by 3]
X
= 6
3x= 18
3x ^18
3 3
x = 6
Example 2.27
Solve : ym + m =
Solution
m + m = 22
Verification: LHS = 2x + 5 = 2 (6) + 5 = 17,
RHS = 23-x = 23-6 = 17.
[Dividing both the sides by 3]
22
2
9m + 2m
= 22
11m _
m =
m = 4
Example 2.28
Solve: --4- =
Solution
22
22X2
11
1
9
[Taking LCM on LHS]
[By cross multiplication]
Verification:
LHS
Q
= ym + m =
f (4) + 4
= 18 + 4 = 22
= RHS
2__ J_
6-5
3x
X
3x
1
9
1
9
[Taking LCM on LHS]
OA Zj J A — _ J A lJ.
Verification:
LHS
^ 2__
5
3x
_ 2
3
5
3(3)
. 2 5
" 3 9
_ 6-
9
5 _ 1
9
= RHS
Algebra
Example 2.29
Find the two consecutive positive odd integers whose sum is 32.
Solution
Let the two consecutive positive odd integers be x and {x + 2).
Then, their sum is 32.
.-. (x) + {x + 2) =32
2x + 2 = 32
2x = 32-2
2x = 30
X = 3(1 = 15
Verification:
15 + 17 = 32
2
Since x = 15, then the other integer, x + 2 = 15 + 2= 17
.-. The two required consecutive positive odd integers are 15 and 17.
Example 2.30
One third of one half of one fifth of a number is 15. Find the number.
Solution
Let the required number be x.
. , . Verification:
Then, ^of^of^ofx = 15.
3 2 5 111
. Ill ir LHS = ^X^X-^XX
I.e. ^XyX^X:^: =15 3 2 5
X = 15x3x2x5 = ixyXyX450
= 15 = RHS
X = 45 X 10 = 450
Hence the required number is 450.
Example 2.31
5 7
A rational number is such that when we multiply it by ^ and add ^ to the
product we get ^^. What is the number?
Solution
Let the rational number be x.
When we multiply it by y and add ^ to the product we get ^^.
i.e.,
5 2
xxf + f
- -1
12
Sx
- -7 2
2
12 3
- -7-8
12
_ -15
12
Chapter 2
- -15y2
12 ^5
= ^1
2
1
Verification:
LHS =
2 ><2 + 3 -
-5 + 2
4 ^ 3
=
-15 + 8 _ -7
12 12
= RHS.
Verification:
Hence the required number is
Example 2.32
Arun is now half as old as his father. Twelve years ago the father's age was
three times as old as Arun. Find their present ages.
Solution
Let Arun's age be x years now.
Then his father's age = 2x years
12 years ago, Arun's age was (x - 12) years and
his faher's age was (2x - 12) years.
Given that, (2x-12) = 3(x-12)
2x-12 = 3x-36
36-12 = 3x-2x
X = 24
Therefore, Arun's present age = 24 years.
His father's present age = 2 (24) = 48 years.
Example 2.33
By selling a car for ? 1,40,000, a man suffered a loss of 20%. What was the cost
price of the car?
Solution
Let the cost price of the car be x.
Loss of 20% = ^ of X = ^xx = f
We know that.
Cost price - Loss = Selling price = ^^x 175000
Arun's age
Father's age
Now : 24
48
12 years ago
48 - 12 = 36
24-12 = 12
36 = 3 (Arun's age)
= 3(12) = 36
x-^ = 140000
5^-^ = 140000
^ = 140000
X = 140000x4-
4
X = 175000
Hence the cost price of the car is ? 1,75,000.
Verification:
Loss = 20% of 175000
= -20_
100
= ? 35,000
S.P = C.P-Loss
= 175000-35000
= 140000
Algebra
EXERCISE 2.6
1. Solve the following equations:
(i)3x + 5 = 23 (ii)17=10->'
(iv) 6x = 72
(v)3^=-7
(iii) 2y-l = 1
(vi)3(3jc-7) = 5(2jc-3)
(vii) 4(2;c - 3) + 5(3jc - 4) = 14 (viii)
2x + 5 _ 3
(ix)
1
3x + l
^""M + 4 ~ 2
2. Frame and solve the equations for the following statements:
(i) Half of a certain number added to its one third gives 15. Find the number.
(ii) Sum of three consecutive numbers is 90. Find the numbers.
(iii) The breadth of a rectangle is 8 cm less than its length. If the perimeter is
60 cm, find its length and breadth.
(iv) Sum of two numbers is 60. The bigger number is 4 times the smaller one.
Find the numbers.
(v) The sum of the two numbers is 21 and their difference is 3. Find the
numbers. (Hint: Let the bigger number be x and smaller number be x - 3)
(vi) Two numbers are in the ratio 5 : 3. If they differ by 18, what are the
numbers?
(vii) A number decreased by 5% of it is 3800. What is the number?
(viii) The denominator of a fraction is 2 more than its numerator. If one is added
to both the numerator and their denominator the fraction becomes ^. Find
the fraction.
(ix) Mary is 3 times older than Nandhini. After 10 years the sum of their ages
will be 80. Find their present ages.
(x) Murali gives half of his property to his wife, two third of the remainder to
his son and the remaining ^ 50,000 to his daughter. Find the shares of his
wife and son.
Chapter 2
/
Concept Summary
^ Monomial: An Algebraic expression that contains only one term is called
a monomial.
"V Binomial: An expression that contains only two terms is called a binomial.
^ Trinomial: An expression that contains only three terms is called a
trinomial.
'^ Polynomial: An expression containing a finite number of terms with
non-zero coefficient is called a polynomial.
^ Degree of the Polynomial: The highest power of the term is called the
degree of the polynomial.
Like terms contain the same variables with same powers.
• Only like (or) similar terms can be added or subtracted.
• Products of monomials are also monomials.
• The product of a monomial by a binomial is a binomal.
Identities
(a
+ by--
= a- + lab + b-
{a
-by--
= a' - lab + b-
a'
-b' =
{a + b)ia-b)
ix + a){x + b)
= x' + ia + b)x
+ ab
^ Factorization: The process of expressing any polynomial as a product of
its factor is called factorization.
^ Linear Equation: An equation involving one or more variables each with
power 1 is called a Linear equation.
ax + b = is the general form of linear equation in one variable, where
a ^ 0,a,b are constants and x is the variable.
A Linear equation in one variable has a unique solution.
Algebra
Mathematics Club Activity
Algebraic Comedy
Dear students,
By using certain tricks, we can prove 2 = 3. Is it not unheard of? Yes, But can you
find out where the mistake is?
We begin with the unquestionable equahty.
4_10 = 9-15
We add 6-j-to both sides of the equation
4-10 + 6^ = 9-15 + 6^
4 4
We can write it as T - 2(2)(|-) + (|-)^ = 3'- 2(3)(|-) + {^Y
Taking square root on both sides, we get (2 — — ) = (3 — -^)
Adding — to both sides.
2.1 + 1
2 2
3-^ + ^
2 2
We arrive at 2=3
Now we have proved that 2 is equal to 3.
Where is the mistake?
Let us discuss in detail. An error slipped in when we concluded that from
(2 - ^J = (3 - yJ it follows that
2 ^2
From the fact, the squares of number are equal, it does not follow the numbers are
equal.
that
For example, (- 5)' = 5' [•.• (- 5)(- 5) = (5) (5) = 25]
does not imply -5 = 5.
In the above problem we arrived I -^^ \ = ( ^ ) from this we should not conclude
2 T'
Have you got it now?
Life Mathematics
3.1
Introduction
3.2
Revision - Profit, Loss and Simple Interest
3.3
Application of Percentage, Profit and Loss,
Overhead Expenses, Discount and Tax
3.4
Compound Interest
3.5
Difference between S.I. and C.I.
3.6
Fixed Deposits and Recurring Deposits
3.7
Compound Variation
3.8
Time and Work
3.1 Introduction
Every human being wants to reach the height of 'WIN-WIN'
situation throughout his life. To achieve it effectively, he allocates his
time to work, to earn more wealth and fame.
From stone age to present world, from material exchange to money
transaction, for his produce and land, man has been applying the idea
of ratio and proportion. The monumental buildings like the Taj Mahal
and the Tanjore Brihadisvara Temple, known for their aesthetic looks,
also demonstrate our ancestors' knowledge of skill of using right kind of
ratio to keep them strong and wonderful.
Many of the existing things in the world are connected by cause
and effect relationship as in rain and harvest, nutrition and health, income
and expenditure, etc. and hence compound variation arises.
In our effort to survival and ambition to grow, we borrow or deposit
money and compensate the process preferably by means of compound
interest. The government bears the responsibility on the sectors like
security, health, education and other amenities. To deliver these to all
citizen, we pay various taxes from our income to the government.
This chapter covers the topics which are interwoven in our lives.
Roger Bacon
[1214-1294]
English
philosopher,
Wonderful teacher
emphasised on
empirical methods.
He became a
master at Oxford.
He stated:
"Neglect of
mathematics
works injury to all
knowledge".
He said, "The
importance of
mathematics for
a common man
to underpinned
whenever he
visits banks,
shopping malls,
railways, post
offices, insurance
companies,
or deals with
transport, business
transaction,
imports and
exports, trade and
commerce".
Life Mathematics
3.2 Revision: Profit, Loss and Simple Interest
We have already learnt about profit and loss and simple interest in our previous
class. Let us recall the following results:
RESULTS ON PROFIT, LOSS AND SIMPLE INTEREST
(i) Profit or Gain
(ii) Loss
(iii) Profit %
(iv) Loss %
(v) Simple interest (I)
(vi) Amount
Selling price - Cost price
Cost price - Selling price
C.P.
Loss
C.P.
Principal X Time X Rate _ Pnr
XlOO
100
Principal + Interest
100
3.3 Application of Percentage, Profit and Loss, Overhead Expenses,
Discount and Tax
3.3.1. Application of Percentage
We have already learnt percentages in the previous classes. We present these
ideas as follows:
(i)
Two percent = 2% =
100
(ii) 8% of 600 kg =
(ii) 125%
8
X 600 = 48 kg
5-1 1
100
125
100 4 ^4
Now, we learn to apply percentages in some problems.
Example 3.1
What percent is 15 paise of 2 rupees 70 paise?
Solution
2 rupees 70 paise == (2 x 100 paise + 70 paise)
= 200 paise + 70 paise
= 270 paise
Required percentage
(Do you knowTy
50%
25%
75%
^XlOO = ^ = 5|-%.
.95
Chapter 3
Example 3.2
Find the total amount if 12% of it is ? 1080.
Solution
Let the total amount be x .
Given : 12% of the total amount = ? 1080
^^ XX = 1080
100
X = 1080x100=^9000
12
•■■ The total amount = ? 9000.
Example 3.3
72% of 25 students are good in Mathematics. How many are not good in
Mathematics?
Solution
Percentage of students good in Mathematics = 72%
Number of students good in Mathematics = 72% of 25 students
= j^ X 25 = 18 students
Number of students not good in Mathematics = 25 - 18 = 7.
Example 3.4
Find the number which is 15% less than 240.
Solution
15% of 240 = 3^X240 =36
•'■ The required number = 240 - 36 = 204.
Example 3.5
The price of a house is decreased from Rupees Fifteen lakhs to Rupees Twelve
lakhs. Find the percentage of decrease.
Solution
Original price = ? 15,00,000
Change in price = ? 12,00,000
Decrease in price = 15,00,000 - 12,00,000 = 3,00,000
■■■ Percentage of decrease = ^tSS^x 100 = 20%
Remember
Percentage of increase = Increase in amount ^ ;^00
Original amount
Percentage of decrease = Decrease in amount ^ ^^OO
Original amount
Life Mathematics
^
'(0^
\t<t tAea^
15 sweets are divided between Sliaratli and Bliaratli, so tliat tiiey get 20%
and 80% of tliem respectively. Find the number of sweets got by each.
Atrnty
My Grandma says, in her childhood, gold was X 100 per gram.
Read a newspaper to know the price of gold and note down
the price on the first of every month. Find the percentage of
increase every month.
EXERCISE 3.1
1 . Choose the correct answer.
(i) There are 5 oranges in a basket of 25 fruits. The percentage of oranges is
(A) 5% (B) 25% (C) 10% (D) 20%
(ii) ^ = %■
(A) 25 (B) 4 (C) 8 (D) 15
(iii) 15% of the total number of biscuits in a bottle is 30. The total number of biscuits
is .
(B) 200
(A) 100
(C) 150
(D) 300
(iv) The price of a scooter was ? 34,000 last year. It has increased by 25% this year.
Then the increase in price is .
(A)? 6,500 (B)? 8,500 (C)? 8,000 (D)? 7,000
(v) A man saves ? 3,000 per month from his total salary of? 20,000. The percentage
of his savings is .
(A) 15% (B) 5% (C) 10% (D) 20%
2. (i) 20% of the total number of litres of oil is 40 litres.
Find the total quantity of oil in litres.
(ii) 25% of a journey covers 5,000 km. How long is the whole journey?
(iii) 3.5% of an amount is ? 54.25. Find the amount.
(iv) 60% of the total time is 30 minutes. Find the total time.
(v) 4% sales tax on the sale of an article is ? 2. What is the amount of sale?
3. Meenu spends ? 2000 from her salary for recreation which is 5% of her salary.
What is her salary?
4. 25% of the total mangoes which are rotten is 1,250. Find the total number of
mangoes in the basket. Also, find the number of good mangoes.
Chapter 3
5. The marks obtained by Rani in her twelfth standard exams are tabulated
below. Express these marks as percentages.
Subjects
Maximum
marks
Marks obtained
Percentage of
marks (out of 100)
(i) English
200
180
(ii) Tamil
200
188
(iii) Mathematics
200
195
(iv) Physics
150
132
(v) Chemistry
150
142
(vi) Biology
150
140
6. A school cricket team played 20 matches against another school. The hrst school
won 25% of them. How many matches did the hrst school win?
7. Rahim deposited ? 10,000 in a company which pays 18% p.a. simple interest.
Find the interest he gets if the period is 5 years.
8. The marked price of a toy is ? 1,200. The shop keeper gave a discount of 15%.
What is the selling price of the toy?
9. In an interview for a computer firm 1,500 applicants were interviewed. If
12% of them were selected, how many applicants were selected? Also find the
number of applicants who were not selected.
10. An alloy consists of 30% copper and 40% zinc, and the remaining is nickel.
Find the amount of nickel in 20 kilograms of the alloy.
11. Pandian and Thamarai contested for the election to the Panchayat committee
from their village. Pandian secured 11,484 votes which was 44% of the total
votes. Thamarai secured 36% of the votes. Calculate (i) the number of votes
cast in the village and (ii) the number of voters who did not vote for both the
contestants.
12. A man spends 40% of his income for his food, 15% for his clothes and 20% for
house rent and saves the rest. How much percent does he save? If his income is
? 34,400, find the amount of his savings.
13. Jyothika secured 35 marks out of 50 in English and 27 marks out of 30 in
Mathematics. In which subject did she get better marks and by how much?
14. A worker receives ? 11,250 as bonus, which is 15% of his annual salary. What
is his monthly salary?
15. The price of a suit is increased from ? 2,100 to ^ 2,520. Find the percentage of
increase.
Life Mathematics
It^ iS^ede
1 . 40% - 100% - %
2. If 25% of students in a class come to school by walk, 65% of students
come by bicycle and the remaining percentage by school bus, what
percentage come by school bus?
3. In a particular class of students, 30% of them take Hindi, 50% take Tamil
and the remaining take French as their second language. What percent
take French as their second language?
4. In a city, 30% are females, 40% are males and the remaining are children.
What percent are children?
J}0yip/
Amutha buys silk sarees from two different merchants Ganesan and Govindan.
I^Ganesan weaves 200gms of silver thread with lOOgms of bronze thread whereas
Govindan weaves 300gms of silver thread with 200gms of bronze thread for
the sarees.Calculate the percentage of silver thread in each and find who gives a
better quality. [Note: More the silver thread better the quality.]
3.3.2 Applications of Profit and Loss
In this section, we learn to solve problems on applications of profit and loss.
(i) Illustration of the formula for S.P.
Consider the following situation:
Rajesh buys a pen for ? 80 and sells it to his friend.
If he wants to make a profit of 5%, can you say the price for
which he would have sold?
Rajesh bought the pen for ^ 80 which is the Cost
Price (C.R). When he sold, he makes a profit of 5% which is
calculated on the C.R
.-. Profit = 5% of C.R = ^ x 80 = ^ 4
Since there is a profit, S.P. > C.R
S.P. = C.P. + Profit
- 80 + 4 = ^84.
•"• The price for which Rajesh would have sold = ? 84.
The same problem can be done by formula method:
Selling price (S.P.) =
(100 + Profit%)
100
^ (100 + 5)
100
XC.P.
XSO = ISI-XSO =?84.
100
Chapter 3
(ii) Illustration of the formula for C.P.
Consider the following situation:
Suppose a shopkeeper sells a wrist watch for ? 540
making a profit of 5 %, then what would have been the cost
of the watch?
The shopkeeper had sold the watch at a profit of 5 %
on the C.P. Since C.P. is not known, let us take it as ? 100.
Profit of 5% is made on the C.P.
••■ Profit = 5% of C.P.
X 100
100
= ?5.
We know, S.P. = C.P. + Profit
= 100 + 5
= ?105.
Here , if S.P. is ? 105, then C.P is ? 100.
When S.P. of the watch is ? 540, C.P. = ^^^^Qg^^^ = ^ 514.29
•'■ The watch would have cost ^ 514.29 for the shopkeeper.
The above problem can also be solved by using the formula method.
100
C.P. =
100 + profit %
XS.P.
100
100 + 5
X540
= ? 514.29.
We now summarize the formulae to calculate S.P. and C.P. as follows:
1. When there is a profit
(i) C.P =
100
100 + profit%
XS.P.
1. When there is a loss
2. When there is a profit
(i)S.P. = ( "'» + Pf"°''° lxc.P.
2. When there is a loss.
(ii)S.P. = (
100-loss%
100
XC.P.
Life Mathematics
Example 3.6
Hameed buys a colour T.V set for ^ 15,200 and sells it at a loss of 20%. What is
the selling price of the T.V set?
Solutio
n
Raghul
used this method:
Loss is
20%oftheC.P.
- 100 X 1^200
= ?3040
S.P.
= C.P. - Loss
= 15,200-3,040
= ? 12,160
Roshan used the formula method:
C.P. = ? 15, 200
Loss - 20%
OR
S.P
100 - Loss%
100
100-20
xC.P.
100
X 15200
80
X 15200
100
? 12,160
Both Raghul and Roshan came out with the same answer that the selling price
ofthe T.V. set is? 12,160.
Example 3.7
A scooty is sold for ? 13,600 and fetches a loss of 15%. Find the cost price of
the scooty.
Revathi used the formula method:
Loss = 15%.
S.R = ? 13,600
OR
Devi used this method:
Loss of 15% means,
if C.R is ? 100, Loss - ? 15
Therefore, S.P. would be
(100-15) = ?85
If S.P is? 85, C.P is? 100
When S.R is ?13,600, then
C.P
100
100 - Loss%
xS.P
100
100- 15
X 13600
C.R =
100X13600
85
? 16,000
= J^ X 13600
= ? 16,000
Hence the cost price of the scooty is ? 16,000.
Items
Cost price in ?
Profit/Loss
Selling
price in ?
Washing Machine
16,000
9% Profit
Microwave Oven
13,500
12% Loss
Wooden Shelf
13% Loss
6,786
Sofa set
12'/2% Profit
7,000
Air Conditioner
32,400
7% Profit
Chapter 3
Example 3.8
The cost price of 11 pens is equal to the selhng price of 10 pens. Find the loss
or gain percent.
Solution
Let S.P. of each pen be x.
S.P. oflOpens = ? lOx
S.P. of 11 pens = ?llx
Given: C.P. of 11 pens = S.P. of 10 pens = ^ lOx
Here, S.P. > C.P.
•■■ Profit = S.P. - C.P.
= llx - lOx = X
Profit % = ^^ X 100 = ^ X 100 = 10%.
C.P.
\0x
Example 3.9
A man sells two wrist watches at ? 594 each. On one he gains 10% and on the
other he loses 10%. Find his gain or loss percent on the whole.
Solution
Given : S.P. of the first wrist watch = ? 594, Gain% = 10%
100
•. C.P. of the first wrist watch =
100 + profit%
100
XS.P.
(100+ 10)
X594
100
110
X 594 = ? 540.
Similarly, C.P. of the second watch on which he loses 10% is
100 c p
(100 - Loss%) ■ ■
= (10^10) ^='* = W^^'*=^«™-
To say whether there was an overall Profit or Loss, we need to find the combined
C.P. and S.P.
Total C.P. of the two watches = 540 + 660 =? 1,200.
Total S.P. of the two watches = 594 + 594 =? 1,188.
Net Loss = 1,200 - 1,188 = ? 12.
Loss% = ^^XlOO
= ^X 100 = 10/0.
Life Mathematics
3.3.3. Application of Overhead Expenses
Maya went with her father to purchase an ^m"
Air cooler. They bought it for ? 18,000. The shop of ^ /"^^^B
purchase was not nearer to their residence. So they have ?*|f^^» ^^
to arrange some conveyance mode to take the air cooler
to their residence. The conveyance charges came to
^ 500. Hence the C.P. of the air cooler is not only
^ 18,000 but it also includes the Conveyance Charges (Transportation charges)
? 500 which is called as Overhead Expenses .
Now,
C.P. of the air cooler = Real cost + Conveyance charges
= 18,000 + 500-^18,500
Consider another situation, where Kishore's father buys an old Maruti car from
a Chennai dealer for ? 2,75,000 and spends ? 25,000 for painting the car. And then
he transports the car to his native village for which he spends again ? 2,000. Can you
answer the following questions:
(i) What is the the overall cost price of the car?
(ii) What is the real cost price of the car?
(iii) What are the overhead expenses referred here?
In the above example the painting charges and the transportation charges are
the overhead expenses.
.•. Cost price of the car = Real cost price + Overhead expenses
= 2,75,000 + (25,000 + 2,000)
2,75,000 + 27,000 = ? 3,02,000
Thus, we come to the conclusion that.
Sometimes when an article is bought or sold, some additional expenses are made
while buying or before selling it. These expenses have to be included in the cost
price. These expenses are referred to as Overhead Expenses. These may include
expenses like amount spent on repairs, labour charges, transportation, etc..
Example 3.10
Raju bought a motorcycle for ? 36,000 and then bought some extra fittings to
make it perfect and good looking. He sold the bike at a profit of 10% and he got a
selling price of ^ 44,000. How much did he spend for the extra fittings made for the
motorcycle?
Chapter 3
Solution
Let the C.R be ? 100.
Profit = 10%, S.P. = ? 110
If S.P. is ? 110, then C.P. is ? 100.
When S.P is Rs 44,000
C.P = 4400Q X 100 = ^ 40,000
.-. Amount spent on extra fittings = 40,000 - 36,000 = ? 4,000.
EXERCISE 3.2
1. Find the Cost price / Selling price.
Cost price
Selling price
Profit
Loss
(i) ? 7,282
?208
(ii)
?572
?72
(iii) ? 9,684
?684
(iv)
? 1,973
?273
(v) ? 6,76,000
? 18,500
2. Fill up the appropriate boxes and leave the rest.
C.P.
S.P.
Profit & Profit %
Loss & Loss%
(i) ? 320
?384
(ii) ? 2,500
? 2,700
(iii) ^ 380
?361
(iv) ? 40
^ 2 Loss
(v) ? 5,000
? 500 Profit.
3. Find the S.P. if a profit of 5% is made on
(i) a bicycle of ? 700 with ? 50 as overhead charges.
(ii) a computer table bought at ? 1,150 with ? 50 as transportation charges.
(iii) a table-top wet grinder bought for ? 2,560 and an expense of ? 140 made on
its repairs.
Life Mathematics
4. By selling a table for ^ 1,320, a trader gains 10%. Find the C.P. of the table.
5. The cost price of 16 note books is equal to the selling price of 12 note books.
Find the gain percent.
6. A man sold two articles at ? 375 each. On the first article, he gains 25% and
on the other, he looses 25%. How much does he gain or lose in the whole
transaction? Also, find the gain or loss percent in the whole transaction.
7. Anbarasan purchased a house for ? 17,75,000 and spent ? 1,25,000 on its
interior decoration. He sold the house to make a profit of 20%. Find the S.P. of |
the house.
8. After spending Rupees sixty thousand for remodelling a house, Amla sold a
house at a profit of 20%. If the selling price was Rupees forty two lakhs, how
much did she spend to buy the house?
9. Jaikumar bought a plot of land in the outskits of the city for ? 2 1 ,00,000. He
built a wall around it for which he spent ^ 1,45,000. And then he wants to sell it
at ? 25,00,000 by making an advertisement in the newspaper which costs him
? 5,000. Now, find his profit percent.
10. A man sold two varieties of his dog for ? 3,605 each. On one he made a gain of
15% and on the other a loss of 9%. Find his overall gain or loss.
[H/m.- Find C.R of each]
3.3.4 Application of Discounts
Yesterday Pooja went to a shop with her parents to
purchase a dress for Pongal festival. She saw in the shops
many banners, the content of which she didn't understand.
Having this in mind, she entered the shop and
purchased a frock.
The price labelled on the frock was ^ 550 called
as Marked Price (abbreviated as M.P.) and she gave the
shopkeeper ? 550. But the shopkeeper returned the balance
amount and informed her that there is a discount of 20%.
Here, 20%) discount means, 20%) discount on the Marked
Price.
Discount
^x 550 = ? 110.
Discount is the reduction on
the Marked Price or List Price of
the article.
The normal price attached to the article
before the discount made is called as
Marked Price (M.P.) or List Price of the
article.
Chapter 3
Amount paid by Pooja to the shop keeper is ^ 440
= ? 550 - ? 110
= Marked Price-Discount
Hence we conclude the following:
During festival seasons
Discount = Marked Price - Selling Price and in the Tamil month
Selling Price = Marked Price - Discount , ^ i i no/ ono/
° rebates of 10%, 20%,
Marked Price = Selling Price + Discount 30%, etc., are offered to
attract customers by Co-
optex, Khadi and other
shops for various items to
promote sales.
Example 3.11
A bicycle marked at ^ 1,500 is sold for ^ 1,350. What is the percentage of
discount?
Solution
Given : Marked Price = ? 1500, Selling Price = ? 1350
Amount of discount = Marked Price - Selling Price
= 1500 - 1350
= ?150
Discount for ^ 1500 = ? 150
Discount for ? 100
_ 150
XlOO
1500
Percentage of discount = 10%.
Example 3.12
The list price of a frock is ^ 220. A discount of 20% on sales is announced.
What is the amount of discount on it and its selling price?
Solution
Given : List (Marked) Price of the frock = ? 220, Rate of discount = 20%
20
Amount of discount = Jqq X 220
= ?44
.•. Selling Price of the frock = Marked Price - Discount
= 220-44
= f 176.
Life Mathematics
Example 3.13
An almirah is sold at ^ 5,225 after allowing a discount of 5%. Find its
marked price.
Vignesh used the formula method:
S.P. = Rs5225
Discount === 5%
M.R = ?
100
Solution
Krishna used this method:
The discount is given in percentage.
Hence, the M.R is taken as
?100.
Rate of discount =
5%
Amount of discount =
5
100
X 100 [^
=
?5.
Selling Price =
M.R
- Discount
=
100-
- 5 = ? 95
If S.R is ? 95, then M.R is
^100.
When S.R is ^ 5225,
M.R -
100
X 5225
■'■ The M.R of the almirah
= ? 5,500.
M.R
'-(i
00
Discount%
100
XS.P.
VlOO-5/
100
100
95
? 5,500.
5
X5225
X5225
Example 3.14
A shopkeeper allows a discount of 10% to his customers and still gains 20%.
Find the marked price of an article which costs ^ 450 to the shopkeeper.
Solution
Vanitha used this method:
Let M.R be ? 100.
Vimal used the formula method
Discount = 10% of M.R
Discount = 10%, Gain = 20%,
- 1^ of MP - ^^ xlOO
100 100
C.R - ? 450, M.R = ?
= ^10
S.P. = M.R - Discount [o
R] 100 - Discount%
= 100 - 10 = ? 90
Gain =20%ofC.R
-(100 + 20)^
(100-10)^^^^
= ^X450=?90
= ^X450
S.R = C.R + Gain
= 450 + 90 = ? 540.
= ?600
If S.R is ^ 90, then M.R is ? 100.
When S.R is ^ 540,
M.R- 540X^100 -^600
.-. The M.R of an article = ? 600
Chapter 3
Example 3.15
A dealer allows a discount of 10% and still gains 10%. What is the cost price
of the book which is marked at ^ 220?
Solution
Sugandan used this method:
M.P.
Discount
= f 220.
= 10% of M.P.
_ 10
X220 = ?22
Mukundan used the formula method:
Discount = 10%
Gain = 10%
100
S.P. = M.P. - Discount
= 220 - 22 = ? 198
Let C.P. be ? 100.
M.P.
C.P.
[OR]
Gain
= 10% of C. P.
= ?220
= 100 - Discount%
100 + Gain%
= 100- 10
XM.P.
100+ 10
X220
_ 10
X 100 = ? 10
S.P.
100
= C.P. + Gain
= 100 + 10
= ? 110.
If S.P. is ? 110, then C.P. is ? 100.
When S.P is ? 198,
C.P =
90
110
X 220 = ? 180.
198x100
110
= ? 180.
Example 3.16
A television set was sold for ? 14,400 after giving successive discounts of 10%
and 20% respectively. What was the marked price?
Solution
Selling Price = ^14,400
Let the M.P be ? 100.
First discount = 10% = ^x 100 =? 10
S.P after the first discount = 100 - 10 = ? 90
Second discount = 20% =
20
100
X90 =?18
Selling Price after the second discount =90 - 18 = ^ 72
If S.P. is ? 72, then M.P is ? 100.
When S.P. is ^ 14,400,
M.P = 14400X100 =^20,000
M.P = ? 20,000.
Life Mathematics
Example 3.17
A trader buys an article for ? 1,200 and marks it 30% above the C.P. He then
sells it after allowing a discount of 20%. Find the S.R and profit percent.
Solution:
Let C.P. of the article be ? 100
M.P. = 30% above C.P = ? 130
If C.P. is ? 100, then M.P. is ? 130.
When C.P is ? 1200, M.P
Discount = 20% of 1560
1200x130 =?is6n
100
^X 1560 =^312
S.P = M.P - Discount
= 1560-312 = ^1248
Profit = S.P -C.P.
= 1248-1200 = ^48.
.-. Profit % = I#txl00
C.P.
48
1200
Xl00=4%
ifi^^
A shop gives 20% discount. What will be the S.P. of the following?
(i)
A dress marked at ? 120
(ii)
A bag marked at ? 250
(iii)
A pair of shoes marked at ? 750.
3.3.5 Application of Tax
Children! Very often we find advertisements in
newspapers and on television requesting people to pay their
taxes in time. What is this tax? Why does the Government
collect the tax from the common people?
Tax is paid to the Government to create better
infrastructure facilities like roads, railways, irrigation
facility, water supply, electricity, etc., to build a better nation. For all these, the money
needed by the Government is collected in the form of tax from the people.
Chapter 3
Taxes are of two types:
1. Direct Tax
Tax which is collected in the form of Income Tax, Property Tax, Professional
Tax, Water Tax, etc., is called as Direct Tax. These are paid directly to the government
by the public.
2. Indirect Tax
Some of the taxes which are not paid directly to the government are Indirect
Taxes and are explained below.
Excise Tax
This tax is charged on some items which are manufactured in the country. This
is collected by the Government of India.
Service Tax
Tax which is charged in Hotels, Cinema theatres, for service of Chartered
Accountants, Telephone Bills, etc., comes under Service Tax. This tax is collected by
the service provider from the user and deposited to the Government.
Income Tax
This is the most important source of revenue for the Government which
is collected from every citizen who is earning more than a minimum stipulated
income annually. As true citizens of our country, we should be aw are of our duty
and pay the tax on time.
Sales Tax / Value Added Tax
Sales Tax
Sales Tax is the tax levied on the sales made
by a seller at the time of selling the product. When the
buyer buys the commodity the sales tax is paid by him
together with the price of the commodity.
This sales tax is charged by the Government
on the selling price of an item and is added to the
value of the bill.
These days, however, the prices include the
tax known as Value Added Tax (VAT). This means
that the price we pay for an item is added with VAT.
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Life Mathematics
Can you find the prevaihng rate of
Sales Tax for some commodities in <
the year 2011.
1. Electrical instruments %
2. Petrol % 3. Diesel '
4. Domestic appliances %
5. Chemicals %
The Government gives exemption
of Sales Tax for some commodities
like rice, sugar, milk, salt, pen,
pencils and books.
Calculation of Sales Tax
Amount of Sales tax
Rate of Sales tax
Bill amount
Rate of Sales tax ^ ^^^^^ ^f ^^^ -^^^
Amount of Sales tax ., ,qq
Cost of the item
Cost of the item + Amount of Sales tax
Example 3.18
Vinodh purchased musical instruments for ? 12,000. If the rate of sales tax is
8%, find the sales tax and the total amount paid by him.
Solution
Value of the musical instruments = ^ 12,000
Rate of Sales Tax = 8%
Amount of Sales Tax = y^x 12000
= ?960
Total amount paid by Vinodh including Sales Tax = 12,000 + 960
- ? 12,960
Example 3.19
A refrigerator is purchased for ? 14,355, including sales tax. If the actual cost
price of the refrigerator is ^ 13,050, find the rate of sales tax.
Solution
Given: For the refrigerator, bill amount = ? 14,355, Cost price = ? 13,050.
Sales tax = Bill amount - Cost of the item
- 14,355-13,050 = ^1,305
Rate of Sales Tax = Amount of Sales Tax ^ ^qq
Cost ot the item
= J||xioo^lO%
Chapter 3
Example 3.20
Priya bought a suitcase for ? 2,730. The VAT for this item is 5%. What was the
price of the suitcase before VAT was added? Also state how much is the VAT.
Solution
Given : VAT is 5%.
If the price without VAT is ? 100, then the price including VAT is ? 105.
Now, when price including VAT is ^ 105, original price is ^ 100.
When price including VAT is ^ 2,730, the original price of the suitcase
= 1^X2730 =? 2,600
The original price of the suitcase = ? 2,600
.-. VAT = 2,730 - 2,600 = ? 130
1. Find the buying price of each of the following when 5% Sales Tax is
added on the purchase of:
(i) A pillow for ? 60 (ii) Two bars of soap at ? 25 each.
2. If 8% VAT is included in the prices, find the original price of :
(i) An electric water heater bought for ? 14,500 (ii) A crockery set
bought for ? 200.
EXERCISE 3.3
1. Choose the correct answer:
(i) The discount is always on the .
(A) Marked Price (B) Cost Price (C) Selling Price
(ii) If M.P - ? 140, S.P - ? 105, then Discount = .
(A) ? 245 (B) ? 25 (C) ? 30
(iii) = Marked Price - Discount.
(A) Cost Price (B) Selling Price (C) List Price
(iv) The tax added to the value of the product is called
(A) Sales Tax (B) VAT (C) Excise Tax
(D) Interest
(D) ? 35
(D) Market price
Tax.
(D) Service Tax
(v) If the S.P. of an article is ? 240 and the discount given on it is ? 28, then the M.P.
is .
(A) ^212 (B)?228 (C) ? 268 (D) ? 258
2. The price marked on a book is ^ 450. The shopkeeper gives 20% discount on it in|
a book exhibition. What is the Selling Price?
3. A television set was sold for ^ 5,760 after giving successive discounts of 10% anc
20% respectively. What was the Marked Price?
Life Mathematics
4. Sekar bought a computer for ^ 38,000 and a printer for ^ 8,000. If the rate of
sales tax is 7% for these items, find the price he has to pay to buy these two
items.
5. The selling price with VAT, on a cooking range is ? 19,610. If the VAT is 6%,
what is the original price of the cooking range?
6. Richard got a discount of 10% on the suit he bought. The marked price was
? 5,000 for the suit. If he had to pay sales tax of 10% on the price at which he
bought, how much did he pay?
7. The sales tax on a refrigerator at the rate of 9% is ? 1,170. Find the actual sale
price.
8. A trader marks his goods 40% above the cost price. He sells them at a discount
of 5%. What is his loss or gain percentage?
9. AT.V with marked price ? 11,500 is sold at 10% discount. Due to festival
season, the shop keeper allows a further discount of 5%. Find the net selling
price of the TV
10. A person pays ? 2,800 for a cooler listed at ^ 3,500. Find the discount percent
offered.
11. Deepa purchased 15 shirts at the rate of ^ 1,200 each and sold them at a profit
of 5%. If the customer has to pay sales tax at the rate of 4%, how much will one
shirt cost to the customer?
12. Find the discount, discount percent, selling price and the marked price.
SI. No
Items
M,P
Rate of
Discount
Amount of
Discount
S.P
(i)
Saree
? 2,300
20%
(ii)
Pen set
?140
?105
(iii)
Dining table
20%
? 16,000
(iv)
Washing
Machine
? 14,500
? 13,775
(V)
Crockery set
? 3,224
121/2%
il^rii. -r- ,
Which is a better offer? Two successive discounts of 20% and
5% or a single discount of 25%. Give appropriate reasons.
Chapter 3
3.4. Compound Interest
In class VII, we have learnt about Simple Interest and the formula for calculating
Simple Interest and Amount. In this chapter, we shall discuss the concept of Compound
Interest and the method of calculating Compound Interest
and Amount at the end of a certain specified period.
Vinay borrowed ^ 50,000 from a bank for a fixed
time period of 2 years, at the rate of 4% per annum.
Vinay has to pay for the first year.
Simple interest = ^^^ ^
= 50000x1x4 =tn()nn
100
Suppose he fails to pay the simple interest ^ 2,000 at the end of first year, then
the interest ^ 2,000 is added to the old Principal ^ 50,000 and now the sum
P + I = ? 52,000 becomes the new Principal for the second year for which the interest
is calculated.
Now in the second year he will have to pay an interest of
J. . _ PX^Xr
= 52000X1X4 -f TOfto
100
Therefore Vinay will have to pay more
interest for the second year.
This way of calculating interest is called
Compound Interest.
Generally in banks, insurance companies,
post offices and in other money lending and
deposit taking companies compound interest is
followed to find the interest.
Example 3.21
Ramlal deposited ? 8,000 with a finance company for 3 years at an interest of
15% per annum. What is the compound interest that Ramlal gets after 3 years?
Solution
Step 1: Principal for the first year
Interest for the first year
vDo you know?
When the interest is paid
on the Principal only, it is
called Simple Interest. But
if the interest is paid on the
Principal as well as on the
accrued interest, it is called
Compound Interest.
? 8,000
VXnXr
Amount at the end of first year
100
8000x1x15 =^1200
100
P + I = 8,000 + 1,200 = ? 9,200
Life Mathematics
Step 2: The amount at the end of the first year becomes the Principal for the
second year.
Principal for the second year = ^ 9,200
P X n X r
Interest for the second year = ^^
_ 9200x1x15 - ^ 1 oon
100 ^ ^'^^°
Amount at the end of second year = P + I = 9,200 + 1,380 = ^ 10,580
Step 3: The amount at the end of the second year becomes the Principal for the
third year.
Principal for the third year = ? 10,580
Interest for the third year = tqq
_ 10580x1x15 - ? 1 ro7
100 ^^'^^^
Amount at the end of third year = P + I
= 10,580 + 1,587 = ? 12,167
Hence, the Compound Interest that Ramlal gets after three years is
A-P = 12,167-8,000 =? 4,167.
Deduction of formula for Compound Interest
The above method which we have used for the calculation of Compound
Interest is quite lengthy and cumbersome, especially when the period of time is very
large. Hence we shall obtain a formula for the computation of Amount and Compound
Interest.
If the Principal is P, Rate of interest per annum is r % and the period of time or
the number of years is n, then we deduce the compound interest formula as follows:
Step 1 : Principal for the first year = P
VXnXr
Interest for the first year =
100
P X 1 X r _ Pr
100 100
Amount at the end of first year = P + I
= -C + ISo)
Chapter 3
Step 2 : Principal for the second year = P^l + j?^)
Interest for the second year
= P
100
(using the S.I.formula)
IX- '•
V 100/ 100
Amount at the end of second year = P + I
= P 1 +
100
r V
Step 3 : Principal for the third year = P[l +
100
Interest for the third year =
i'^m)
p 1 +
X Ixr
100
(using the S.I.formula)
= p(i + T^Yx '
100
100
Amount at the end of third year = P + I
lOO") V
100 / 100
= p{i + ^Y(i+ '
100
= "C+ikr
r v
Similarly, Amount at the end of n* year is A = P^l + t^)'
and C. I. at the end of 'n' years is given by A - P
(i. e.) C. I. = P(
1 +
r v
100
)"
To Compute Compound Interest
Case 1: Compounded Annually
When the interest is added to the Principal at the end of each year, we say that
the interest is compounded annually.
Here A = Ph + ^)" and C.I. = A - P
Life Mathematics
Case 2: Compounded Half - Yearly (Semi - Annually)
When the interest is compounded Half - Yearly, there are two conversion periods
in a year each after 6 months. In such situations, the Half - Yearly rate will be half of |
the annual rate, that is (^).
'-lim)
and C.I. = A- P
In this case, A = P
Case 3: Compounded Quarterly
When the interest is compounded quarterly, there are four conversion periods in
a year and the quarterly rate will be one-fourth of the annual rate, that is (^].
In this case, A = P[l + ^(
100
and C.I. = A - P
Case 4: Compounded when time being fraction of a year
When interest is compounded annually but time being a fraction.
In this case, when interest is compounded annually but time being a fraction of
a year, say 5^ years, then amount A is given by
i I
for 5 years for Va of year
Example 3.22
Find the C.I. on ^ 15,625 at 8% p. a. for 3 years compounded annually.
Solution
We know.
Amount after 3 years = v(l+ t^T
= 15625(1 +t|o/
= 15625(1+^/
= 15625 (§/
= 15625 x^X^X^
25 25 25
= ? 19,683
Now, Compound interest = A - P = 19,683 - 15,625
= ? 4,058
Chapter 3
To find the C.I. when the interest is compounded annually or half-yearly
Let us see what happens to ? 100 over a period of one year if an interest is
compounded annually or half-yearly.
Is.No
nnually
P = ? 100 at 10% per annum
compounded annually
Half-yearl
P = ? 100 at 10% per annum
compounded half-yearly
The time period taken is 1 year
The time period is 6 months or Vi year.
. lOOXlOXl - ^.n
100
I
lOOXlOX
100
?5
A =100 + 10 = ? 110
A= 100 + 5 = ? 105
For the next 6 months, P = ? 105
So, 1 =
105 X 10 X
'2 -
100
= ?5.25
and A =105 + 5.25 = ? 110.25
A = ?110
A = ? 110.25
Thus, if interest is compounded half - yearly, we compute the interest two times
and rate is taken as half of the annual rate.
Example 3.23
Find the compound interest on ? 1000 at the rate of 10% per annum for 18
months when interest is compounded half-yearly.
Solution
Here, P = ? 1000, r = 10% per annum
and n = 18 months = Ty years = y years = ^ y years
1
.•. Amount after 18 months = P 1 + y[
r 1 , 1 A N n2X^
= 1000
100
2V100
= 1000(1+^/
= 1000 ( 20 )
= ? 1157.625
= ? 1157.63
C. I. = A-P
= 1157.63 - 1000 = ? 157.63
A sum is taken for
one year at 8% p. a. If
interest is compounded
after every three
months, how many
times will interest be
charged in one year?
Life Mathematics
Example 3.24
Find the compound interest on ? 20,000 at 15% per annum for 2— years.
Solution
Here, P = ? 20,000, r = 15% p. a. and n = 2^ years.
Amount after 2 3- years = A = ^{^ + j^J[^ + j{j^)
= 20000 1 +
15
I 100 i
20000 (i+4r)'(i
100
3 ^'
20
[iH(iw)
+
20
23\^/21
20000 (tnt)
= 20000x§X§xf
= ? 27, 772.50
C.I. = A-P
= 27,772.50-20,000
= ? 7,772.50
r \"
Inverse Problems on Compound Interest
We have already learnt the formula, A = P[l + . ^.p, 1 ,
where four variables A, P, r and n are involved. Out of these four variables, if
any three variables are known, then we can calculate the fourth variable.
Example 3.25
At what rate per annum will ^ 640 amount to ^ 774.40 in 2 years, interest being
compounded annually?
Solution:
Given: P = ? 640, A = ? 774.40, n = 2 years, r = ?
We know.
A
= P(
1 +
r \n
100
774.40 = 640(1+ '' ^'
774.40
= (1 + 100
64000 ^ V-~^
100
r V
640
77440 _ /-, , r V
121
100
= (
1 +
100
r \2
100
Chapter 3
11
(w) = (1 + ik)'
11
10
r
100
r
100
r
100
1 +
r
100
11
10
- 1
11-
-10
r =
10
X
10
100
10
Rate r = 10% per annum.
Example 3.26
In how much time will a sum of ? 1600 amount to ? 1852.20 at 5% per annum
compound interest.
Solution
Given: P = ? 1600, A = ? 1852.20, r = 5% per annum, n = ?
We know, A = Pfl + '"
I 100 i
1852.20 = 1600(1+^)"
1852.20 /105
1600
185220
160000
9261
8000
Find the time period
and rate for each of the cases
given below :
1. A sum taken for 2 years at
8% p. a. compounded
half - yearly.
2. A sum taken for IVi years at
4% p. a. compounded half -
yearly.
~ liooi
- l20i
l20i
i20i '20i
.•. n =3 years
3.5 Difference between Simple Interest and Compound Interest
When P is the Principal, n = 2 years and r is the Rate of interest.
Difference between C. I. and S. I. for 2 years = Pfj^Y
Example 3.27
Find the difference between Simple Interest and Compound Interest for a sum
of ? 8,000 lent at 10% p. a. in 2 years.
Solution
Here, P = ? 8000, n = 2 years, r = 10% p. a.
Life Mathematics
r y
Difference between Compound Interest and Simple Interest for 2 years = p(t^)
= 8000 (^;
= 8000
(A
= 8000x^X^=?80
EXERCISE 3.4
1. Find the Amount and Compound Interest in the following cases:
10.
11.
SI. No.
Principal in Rupees
Rate% per annum
Time in years
(i)
1000
5%
3
(ii)
4000
10%
2
(iii)
18,000
10%
4
Sangeetha borrowed ? 8,000 from Alex for 2 years at Y1V2% per annum. What
interest did Sangeetha pay to Alex if the interest is compounded annually.
Maria invested ? 80,000 in a business. She would be paid interest at 5% per
annum compounded annually. Find (i) the amount standing to her credit at the
end of second year and (ii) the interest for the third year.
Find the compound interest on ^ 24,000 compounded semi - annually for IVi
years at the rate of 10% per annum.
Find the amount that Dravid would receive if he invests ? 8,192 for 18 months at
12V2% per annum, the interest being compounded half - yearly.
Find the compound interest on ^ 15,625 for 9 months, at 16% per annum
compounded quarterly.
Find the Principal that will yield a compound interest of ^ 1,632 in 2 years at 4%
rate of interest per annum.
Vicky borrowed ? 26,400 from a bank to buy a scooter at the rate of 15% p. a.
compounded yearly. What amount will he pay at the end of 2 years and 4 months
to clear the loan.
Arif took a loan of ? 80,000 from a bank. If the rate of interest is 10% p. a., find
the difference in amounts he would be paying after IVi years if the interest is
(i) compounded annually and (ii) compounded half - yearly.
Find the difference between simple interest and compound interest on ^ 2,400 at
2 years at 5% per annum compounded annually.
Find the difference between simple interest and compound interest on ^ 6,400
for 2 years at 6 V\% p. a. compounded annually.
Chapter 3
12. The difference between C. I. and S. I. for 2 years on a sum of money lent at
5% p. a. is ^ 5. Find the sum of money lent.
13. Sujatha borrows ? 12,500 at 12% per annum for 3 years at simple interest and
Radhika borrows the same amount for the same time period at 10% per annum
compounded annually. Who pays more interest and by how much?
14. What sum invested for IVi years compounded half -yearly at the rate of 4% p. a.
amounts to ? 1,32,651?
15. Gayathri invested a sum of ^ 12,000 at 5% p. a. at compound interest. She
received an amount of ? 13,230 after 'n' years. Find the value of 'n'.
16. At what rate percent compound interest per annum will ^ 640 amount to
? 774.40 in 2 years.
17. Find the rate percent per annum, if ^ 2,000 amount to ^ 2,315.25 in an year and a
half, interest being compounded six monthly.
3.5.1 Appreciation and Depreciation
a) Appreciation
In situations like growth of population, growth of
bacteria, increase in the value of an asset, increase in price of
certain valuable articles, etc., the following formula is used.
A= P('l +
100
b) Depreciation
In certain cases where the cost of machines, vehicles,
value of some articles, buildings, etc., decreases, the following
formula can be used.
A
= P(
1
100
World Population
Year Population
1700 600,000,000
1800 900,000,000
1900 1,500,000,000
2000 6,000,000,000
In 3 centuries, population
has multiplied 10 fold.
Example 3.28
The population of a village increases at the rate of 7% every year. If the present
population is 90,000, what will be the population after 2 years?
Solution
Present population P = 90,000, Rate of increase r = 7%,
Number of years n = 2.
The population after 'n' years = P(l + t^Y
/ 7 \^
.•. The population after two years = 90000 (1 + ttw^^)
Life Mathematics
= 90000
107 f
viooi
= 103041
The population after two years = 1,03,041
Example 3.29
The value of a machine depreciates by 5% each year. A man pays ? 30,000 for
the machine. Find its value after three years.
Solution
Present value of the machine P = ? 30,000, Rate of depreciation r = 5%,
Number of years n = 3
The value of the machine after 'n' years = P[l
r
100
.•. The value of the machine after three years = 30000(l
5
100
= 30000/
95
liooi
= 30000 X^X^X ^^
100 100 100
= 25721.25
The value of the machine after three years = ^ 25,721.25
Example 3.30
The population of a village has a constant growth of 5% every year. If its present
population is 1,04,832, what was the population two years ago?
Solution
Let P be the population two years ago.
•■■^(l+lio/ = 104832
^(loo) = 104832
p ^ 104832x100x100
105X105
= 95085.71
= 95,086 (rounding off to the nearest whole number)
.•. Two years ago the population was 95,086.
Chapter 3
EXERCISE 3.5
1. The number of students enrolled in a school is 2000. If the enrollment increases
by 5% every year, how many students will be there after two years?
2. A car which costs ^ 3,50,000 depreciates by 10% every year. What will be the
worth of the car after three years?
3. A motorcycle was bought at ^ 50,000. The value depreciated at the rate of 8%
per annum. Find the value after one year.
4. In a Laboratory, the count of bacteria in a certain experiment was increasing at
the rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was
initially 5,06,000.
5. From a village people started migrating to nearby cities due to unemployment
problem. The population of the village two years ago was 6,000. The migration is
taking place at the rate of 5% per annum. Find the present population.
6. The present value of an oil engine is ^ 14,580. What was the worth of the engine
3 years before if the value depreciates at the rate of 10% every year?
7. The population of a village increases by 9% every year which is due to the job
opportunities available in that village. If the present population of the village is
11,881, what was the population two years ago?
3.6 Fixed Deposits and Recurring Deposits
Banks, post offices and many other financial institutions
accept deposits from public at varying rates of interest. People
save in these institutions to get regular periodical income.
Different saving schemes are offered by these financial
institutions. Few of those schemes are
(i) Fixed Deposit and (ii) Recurring Deposit
(i) Fixed Deposit
In this type of deposit, people invest a quantum of money for specific periods.
Such a deposit is called Fixed Deposit (in short form, F.D.)
Note: Deposits can either be for a short term or long term. Depending on the
period of deposits, they offer a higher rate of interest.
(ii) Recurring Deposit
Recurring Deposit (in short form, R.D.) is entirely different from Fixed Deposit.
In this scheme, the depositor has the freedom to choose an amount according
to his saving capacity, to be deposited regularly every month over a period of years in
the bank or in the post office.
Life Mathematics
The bank or post office repays the total amount
deposited together with the interest at the end of the
period. This type of Deposit is known as Recurring
Deposit.
Note: The interest on Recurring Deposit is calculated
using simple interest method.
Do you know?
kThe monthly instalments
can be paid on any day
within the month for R.D.
To find the formula for calculating interest and the maturity amount for
R.D
Let r % be the rate of interest paid and 'P' be the monthly instalment paid for
'n' months.
n{n + 1)'
Interest = ^r^, where N = -^
Total Amount due at maturity is A = Pn -f
years
PNr
100
Example 3.31
Tharun makes a deposit of Rupees two lakhs in a bank for 5 years. If the rate of
interest is 8% per annum, find the maturity value.
Solution
Principal deposited P = ? 2,00,000, n = 5 years, r = 8% p. a.
Interest = ^= 200000 x 5 Xy^
= ? 80, 000
.-. Maturity value after 5 years = 2,00,000 + 80,000 = ? 2,80,000.
Example 3.32
Vaideesh deposits ^ 500 at the beginning of every month for 5 years in a post
office. If the rate of interest is 7.5%, find the amount he will receive at the end of 5
years.
Solution
Amount deposited every month, P = ? 500
Number of months, n = 5 x 12 = 60 months
Rate of interest, r = 7!-% = i|-%
Total deposit made = Pn = 500 x 60
= ? 30,000
Period for recurring deposit, N =
12
n{n + 1)
2
years
=^X 60x61 = ^ years
24
Chapter 3
Interest, I
_ PNr
100
= 5««xTx 2x100
= ? 5,718.75
Total amount due
= ^'-lf
= 30,000 + 5,718.75
= ? 35,718.75
Example 3.33
Vishal deposited ? 200 per month for 5 years in a recurring deposit account in
a post office. If he received ? 13,830 find the rate
of interest.
Solution
Maturity Amount, A = ? 13,830, P = ? 200
, n = 5 X 12 = 60 months
Period, N
_ 1
12
n{n+ 1)
[ 2 J
years
1 v^nv 61 305
_x60x^ = ^years
Amount Deposited = Pn = 200 x 60 = ? 12, 000
Maturity Amount = Pn +
PNr
100
13830 = 12000 + 200 x ^ X
13830 - 12000 = 305 X r
100
1830
.•. r =
= 305 X r
1830
305
= 6%
3.6.1 Hire Purchase and Instalments
Banks and financial institutions have introduced a scheme called hire purchase
and instalment to satisfy the needs of today's consumers.
Hire purchase: Under this scheme, the article will not be owned by the buyer
for a certain period of time. Only when the buyer has paid the complete price of the
article purchased, he/she will become its owner.
Instalment: The cost of the article along with interest and certain other charges
is divided by the number of months of the loan period. The amount thus got is known
as the instalment.
Life Mathematics
Equated Monthly Instalment ( E.M.I. )
Equated Monthly Instalment is also as equivalent as the instalment scheme but
with a dimnishing concept. We have to repay the cost of things with the interest along
with certain charges. The total amount received should be divided by the period of
months. The amount thus arrived is known as Equated Monthly Instalment.
Principal + Interest
Number of months
Different schemes of Hire purchase and Instalment scheme
1. 0% interest scheme: Companies take processing charge and 4 or 5 months
instalments in advance.
2. 100% Finance: Companies add interest and the processing charges to the
cost price.
3. Discount Sale: To promote sales, discount is given in the instalment schemes.
4. Initial Payment: A certain part of the price of the article is paid towards the
purchase in advance. It is also known as Cash down payment.
Example 3.34
The cost price of a washing machine is ^ 18,940. The table given below illustrates
various schemes to purchase the washing machine through instalments. Choose the
best scheme to purchase.
SI.
No
Different
schemes
S. P.
in?
Initial
payments
Rate of
interest
Processing
fee
Period
(i)
75%
Finance
18,940
25%
12%
1%
24
months
(ii)
100%
Finance
18,940
Nil
16%
2%
24
months
(iii)
0%
Finance
18,940
4E. M. I.
in
advance
Nil
2%
24
months
Calculate the E. M. I. and the total amount for the above schemes.
Solution
(i) 75% Finance
P = ? 18, 940, Initial payment = 25%, Rate = 12%, Processing fee = 1%
Processing fee = 1% of? 18, 940
= j^-r X 18940 = ? 189.40 = ? 189
Initial payment = 25% of? 18,940
= ^X 18940 =?4, 735
Chapter 3
Loan amount
=
18,940 - 4,735 = ? 14,205
Interest
=
14205X12X2
100
=
? 3,409. 20 ~ ? 3,409
E. M. I.
=
Loan amount + Interest
Number of months
=
14205 + 3409 _ 17614
24 24
=
? 733.92 ~ ? 734
.•. Total amount to be paid
=
4,735 + 14,205 + 3,409 + 189
=
? 22,538
(ii) 100% finance
Processing fee
=
2% of ? 18,940
=
^2^X18940 -? 378.80 ~ ? 379
Rate of Interest
Interest
:
16%
18940 X 100 ^^
=
? 6060.80 ~ ? 6,061
E. M. I.
=
Loan amount + Interest
Number of months
=
18940 + 6061 _ 25001
24 24
=
^ 1,041.708 ~ ? 1,041.71
=
? 1,042
Total amount to be paid
=
6,061 + 18,940 + 379 = ? 25,380
(iii) 0% interest scheme
Processing fee
=
2% of ? 18,940
=
2 X 18940 - ? 378.80 ~ ? 379
E. M. I.
=
Loan amount + Interest
Number of months
=
18940 + _ 18940
24 24
=
? 789.166 ~ ? 789
Total amount to be paid
=
18,940 + 3,156 + 379 = ? 22,475
Advance E. M. I. paid
=
? 789 X 4 = ? 3,156
Hence, 0% interest scheme i
s the best scheme.
Life Mathematics
Example 3.35
The cost of a computer is ? 20,000. The company offers it in 36 months, but
charges 10% interest. Find the monthly instalment the purchaser has to pay.
Solution
Cost of computer = ? 20,000, Interest = 10% p. a.. Period = 36 months (3 years)
Total Interest = 20000X^X3
= ? 6,000
.-. Total amount to be paid = 20,000 + 6,000
= ? 26,000
Monthly Instalment
_ Total amount
~ Number of months
= 26000
36
= ? 722. 22
~ ?722
EXERCISE 3.6
Swetha makes a fixed deposit of ^ 25,000 in a bank for 2 years. If the rate of
interest is 4% per annum, find the maturity value.
Nitin makes a fixed deposit of ? 75,000 in a bank for 3 years. If the rate of
interest is 5% per annum, find the maturity value.
Imran deposits ? 400 per month in a post office as R.D. for 2 years. If the rate of
interest is 12%, find the amount he will receive at the end of 2 years.
Ramesh deposited a certain amount every month for 6 years in a post office at
the rate of 8% and receives ^ 17,904 at the end of 6 years. Find his monthly
deposit.
Ethen deposited ^ 700 per month in a bank as R.D. for 6 years. Calculate the rate
of interest if he received ? 64,197 at the end of 6 years.
The cost of a microwave oven is ^ 6,000. Poorani wants to buy it in 5
instalments. If the company offers it at the rate of 10% p. a. Simple Interest, find
the E.M.I, and the total amount paid by her.
The cost price of a refrigerator is ^ 16,800. Ranjith wants to buy the refrigerator
at 0% finance scheme paying 3 E.M.I, in advance. A processing fee of 3% is also
collected from Ranjith. Find the E.M.I, and the total amount paid by him for a
period of 24 months.
The cost of a dining table is ? 8,400. Venkat wants to buy it in 10 instalments. If
the company offers it for a S.I. of 5% p. a., find the E.M.I, and the total amount
paid by him.
Chapter 3
3.7 Compound Variation
In the earUer classes we have already learnt about Direct and Inverse Variation.
Let us recall them.
Direct Variation
If two quantities are such that an increase or decrease in one leads to a
corresponding increase or decrease in the other, we say they vary directly or the
variation is Direct.
Examples for Direct Variation:
1. Distance and Time are in Direct Variation, because more the distance
travelled, the time taken will be more( if speed remains the same).
2. Principal and Interest are in Direct Variation, because if the Principal is
more the interest earned will also be more.
3. Purchase of Articles and the amount spent are in Direct Variation, because
purchase of more articles will cost more money.
Indirect Variation or Inverse Variation:
If two quantities are such that an increase or decrease in one leads to a
corresponding decrease or increase in the other, we say they vary indirectly or the
variation is in inverse.
Examples for Indirect Variation:
1. Work and time are in Inverse Variation, because more the number of
the workers, lesser will be the time required to complete a job.
2. Speed and time are in Inverse Variation, because higher the speed, the
lower is the time taken to cover a distance.
3. Population and quantity of food are in Inverse Variation, because if the
population increases the food availabilty decreases.
Compound Variation
Certain problems involve a chain of two or more variations, which is called as
Compound Variation.
The different possibilities of variations involving two variations are shown in
the following table:
Life Mathematics
Variation I
Variation II
Direct
Direct
Inverse
Inverse
Direct
Inverse
Inverse
Direct
Let us work out some problems to illustrate compound variation.
Example 3.36
If 20 men can build a wall 112 meters long in 6 days, what length of a similar
wall can be built by 25 men in 3 days?
Solution:
Method 1: The problem involves set of 3 variables, namely- Number of men.
Number of days and length of the wall.
Number of Men
Number of days
Length of the wall in metres
20
6
112
25
3
X
Step 1 : Consider the number of men and the length of the wall. As the number
of men increases from 20 to 25, the length of the wall also increases. So it is in Direct
Variation.
Therefore, the proportion is 20 : 25 :: 112 : x
(1)
Step 2: Consider the number of days and the length of the wall. As the number
of days decreases from 6 to 3, the length of the wall also decreases. So, it is in Direct
Variation.
Therefore, the proportion is 6 : 3 :: 112 : x
Combining (1) and (2), we can write
(2)
We know. Product of Extremes = Product of Means.
Extremes Means Extremes
20 : 25::112 : x
6 :
Chapter 3
So,
Method 2
20x6 XX
X
25X3X112
25x3x112 -
20x6
70 meters.
Number of Men
Number of days
Length of the wall in metres
20
6
112
25
3
X
Step 1: Consider the number of men and length of the wall. As the number
of men increases from 20 to 25, the length of the wall also increases. It is in direct
variation.
The multiplying factor = 25_
Step 2: Consider the number of days and the length of the wall. As the number
of days decreases from 6 to 3, the length of the wall also decreases. It is in direct
variation.
_ 3
The multiplying factor - ^.
•■• ^ = il^xj-x 112 = 70meters
Example 3.37
Six men working 10 hours a day can do a piece of work in 24 days. In how many
days will 9 men working for 8 hours a day do the same work?
Solution
Method 1: The problem involves 3 sets of variables, namely - Number of men.
Working hours per day and Number of days.
Number of Men
Number of hours per day
Number of days
6
10
24
9
8
X
Step 1: Consider the number of men and the number of days. As the number of
men increases from 6 to 9, the number of days decreases. So it is in Inverse Variation.
Therefore the proportion is 9 : 6 :: 24 : x (1)
Step 2: Consider the number of hours worked per day and the number of days.
As the number of hours working per day decreases from 10 to 8, the number of days
increases. So it is inverse variation.
Therefore the proportion is 8 : 10 :: 24 : x (2)
Combining (1) and (2), we can write as
9:6
8:10
::24:x
Life Mathematics
We know, Product of extremes = Product of Means.
Extremes Means Extremes
9 : 6 : : 24 : x
So,
9x8xx = 6x10x24
= 6X10X24 -
9x8
X
20 days
Note: 1. Denote the Direct variation as i (Downward arrow)
2. Denote the Indirect variation as t (Upward arrow)
3. Multiplying Factors can be written based on the arrows. Take the
number on the head of the arrow in the numerator and the number on
the tail of the arrow in the denominator.
For method two, use the instructions given in the note above .
Method 2 : (Using arrow marks)
Number of Men
Number of hours per day
Number of days
6
10
24
9
8
X
Step 1 : Consider men and days. As the number of men increases from 6 to 9,
the number of days decreases. It is in inverse variation.
The multiplying factor = ^
Step 2 : Consider the number of hours per day and the number of days. As the
number of hours per day decreases from 10 to 8, the number of days increases. It is
also in inverse variation.
The multiplying factor
= 10.
8
X =
|-xi?-x24 = 20 days.
9 8 -^
EXERCISE 3.7
Twelve carpenters working 10 hours a day complete a furniture work in 18 days.
How long would it take for 15 carpenters working for 6 hours per day to complete
the same piece of work.
Eighty machines can produce 4,800 identical mobiles in 6 hours. How many
mobiles can one machine produce in one hour? How many mobiles would 25
machines produce in 5 hours?
If 14 compositors can compose 70 pages of a book in 5 hours, how many
compositors will compose 100 pages of this book in 9 hours?
If 2,400 sq.m. of land can be tilled by 12 workers in 10 days, how many workers
are needed to till 5,400 sq.m. of land in 18 days?
Working 4 hours daily, Swati can embroid 5 sarees in 18 days. How many days
will it take for her to embroid 10 sarees working 6 hours daily?
A sum of ^ 2,500 deposited in a bank gives an interest of ^ 100 in 6 months.
What will be the interest on ? 3,200 for 9 months at the same rate of interest?
Chapter 3
3.8 Time and Work
When we have to compare the work of several persons, it is necessary to
ascertain the amount of work each person can complete in one day. As time and work
are of inverse variation and if more people are joined to do a work, the work will be
completed within a shorter time.
In solving problems here, the following points should be remembered:
1. If a man finishes total work in 'n' days, then in one day he does ' 'of
the total work. For example, if a man finishes a work in 4 days, then in
one day he does ^ of the work.
2. If the quantity of work done by a man in one day is given, then the
total number of days taken to finish the work = l/(one day's work). For
example, if a man does yk of the work in 1 day, then the number of days
taken to finish the work
10
(ll))
ix
10.
1
10 days.
Example 3.38
A can do a piece of work in 20 days and B can do it in 30 days. How long will
they take to do the work together?
Solution
Work done by A in 1 day = 2(75 Work done by B in 1 day = wk
Work done by A and B in 1 day
20 30
3 + 2
60
_5_
60
J_
12
of the work
Total number of days required to finish the work by A and B = ^
12 days.
Example 3.39 ^^^
A and B together can do a piece of work in 8 days, but A alone can do it 12
days. How many days would B alone take to do the same work?
Solution
Work done by A and B together in 1 day =
Work done by A in 1 day =
Work done by B in 1 day =
of the work
Number of days taken by B alone to do the same work =
Y2^ of the work
J_= 3-2
12
1
1
8
1,
24
24 days
24
'24
Life Mathematics
isjgaa g I't^
Example 3.40
Two persons A and B are engaged in a
work. A can do a piece of work in 12 days and
B can do the same work in 20 days. They work
together for 3 days and then A goes away. In
how many days will B finish the work?
Solution
Work done by A in 1 day =
Work done by B in 1 day =
Work done by A and B together in 1 day =
Work done by A and B together in 3 days =
Remaining Work =
\i^ t^eAc
While A, B and C working
individually can complete a
job in 20,5,4 days respectively.
If all join together and work,
find in how many days they
will finish the job?
12
J_
20
12 20
5+3 ^8^2
60 60 15
Number of days taken by B to finish the remaining work
5
1
- 5 _ 3 ., 20
J_~5 ^ 1
20
= 12 days.
Example 3.41
A and B can do a piece of work in 12 days, B and C in 15 days, C and A in 20
days. In how many days will they finish it together and separately?
Solution
Work done by A and B in 1 day =
X
12
Work done by B and C in 1 day =
Work done by C and A in 1 day =
Work done by (A+B)+(B+C)+(C+A) in 1 day =
Work done by (2A + 2B + 2C) in 1 day =
Work done by 2(A + B + C) in 1 day =
Work done by A, B and C together in 1 day =
X
15
X
20
X + X + J_
12 15 20
5 + 4 + 3
60
12_
60
lyI2_ = J-
2 ^ 60 10
Chapter 3
.-. A,B and C will finish the work in 10
days.
Work done by A in 1 day
(i.e.) [(A + B + C)'s work - (B + C)'s work] =
1
10
1
15
3-2
30
1
30
.-.A will finish the work in 30 days.
Workdone by B in 1 day
(i.e.) [(A + B + C)'s work - (C + A)'s
work] =
1
10
1
20
2-1
20
1
20
.-. B will finish the work in 20 days.
Work done by C in 1 day
(i.e.) [(A + B + C)'s work - (A + B)'s
work] =
1
10
1
12
6-5
60
1
60
.-. C will finish the work in 60 days.
Example 3.42
A can do a piece of work in 10 days and B can do it
each of them get if they finish the work and earn ? 1500?
in 15
days
. How much does
Solution
Work done by A in 1 day
1
~ 10
Work done by B in 1 day
1
~ 15
Ratio of their work
1 . 1
- 10 ■ 15 -
3:2
Total Share
= ?1500
A's share
= ^Xl500
= ?900
B's share
= ^Xl500
= ?600
Example 3.43
Two taps can fill a tank in 30 minutes and 40 minutes. Another tap can empty
it in 24 minutes. If the tank is empty and all the three taps are kept open, in how much
time the tank will be filled?
Solution
Quantity of water filled by the first tap
in one minute
=
1
30
Quantity of water filled by the second 1
tap in one minute
=
1
40
Quantity of water emptied by the third
tap in one minute
=
1
24
Life Mathematics
Quantity of water filled in one minute,")
when all the 3 taps are opened j
J_ + J__J_
30 40 24
4 + 3-5
120
7-5
120
Time taken to fill the tank =
2
120
1
60
= 60 minutes
/60
1 hour
EXERCISE 3.8
1. A man can complete a work in 4 days, whereas a woman can complete it in only
12 days. If they work together, in how many days, can the work be completed?
2. Two boys can finish a work in 10 days when they work together. The first boy
can do it alone in 15 days. Find in how many days will the second boy do it
all by himself?
3. Three men A, B and C can complete a job in 8, 12 and 16 days respectively.
A and B work together for 3 days; then B leaves and C joins. In how many
days, can A and C finish the work?
4. A tap A can fill a drum in 10 minutes. A second tap B can fill in 20 minutes.
A third tap C can empty in 15 minutes. If initially the drum is empty, find
when it will be full if all taps are opened together?
5. A can finish a job in 20 days and B can complete it in 30 days. They work
together and finish the job. If ^ 600 is paid as wages, find the share of each.
6. A, B and C can do a work in 12, 24 and 8 days respectively. They all work for
one day. Then C leaves the group. In how many days will A and B complete
the rest of the work?
7. A tap can fill a tank in 15 minutes. Another tap can empty it in 20 minutes.
Initially the tank is empty. If both the taps start functioning, when will the
tank become full?
Abbreviation:
C.P
. = Cost Price,
S.P. = Selhng Price,
M.P
= Marked Price,
P =
Principal, r =
Rate of interest, n =
time
period.
A =
Amount, C. I
. = Compound Interest.
Chapter 3
Concept Summary
^ Percent means per hundred. A fraction with its denominator 100 is called
a percent.
'V In case of profit, we have
Profit = S.P. - C.P.; Profit percent = ^^^^ x 100
gp _ / 100 + Profit% \ pp . CV - ( ^QQ \y^.V
^■^■~[ 100 J^^-^-' ^•^•~llOO + Profit%)^^-^-
'^ In case of Loss, we have
Loss = C.P - S.P.; Loss percent = ^^°^x 100
S p ^ / 100-Loss% X^C.P.; C.P. = ( 100 ]xS.P.
V 100 / ' V100-Loss%/
'^ Discount is the reduction given on the Marked Price.
'^ Selling Price is the price payable after reducing the Discount from the
Marked Price.
%■ Discount = M.P - S.P
% M.P. = ^ XS.P.; S.P = 100-Discount% ^j^p
100 - Discount% 100
^ rv>- 100 ~ Discount% ^ jy^^ p . j^| p _ 100 + Profit% n^ (^ p
100 + Profit% " ■ ' ■ ■ 100 - Discount%
% Discount Percent = Discount x 100.
M.P.
^ When the interest is
(i) compounded annually, A — Pn + t^)
i2n
100
1/ r
(ii) compounded half- yearly, A = p[l + yfy^)
(iii) compounded quarterly, A = p[l + yfj^)
^ Appreciation, A = P(^l + y^)"; Depreciation, A = P('l - -^f
^ The difference between C. I. and S. I. for 2 years — ^{jT^j
^ One day's work of A = , , ,
Number ot days taken by A
^ Work completed in 'x' days = One day's work x x.
Measurements
4.1 Introduction
4.2 Semi Circles and Quadrants
4.3 Combined Figures
4.1 Introduction
Measuring is a skill. It is required for every individual in his / her
life. Everyone of us has to measure something or the other in our daily
life. For instance, we measure
//V-/-/-V--/I
V,
n,
T^M ^'W ^X
Fig. 4.1
(i) the length of a rope required for drawing water from a well,
(ii) the length of the curtain cloth required for our doors and
windows,
(iii) the size of the floor in a room to be tiled in our house and
(iv) the length of cloth required for school uniform dress.
In all the above situations, the idea of 'measurements' comes in.
The branch of mathematics which deals with the measure of lengths,
angles, areas, perimeters in plane figures and surface areas, volumes
in solid figures is called 'measurement and mensuration'.
Recall
Let us recall the following definitions which we have learnt in class VII.
(i) Area
Area is the portion inside the closed figure in
a plane surface. C The word 'peri' in Greek
means 'around' and 'meter'-
(ii) Perimeter ^> ^^^^^ measure'.
The perimeter of a closed figure is the total
measure of the boundary.
Thus, the perimeter means measuring around a figure or measuring along a
curve.
Can you identify the shape of the following objects?
Fig. 4.2
The shape of each of these objects is a 'circle'.
(iii) Circle
Let 'O' be the centre of a circle with radius 'r' units (OA).
Area of a circle, A = zr^ sq.units.
Perimeter or circumference of a circle,
P = 27rr units, I O- 1 A
where tt ~ 4^ or 3. 1 4. %---^^ Fig. 4.4
Fig. 4.3
cumfe^
c^pmty
Take a cardboard
and draw circles of
different radii. Cut the
circles and find their
areas and perimeters.
S.No.
Radius
Area
Perimeter
1.
2.
3.
Measurements
4.2 Semi circles and Quadrants
4.2.1 Semicircle
Have you ever noticed the sky during night time after 7 days of new moon day
or full moon day?
What will be the shape of the moon?
It looks like the shape of Fig. 4.6.
How do you call this? ^'^- ^-^
This is called a semicircle. [Half part of a circle]
The two equal parts of a circle divided by its diameter are called semicircles.
(a) Perimeter of a semicircle
Perimeter, P = y x (circumference of a circle) + 2 x r units a /• o / b
P = Trr + 2r = {7r + 2)r units
(b) Area of a semicircle
_ 1
Area, A = y x (Area of a circle)
_ 1
XTir
A = ^^ sq. units.
4.2.2 Quadrant of a circle
Cut the circle through two of its perpendicular diameters. We
get four equal parts of the circle. Each part is called a quadrant of a;
the circle. We get four quadrants OCA, OAD, ODB and OBC while
cutting the circle as shown in the Fig. 4.11.
A, fi Fig. 4.11
Note: The central angle of the quadrant is 90°.
Chapter 4
(a) Perimeter of a quadrant
Perimeter, P = ^ x (circumference of a circle) + 2r units
4
- Kr
+ 2r =
=(f + 2)
r units
(b) Area of a quadrant
Area, A = ^ x (Area of a circle)
A = 4- X Trr^ sq.units
Example 4.1
Find the perimeter and area of a semicircle whose radius is 14 cm.
Solution 14 cm
Given: Radius of a semicircle, r = 14 cm
Perimeter of a semicircle, P = (tt + 2) r units ^. ^ ^^
Fig. 4.15
.-. P = (^ + 2)X14
= (^^^J^)X14=^X14 = 72
Perimeter of the semicircle = 72 cm.
Area of a semicircle, A = -^^ sq. units
. A = ^xl4|14=308cm\
Example 4.2
The radius of a circle is 21 cm. Find the perimeter and area of
a quadrant of the circle.
Solution
Given: Radius of a circle, r = 21 cm
Perimeter of a quadrant, P = (^ + 2\r units
Fig. 4.16
22
22
^+2x21 = (fA + 2)x21
_ /22 + 28
p = ( ^^7^^° ]x21 =^X21
14
= 75 cm.
2
Area of a quadrant, A = ^^ sq. units
A = 22 y^ 21x21
_ 50^
14
7 4
= 346.5 cm\
Measurements
Example 4.3
The diameter of a semicircular grass plot is 14 m. Find
the cost of fencing the plot at ? 10 per metre .
Solution
Given: Diameter, d = 14 m.
.". Radius of the plot, r = ^ = 7m.
14 m
Fig. 4.17
To fence the semicircular plot, we have to find the perimeter of it.
Perimeter of a semicircle, P = (7r + 2)xr units
- /22
_ / 22 + 14
X7
X7
\ 7
P = 36 m
Cost of fencing the plot for 1 metre = ? 10
.•. Cost of fencing the plot for 36 metres = 36 x 10 = ? 360
Example 4.4
The length of a chain used as the boundary of a '"\
semicircular park is 36 m. Find the area of the park.
Solution
Given:
Length of the boundary = Perimeter of a semicircle
A.
O
-• -,
B
Fig. 4.18
.■.{TT + iy
= 36 m
(f.2)xr
= 36
22+")xr
= 36
36 xr
= 36 ^ r = 7 m
ea of the park
= Area of the semicircle
A =
Tir
sq. units
Area of the park
22 y 7 X 7
7 ^
77 m'.
^ X^^= 77m'
Chapter 4
EXERCISE 4.1
1
(i
(ii
(iii
(iv
(V
(vi
(vii
(viii
(ix
(X
times the area of the circle.
(C) one-half (D) one-quarter
Choose the correct answer:
Area of a semicircle is
(A) two (B) four
Perimeter of a semicircle is
(A) ( ^'^'^ ) r units (B) (;r + 2) r units (C) 2r units (D) (?: + 4)r units
If the radius of a circle is 7 m, then the area of the semicircle is
(A) 77 m' (B) 44 m' (C) 88 m' (D) 154 m-
If the area of a circle is 144 cm\ then the area of its quadrant is
(A) 144 cm' (B) 12 cm' (C) 72 cm' (D) 36 cm'
The perimeter of the quadrant of a circle of diameter 84 cm is
(A) 150 cm (B) 120 cm (C) 21 cm
The number of quadrants in a circle is
(A) 1 (B) 2 (C) 3
Quadrant of a circle is of the circle.
(A) one-half (B) one-fourth (C) one-third
The central angle of a semicircle is .
(A) 90° (B) 270°
The central angle of a quadrant is _
(A) 90° (B) 180°
(C) 180^
(D) 42 cm
(D)4
(D) two-thirds
(D) 360°
(D)0°
(C) 270°
If the area of a semicircle is 84 cm', then the area of the circle is
(A) 144 cm' (B) 42 cm' (C) 168 cm' (D) 288 cm'
Find the perimeter and area of semicircles whose radii are,
(i) 35 cm (ii) 10.5 cm (iii) 6.3 m (iv) 4.9 m
Find the perimeter and area of semicircles whose diameters are,
(i) 2.8 cm (ii) 56 cm (iii) 84 cm (iv) 112 m
Calculate the perimeter and area of a quadrant of the circles whose radii are,
(i) 98 cm (ii) 70 cm (iii) 42 m (iv) 28 m
5. Find the area of the semicircle ACB and the quadrant BOC in the
given figure.
6. A park is in the shape of a semicircle with radius 21m. Find the
cost of fencing it at the cost of ? 5 per metre.
Measurements
4.3 Combined Figures
VJ
r^
(
)
C~)
A
(a)
(b)
(c)
Fig. 4.19
What do you observe from these figures?
In Fig. 4.19 (a), triangle is placed over a
semicircle. In Fig. 4.19 (b), trapezium is placed
over a square etc.
Two or three plane figures placed
adjacently to form a new figure. These are
'combined figures'. The above combined
figures are Juxtaposition of some known figures;
Can we see some examples?
(d)
(e)
Some combinations of plane
figures placed adjacently, with
one side equal in length to a
side of the other is called a
Juxtaposition of figures.
triangle, rectangle, semi-circle, etc.
S.No,
1.
2.
3.
Plane figures
Two scalene triangles
Two right triangles and
a rectangle
Six equilateral triangles
Juxtaposition
Quadrilateral
D C
Trapezium
i
A E F B
Hexagon
(a) Polygon
A polygon is a closed plane figure formed by 'n'
line segments.
A plane figure bounded by straight line segments is
a rectilinear figure.
A rectilinear figure of three sides is called a
triangle and four sides is called a Quadrilateral.
Polygon of Polygon of
4 line segments 6 line segments
A
V
Fig. 4.20
Chapter 4
(b) Regular polygon
If all the sides and angles of a polygon are equal, it is called a regular polygon.
For example,
(i) An equilateral triangle is a regular polygon with three
sides.
(ii) Square is a regular polygon with four sides.
3 — I— q Fig. 4.21
Fig. 4.22
(c) Irregular polygon
Polygons not having regular geometric shapes are called irregular polygons.
(d) Concave polygon
A polygon in which atleast one angle is more than 180°, is called
a concave polygon.
(e) Convex polygon
A polygon in which each interior angle is less than 180°, is
called a convex polygon.
Polygons are classified as follows. P^9- ^-24
Fig. 4.23
Number of sides
Name of the polygon
3
Triangle
4
Quadrilateral
5
Pentagon
6
Hexagon
7
Heptagon
8
Octagon
9
Nonagon
10
Decagon
Most of the combined figures are irregular polygons. We divide them into
known plane figures. Thus, we can find their areas and perimeters by applying the
formulae of plane figures which we have already learnt in class VII. These are listed
in the following table.
Measurements
No.
Name of the
Figure
Figure
Area (A)
(sq. units)
Perimeter (P)
(units)
Triangle
— XbXh
2
AB + BC + CA
B b C
2.
Right triangle
— XbXh
2
(base + height +
hypotenuse)
B base (ft) C
Equilateral
triangle
3 2
-,— fl where
(V3 ~ 1.732)
AB+BC+CA = 3a ;
Altitude, h =
units
Isosceles triangle
h X y<3^ - h^
2a +2 -J a^ - }?
Scalene triangle
\j s{s - a){s - b) {s — c)
where s=
a + h + c
AB + BC + CA
= {a + h + c)
Quadrilateral
■XdX{hi + hi)
AB + BC + CD + DA
Parallelogram
D
h
/
h
/
/
1
/
bx h
2x(a + b)
Rectangle
I X b
2x(l + b)
9.
Trapezium
■XhX(a+b)
AB + BC + CD + DA
10.
Rhombus
a
</
^
a
\
d.
y
a
N
1/
"a
— XdiX di where
2
(ii,fi?2 are diagonals
4a
11.
Square
4a
A a B
Chapter 4
Fig. 4.26 Fig. 4.27
Solution
(i) It is a combined figure made up of a square ABCD
and a semicircle DEA. Here, arc DEA is half the
circumference of a circle whose diameter is AD.
Given: Side of a square = 7 cm
.-. Diameter of a semicircle = 7 cm
.-. Radius of a semicircle, r = y cm
Perimeter of the combined figure
P =
AB + BC + CD + DEA
7 + 7 + 7 + yX (circumference of a circle)
21 + ^XlTvr = 21+ ^X^
P = 21 + 11 = 33 cm
Perimeter of the combined figure = 33 cm.
Area of the combined figure = Area of a semicircle + Area of a square
A = ^ + a'
2
22
X
7X7
+ T
- 77
7X2 2X2 ■ 4
•. Area of the given combined figure = 19.25 +49 = 68.25 cm'
+ 49
Measurements
(ii) The given combined figure is made up of a square ABCD and
an equilateral triangle DEA.
Given: Side of a square = 4 cm
.-. Perimeter of the combined figure = AB + BC + CD + DE + EA
= 4 + 4 + 4 + 4 + 4 = 20 cm
.-. Perimeter of the combined figure = 20 cm.
Area of the given combined figure = Area of a square +
B
Area of an equilateral triangle
= a' +
= 4x4 +
a
1.732
X4x4
= 16 + 1.732 X 4
Area of the given combined figure = 16 + 6.928 = 22.928
Area of the given figure ~ 22. 93 cm-.
Example 4.6
Find the perimeter and area of the shaded portion
A D
(ii)
B
4cm
Fig. 4.28
c
Solution p.g ^29
(i) The given figure is a combination of a rectangle ABCD and two semicircles
AEB and DEC of equal area. ^ ^
Given: Length of the rectangle, / = 4 cm
Breadth of the rectangle, b = 2 cm
Diameter of a semicircle = 2 cm
.-. Radius of a semicircle, r
_ 2 _
1 cm
•. Perimeter of the given figure = AD+BC+ AEB + DFC
= 4+ 4+ 2 X y X (circumference of a circle)
= 8 + 2 X ^X27rr
= 8 + 2 X ^Xl
= 8 + 2x3.14
= 8 + 6.28
Chapter 4
Perimeter of the given figure = 14.28 cm.
Area of the given figure = Area of a rectangle ABCD +
2 X Area of a semicircle
= /x b + 2 X
7tr
= 4x2 + 2 X 22X1X1
7x2
.-. Total area = 8 + 3. 14 = 11. 14 cm^
(ii) Let ADB, BEC and CFA be the three semicircles I, II and III respectively.
Given:
Radius of a semicircle I, r^ = A^ = 5 cm
Radius of a semicircle II, r = -i- = 4 cm
Radius of a semicircle III, r
1 2
= 1
2 2
_ 6
= 3 cm
3 2
Perimeter of the shaded portion = Perimeter of a semicircle I +
Perimeter of a semicircle II +
Perimeter of a semicircle III
= (7r + 2)x5 + (7r + 2)x4 + (7r + 2)x3
= (7r + 2)(5 + 4 + 3) = (7r+2)xl2
= ^ 22 + 14 j^^2 = J^xl2 = 61.714
Perimeter of the shaded portion ~ 61.71cm.
Area of the shaded portion, A = Area of a semicircle I +
Area of a semicircle II +
Area of a semicircle III
A =
Ttrf , 7rr{ , Tin
111
^X5X5 + ^X4X4 + ^X3X3
A =
275 I 176 . 99 ^ 550
111 1
= 78.571cm'
Area of the shaded portion ~ 78.57 cm'
In this example we observe that.
Area of semicircle BEC + Area of semicircle CFA = Area of semicircle ADB
Measurements
A 28m E
Fig. 4.30
Example 4.7
A horse is tethered to one corner of a rectangular
field of dimensions 70 m by 52 m by a rope 28 m long
for grazing. How much area can the horse graze inside?
How much area is left ungrazed?
Solution
Length of the rectangle, / = 70 m
Breadth of the rectangle, b = 52 m
Length of the rope = 28 m
Shaded portion AEF indicates the area in which the horse can graze. Clearly, it
is the area of a quadrant of a circle of radius, r = 28 m
Area of the quadrant AEF = -^xirr sq. units
= ^X^X 28x28 =616m^
.-. Grazing Area = 616 m-.
Area left ungrazed = Area of the rectangle ABCD -
Area of the quadrant AEF
Area of the rectangle ABCD = I x b sq. units
= 70 X 52 = 3640 m^
.-. Area left ungrazed = 3640 - 616 = 3024 m^
Example 4.8 ~7^
In the given figure, ABCD is a square of side 14 cm. Find the
area of the shaded portion.
Solution
Side of a square, a = 14 cm
n
Radius of each circle, r - -^ cm
Eig. 4.31
Area of the shaded portion = Area of a square - 4 x Area of a circle
D 14cm C
r
= a'
4 {Trr')
= 14x 14-4X ^X^X^
7
= 196-154
Area of the shaded portion = 42 cm'.
Eig. 4.32 B
Chapter 4
Fig. 4.33
Example 4.9
A copper wire is in the form of a circle with radius 35 cm. It is bent into a
square. Determine the side of the square.
Solution
Given: Radius of a circle, r = 35 cm.
Since the same wire is bent into the form of a square,
Perimeter of the circle = Perimeter of the square
Perimeter of the circle = 27tr units
= 2x^x35cm
P = 220 cm.
Let 'a' be the side of a square.
Perimeter of a square = 4a units
4a = 220
a = 55 cm
.-. Side of the square = 55 cm.
Fig. 4.34
Example 4.10
Four equal circles are described about four corners of
a square so that each touches two of the others as shown in the
Fig. 4.35. Find the area of the shaded portion, each side of the
square measuring 28 cm.
Solution
Let ABCD be the given square of side a.
.-. a = 28cm
- 28
Radius of each circle, r
2
= 14 cm
Fig. 4.35
Area of the shaded portion = Area of a square - 4 x Area of a quadrant
= a- - 4 x4-X7rr
4
= 28x28-4x4x4?-xl4xl4
= 784-616
•. Area of the shaded portion = 168 cm^
Measurements
Example 4.11
A 14 m wide athletic track consists of two straight
sections each 120 m long joined by semi-circular ends
with inner radius is 35 m. Calculate the area of the track.
Solution
Given: Radius of the inner semi circle, r
Width of the track =
.-. Radius of the outer semi circle, R =
R =
120 m
14m
= 35 m
14 m
35 + 14 = 49 m
120 m G
Fig. 4.36
49 m
Area of the track is the sum of the areas of the semicircular tracks and the areas
of the rectangular tracks.
Area of the rectangular tracks ABCD and EFGH = 2 x (/ x b)
= 2 X 14 X 120 = 3360 m^
Area of the semicircular tracks = 2 x (Area of the outer semicircle -
Area of the inner semicircle)
= 2x(i
ttR'
1
jrr
2x^X7r{R'-r')
22
7
22
X(49'-35')
(49 + 35)(49-35)
- 2^X84X14
= 3696 m^
.-. Area of the track = 3360 + 3696
= 7056 m^ p^
Example 4.12
^\
In the given Fig. 4.37, PQSR represents a flower bed. If g
R
^^\ \
OP = 21 m and OR = 14 m, find the area of the shaded portion. ""^
a
\ \
Solution
-i \ \
C
) s c
Given : OP = 21 m and OR = 14 m ^'^- "^'^^
Area of the flower bed = Area of the quadrant OQP -
Area of the quadrant OSR
= ^TTXOP'-^TTXOR'
4 4
Chapter 4
= 4-X TTX 21- -^-XTTX 14'
4 4
= ^x;rx(2r-14')
= ^X^X(21+14)X(21-14)
.-. Area of the flower bed = ^x^x35x 7 = 192. 5 m^
Example 4.13
Find the area of the shaded portions in the Fig. 4.38, where
ABCD is a square of side 7 cm.
Solution
Let us mark the unshaded portions by I, II, III and IV as
shown in the Fig. 4.39.
Let P,Q,R and S be the mid points of AB, BC,CD and DA
respectively.
7 cm
Fig. 4.38
Side of the square, a - 1 cm
n
Radius of the semicircle, ^--t^ cm
Area of I + Area of III = Area of a square ABCD -
Area of two semicircles
with centres P and R
= a'-2XyX7rr
= 7x7-2x^X^X^X^
.-. Area of I + Area of III = (49 - ^) cm- = ^ cm^
Similarly, we have
Area of II + Area of IV = (49 - ^) cm' = ^ cm^
Area of the shaded portions = Area of the square ABCD - (Area of I +
Area of II + Area of III + Area of IV)
= 49-(f + f)
= 49 - 21= 28 cm'
.-. Area of the shaded portions = 28 cm'.
Example 4.14
A surveyor has sketched the measurements of a land as below.
Find the area of the land.
Solution
Let J, K, L, M be the surveyor's marks from A to D. pig^ 44Q
Metres
ToD
20
15
lOtoC
9toE
12
10
6toB
7toF
5
From A
Measurements
Given: AJ = 5 m , JF = 7 m,
D
KB = 6 m, LE = 9 m , MC = 10 m,
/ Sm
^^^
AK = 10 m, AL = 12 m,
AM = 15 m and AD = 20 m v/--
/ M
3m
9m
10m^\^
J "7^
L /
XiiVX Xi_l 111 ClllU. XiX-/ £—\J 111* tf"""""
'2m[l
6m /
The given land is the combination of the trapezium \
K
5m
J "p
KBCM, LEFJ and right angled triangles ABK, MCD, DEL )v;
and JFA.
7m
T /
\" L
\„^^ 5m
/
Let A^ denote the area of the trapezium KBCM.
t
^
A^ = ^X(KB + MC)XKM
(".■ parallel sides are KB
MC and height is KM
= yX(6+ 10)X5
KB = 6 m, MC = 10 m,
KM=AM-AK
A^ = ^xl6x5 = 40m^
= 15 - 10 = 5 m)
Let A^ denote the area of the trapezium LEFJ.
(".■ parallel sides are LE
A^ = ^X(JF + LE)XJL
JF and height is JL
JF = 7 m, LE = 9 m,
= ^X(7 + 9)X7
JL = AL-AJ
= 12 - 5 = 7 m)
A^ = ^x 16x7 = 56m'
Let A3 denote the area of the right angled triangle ABK.
A3 = ^XAKXKB
A3 = ^Xl0x6 = 30m'
Let A^ denote the area of the right angled triangle MCD.
A^ = ^xMCxMD.
= ^X10X5
A, = f = 25m^
Let Aj. denote the area of the right angled triangle DEL.
A^ = ^XDLXLE
= ^X(AD-AL)XLE
= i^(20-12)x9
Aj_ = ^ X 8 X 9 = 36 m'
•
Chapter 4
Let A denote the area of the right angled triangle JFA.
A.
XAJXJF
6 2
= 1x5x7 = ^= 17.5m\
Area of the land = Ai + A2 + A3 + A4 + As + As
= 40 + 56 + 30 + 25 + 36 + 17.5
.-. Area of the land = 204.5 m^.
EXERCISE 4.2
1. Find the perimeter of the following figures
(i)
(ii)
8cm
4cm
(iii) 4cm
, 4cm
4cm
6cm
10 cm
(iv)
(V)
2. Find the area of the following figures
12cm
(i)
14cm
6 cm
6 cm
10 cm
2m a
(111) ^
14 cm
(iv)
6 cm
6 cm
(V)
Measurements
3. Find the area of the coloured regions
2m
(i) a
20cm
6m
(ii)
2m
6m
7 cm
6cm
(V) X
(vi)
3.5cm 3.5cm
4. In the given figure, find the area of the shaded portion if
AC = 54 cm, BC = 10 cm, and O is the centre of bigger
circle.
5. A cow is tied up for grazing inside a rectangular field of dimensions 40 m x 36 m
in one corner of the field by a rope of length 14 m. Find
the area of the field left ungrazed by the cow.
A square park has each side of 100 m. At each corner of
the park there is a flower bed in the form of a quadrant of
radius 14 m as shown in the figure. Find the area of the
remaining portion of the park.
100 m
1 cm
Find the area of the shaded region shown in the figure. The
four corners are quadrants. At the centre, there is a circle
of diameter 2 cm.
4 cm
8. A paper is in the form of a rectangle ABCD in which AB - 20 cm and BC = 14
cm. A semicircular portion with BC as diameter is cut off. Find the area of the
remaining part.
Chapter 4
9. On a square handkerchief, nine circular designs each of radius
7 cm are made. Find the area of the remaining portion of the
handkerchief.
10. From each of the following notes in the field book of a surveyor, make a rough
plan of the field and find its area.
(i)
Metres
ToD
240
210
30 C
40
170
60
130
70
50toB
From A
(h)
300 M
To J
1000 K
6001
300 H
From G
800 L
i|ctmty
Can you help the ant?
An ant is moving around a few food
pieces of different shapes scattered on the
floor. For which food-piece would the ant
have to take a shorter round and ^^^ J^y2,r-
i^
longer round?
1.4 cm
i^^t
%& inu' t^ea^
Which is smaller? The perimeter of a square or the perimeter of
a circle inscribed in it?
14cm
Measurements
Concept Summary
% The central angle of a circle is 360°.
% Perimeter of a semicircle = (tt + 2) x r units.
% Area of a semicircle = ^^ sq . units.
% The central angle of a semicircle is 180°.
% Perimeter of a quadrant = (^ + 2)xr units.
^ Area of a quadrant = ^^ sq . units.
% The central angle of a quadrant is 90°.
% Perimeter of a combined figure is length of its boundary.
•V A polygon is a closed plane figure formed by 'n' line segments.
"^ Regular polygons are polygons in which all the sides and angles are
equal.
^ Irregular polygons are combination of plane figures.
Geometry
5.1 Introduction
5.2 Properties of Triangle
5.3 Congruence of Triangles
5.4 Concurrency in Triangle
5.5 Pythagoras Theorem
5.6 Circles
5.1 Introduction
Geometry was developed by Egyptians more than 1000 years
before Christ, to help them mark out their fields after the floods from the
Nile. But it was abstracted by the Greeks into logical system of proofs
with necessary basic postulates or axioms.
Geometry plays a vital role in our life in many ways. In nature,
we come across many geometric shapes like hexagonal bee-hives,
spherical balls, rectangular water tanks, cylindrical wells and so on. The
construction of Pyramids is a glaring example for practical application
of geometry. Geometry has numerous practical applications in many
fields such as Physics, Chemistry, Designing, Engineering, Architecture
and Forensic Science.
The word 'Geometry' is derived from two Greek words 'Geo'
which means 'earth' and 'metro' which means 'to measure'. Geometry
is a branch of mathematics which deals with the shapes, sizes, position
and other properties of the object.
In class VII, we have learnt about the properties of parallel lines,
transversal lines, angles in intersecting lines, adjacent and alternative
angles. Moreover, we have also come across the angle sum property of
a triangle.
Euclid
Father of Geometry |
"Euclid
was a great
Mathematician
who gave birth to
logical thinking
in geometry".
Euclid collected
the various
information on
geometry around
300B.C. and
published them
in the form of
13 books in a
systematic manner.
These books are
called Euclid
Elements.
Euclid said :
"The whole is
greater with any of
its parts".
Geometry
Let us recall the results through the following exercise.
1. In Fig.5.1, x° = 128°. Findy°.
c
REVISION EXERCISE
2. Find ZBCE and ZECD in the
Fig.5.2, where ZACD = 90°
c
A
x°+10°
A O
Fig. 5.1
B
D
Fig. 5.2
3. Two angles of a triangle are 43° and 27°. Find the third angle.
4. Find x° in the Fig.5.3, if PQ || RS. 5. In the Fig.5.4, two Unes AB and CD
intersect at the point O. Find the
value of x° and y°.
A ^D
Fig. 5.4
6. In the Fig. 5.5 AB || CD. Fill in the blanks.
(i) ZEFB and ZFGD are angles.
(ii) ZAFG and ZFGD are angles.
(iii) ZAFE and ZFGC are angles.
5.2 Properties of Triangles
A triangle is a closed figure bounded by three line segments
in a plane.
Triangle can be represented by the notation 'A'.
In any triangle ABC, the sides opposite to the vertices
A, B, C can be represented by a, b, c respectively.
Chapter 5
1 5.2.1. Kinds of Triangles
Triangles can be classified into two types based on sides and angles.
I Based on sides:
|(a) Equilateral Triangle (b) Isosceles Triangle (c) Scalene Triangle
3 cm
Three sides
are equal
Based on angles:
(d) Acute Angled
Triangle
6 cm
Two sides
are equal
(e) Right Angled
Triangle
4 cm
All sides are
different
(f ) Obtuse Angled
Triangle
Three acute angles
One right angle
5.2.2 Angle Sum Property of a Triangle
Theorem 1
One obtuse angle
A
X
<f-<-
The sum of the three angles of a triangle is 180^^
Given
To Prove
Construction
Proof
ABC is a Triangle.
ZABC + ZBCA + ZCAB = 180°
Through the vertex A draw XY parallel to BC.
Fig. 5.7
Statement
Reason
(i) BC 1 XY and AB is a transversal
.-. ZABC = ZXAB
Alternate angles.
(ii) and ZBCA = ZYAC
Alternate angles.
(iii) ZABC + ZBCA = ZXAB + ZYAC
By adding (i) and (ii).
(iv) (ZABC + ZBCA) + ZCAB =
(ZXAB + ZYAC) + ZCAB
By adding ZBACon both sides.
(v) .-. ZABC + ZBCA + ZCAB =180°
The angle of a straight line is 180°.
Geometry
(i) Triangle is a polygon of three sides.
(ii) Any polygon could be divided into triangles by joining the diagonals,
(iii) The sum of the interior angles of a polygon can be given by
the formula (n - 2) 180°, where n is the number of sides.
Illustration
Figure
/\
f)
Number of sides
3
4
5
Classification
Triangle
Quadrilateral
Pentagon
Sum of angles
(3 - 2) 180° = 180°
(4 -2)180° = 360°
(5 - 2) 180° = 540°
Theorem 2
If a side of a triangle is produced, tlie
exterior angle so formed, is equal to the sum of
the two interior opposite angles.
Given
To Prove
Proof
ABC is a triangle.
BC is produced to D.
ZACD = ZABC + ZCAB
Fig. 5.8
Statement
Reason
(i) InAABC,ZABC + ZBCA + ZCAB =180°
Angle sum property of a triangle
(ii) ZBCA + ZACD = 180°
Sum of the adjacent angles of a straight
(iii) ZABC + ZBCA + ZCAB =
line
ZBCA + ZACD
Equating (i) and (ii)
(iv) .-. ZABC + ZCAB=ZACD
Subtracting ^BCA on both sides of (iii)
(v) The exterior angle ZACD is equal to the
Hence proved.
sum of the interior opposite angles
ZABC and ZCAB.
Chapter 5
(i) In a traingle the angles opposite to equal sides are equal,
(ii) In a traingle the angle opposite to the longest side is largest.
Fig. 5.9
Example 5.1
In AABC, ZA = 75°, ZB = 65° find ZC.
Solution
We know that in AABC,
ZA + ZB + ZC = 180°
75° + 65° + ZC = 180°
140° + ZC = 180°
ZC = 180° -140°
.-. ZC = 40°.
Example 5.2
In ZiABC, given that ZA = 70° and AB = AC. Find the other angles of A ABC.
Solution
Let ZB = x° and ZC = y°.
Given that AABC is an isosceles triangle.
AC = AB
ZB = ZC [Angles opposite to equal sides are equal]
A
x" = y"
In AABC, Z A + ZB + ZC = 180°
70" + x° + / = 180°
70" + x° + x" = 180° [■:x° =y]
2x° = 180° -70°
2x° = 110°
Fig. 5.10
x° = i^ = 55°. Hence ZB = 55° and ZC = 55°.
Example 5.3
The measures of the angles of a triangle are in the ratio 5:4:3. Find the angles
of the triangle.
Solution
Given that in a AABC, ZA : ZB : ZC = 5 : 4 : 3.
Let the angles of the given triangle be 5 x°, 4 x° and 3 x°.
Geometry
We know that the sum of the angles of a triangle is 180'^
5 x° + 4 x° + 3x° = 180° ^ 12 x° = 180°
= 15°
.o - 180"
X" =
12
So, the angles of the triangle are 75° 60° and 45°.
Example 5.4
Find the angles of the triangle ABC, given in Fig. 5.11.
Solution
BD is a straight line.
We know that angle in the line segment is 180°.
x°+ 110° = 180°
x° = 180° - 110°
x° = 70°
We know that the exterior angle is equal to the sum of the two interior opposite
angles.
x° + y° = 110°
70°+y° = 110°
yO = 110° -70° = 40°
Hence, x° = 70°
andy° = 40°.
Example 5.5
Find the value of ZDEC from the given Fig. 5.12.
Solution
We know that in any triangle, exterior angle is equal
to the sum of the interior angles opposite to it.
In AABC, ZACD = ZABC + ZCAB
.-. ZACD = 70° + 50° = 120°
Also, ZACD = ZECD = 120°.
Considering AECD,
ZECD + ZCDE + ZDEC =180" [Sum of the angles of a triangle]
120" + 22" + ZDEC = 180"
ZDEC = 180° -142°
ZDEC = 38°
Fig. 5.12
Chapter 5
jMvi^
Draw all the types of triangles T^, T^, T^, T^, T^and Tg. Let us name the triangles
as ABC. Let a, b, c be the sides opposite to the vertices A, B, C respectively.
Measure the sides and arrange the data as follows:
Serial
No.of A
a
(cm)
b
(cm)
c
(cm)
(c+a) > b
True / False
(a + b)> c
True / False
(b + c)> a
True / False
T,
T,
T.
T.
T.
T.
JWMtjdcryou observe from this table ?
Theorem 3
(This is known as Triangle Inequality)
Verification :
Consider the triangle ABC such that BC = 12 cm, AB = 8 cm, AC = 9 cm.
(i) AB = 8 cm , AB + BC = 20 cm
(ii) BC = 12 cm , BC + CA = 21 cm
(iii) CA = 9 cm, CA + AB = 17 cm
Now clearly ,
(i) AB + BC> CA
(ii) BC + CA > AB
(iii) CA + AB > BC
In all the cases, we find that the sum of any two sides of a triangle is greater
than the third side.
Example 5.6
Which of the following will form the sides of a triangle?
(i) 23cm, 17cm, 8 cm (ii) 12 cm, 10 cm, 25 cm (iii) 9 cm, 7 cm, 16 cm
Solution
(i) 23 cm, 17cm, 8 cm are the given lengths.
Here 23 + 17 > 8, 17 + 8 > 23 and 23 + 8 > 17.
.-. 23cm, 17cm, 8cm will form the sides of a triangle,
(ii) 12cm, 10cm, 25cm are the given lengths.
Geometry
Here 12 + 10 is not greater than 25. ie, [12 + 10 > 25]
.-. 12cm, 10cm, 25cm will not form the sides of a triangle,
(iii) 9 cm, 7 cm, 16 cm are given lengths. 9 + 7 is not greater than 16.
ie, [9 + 7 = 16,9 + 7^16]
.•.9 cm, 7cm and 16cm will not be the sides of a triangle.
^i) '
c + a> b
— ^
b < c + a
— ^
b-
- c < a
(ii)
b + c > a
=^
a < b + c
— ^
a -
-b<c
(iii)
a + b> c
— ^
c <a + b
— >
c -
-a<b
From the above result we observe that in any triangle the difference between
the length of any two sides is less than the third side.
EXERCISE 5.1
1. Choose the correct answer:
(i) Which of the following will be the angles of a triangle?
(A) 35°, 45°, 90° (B) 26°, 58°, 96°
(C) 38°, 56°, 96° (D) 30°, 55°, 90°
(ii) Which of the following statement is correct ?
(A) Equilateral triangle is equiangular.
(B) Isosceles triangle is equiangular.
(C) Equiangular triangle is not equilateral.
(D) Scalene triangle is equiangular
(iii) The three exterior angles of a triangle are 130°, 140°, x° then x° is
(A) 90° (B) 100° (C) 110° (D) 120°
(iv) Which of the following set of measurements will form a triangle?
(A) 11 cm, 4 cm, 6 cm (B) 13 cm, 14 cm, 25 cm
(C) 8 cm, 4 cm, 3 cm (D) 5 cm, 16 cm, 5 cm
(v) Which of the following will form a right angled triangle, given that the
two angles are
(A) 24°, 66° (B) 36°, 64°
(C) 62°, 48° (D) 68°, 32°
2. The angles of a triangle are (x - 35)°, (x - 20)° and (x + 40)°.
Find the three angles.
3. In AABC, the measure of ZA is greater than the measure of ZB by 24°. If
exterior angle ZC is 108°. Find the angles of the AABC.
Chapter 5
4. The bisectors of ZB and ZC of a AABC meet at O.
Show that ZBOC = 90° +
ZA
5. Find the value of x° andy° from the following figures:
A
A
6. Find the angles x°, y° and z° from the given figure
5.3 Congruence of Triangles
We are going to learn the important geometrical idea "Congruence".
To understand what congruence is, we will do the following activity:
cover tne otner completely ana exactly.
From the above activity we observe that the figures are of the same shape and
the same size.
In general, if two geometrical figures are identical in shape and size then they
are said to be congruent.
ijcthdiy
Check whether the following objects are congruent or not
(a) Postal stamps of same denomination.
(b) Biscuits in the same pack.
(c) Shaving blades of same brand.
Geometry
Now we will consider the following plane figures.
F u
D
Fig. 5.13 Fig. 5.14
Observe the above two figures. Are they congruent? How to check?
We use the Method of Superposition.
Step 1 : Take a trace copy of the Fig. 5.13. We can use Carbon sheet.
Step 2 : Place the trace copy on Fig. 5.14 without bending, twisting and
stretching.
Step 3 : Clearly the figure covers each other completely.
Therefore the two figures are congruent.
Congruent: Two plane figures are Congruent if each when superposed on the other
covers it exactly. It is denoted by the symbol "=".
5.3.1 (a) Congruence among Line Segments
Two line segments are congruent, if they have the same length
3cm
D
Here, length AB = length CD. Hence AB = CD
(b) Congruence of Angles
Two angles are congruent, if they have the same measure.
Here the measures are equal. Hence ZMON = ZPQR.
Chapter 5
(c) Congruence of Squares
Two squares having same sides are congruent to each other.
D 2 cm c S 2 cm r
a
o
A 2 cm B P 2 cm Q
Here, sides of the square ABCD = sides of the square PQRS.
.-. Square ABCD = Square PQRS
(d) Congruence of Circles
Two circles having the same radius are congruent.
In the given figure, radius of circle Ci = radius of circle C2 .
.-. Circle Ci = Circle C2
The above four congruences motivated us to learn about the congruence of
triangles.
Let us consider the two triangles as follows:
B 8 cm C Q 8 cm R
If we superpose AABC on APQR with A on P, B on Q and C on R such that
the two triangles cover each other exactly with the corresponding vertices, sides and
angles.
We can match the corresponding parts as follows:
Corresponding Vertices
Corresponding Sides
Corresponding Angles
A — P
AB = PQ
ZA = ZP
B^Q
BC = QR
ZB = ZQ
C — R
CA=RP
ZC = ZR
Geometry
5.3.2. Congruence of Triangles
Two triangles are said to be congruent, if the three sides and the three angles
of one triangle are respectively equal to the three sides and three angles of the other.
Note: While writing the congruence condition between two triangles the order of
the vertices is significant.
5 cm
5 cm
If A ABC = APQR, then the congruence could be written as follows in different orders
ABAC = AQPR, ACBA = ARQP and so on. We can also write in anticlockwise
direction.
5.3.3. Conditions for Triangles to be Congruent
We know that, if two triangles are congruent, then six pairs of their corresponding
parts (Three pairs of sides, three pairs of angles) are equal.
But to ensure that two triangles are congruent in some cases, it is sufficient to
verify that only three pairs of their corresponding parts are
, , . , . . Axiom: The simple
equal, which are given as axioms. , ,
properties which are
There are four such basic axioms with different ^^^^ without actually
combinations of the three pairs of corresponding parts. These proving them.
axioms help us to identify the congruent triangles.
If 'S' denotes the sides. A' denotes the angles, 'R' denotes the right angle and
'H' denotes the hypotenuse of a triangle then the axioms are as follows:
(i) SSS axiom (ii) SAS axiom (iii) ASA axiom (iv) RHS axiom
(i) SSS Axiom (Side- Side- Side axiom)
If three sides of a triangle are respectively equal to the three sides of another
triangle then the two triangles are congruent.
Chapter 5
We consider the triangles ABC and PQR such that,
AB = PQ, BC = QR and CA= RP.
Take a trace copy of AABCand superpose on APQR such that
AB on PQ , BC on QR and AC on PR
Since AB = PQ ^ A Ues on P, B Ues on Q
Similarly BC = QR ^ C lies on R
Now, the two triangles cover each other exactly.
.•.AABC = APQR
Think it!-
Here, we observe that AB = PQ , BC = QR , CA =
It can be written as ^ = ^ = ^ = 1.
PQ QR RP
RP?j
What will happen
when the ratio is not
equal to 1?
Example 5.7
From the following figures, state whether the given pairs of triangles are
congruent by SSS axiom.
Q 4.5 cm K ^3 cm
Solution
Compare the sides of the APQR and AXYZ
PQ = XY = 5cm, QR = YZ = 4.5cm and RP = ZX = 3cm.
If we superpose A PQR on A XYZ .
P lies on X, Q lies on Y, R lies on Z and APQR covers AXYZ exactly.
.-. A PQR = AXYZ [by SSS axiom]. ?
Example 5.8
In the figure, PQSR is a parallelogram.
PQ = 4.3 cm and QR = 2.5 cm. Is APQR = APSR?
Solution
4.3 cm
4.3 cm
Consider APQR and APSR. Here, PQ = SR = 4.3cm
andPR=QS = 2.5cm. PR = PR [common side]
.-. APQR = ARSP [by SSS axiom]
.-. APQR ^ APSR [ARSP and APSR are of different order]
Geometry
(ii) SAS Axiom (Side- Angle- Side Axiom)
If any two sides and the included angle of a triangle are respectively equal
to any two sides and the included angle of another triangle then the two triangles
are congruent. a
Q R
We consider two triangles, AABC and APQR such that AB = PQ, AC = PR
and included angle BAG = included angle QPR.
We superpose the trace copy of AABC on APQR with AB along PQ and AC
along PR.
Now, A lies on P and B lies on Q and C lies on R. Since,AB = PQ and AC = PR,
B lies on Q and C lies on R. BC covers QR exactly.
.-. AABC covers APQR exactly.
Hence, AABC = APQR
(iii) ASA Axiom (Angle -Side -Angle Axiom)
If two angles and a side of one triangle are respectively equal to two
angles and the corresponding side of another triangle then the two triangles are
congruent.
B C
Consider the triangles, AABC and APQR .
Here, BC = QR, ZB = ZQ, ZC = ZR.
By the method of superposition, it is understood that ZABC covers ZPQR
exactly and ZBCA covers ZQRP exactly.
So, B lies on Q and C lies on R. Hence A lies on P.
.-. AABC covers APQR exactly. Hence, AABC = APQR.
As the triangles are congruent, we get remaining corresponding parts are also
equal, (i.e.) AB = PQ, AC = PR and ZA = ZP
Chapter 5
Representation: The Corresponding Parts of Congruence Triangles are Congruent
is represented in short form as c.p.c.t.c. Hereafter this notation will be used in the
problems.
Example 5.9
AB and CD bisect each other at O. Prove that AC = BD.
Solution
Given : O is mid point of AB and CD.
.-. AO = OB and CO = OD
To prove : AC = BD
Proof : Consider AAOC and ABOD
AO = OB [Given]
CO = OD [Given]
ZAOC = ZBOD [Vertically Opposite angle]
AAOC = ABOD [by SAS axiom]
Hence we get, AC = BD [by c.p.c.t.c]
Example 5.10
In the given figure, ADAB and ACAB are on
the same base AB. Prove that ADAB = ACAB
Solution
Consider ADAB and ACAB
ZDAB = 35° + 20° = 55° = ZCBA [Given]
ZDBA = ZCAB = 20° [Given]
AB is common to both the triangles.
. • . ADB A = ACAB [by AS A axiom]
Hypotenuse
Do you know what is meant by hypotenuse ?
Hypotenuse is a word related with right angled triangle.
Q B
Fig. 5.13
hypotenuse
Consider the right angled triangle ABC. ZB is a right angle.
The side opposite to right angle is known as the hypotenuse.
Here AC is hypotenuse.
Geometry
(iv) RHS Axiom (Right angle - Hypotenuse - Side)
If the hypotenuse and one side of the right angled triangle are respectively
equal to the hypotenuse and a side of another right angled triangle, then the two
triangles are congruent.
A D
B C E F
Consider AABC and ADEF where, ZB = ZE = 90°
Hypotenuse AC = Hypotenuse DF [Given]
SideAB = Side DE [Given]
By the method of superposing, we see that AABC = ADEF .
5.3.4 Conditions which are not sufficient for congruence of triangles
(i) AAA (Angle - Angle - Angle)
It is not a sufficient condition for congruence of triangle. Why?
Let us find out the reason. Consider the following triangles.
P
In the above figures,
ZA = ZP, ZB = ZQ and ZC = ZR
But size of AABC is smaller than the size of APQR.
.•. When AABC is superposed on the APQR, they will not cover each other
exactly .-. AABC ^ APQR.
(ii) SSA (Side-Side-Angle)
We can analyse a case as follows:
Construct AABC with the measurements ZB = 50°, AB = 4.7 cm and
AC = 4 cm. Produce BC to X. With A as centre and AC as radius draw an arc of 4 cm.
It will cut BX at C and D.
Chapter 5
.-. AD is also 4cm [•.• AC and ad are the radius of A
the same circle] / \
Consider AABC and AABD . ^/ \
ZB is common. v/ /^ \^
AB is common and AC = AD = 4cm / ht- \
[by construction] /\ "--^1 \^
^
Side AC, side AB and ZB of AABC and side ^ ^^ ^°
X
AD , side AB and ZB of AABD are respectively
congruent to each others. But BC and BD are not equal.
.-. AABC ^ AABD.
Example 5.11
Prove that the angles opposite to equal side of a triangle are equal.
Solution
ABC is a given triangle with, AB = AC. ^
To prove : Angle opposite to AB = Angle /
\
opposite to AC (i.e.) ZC = ZB . /
\
Construction : Draw AD perpendicular to BC. /
\
.-. ZADB = ZADC =90° / ^ ^
/ n
c
Proof : B D
Condiser AABD and AACD.
AD is common
AB = AC [AABC isanisosecles]
ZADB = ZADC = 90° [by construction]
.-. AADB = AADC [by RHS axiom]
Hence ZABD = ZACD [by c.p.c.t.c]
(or)ZABC =ZACB.
ZB = ZC . Hence the proof.
This is known as Isosceles triangle theorem.
Example 5.12
Prove that the sides opposite to equal angles of a triangle are equal.
Solution
Given : In a AABC, ZB = ZC.
To prove : AB = AC.
Construction : Draw AD perpendicular to BC.
Geometry
[given]
[given]
[by SAS ,
J
Proof :
ZADB = ZADC = 90° [by construction]
ZB = ZC [given]
AD is common side.
.-. AADB = AADC (by AAS axiom)
Hence, AB = AC. [by c.p.c.t.c] b d
So, the sides opposite to equal angles of a triangle are equal.
This is the converse of Isosceles triangle theorem.
Example 5.13
In the given figure AB = AD and ZBAC = ZDAC. Is AABC = AADC?
If so, state the other pairs of corresponding parts.
Solution
In AABC and AADC, AC is common.
ZBAC = ZDAC
AB = AD
.-. AABC = AADC
So, the remaining pairs of corresponding parts are
BC = DC, ZABC = ZADC, ZACB = ZACD.
Example 5.14
APQR is an isosceles triangle with PQ = PR, QP is produced to S and PT
bisects the extension angle 2x°. Prove that ZQ = x° and hence prove that PT || QR.
Solution
Given : APQR is an isosceles triangle with PQ = PR .
Proof : PT bisects exterior angle ZSPR and therefore ZSPT = ZTPR = x°.
.". ZQ = ZR . [Property of an isosceles triangle]
Also we know that in any triangle,
exterior angle = sum of the interior opposite angles.
.-. In APQR, Exterior angle ZSPR = ZPQR + ZPRQ
2x° = ZQ + ZR
= ZQ + ZQ
2x" = 2ZQ
x" = ZQ
Hence ZQ = x°.
[by c.p.c.t.c]
Chapter 5
To prove : PT || QR
Lines PT and QR are cut by the transversal SQ. We have ZSPT=x°.
We already proved that ZQ = x".
Hence, ZSPT and ZPQR are corresponding angles..-. PT || QR.
EXERCISE 5.2
1. Choose the correct answer:
(i) In the isosceles AXYZ, given XY = YZ then which of the following angles are
equal?
(A) ZX and ZY (B) ZY and ZZ (C)ZZ and ZX (D) ZX, ZY and ZZ
(ii) In AABC and ADEF, ZB = ZE, AB = DE, BC = EF. The two triangles are
congruent under axiom
(A)SSS (B)AAA (C)SAS (D) ASA
(iii) Two plane figures are said to be congruent if they have
(A) the same size (B) the same shape
(C) the same size and the same shape (D) the same size but not same shape
(iv) In a triangle ABC, ZA = 60° and AB ^ AC, then ABC is triangle.
(A) a right angled (B) an equilateral (C) an isosceles (D) a scalene
(v) In the triangle ABC, when ZA = 90° the hypotenuse is
(A)AB (B)BC (C)CA (D) None of these
(vi) In the APQR the angle included by the sides PQ and PR is
(A) ZP (B) ZQ
(C) ZR (D) None of these
(vii) In the hgure, the value of x° is
(A) 80° (B) 100°
(C) 120° (D) 200°
In the figure, ABC is a triangle in
which AB = AC. Find x° and y°.
3. In the figure. Find x°.
Geometry
4. In the figure APQR and ASQR
are isosceles triangles. Find x°.
p
5. In the figure, it is given that BR = PC|
and ZACB = ZQRPand AB || PQ.
Prove that AC = QR.
6. In the figure, AB - BC = CD , ZA = x°. 7. Find x°, y°, z° from the figure.
Prove that ZDCF = 3ZA. where AB - BD, BC - DC and
ZDAC = 30°.
8. In the figure, ABCD is a paraUelogram. 9. In figure, BO bisects ZABC of AB
is produced to E such that AB - BE. AABC. P is any point on BO. Prove
AD produced to F such that AD = DF. that the perpendicular drawn from P
Show that AFDC = ACBE . to BA and BC are equal
F
10. The Indian Navy flights fly in a formation
that can be viewed as two triangles with
common side. Prove that ASRT = A QRT,
if T is the midpoint of SQ and SR - RQ.
A
Chapter 5
1 5.4 Concurrency in Triangles
Draw three or more hnes in a plane. What are the possible ways?
The possibilities may be as follows:
A
C
B
D
E
F
(a)
(b)
(c)
(d)
In fig (a), AB ,CD and EF are parallel so they are not intersecting.
In fig (b), AB and CD intersect at P, AB and EF intersect at Q. So P, Q are
two points of intersection.
In fig (c), P, Q, R are three point of intersection.
But in fig (d), P is the only point of intersection. Here AB, CD, EF passing
through the same point P. These lines are called as concurrent lines. The point P is
called the point of concurrency.
In a triangle there are some special points of concurrence, which are Centroid
of a triangle, Orthocentre of a triangle, Incentre of a triangle and Circumcentre of a
triangle. Now we are going to study how to obtain these points in a triangle.
5.4.1 Centroid of a Triangle
In the adjacent figure, ABC is a triangle.
D is mid point of BC. Join AD .
Here AD is one of the medians of A ABC.
A median of a triangle is the line segment joining a vertex
and the midpoint of the opposite side.
Now consider the adjacent figure, in which AD, BE, CF are
the three medians of A ABC.
They are concurrent at G. This point is called as centroid.
The three medians of a triangle are concurrent and the point
of concurrency is known as Centroid. It is denoted by 'C.
Note : (i) The Centroid divides each of the median in the ratio 2
(ii) The Centroid would be the physical centre of gravity.
Geometry
4 cm Dl'^™C
5.4.2 Orthocentre of a Triangle
In the adjacent figure, ABC is a triangle .
From A, draw a perpendicular to BC ,
AD is perpendicular to BC .
ZADB = ZADC = 90°. Here D need not be the B
mid point. Here AD is an altitude from vertex A.
Altitude of a triangle is a perpendicular line
segment drawn from a vertex to the opposite side. Now
consider the figure, the triangle ABC in which AD, BE,
CF are the three altitudes.
They are concurrent at H. This point in known as
Orthocentre.
The three altitudes of a triangle are concurrent and the point of concurrency is
known as Orthocentre.
Different positions of orthocentre
(a) (b) (c)
Case (i) : In fig (a), ABC is an acute angled triangle .
Here orthocentre lies inside the AABC .
Case (ii) : In fig (b), ABC is a right angled triangle .
Here orthocentre lies on the vertex at the right angle.
Case (iii) : In fig (c), ABC is an obtuse angled triangle.
Here orthocentre lies outside the AABC .
5.4.3 Incentre of a Triangle
In the adjacent figure , ABC is a triangle.
The ZA is bisected into two equal parts by AD.
Therefore ZBAD = ZDAC.
Here AD is said to be the angle bisector of z A.
Chapter 5
Angle bisector of a triangle is a line segment which
bisects an angle of a triangle.
Now consider the figure in which AD, BE, CF are
three angle bisectors of AABC.
They are concurrent at I.
This point is known as incentre of the triangle.
The three angle bisectors of a triangle are concurrent
and the point of concurrence is called the Incentre .
5.4.4 Circumcentre of a Triangle
We have learnt about perpendicular bisector in previous class.
What is a perpendicular bisector in a triangle?
Refer the following figures:
4 cm D 1 '=™ C B 3 cm D 3 cm C B 3 cm D 3
cm
(a) (b) (c)
In fig (a): AD is perpendicular from A to EC but not bisecting BC .
In fig (b): AD bisects BC. Hence BD = DC and AD is perpendicular to BC.
In fig (c): DX is perpendicular to BC and DX also bisecting BC. BD = DC but
DX need not passes through the vertex A'.
The perpendicular bisector of the side of a triangle is the line that is perpendicular
to it and also bisects the side.
Q-h
Now, consider the above figure.
Geometry
Here PQ, RS, and MN are the three perpendicular bisectors of BC, AC and AB
concurrent at O.
O is known as the circumcentre.
The three perpendicular bisectors of a triangle are concurrent and the point of
concurrence is known as circumcentre .
Note : (i) In any triangle ABC , Circumcentre (O) , Centroid (G) and Orthocentre
(H) are always lie on one straight line, which is called as Euler Line,
and OG : GH = 1 : 2.
(ii) In particular for equilateral triangle, Circumcentre (O), Incentre (I),
Orthocentre (H) and Centroid (G) will coincide.
5.5 Pythagoras Theorem
Pythagoras (582 - 497 B.C) was one of the foremost Mathematicians of all
times. He was perhaps best known for the right angled triangle relation which bears
his name.
5.5.1 Pythagoras Theorem
In a right angled triangle the square of the hypotenuse is equal to the sum
of the squares of the other two sides.
Let us consider AABC with ZC = 90°.
BC = a, CA = b and AB = c.
Then, a^ + b' = c^
This was proved in number of ways by different
Mathematicians. ^
We will see the simple proof of Pythagoras
Theorem.
Now, we construct a square of side (a + b) as shown in the figure,
and using the construction we prove
Pythagoras theorem. That is, we prove a^ + b^ = c\
We know that Area of any square is square
of its side.
Area of a square of side (a + b) = (a + b)^
From the figure.
Area of the square of side (a + b) is = (a + b)^
= sum of the area of the
triangles I, II, III and IV +
the area of the square PQRS
b
R
a
IV y/
/\
^\ "^
a
/ c
c^^
S
^
b
N. c
c /
I \
\/
/ II
Chapter 5
i.e., (a + b)' =4 (Area of right angled A) + (Area of the square PQRS)
(a + b)' = 4(i-xaxb) + e
a' + b' + 2ab = 2ab + e
.-.a^ + b^ = e
Hence we proved Pythagoras theorem.
^^ E
B
N
M
R
Atmiy
Pythagoras Theorem
Draw a right angled triangle
ABCsuch that ZC = 90°, AB = 5 cm,
AC = 4 cm and EC = 3 cm.
Construct squares on the three
sides of this triangle. P'
Divide these squares into small
squares of area one cm^ each.
By counting the number of small
squares, pythagoras theorem can be
proved.
Number of squares in ABPQ = 25
Number of squares in BCRS = 9
Number of squares in ACMN = 16
.-.Number of squares in ABPQ = Number of squares in BCRS +
Number of squares in ACMN.
The numbers which are satisfying the Pythagoras theorem are called the
Pythagorian Triplets.
Example 5.15
In AABC,ZB = 90°, AB = 18cm and BC = 24cm. Calculate the length of AC.
Solution
= AB' + BC
= 18' + 24'
= 324 + 576
= 900
.-. AC = /900 = 30cm
Example 5.16
A square has the perimeter 40 cm. What is the sum of the diagonals?
By Pythagoras Theorem, AC'
Geometry
Solution
Let 'a' be the length of the side of the square. AC is a diagonal.
Perimeter of square ABCD = 4 a units
4a = 40cm [given]
a = 4^ = lOcm
4
We know that in square each angle is 90" and the diagonals are equal.
In AABC, AC = AB' + BC'
= 10^ + 10^ = 100 + 100
= 200
.-.AC = /200
= 72x100 =10/2
= 10 X 1.414 = 14.14cm
Diagonal AC = Diagonal BD
Hence, Sum of the diagonals = 14.14 + 14.14 = 28.28 cm.
Example 5.17
p
From the figure PT is an altitude
of the triangle PQR in which PQ = 25cm,
PR = 17cm and PT = 15 cm. If QR = x cm.
Calculate x.
Solution From the figure, we have
QR = QT + TR.
To find : QT and TR.
In the right angled triangle PTQ , "^^
ZPTQ = 90° [PT is attitude] ^
xcm
By Pythagoras Theorem, PQ' = FT' + QT' .-. PQ' - PT' =
... QT^ = 25' - 15' = 625 - 225 = 400
QT'
QT = 7400 = 20 cm
Similarly, in the right angled triangle PTR,
by Pythagoras Theorem, PR' = PT' + TR'
.-. TR' = PR'-PT'
= 17' -15'
= 289 - 225 = 64
TR = /64 = 8 cm
Form (1) and (2) QR = QT + TR = 20 + 8 = 28 cm.
.(1)
(2)
Chapter 5
Example 5.18
A rectangular field is of dimension 40 m by 30 m. What distance is saved by
walking diagonally across the field?
Solution
Given: ABCD is a rectangular field of Length = 40m, Breadth = 30m, ZB = 90°
In the right angled triangle ABC, A D
By Pythagoras Theorem,
AC' = AB' + BC S
= 30' + 40' = 900 + 1600
= 2500
.-. AC = 72500 = 50 m
Distance from A to C through B is
= 30 + 40 = 70 m
Distance saved = 70 - 50 = 20 m.
40 m
EXERCISE 5.3
1. Choose the correct answer
(i) The point of concurrency of the medians of a triangle is known as
(A) incentre (B) circle centre (C) orthocentre (D) centroid
(ii) The point of concurrency of the altitudes of a triangle is known as
(A) incentre (B) circle centre (C) orthocentre (D) centroid
(iii) The point of concurrency of the angle bisectors of a triangle is known as
(A) incentre (B) circle centre (C) orthocentre (D) centroid
(iv) The point of concurrency of the perpendicualar bisectors of a triangle is known as
(A) incentre (B) circumcentre (C) orthocentre (D) centroid
In an isosceles triangle AB - AC and ZB = 65°. Which is the shortest side?
PQR is a triangle right angled at P. If PQ = 10 cm and PR = 24 cm, find QR.
Check whether the following can be the sides of a right angled triangle
AB - 25 cm, BC - 24 cm, AC - 7cm.
5. Angles Q and R of a triangle PQR are 25° and 65°. Is APQR a right angled
triangle? Moreover PQ is 4cm and PR is 3 cm. Find QR.
6. A 15 m long ladder reached a window 12 m high from the ground. On placing it
against a wall at a distance x m. Find x.
7. Find the altitude of an equilateral triangle of side 10 cm.
8. Are the numbers 12, 5 and 13 form a Pythagorian Triplet?
2.
3.
4.
Geometry
A painter sets a ladder up to reach the bottom of
a second story window 16 feet above the ground.
The base of the ladder is 12 feet from the house.
While the painter mixes the paint a neighbour's
dog bumps the ladder which moves the base 2
feet farther away from the house. How far up side
of the house does the ladder reach?
5.6 Circles
You are familiar with the following objects. Can you say the shape of the
following?
(a) Cycle wheel
(b) Ashoka chakra in our National Emblem
(c) Full moon
Sure, your answer will be circle. You know that a circle is described when a
point P moves in a plane such that its distance from a fixed point in the plane remains
constant.
Definition of Circle
A circle is the set of all points in a plane at a constant distance from a fixed point
in that plane.
The fixed point is called the centre of the circle.
The constant distance is known as the radius of the
circle.
In the figure 'O' is centre and OA, OB, OC are radii
of the circle.
Here, OA = OB = OC = r
Note: All the radii of the circle are equal.
Chord
A chord is a line segment with its end points lying
on a circle.
In figure, CD, AB and EF are chords.
Here AB is a special chord passes through the
centre O.
Chapter 5
Diameter
A diameter is a chord that passes through the centre of the circle and
diameter is the longest chord of a circle.
In the figure, AOB is diameter of the circle.
O is the mid point of AB and 0A= OB = radius of the circle
Hence, Diameter = 2 x radius (or) Radius = ( diameter )^ 2
Note : (i) The mid-point of every diameter of the circle is the centre of the circle.
(ii) The diameters of a circle are concurrent and the point of concurrency is
the centre of the circle.
Secant of a Circle
A line passing through a circle and intersecting the circle at two points is called
the secant of the circle.
In the given figure, line AB is a Secant.
It cuts the circle at two points A and B .
Now, let us move the secant AB
downwards. Then the new positions are A^ B^,
A^ B^, .... etc..
While secant AB moves down, the
points A and B are moving closer to each other.
So distance between A and B is
gradually decreases.
At one position the secant AB touches the circle at only one point L. At this
position, the line LM is called as tangent and it touches the circle at only one point.
Tangent
Tangent is a line that touches a circle at exactly one point, and the point is
known as point of contact.
Arc of a Circle
In the figure AB is a chord. The chord AB divides the
circle into two parts.
The curved parts ALB and AMB are known as Arcs.
Arcs will be denoted by the symbol ' '.
The smaller arc ALB is the minor arc.
The greater arc AMB is the major arc.
Geometry
Segment of a Circle
A chord of a circle divides the circular region
into two parts. Each part is called as segment of
the circle.
The segment containing minor arc is called
the minor segment.
The segment containing major arc is called ^
the major segment.
Sector of a Circle
The circular region enclosed by an arc of a circle and
the two radii at its end points is known as Sector of a circle.
The smaller sector OALB is called the minor sector.
The greater sector OAMB is called the major sector.
EXERCISE 5.4
1. Choose the correct answer:
(i) The of a circle is the distance from the centre to the circumference.
(A) sector (B) segment (C) diameters (D) radius
(ii) The relation between radius and diameter of a circle is
(A) radius = 2 x diameters (B) radius - diameter + 2
(C) diameter = radius + 2 (D) diameter = 2 (radius)
(iii) The longest chord of a circle is
(A) radius (B) secant (C) diameter (D) tangent
2. If the sum of the two diameters is 200 mm, find the radius of the circle in cm.
3. Define the circle segment and sector of a cirle.
4. Define the arc of a circle.
5. Define the tangent of a cirle and secant of a cirle.
Chapter 5
Concept Summary
% The sum of the three angles of a triangle is 180°.
^ If the sides of a triangle is produced, the exterior angle so formed, is
equal to the sum of the two interior opposite angles.
^ Any two sides of a triangle together is greater than the third side.
^ Two plane figures are Congruent if each when superposed on the other
covers it exactly. It is denoted by the symbol "=".
^ Two triangles are said to be congruent, if three sides and the three
angles of one triangle are respectively equal to three sides and three
angles of the other
% SSS Axiom: If three sides of a triangle are respectively equal to the
three sides of another triangle then the two triangles are congruent.
^ SAS Axiom: If any two sides and the included angle of a triangle are
respectively equal to any two sides and the included angle of another
triangle then the two triangles are congruent.
^ ASA Axiom: If two angles and a side of one triangle are respectively
equal to two angles and the corresponding side of another triangle then
the two triangles are congruent.
% RHS Axiom: If the hypotenuse and one side of the right angled triangle
are respectively equal to the hypotenuse and a side of another right
angled triangle, then the two triangles are congruent.
^ Centroid : Point of concurrency of the three Medians.
^ Orthocentre : Point of concurrency of the three Altitudes.
^ I ncentre : Point of concurrency of the three Angle Bisectors.
^ Circumcentre : Point of concurrency of the Perpendicular Bisectors of
the three sides.
V Circle : A circle is the set of all points in a plane at a constant distance
from a fixed point in that plane .
Geometry
^ Chord : A chord is a line segment with its end points lying on a circle.
% Diameter : A diameter is a chord that passes through the centre of the
circle.
^ A line passing through a circle and intersecting the circle at two points
is called the secant of the circle.
% Tangent is a line that touches a circle at exactly one point, and the
point is known as point of contact.
^ Segment of a circle : A chord of a circle divides the circular region
into two parts.
■^ Sector of a circle : The circular region enclosed by an arc of a circle
and the two radii at its end points is known as Sector of a circle.
^
^
Mathematics Club Activity
^ r^
The Importance of Congruency
In our daily life, we use the concept of congruence in many ways. In our home,
we use double doors which is congruent to each other. Mostly our house double gate is
congruent to each other. The wings of birds are congruent to each other. The human
body parts like hands, legs are congruent to each other. We can say many examples like
this.
Birds while flying in the sky, they fly in the formation of
a triangle. If you draw a median through the leading bird you
can see a congruence. If the congruency collapses then the
birds following at the end could not fly because they losses their
stability.
Now, try to identify the congruence structures in the
nature and in your practical life.
Practical Geometry
6.1
Introduction
6.2
Quadrilateral
6.3
Trapezium
6.4
Parallelogram
6.5
Rhombus
6.6
Rectangle and Square
6.7
Concentric Circles
6.1 Introduction
Ancient Egyptians demonstrated practical knowledge of geometry
through surveying and construction of projects. Ancient Greeks practised
experimental geometry in their culture. They have performed variety of
constructions using ruler and compass.
Geometry is one of the earliest branches of Mathematics.
Geometry can be broadly classified into Theoretical Geometry and
Practical Geometry. Theoretical Geometry deals with the principles of
geometry by explaining the construction of figures using rough sketches.
Practical Geometry deals with constructing of exact figures using
geometrical instruments.
We have already learnt in the previous classes, the definition,
properties and formulae for the area of some plane geometrical figures.
In this chapter let us learn to construct some specific plane geometrical
figures.
Guass
[1777-1855 A.D.]
Guass was a
German Math-
ematician. At the
age of seventeen
Gauss investigated
the constmctibility
of regular 'p-gons'
(polygons with
p-sides) where p
is prime number.
The construction
was then known
only for
p = 3 and p = 5.
Gauss discovered
that the regular
p-gon is con-
structible if and
only if p is prime
"Fermat Number"
(i.e.) p = 2'" + 1
Practical Geometry
6.2 Quadrilateral
6.2.1 Introduction
We have learnt in VII standard about
quadrilateral and properties of quadrilateral.
Let us recall them.
In Fig. 6.1, A, B, C, D are four points in a
plane. No three points lie on a line.
AB, BC, CD, DA intersect only at the
vertices. We have learnt that quadrilateral is a
four sided plane figure. We know that the sum
of measures of the four angles of a quadrilateral
is 360°.
Fig. 6.1
(AB,AD), (AB,BC), (BC,CD), (CD , DA) are adjacent sides. AC and
BD are the diagonals.
ZA, ZB, ZC and ZD (or ZDAB, ZABC, ZBCD, ZCDA) are the angles
of the quadrilateral ABCD.
.-. ZA+ ZB+ ZC+ ZD = 360°
Note : (i) We should name the quadrilateral in cyclic ways such as ABCD
and BCDA.
(ii) Square, Rectangle, Rhombus, Parallelogram, Trapezium are all
Quadrilaterals.
(iii) A quadrilateral has four vertices, four sides, four angles and two
diagonals.
6.2.2 Area of a Quadrilateral
Let ABCD be any quadrilateral with BD as
one of its diagonals.
Let AE and FC be the perpendiculars drawn
from the vertices A and C on diagonal BD .
From the Fig. 6.2
Area of the quadrilateral ABCD
= Area of A ABD + Area of A BCD
yXBDxAE+ ^XBDXCF
^ X BD X ( AE + CF) = ^ X d X (h^ + h^) sq. units.
Chapter 6
where BD = d, AE = h, and CF = /12.
Area of a quadrilateral is half of the product of a diagonal and the sum of the
altitudes drawn to it from its opposite vertices. That is,
A = y d (h^ + h^) sq. units, where 'd' is the diagonal; 'h^' and 'h^' are the
altitudes drawn to the diagonal from its opposite vertices.
rAm^
_ 1
ly By using paper folding technique, verify A = y d (h^ + h^)
6.2.3 Construction of a Quadrilateral
In this class, let us learn how to construct a quadrilateral.
To construct a quadrilateral first we construct a triangle from the given data.
Then, we find the fourth vertex.
To construct a triangle, we require three independent measurements. Also
we need two more measurements to find the fourth vertex. Hence, we need five
independent measurements to construct a quadrilateral.
We can construct, a quadrilateral, when the following measurements are given:
(i) Four sides and one diagonal
(ii) Four sides and one angle
(iii) Three sides, one diagonal and one angle
(iv) Three sides and two angles
(v) Two sides and three angles
6.2.4 Construction of a quadrilateral when four sides and one diagonal are given
Example 6.1
Construct a quadrilateral ABCD with AB = 4 cm, EC = 6 cm, CD = 5.6 cm
DA = 5 cm and AC = 8 cm. Find also its area.
Solution
Given: AB = 4 cm, EC = 6 cm, CD = 5.6 cm
DA = 5 cm and AC = 8 cm.
To construct a quadrilateral
Steps for construction
Step 1 : Draw a rough figure and mark the given
measurements.
Draw a line segment AB = 4 cm.
Rough Diagram
5.6 cm
Step 2
Step 3
With A and B as centres draw arcs of radii
8 cm and 6 cm respectively and let them cut at C.
4 cm B
Fig. 6.3
Practical Geometry
Step 4 : Join AC and BC .
Step 5 : With A and C as centres draw arcs of radii 5 cm, and 5.6 cm
respectively and let them cut at D.
Join AD and CD.
ABCD is the required quadrilateral.
From B draw BE 1 AC and from D draw DF 1 AC, then measure
the lengths of BE and DF. BE = h^ = 3 cm and DF = h^ = 3.5 cm.
AC = d = 8 cm.
Calculation of area:
In the quadrilateral ABCD, d = 8 cm, ^^ = 3 cm and h^ =3.5 cm.
Step 6
Step?
Area of the quadrilateral ABCD
_ 1
d (h. + h,)
= |(8)(3 + 3.5)
1
X 8x6.5
2
= 26 cm^
6.2.5 Construction of a quadrilateral when four sides and one angle are given
Example 6.2
Construct a quadrilateral ABCD with AB = 6 cm, BC = 4 cm, CD = 5 cm,
DA = 4.5 cm, ZABC = 100° and find its area.
Solution
Given:
AB = 6 cm, BC = 4 cm,CD = 5 cm, DA = 4.5 cm ZABC = 100°.
Chapter 6
To construct a quadrilateral
X
Rough Diagram
4 cm
Fig. 6.6
Steps for construction
Step 1 : Draw a rough diagram and mark the given measurments.
Draw a hne segment BC = 4 cm.
At B on BC make Z CBX whose measure is 100°.
With B as centre and radius 6 cm draw an arc. This cuts BX at A.
Join CA
With C and A as centres, draw arcs of radii 5 cm and 4.5 cm
respectively and let them cut at D.
Join CD and AD.
ABCD is the required quadrilateral.
From B draw BF 1 AC and from D draw DE 1 AC . Measure the
lengths of BF and DE. BF = h^ = 3 cm, DE = h^=2.7 cm and
AC = d = 7.8 cm.
Calculation of area:
In the quadrilateral ABCD, d = 7.8 cm, h^= 3 cm and h2= 2.7 cm.
Area of the quadrilateral ABCD =^ d{\ + h^)
= ^(7.8) (3 + 2.7)
Step 2
Step 3
Step 4
Steps
Step 6
Step 7
1
X 7.8x5.7 = 2.23 cm\
Practical Geometry
6.2.6 Construction of a quadrilateral when three sides, one diagonal and one
angle are given
Example 6.3
Construct a quadrilateral PQRS with PQ = 4 cm, QR = 6 cm, PR = 7 cm,
PS = 5 cm and Z PQS = 40° and find its area.
Solution
Given: PQ = 4 cm, QR = 6 cm, PR= 7 cm.
PS=5cmand ZPQS = 40°.
To construct a quadrilateral
Rough Diagram
p 4 cm Q
Fig. 6.7
Fig. 6.8
Steps for construction
Step 1 : Draw a rough diagram and mark the given measurements.
Step 2 : Draw a line segment PQ = 4 cm.
Step 3 : With P and Q as centres draw arcs of radii 7 cm and 6 cm respectively
and let them cut at R.
Step 4 : Join PR and QR.
Step 5 : At Q on PQ make |PQT whose measure is 40°.
Step 6 : With P as centre and radius 5 cm draw an arc. This cuts QT at S.
Step 7 : Join PS.
PQRS is the required quadrilateral.
Step 8 : From Q draw QX 1 PR and from S draw SY 1 PR. Measure the
lengths QX and SY. QX = h^ = 3.1 cm, SY = h^ = 3.9 cm.
PR = d = 7 cm.
Chapter 6
Calculation of area:
In the quadrilateral PQRS, d — 7 cm, h^ = 3.1 cm and h^ — 3.9 cm.
Area of the quadrilateral PQRS = ^ d{\ + h^)
= i-(7)(3.1 + 3.9)
= yX7X7
= 24.5 cm^.
6.2.7 Construction of a quadrilateral when three sides and two angles are given
Example 6.4
Construct a quadrilateral ABCD with AB = 6.5 cm, AD = 5 cm, CD = 5 cm,
ZBAC = 40° and ZABC = 50°, and also find its area.
Solution
Given:
AB = 6.5 cm, AD = 5 cm, CD = 5 cm,
ZBAC = 40° and ZABC = 50°.
To construct a quadrilateral
\X40° 50° X
6.5 cm B
Fig. 6.9
Rough Diagram
B
6.5 cm
Fig. 6.10
Steps for construction
Step 1 : Draw a rough diagram and mark the given measurements.
Step 2 : Draw a line segment AB = 6.5 cm.
Step 3 : At A on AB make Z BAX whose measure is 40° and at B on AB
make ZABY whose measure is 50°. They meet at C.
Practical Geometry
Step 4
Steps
Step 6
With A and C as centres draw arcs of radius 5 cm and 5 cm
respectively and let them cut at D.
Join AD and CD.
ABCD is the required quadrilateral.
From D draw DE 1 AC and from B draw EC 1 AC . Then measure
the lengths of BC and DE. EC = \ =4.2 cm, DE = h^ =4.3 cm and
AC = d = 5 cm.
Calculation of area:
In the quadrilateral ABCD, d = 5 cm, BC = h^ = 4.2 cm and h^ = 4.3 cm.
Area of the quadrilateral ABCD = y d (h, + h,)
= I (5) (4.2 + 4.3)
= ^X 5x8.5 = 21.25 cml
6.2.8 Construction of a quadrilateral when two sides and three angles are given
Example 6.5
Construct a quadrilateral ABCD with AB = 6 cm, AD = 6 cm, Z ABD = 45°,
Z BDC = 40° and Z DBC = 40°. Find also its area.
Solution
Given: AB = 6 cm, AD = 6 cm, ZABD = 45°,
ZBDC = 40° and ZDBC = 40°.
To construct a quadrilateral
Rough Diagram
D
^^^40^ ■
i
^v
^0°
45°\
A
6 cm
Fig. 6.11
Chapter 6
Steps for construction
Stepl
Draw a rough diagram and mark the given measurements.
Step 2
Draw a line segment AB = 6 cm.
Step 3
At B on AB make ZABX whose measure is 45°.
Step 4
*-
With A as centre and 6 cm as radius draw an arc. Let it cut BX at D.
Steps
Join AD .
Step 6
At B on BD make ZDBY whose measure is 40°.
Step?
At D on BD make Z BDZ whose measure is 40°.
Steps
Let BY and DZ intersect at C.
ABCD is the required quadrilateral.
Step 9 : From A draw AE 1 BD and from C draw CF 1 BD . Then measure
the lengths of AE and CF. AE = h^ = 4.2 cm, CF = h^ = 3.8 cm and
BD = d = 8.5 cm.
Calculation of area:
In the quadrilateral ABCD, d = 8.5 cm, \= 4.2 cm and h^= 3.8 cm.
Area of the quadrilateral ABCD =^d(\ + h^)
= ^(8.5) (4.2 + 3.8)
= i^X 8.5X8 = 34cm2.
EXERCISE 6.1
Draw
f quadrilateral ABCD with
the following measurements. Find also its area. 1
1.
AB = 5 cm, BC = 6 cm, CD = 4 cm, DA= 5.5 cm and AC = 7 cm.
2.
AB = 7 cm, BC = 6.5 cm, AC = 8 cm, CD= 6 cm and DA = 4.5 cm.
3.
AB = 8 cm, BC = 6.8 cm, CD = 6 cm, AD= 6.4 cm and Z B = 50°.
4.
AB = 6 cm, BC = 7 cm, AD = 6 cm, CD= 5 cm, and Z BAC = 45°.
5.
AB = 5.5 cm, BC = 6.5 cm, BD = 7 cm, AD= 5 cm and Z BAC= 50°.
6.
AB = 7 cm, BC = 5 cm, AC = 6 cm, CD= 4 cm, and Z ACD = 45°.
7.
AB = 5.5 cm, BC = 4.5cm, AC = 6.5 cm, Z CAD = 80° and Z ACD = 40°.
8.
AB = 5 cm, BD = 7 cm, BC = 4 cm, Z BAD = 100° and Z DBC = 60.
9.
AB = 4 cm, AC = 8 cm, Z ABC = 100°, Z ABD = 50° and
Z CAD = 40°.
10.
AB = 6 cm, BC = 6 cm, Z BAC = 50°, Z ACD = 30° and Z CAD = 100°.
Practical Geometry
6.3 Trapezium
6.3.1 Introduction
In the class VII we have learnt special quadrilaterals such as trapezium and
isosceles trapezium. We have also learnt their properties. Now we recall the definition
of a trapezium.
A quadrilateral in which only one pair of opposite sides are parallel is
called a trapezium.
6.3.2 Area of a trapezium
Let us consider the trapezium EASY
b
Fig. 6.13
We can partition the above trapezium into two triangles by drawing
a diagonal YA.
One triangle has base EA ( EA = a units )
The other triangle has base YS ( YS = b units )
We know EA 1 1 YS
YF = HA = h units
Now, the area of A EAY is y ah. The area of A YAS is y bh.
Hence,
the area of trapezium EASY = Area of A EAY + Area of A YAS
= ^ah + ^bh
= y h (a + b) sq. units
= y X height X (Sum of the parallel sides) sq. units
Area of Trapezium
A = y fi (a + fcj) sq. units where 'a' and 'b' are the lengths of the parallel sides
and 'h' is the perpendicular distance between the parallel sides.
Chapter 6
6.3.3 Construction of a trapezium
In general to construct a trapezium, we take the parallel sides which has
greater measurement as base and on that base we construct a triangle with the given
measurements such that the triangle lies between the parallel sides. Clearly the vertex
opposite to the base of the triangle lies on the parallel side opposite to the base. We
draw the line through this vertex parallel to the base. Clearly the fourth vertex lies on
this line and this fourth vertex is fixed with the help of the remaining measurement.
Then by joining the appropriate vertices we get the required trapezium.
To construct a trapezium we need four independent data.
We can construct a trapezium with the following given information:
(i) Three sides and one diagonal
(ii) Three sides and one angle
(iii) Two sides and two angles
(iv) Four sides
6.3.4 Construction of a trapezium when three sides and one diagonal are given
Example 6.6
Construct a trapezium ABCD in which AB is parallel to DC , AB = 10 cm,
BC = 5 cm, AC = 8 cm and CD = 6 cm. Find its area.
„ , . Rough Diagram
Solution
Given:
AB is parallel to DC, AB = 10 cm,
BC = 5 cm, AC = 8 cm and CD = 6 cm.
To construct a trapezium
10 cm
Fig. 6.14
X<
10 cm
Fig. 6.15
Practical Geometry
Steps for construction
Step 1 : Draw a rough diagram and mark the given measurements.
Draw a line segment AB = 10 cm.
With A and B as centres draw arcs of radii 8 cm and 5 cm respectively
and let them cut at C.
Step 2
Step 3
Step 4 : Join AC and BC.
Step 5 : Draw CX parallel to BA.
Step 6 : With C as centre and radius 6 cm draw an arc cutting CX at D.
Step 7 : Join AD.
ABCD is the required trapezium.
Step 8 : From C draw CEl AB and measure the length of CE.
CE = h = 4 cm.
AB = a = 10 cm, DC = b = 6 cm.
Calculation of area:
In the trapezium ABCD, a = 10 cm, b = 6 cm and h = 4 cm.
Area of the trapezium ABCD
= ^(4) (10 + 6)
= ^X4X16
= 32 cml
6.3.5 Construction of a trapezium when three sides and one angle are given
Example 6.7
Construct a trapezium PQRS in which PQ is parallel to SR, PQ = 8 cm
Z PQR = 70°, QR = 6 cm and PS = 6 cm. Calculate its area.
Solution
Given:
PQ is parallel to SR, PQ = 8 cm, ZPQR = 70°,
QR = 6 cm and PS = 6 cm.
Rough Diagram
S . R
8 cm
Fig 6.16
Chapter 6
To construct a trapezium x
\
^^S 3.9 cm \5^
Y < "^
^\ ' ^\
<c /
o \ '^i
^ \
1 r
70°A
^. L A
P T o ' Q
^ 8 cm ^
Fig. 6.17
Steps for construction
Stepl
Draw a rough diagram and mark the given measurements.
Step 2
Draw a line segment PQ = 8 cm.
Step 3
At Q on PQ make Z PQX whose measure is 70°.
Step 4
With Q as centre and 6 cm as radius draw an arc. This cuts QX at R.
Steps
Draw RY parallel to QP.
Step 6
With P as centre and 6 cm as radius draw an arc cutting RY at S.
Step?
Join PS.
PQRS is the required trapezium.
Steps
From S draw ST 1 PQ and measure the length of ST.
ST = h = 5.6 cm,
RS = fa = 3.9 cm. PQ = a = 8 cm.
Calculatio
n of area:
In the trapezium PQRS, a = 8 cm, b = 3.9 cm and h = 5.6 cm. 1
Area of t
he trapezium PQRS =^h{a + b)
= i^(5.6)(8 + 3.9)
= ^X 5.6X11.9
^^^^
= 33.32 cm2.
Practical Geometry
6.3.6. Construction of a trapezium when two sides and two angles are given
Example 6.8
Construct a trapezium ABCD in which AB is parallel to DC, AB = 7 cm,
BC = 6 cm, ZBAD = 80° and ZABC = 70° and calculate its area.
Solution
Given:
Rough Diagram
D . C
AB is parallel to DC, AB = 7 cm,
BC = 6 cm, ZBAD = 80° and ZABC = 70°.
To construct a trapezium
X
4 cm
>—
Steps for
Stepl
Step 2
Step 3
Step 4
Steps
Step 6
Step?
7 cm
Fig. 6.19
construction
Draw a rough diagram and mark the given measurements.
Draw a line segment AB = 7 cm.
On AB at A make ZBAX measuring 80°.
On AB at B make ZABY measuring 70°.
With B as centre and radius 6 cm draw an arc cutting BY at C.
Draw CZ parallel to AB . This cuts AX at D.
ABCD is the required trapezium.
From C draw CE 1 AB and measure the length of CE.
CE = h = 5.6 cm and CD = b = 4 cm.
Also, AB = a = 7 cm.
Chapter 6
Calculation of area:
In the trapezium ABCD, a-1 cm, b = 4 cm and h - 5.6 cm.
Area of the trapezium ABCD = -y h (a + b)
= 1- (5.6) (7 + 4)
= ^X5.6xll
= 30.8 cml
6.3.7. Construction of a trapezium when four sides are given
Example 6.9
Construct a trapezium ABCD in which AB is parallel to DC, AB = 7 cm,
BC = 5 cm, CD = 4 cm and AD = 5 cm and calculate its area.
, . Rough Diagram
Solution
D 4 cm C
J :>
Given:
AB is parallel to DC, BC = 5 cm,
CD = 4 cm and AD = 5 cm.
To construct a trapezium
X<
"o
4 cm
3 cm
7 cm
Fig. 6.20
Fig. 6.21
Steps for construction
Step 1 : Draw a rough diagram and mark the given measurements.
Draw CE 1 1 DA . Now AECD is a parallelogram.
.-. EC = 5 cm, AE = DC = 4 cm, EB = 3cm.
Draw a line segment AB = 7 cm.
Step 2
Step 3
Mark E on AB such that AE = 4 cm. [•.• DC = 4 cm]
Practical Geometry
Step 4
Steps
Step 6
Step?
Steps
With B and E as centres draw two arcs of radius 5 cm and let them
cut at C.
Join BC and EC .
With C and A as centres and with 4 cm and 5 cm as radii draw two
arcs. Let them cut at D.
Join AD and CD.
ABCD is the required trapezium.
From D draw DF 1 AB and measure the length of DF.
DF = h = 4.8 cm. AB = a = 7 cm, CD = b = 4 cm.
Calculation of area:
In the trapezium ABCD, a = 7 cm, b = 4 cm and h = 4.8 cm.
Area of the trapezium ABCD
= fh{a + b)
= ^(4.8) (7 + 4)
= ^X4.8X11
= 2.4x11
= 26.4 cml
6.3.8 Isosceles trapezium
In Fig. 6.22 ABCD is an isosceles trapezium
In an isosceles trapezium, d
(i) The non parallel sides are
equal in measurement i.e., AD = BC.
(ii) ZA=ZB.
and ZADC = Z BCD a
(iii) Diagonals are equal in length
i.e., AC = BD
Fig. 6.22
(iv) AE = BF, (DB 1 AB, CF 1 BA)
To construct an isosceles trapezium we need only three independent
measurements as we have two conditions such as
(1) One pair of opposite sides are parallel,
(ii) Non - parallel sides are equal.
Chapter 6
6.3.9. Construction of isosceles trapezium
Example 6.10
Construct an isosceles trapezium ABCD in which AB is parallel to
AB = 11 cm, DC = 7 cm, AD = BC = 6 cm and calculate its area.
Solution
Given:
DC,
AB is parallel to DC, AB = 11 cm,
DC = 7 cm, AD = BC = 6 cm.
To construct an isosceles trapezium
D
Rough Diagram
7 cm
C
Vo /
V
L
11 cm
Fig. 6.23
B
7 cm
— ^
7 cm
— ^ —
B
11 cm
Steps for
Stepl
Step 2
Step 3
Step 4
Steps
Step 6
Step?
Steps
Fig. 6.24
construction
: Draw a rough diagram and mark the given measurements.
: Draw a line segment AB = 11 cm.
: Mark E on AB such that AE = 7 cm ( since DC = 7 cm)
: With E and B as centres and (AD = EC = 6 cm) radius 6 cm draw
two arcs. Let them cut at C.
: Join BC and EC.
: With C and A as centres draw two arcs of radii 7 cm and 6 cm
respectively and let them cut at D.
: Join AID and CD.
ABCD is the required isosceles trapezium.
: From D draw DF 1 AB and measure the length of DF.
DF = /j = 5.6 cm. AB = a = 11 cm and CD = b = 7 cm.
Practical Geometry
Calculation of area:
In the isosceles trapezium ABCD, a = 11 cm, b = 7 cm and h = 5.6 cm.
Area of the isosceles trapezium ABCD = ^h (a + b)
= |(5.6) (11 + 7)
= ^X 5.6X18
= 50.4 cml
EXERCISE 6.2
I. Construct trapezium PQRS with the following measurements. Find also its
area.
1. PQ is parallel to SR, PQ = 6.8 cm, QR - 7.2 cm, PR - 8.4 cm and RS - 8 cm.
2. PQ is parallel to SR, PQ - 8 cm, QR - 5 cm, PR - 6 cm and RS - 4.5 cm.
3. PQ is parallel to SR, PQ = 7 cm, Z Q = 60°,QR - 5 cm and RS - 4 cm.
4. PQ is parallel to SR, PQ = 6.5 cm, QR - 7 cm, Z PQR = 85° and PS = 9 cm.
5. PQ is parallel to SR, PQ - 7.5 cm, PS - 6.5 cm, Z QPS = 100° and
Z PQR = 45°.
6. PQ is parallel to SR, PQ - 6 cm, PS = 5 cm, Z QPS = 60° and Z PQR = 100°.
7. PQ is parallel to SR, PQ - 8 cm, QR - 5 cm, RS - 6 cm and SP = 4 cm.
8. PQ is parallel to SR, PQ - 4.5 cm, QR - 2.5 cm, RS -3 cm and SP - 2 cm.
II. Construct isosceles trapezium ABCD with the following measurements
and find its area.
1. AB is parallel to DC, AB = 9 cm, DC = 6 cm and AD = BC = 5 cm.
2. AB is parallel to DC, AB = 10 cm, DC = 6 cm and AD = BC = 7 cm.
It is interesting to note that many of the properties of quadrilaterals were
known to the ancient Indians. Two of the geometrical theorems which are
explicitly mentioned in the Boudhayana Sutras are given below:
i) The diagonals of a rectangle bisect each other. They divide the rectangle
into four parts, two and two.
ii) The diagonals of a Rhombus bisect each other at right angles.
Chapter 6
6.4 Parallelogram
6.4.1. Introduction
In the class VII we have come across parallelogram. It is defined as follows:
A quadrilateral in which the opposite sides are parallel is called a
parallelogram.
Consider the parallelogram BASE given in the Fig. 6.25,
Then we know its properties
(i) BA||ES ; BE||AS
(ii) BA = ES , BE = AS
(iii) Opposite angles are equal in measure.
ZBES= ZBAS; ZEBA= ZESA
Diagonals bisect each other.
OB = OS; OE = OA, but BS # AE .
(iv)
(V)
Fig. 6.25
Sum of any two adjacent angles is equal to 180°.
Now, let us learn how to construct a parallelogram, and find its area.
6.4.2 Area of a parallelogram E
Let us cut off the red portion ( a right
angled triangle EPS ) from the parallelogram
FAME. Let us fix it to the right side of the
figure FAME. We can see that the resulting
figure is a rectangle. See Fig. 6.27.
M
Fig. 6.26
We know that the area of a rectangle
having length b units and height h units is given hy A = bh sq. units.
E
Here, we have actually converted
the parallelogram FAME into a rectangle.
Hence, the area of the parallelogram is
A = bh sq. units where 'b' is the base of the
parallelogram and 'h' is the perpendicular
distance between the parallel sides.
Fig. 6.27
S'(S)
Practical Geometry
6.4.3 Construction of a parallelogram
Parallelograms are constructed by splitting up the figure into suitable triangles.
First a triangle is constructed from the given data and then the fourth vertex is found.
We need three independent measurements to construct a parallelogram.
We can construct a parallelogram when the following measurements are given .
(i) Two adjacent sides, and one angle
(ii) Two adjacent sides and one diagonal
(iii) Two diagonals and one included angle
(iv) One side, one diagonal and one angle.
6.4.4 Construction of a parallelogram when two adjacent sides and one angle are
given
Example 6.11
Construct a parallelogram ABCD with AB = 6 cm, BC = 5.5 cm and
ZABC = 80° and calculate its area.
Solution
Given: AB = 6 cm, BC = 5.5 cm and ZABC = 80°. Ro^g^ Diagram
°\ ^ '^
To construct a parallelogram
6 cm
^—
->■
-^ 6 cm B
Fig. 6.28
6 cm
Fig. 6.29
B
Chapter 6
Steps for construction
Step 1 : Draw a rough diagram and mark the given measurements.
Step 2 : Draw a line segment AB = 6 cm.
Step 3 : At B on AB make ZABX whose measure is 80°.
Step 4 : With B as centre draw an arc of radius 5.5 cm and
let it cuts BX at C.
Step 5 : With Cand A as centres draw arcs of radii 6 cm and 5.5 cm repectively
and let them cut at D.
Step 6
Step?
Join AD and CD.
ABCD is the required parallelogram.
From C draw CE 1 AB and measure the length of CE.
CE = h = 5.4 cm. AB = b = 6 cm.
Calculation of area:
In the parallelogram ABCD, b = 6 cm and h = 5.4 cm.
Area of the parallelogram ABCD = b ^ h = 6 x 5.4
= 32.4 cm^.
6.4.5. Construction of parallelogram when two adjacent sides and one diagonal
are given
Example 6.12
Construct a parallelogram ABCD with AB = 8 cm, AD = 7 cm and BD = 9 cm
and find its area.
Solution
Given: AB = 8 cm, AD = 7 cm and BD = 9 cm.
Rough Diagram
To construct a parallelogram
8 cm
Fig. 6.30
Fig. 6.31
Practical Geometry
Steps for construction
Step 1 : Draw a rough diagram and mark the given measurements.
Step 2 : Draw a line segment AB = 8 cm.
Step 3 : With A and B as centres draw arcs of radii 7 cm and 9 cm respectively
and let them cut at D.
Step 4 : Join AD and BD.
Step 5 : With B and D as centres draw arcs of radii 7 cm and 8 cm respectively
and let them cut at C.
Join CD and BC .
ABCD is the required parallelogram.
Step 6
Step?
From D draw DEI AB and measure the length of DE.
DE = h = 6.7 cm. AB = DC= b = 8 cm
Calculation of area:
In the parallelogram ABCD, b = 8 cm and h = 6.7 cm.
Area of the parallelogram ABCD = b x h
= 8 X 6.7 = 53.6 cml
6.4.6. Construction of a parallelogram when two diagonals and one included angle
are given
Example 6.13
Draw parallelogram ABCD with AC = 9 cm, BD = 7 cm and ZAOB = 120°
where AC and BD intersect at 'O' and find its area.
Solution
Given: AC = 9 cm, BD = 7 cm and Z AOB = 120°.
X
Rough Diagram
D.
,C
Fig. 6.33
Chapter 6
To construct a parallelogram
Steps for construction
Step 1 : Draw a rough diagram and mark the given measurements.
Step 2 : Draw a line segment AC = 9 cm.
Step 3 : Mark 'O' the midpoint of AC.
Step 4 : Draw a Une XY through 'O' which makes ZAOY = 120°.
Step 5 : With O as centre and 3.5 cm as radius draw two arcs on XY on either
sides of AC cutting OX at D and OY at B.
Step 6 : Join AB, BC, CD and DA.
ABCD is the required parallelogram.
Step 7 : From D draw DE 1 AB and measure the length of DE.
DE = h = 4 cm. AB = fa = 7 cm.
Calculation of area:
In the parallelogram ABCD, fa = 7 cm and h = 4 cm.
Area of the parallelogram ABCD = bxh = 7x4 = 28 cm^.
6.4.7. Construction of a parallelogram when one side, one diagonal and one angle
are given
Example 6.14
Construct a parallelogram ABCD, AB = 6 cm, ZABC = 80° and AC = 8 cm
and find its area.
Solution
Given: AB = 6 cm, Z ABC = 80° and AC = 8 cm.
To construct a parallelogram
Rough Diagram
6 cm
Fig. 6.34
6 cm
E B pig^ 6.35
Practical Geometry
Steps for construction
Stepl
Draw a rough diagram and mark the given measurements.
Step 2
Draw a line segment AB = 6 cm
Step 3
At B on AB make ZABX whose measure is 80°.
Step 4
With A as centre and radius 8 cm draw an arc. Let it cut BX at C.
Steps
Join AC .
Step 6
With C as centre draw an arc of radius 6 cm.
Step?
With A as centre draw another arc with radius equal to the length o:
BC. Let the two arcs cut at D.
Steps
Join AD and CD.
ABCD is the required parallelogram.
Step 9
From C draw CEl AB and measure the length of CE.
CE = h = 6.4 cm. AB = b = 6 cm.
Calculatio
n of area:
In the parallelogram ABCD, b = 6 cm and h = 6.4 cm.
Area of
the parallelogram ABCD =h ^h
= 6 X 6.4
Draw parallelo
= 38.4 cm2
EXERCISE 6.3
gram ABCD with the following measurements and calculate its area
1. AB= 7
cm, BC - 5 cm and Z ABC = 60°.
2. AB - 8.5 cm, AD - 6.5 cm and Z DAB = 100°.
3. AB- 6
cm, BD - 8 cm and AD - 5 cm.
4. AB-5
cm, BC = 4 cm, AC - 7 cm.
5. AC - 1(
D cm, BD = 8 cm and Z AOB = 100° where AC and BD intersect
at '0'.
6. AC = 8
cm, BD - 6 cm and Z COD = 90° where AC and BD intersect at '0'.
7. AB = 8
cm, AC - 10 cm and Z ABC = 100°.
8. AB = 5.
5 cm, Z DAB = 50° and BD - 7 cm.
Chapter 6
6.5 Rhombus
6.5.1. Introduction
A parallelogram in which the adjacent sides are equal is called a rhombus.
In rhombus ABCD, see Fig. 6.36.
(i) All sides are equal in measure.
i.e., AB = BC = CD = DE
(ii) Opposite angles are equal in measure.
i.e., ZA= ZCand ZB = ZD
(iii) Diagonals bisect each other at right
angles.
i.e., AO = OC ; BO = OD, ^. _ _ _
Fig. 6.36
At 'O', AC and BD are perpendicular to
each other,
(iv) Sum of any two adjacent angles is equal to 180°.
(v) Each diagonal of a rhombus divides it into two congruent triangles,
(vi) Diagonals are not equal in length.
6.5.2 Area of a rhombus
Let us consider the rectangular sheet of paper JOKE as shown below.
F
Fig. 6.37
Let us mark the mid - points of the sides. (We use the paper folding technique
to find the mid point). The mid-point of JO is F ; the mid-point of OK is A ; the mid-
point of KE is I and the mid-point of EJ is R. Let us join RA and IF. They meet at C.
FAIR is a rhombus.
We have eight congruent right angled triangles. The area of the required rhombus
FAIR is the area of four right angled triangles.
In other words, we can say that the area of the rhombus FAIR is half of the
rectangle JOKE.
Practical Geometry
We can clearly see that JO , the length of rectangle becomes one of the diagonals
of the rhombus (RA). The breadth becomes the other diagonal (IF) of the rhombus.
Area of rhombus FAIR = y di X d2
Area of rhombus A = y X di X di sq. units
where d^ and d^are the diagonals of the rhombus.
6.5.3 Construction of a rhombus
Rhombus is constructed by splitting the figure into suitable triangles. First, a
triangle is constructed from the given data and then the fourth vertex is found. We
need two independent measurements to construct a rhombus.
We can construct a rhombus, when the following measurements are given
(i) One side and one diagonal
(11) One side and one angle
(ill) Two diagonals
(Iv) One diagonal and one angle
6.5.4 Construction of Rhombus when one side and one diagonal are given
Example 6.15
Construct a rhombus PQRS with PQ = 6 cm and PR = 9 cm and find its area.
Solution
Given: PQ = 6 cm and PR = 9 cm.
To construct a rhombus
Rough Diagram
S . R
° 6 cm
Fig. 6.38
Fig. 6.39
Chapter 6
Steps for construction
Step 1 : Draw a rough diagram and mark the given measurements.
Step 2 : Draw a line segment PQ = 6 cm.
Step 3 : With P and Q as centres, draw arcs of radii 9 cm and 6 cm respectively
and let them cut at R.
Step 4 : Join PR and QR.
Step 5 : With P and R as centres draw arcs of radius 6 cm
and let them cut at S.
Step 6 : Join PS and RS.
PQRS is the required rhombus.
Step 7 : Measure the length of QS.
QS = d^ = 8 cm. PR = d^ = 9 cm.
Calculation of area:
In the rhombus PQRS, d^ = 9 cm and d^ = 8 cm.
Area of the rhombus PQRS =yXdiXd2=yX9x8=36 cml
6.5.4 Construction of a rhombus when one side and one angle are given
Example 6.16
Construct a rhombus ABCD with AB = 7 cm and
ZA = 60° and find its area.
Solution
Given: AB = 7 cm and Z A = 60°.
To construct a rhombus X
Rough Diagram
Practical Geometry
Steps for construction
Step 1 : Draw a rough diagram and mark the given measurements.
Step 2 : Draw a line segment AB = 7 cm.
Step 3 : At A on AB make Z BAX whose measure 60°.
Step 4 : With A as centre draw an arc of radius 7 cm. This cuts AX at D.
Step 5 : Wth B and D as centres draw arcs of radius 7 cm.
and let them cut at C.
Step 6 : Join BC and DC .
ABCD is the required rhombus.
Step 7 : Measure the lengths AC and BD.
AC = d^ = 12.2 cm and BD = d^ = 7 cm.
Calculation of area:
In the rhombus ABCD, d^ = 12.2 cm and d^ = 7 cm.
Area of the rhombus ABCD = yXdiXdi
= ^X 12.2x7
= 42.7 cm2.
6.5.5 Construction of a rhombus when two diagonals are given
Example 6.17
Cosnstruct a rhombus PQRS with PR = 8 cm and QS = 6 cm and find its area.
Solution
Given: PR = 8 cm and QS = 6 cm.
Steps for construction
Step 1 : Draw a rough diagram and mark the given measurements.
Step 2 : Draw a line segment PR = 8 cm
Step 3 : Draw the perpendicular bisector 5CY to PR. Let it cut PR at "O" .
Step 4 : With O as centre and 3 cm (half of QS) as radius draw arcs on either
side of 'O' which cuts XY at Q and S as shown in Fig. 6.43.
Step 5 : Join PQ, QR, RS and SP.
PQRS is the required rhombus.
Step 6 : We know, PR = d^ = 8 cm and QS = d^ = 6 cm.
Chapter 6
To construct a rhombus
Rough Diagram
Fig. 6.42
Y
Calculation of area:
In the Rhombus PQRS, d^ = 8 cm and d^ = 6 cm.
Area of the rhombus PQRS = y X d, X d
= ^X8X6
= 24 cml
6.5.6 Construction of a rhombus when one diagonal and one angle are given
Example 6.18
Construct a rhombus ABCD with AC = 7.5 cm and ZA = 100°. Find its area.
Solution
Given: AC = 7.5 cm and ZA = 100°.
Practical Geometry
To construct a rhombus
M
Rough Diagram
D, - C
Fig. 6.44
N ^ Fig. 6.45
Steps for construction
Step 1 : Draw a rougli figure and mark tlie given measurements.
Step 2 : Draw a line segment AC = 7.5 cm.
Step 3 : At A draw AX and AY on either side of AC making on angle 50°
with AC .
Step 4 : At C draw CM and CN on either side of CA making an angle 50°
with CA.
Step 5 : Let AX and CM cut at D and AY and CN cut at B.
ABCD is the required rhombus.
Step 6 : Measure the length BD. BD = d^ = 9 cm. AC = d^ = 7.5 cm.
Calculation of area:
In the rhombus ABCD, d^ = 7.5 cm and d^ = 9 cm.
Area of the rhombus ABCD = y X d, X d
= ^ X 7.5 X 9 = 7.5 X 4.5 = 33.75 cml
Chapter 6
EXERCISE 6.4
Draw rhombus BEST with the following measurements and calculate its area.
1. BE - 5 cm and BS - 8 cm.
2. BE = 6 cm and ET - 8.2 cm.
3. BE - 6 cm and Z B = 45°.
4. BE - 7.5 cm and Z E = 65°.
5. BS - 10 cm and ET - 8 cm.
6. BS = 6.8 cm and ET - 8.4 cm.
7. BS = 10 cm and Z B = 60°.
8. ET - 9 cm and Z E = 70°.
6.6 Rectangle and Square
6.6.1 Rectangle
A rectangle is a parallelogram whose one of the angle is a right angle.
Its properties are
(i) The opposite sides are equal.
(ii) All angles are equal.
(iii) Each angle is a right angle.
(iv) The diagonals are equal in length.
(v) The diagonals bisect each other.
Area of a rectangle:
Area of the rectangle ABCD = length x breadth
A = / X £) sq. units.
6.6.2 Construction of a rectangle
We can construct a rectangle, when the following measurements are given:
Fig. 6.46
(i) Length and breadth
(11) A side and a diagonal
Practical Geometry
6.6.3.Construction of a rectangle when length and breadth are given
Example 6.19
Construct a rectangle whose adjacent sides are 6 cm and 4 cm and find its area.
Solution
Given:
Adjacent sides are 6 cm and 4 cm.
To construct a rectangle
Rough Diagram
D u C
Fig. 6.48
Steps for construction
Stepl
Step 2
Step 3
Step 4
Steps
Draw a rough diagram and mark the given measurements.
Draw a line segment AB = 6 cm.
At A, with a compass construct AX 1 AB .
With A as centre draw an arc of radius 4 cm and let it cut AX at D.
With D as centre draw an arc of radius 6 cm above the line segment
AB.
Step 6 : With B as centre draw an arc of radius 4 cm cutting the previous arc
at C. Join BC and CD.
ABCD is the required rectangle.
Step 7 : AB = / = 6 cm and BC = b = 4 cm.
Calculation of area:
In the rectangle ABCD, / = 6 cm and b = 4 cm.
Area of the rectangle ABCD = I x b
= 6 X 4 = 24 cm^.
Chapter 6
6.6.4 Construction of a rectangle when one diagonal and one of a side are given
Example 6.20
Construct a rectangle whose diagonal is 7 cm and length of one of its side is
4 cm. Find also its area.
Solution
Given:
Diagonal = 7 cm and length of one side = 4 cm.
To construct a rectangle
Rough Diagram
D, u ,C
4 cm B
Fig. 6.49
Steps for construction
Step 1 : Draw a rough diagram and mark the given measurements.
Step 2 : Draw a line segmeant AB = 4 cm.
Step 3 : Construct BX*1 AB.
Step 4 : With A as centre, draw an arc of radius 7 cm which cuts BX at C.
Step 5 : With BC as radius draw an arc above AB with A as centre.
Step 6 : With C as centre and 4 cm as radius to cut the previous arc at D.
Step 7 : Join AD and CD . ABCD is the required rectangle.
Step 8 : Measure the length of BC. BC = / = 5.8 cm
Calculation of area:
In the rectangle ABCD, / = 5.8 cm and b = 4 cm.
Area of the rectangle ABCD = / x b = 5.8 x 4 = 23.2 cm^.
Practical Geometry
6.6.5 Construction of a Square
Square
A square is a rectangle, whose adjacent sides are equal in length.
D
The properties of a square are :
(i) All the angles are equal.
(ii) All the sides are of equal length.
(iii) Each of the angle is a right angle.
(iv) The diagonals are of equal length and
(v) The diagonals bisect each other at right angles
Area of a square = side x side if the diagonal is known ,
A = axa
A = a' sq. units
To construct a square we need only one measurement.
We can construct a square when the following measurements are given:
B
Fig. 6.51
(i) one side, (ii) a diagonal
6.6.6 Construction of a square when one side is given
Example 6.21
Construct a square of side 5 cm. Find also its area.
Solution
Given: Side = 5 cm.
To construct a square
Rough Diagram
S, ,R
5 cm
Fig. 6.52
Chapter 6
Steps for construction
Step 1 : Draw a rough diagram and mark the given measurements.
Step 2 : Draw a line segment PQ = 5 cm.
Step 3 : At P using a compass construct PX _L PQ .
Step 4 : With P as centre draw an arc of radius 5 cm cutting PXat S.
Step 5 : With S as centre draw an arc of radius 5 cm above the line segment
PQ.
Step 6 : With Q as centre and same radius, draw an arc, cutting the previous
arc at R.
Step 7 : Join QR and RS.
PQRS is the required square.
Calculation of area:
In the square PQRS, side a = 5 cm
Area of the square PQRS = a x a
= 5 X 5 = 25 cm2.
6.6.7 Construction of a square when one diagonal is given
Example 6.22
Construct a square whose diagonal is 6 cm. Measure the side. Find also its area.
Solution ^ Rough Diagram
Given: Diagonal = 6 cm.
To construct a square
A B
Fig. 6.54
Fig. 6.55
Practical Geometry
Steps for construction
Step 1
Step 2
Step 3
Step 4
Steps
Step 6
Draw the rough diagram and mark the given measures.
Draw a line segment AC = 6 cm.
Construct a perpendicular bisector XY of AC .
XY intersects AC at O. We get OC = AO = 3 cm.
With O as centre draw two arcs of radius 3 cm cutting the line XY
at points B and D.
Join AB, BC, CD and DA.
ABCD is the required square.
Calculation of area:
In the square ABCD, diagonal d = 6 cm
Area of the Square ABCD
_ J^_ 6x6 _
18cm2
EXERCISE 6.5
1. Construct
area.
(i)
(ii)
(iii)
(iv)
2. Construct
area.
(i)
(ii)
(iii)
(iv)
3. Construct
(i)
(iii)
4. Construct
area.
(i)
(iii)
rectangle JUMP with the following measurements. Find also its
JU = 5.4 cm and UM = 4.7 cm.
JU = 6 cm and JP = 5 cm.
JP = 4.2 cm and MP= 2.8 cm.
UM = 3.6 cm and MP = 4.6 cm.
rectangle MORE with the following measurements. Find also its
MO = 5 cm and diagonal MR = 6.5 cm.
MO = 4.6 cm and diagonal OE = 5.4 cm.
OR = 3 cm and diagonal MR = 5 cm.
ME = 4 cm and diagonal OE = 6 cm.
square EASY with the following measurements. Find also its area.
Side 5.1 cm. (ii) Side 3.8 cm.
Side 6 cm (iv) Side 4.5 cm.
square GOLD, one of whose diagonal is given below. Find also its
4.8 cm.
5 cm.
(ii) 3.7 cm.
(iv) 7 cm.
Chapter 6
6.7 Concentric Circles
In this section, we are going to learn about Concentric Circles. Already we are
familiar with Circles.
6.7.1. Motivation
When a small stone was dropped in still water, you might have seen circular
ripples were formed. Which is the centre of these circles? Is it not the place where the
stone was dropped? Yes.
The circles with different measures of radii and with the same centre are called
concentric circles. The centre is known as common centre.
The Concentric Circles
Circles drawn in a plane with a common centre and different radii are called
concentric circles. See Fig. 6.56 and 6.57.
Fig. 6.56
Look at the following two figures:
Fig. 6.57
Fig. 6.58 Fig. 6.59
Fig. 6.58 represents two concentric circles.
In Fig. 6.59 the area between the two concentric circles are shaded with colour.
The coloured area is known as circular ring.
Practical Geometry
Description : Circular Ring
In Fig. 6.60, C^ and C^ are two circles having the
same centre O with different radii n and ri
Circles C^ and C2 are called concentric circles.
The area bounded between the two circles is
known as circular ring.
Width of the circular ring = OB - 0A= ri — n
6.7.2. Construction of concentric circles when the radii are given.
Example 6.23
Draw concentric circles with radii 3 cm and 5 cm and shade the circular ring.
Find its width. Rough Diagram
Solution
Given: The radii are 3 cm and 5 cm.
If '^/ \ ^
|B
To construct concentric circles
Fig. 6.61
Fig. 6.62
Chapter 6
Steps for construction
Step 1 : Draw a rough diagram and mark the given measurements.
Step 2 : Take any point O and mark it as the centre.
Step 3 : With O as centre and draw a circle of radius OA = 3 cm
Step 4 : With O as centre and draw a circle of radius OB = 5 cm.
Thus the concentric circles C^ and C^ are drawn.
Width of the circular ring = OB - OA
= 5-3
= 2 cm.
EXERCISE 6.6
1. Draw concentric circles for the following measurements of radii. Find out the
width of each circular ring.
(i) 4 cm and 6 cm.
(ii) 3.5 cm and 5.5 cm.
(iii) 4.2 cm and 6.8 cm.
(iv) 5 cm and 6.5 cm.
(v) 6.2 cm and 8.1 cm.
(vi) 5.3 cm and 7 cm.
Interesting Information
The golden rectangle is a rectangle which has appeared in art and architecture
through the years. The ratio of the lengths of the sides of a golden rectangle is
approximately 1 : 1.6. This ratio is called the golden ratio. A golden rectangle is
pleasing to the eyes. The golden ratio was discovered by the Greeks about the
middle of the fifth century B.C.
The Mathematician Gauss, who died in 1855, wanted a 17-sided polygon drawn on
his tombstone, but it too closely resembled a circle for the sculptor to carve.
Mystic hexagon: A mystic hexagon is a regular hexagon with all
its diagonals drawn.
Practical Geometry
Concept Summary
%
%
%
%
A quadrilateral is a plane figure bounded by four line segments.
To construct a quadrilateral, five independent measurements are necessary.
A quadrilateral with one pair of opposite sides parallel is called a trapezium.
To construct a trapezium four independent measurements are necessary.
If non-parallel sides are equal in a trapezium, it is called an isosceles
trapezium.
To construct an isosceles trapezium three independent measurements are
necessary.
A quadrilateral with each pair of opposite sides parallel is called a
parallelogram.
To construct a parallelogram three independent measurements are
necessary.
A quadrilateral with each pair of opposite sides parallel and with each pair
of adjacent sides equal is called a rhombus.
To contruct a rhombus two independent measurements are necessary.
A parallelogram in which each angle is a right angle, is called a rectangle.
Square is a rectangle, whose pair of adjecent sides are equal.
Circles drawn in a plane with a common centre and different radii are
called concentric circles.
The area bounded between two concentric circles is known as circular
ring.
'The area of a quadrilateral, A = — d {h^ + h^ sq. units, where 'd' is the
diagonal, '/l^' and 'h^' are the altitudes drawn to the diagonal from its
opposite vertices.
The area of a trapezium, A = — h{a + b) sq. units, where 'a' and 'b' are the
lengths of the parallel sides and '/f is the perpendicular distance between
the two parallel sides.
The area of a parallelogram, A= bh sq. units, where 'b' is the base of the
parallelogram and 'h' is the perpendicular distance between the parallel
sides.
The area of a rhombus, A = — d^d^ sq. units, where d^ and d^ are the two
diagonals of the rhombus.
Graphs
7.1
Introduction
7.2
Introduction to Cartesian Plane with Axes
7.3
Plotting of points for different kinds of
situations
7.4
Drawing Straight Lines and Parallel Lines
to the Coordinate Axes
7.5
Linear Graphs
7.6
Reading Linear Graphs
7.1 Introduction
The story of a fly and the graph
The mathematician who introduced graph was Rene Descartes,
a French Mathematician in early 17th century. Here is an interesting
anecdote from his life.
Rene Descartes, was a sick child
and was therefore, allowed to remain
in bed till quite late in the morning.
Later, it became his nature. One
day when he was lying on the bed,
he saw a small insect(fly) near one
corner of the ceiling. It's movement
led Rene Descartes to think about the
problem of determining its position
on the ceiling. He thought that it was
sufficient to know the eastward and
northward distance of the fly from the corner 'O' of the ceiling (refer
figure). This was the beginning of the subject known as Graphs.
His system of fixing a point with the help of two measurements
one with vertical and another with horizontal is known as 'Cartesian
System'. The word 'Cartes' is taken from Rene Descartes and is named
as 'Cartesian System', in his honour. The two axes x andy are called as
Cartesian axes.
Comer of the ceiling
Graphs
7.2 Introduction to cartesian plane with axes
7.2.1 Location of a point
Look at the Fig.7.1. Can you tell us where
the boy is? Where the church is? Where the
temple is? Where the bag is? and Where the
mosque is? Is it easy? No. How can we locate the
position of the boy, the church, the temple, the
bag and the mosque correctly?
Let us first draw parallel horizontal lines
with a distance of 1 unit from each other. The
bottom line is OX. Now the figure 7.1 will look
like the figure 7.2.
Try to express the position of the boy, the
church, the temple, the bag and the mosque now.
The boy and the church are on the first horizontal
line, (i.e.) Both of them are 1 unit away from the
bottom line OX. Still we are not able to locate
their position exactly. There is some confusion
for us yet. In the same manner, it is difficult for
us to locate the exact positions of the temple, the
bag and the mosque because they lie on different
parallel lines.
To get rid of this confusion, let us now
draw the vertical lines with a distance of 1 unit
from each other in the figure 7.2. The left most
vertical line is OY. Then it will look like the
following figure 7.3.
Now with the help of both the horizontal
and vertical lines, we can locate the given objects.
Let us first locate the position of the boy. He is
1 unit away from the vertical line OY and 1 unit
away from the horizontal line OX. Hence his
position is represented by a point (1, 1).
Similarly the position of the church is
represented by the point (4 , 1), the position of the
temple is (2 , 2), the position of the bag is (4 , 3) and the position of the mosque is (3 , 4).
Fig. 7.2
2 unts
3 units
■it.
'-^m
M.
%
N
■1£k
m—
^x
Fig. 7.3
Chapter 7
7.2.2. Coordinate system
"T"
1
1
1
1
1
1
1
1
1
1
■I ■■
1
,, |,
II 1
Now let us define formally
.......:i
J...
/\
[Tells how many] Tells how many]
. units to move ., units to move ....
l^ to the right J [ to up J
what the coordinate system is.
f
A
)SCi
isa
±^
Let X'OX and Y'OY be
T
T
■■■■/■
■■■■fi
—
P(4,6|) 1
the two number lines intersecting
each other perpendicularly at zero.
They will divide the whole plane
J.
1.
c
4 units
^
J.
u
■■■4
~-cr>
-= -Ordinate
■■■3
■■■2
JO
of the paper into four parts which
■■4-
J
Y' "
-•^
we call quadrants [I, II, III and
<
7-
6-
5-
4-
3-
2-
1.^:
f
1
) ^
■ ii
(i ■"■''^
IV]. See the figure.
1 '
.4-
4..
_2.
^
V
y f
■a
^r
rigir
(0,
The line X'OX is called the x-axis.
Tl
rT
-4
c
0)
TT T
The line Y'OY is called they-axis.
I
[X.
-5
ft
W-
The point '0' is called the Origin.
7
1'
Thus, origin is the point of
. — .> u......
Y-
axis
intersection of x-axis and v-axis.
'1.
1
1
1
1
1
1
1
1
1
1
1
1
„„i
1 1 1,
This is called the Cartesian c
oordinate system.
Note : To mark a point, we always write the x-coordinate (or the number on the
horizontal axis) first and then the y-coordinate (or the number on the vertical
axis). The first number of the pair is called the x-coordinate or abscissa.
The second number of the pair is called the y-coordinate or ordinate.
Observation : Let us consider the point P (4, 6) in the figure. It is 4 units away
from the right side of the y-axis and 6 units above the x-axis. Then the coordinate of
the point P is (4, 6).
7.3 Plotting of Points for different kinds of situations
7.3.1 Plotting a point on a Graph sheet
Example 7.1
Plot the point (4, 5) on a graph sheet. Is it the same as (5, 4) ?
Solution
Draw X'OX and Y'OY and let them cut at the origin at 0.
Mark the units along the x-axis and y-axis with a suitable scale.
The given point is P (4, 5).
liere the x-coordinate of P is 4 and the y-coordinate of P is 5.
Graphs
Scale:
Xaxis 1 cm = 1 unit
Y axis 1 cm = 1 unit
w
L !
yj '
(+,
+)
4 units
.P(4,5)
^r
1
s
.Q(5,
4)
I
(/)
■-»
■1/
k
LO
A
■*
i
5 ^
i
I ^ ^
■yr
And both are positive. Hence the point
P (4, 5), hes in the first quadrant.
To plot, start at the Origin 0(0,0) . Move
4 units to the right along the x-axis. Then turn
and move 5 units up parallel to y-axis. You will
reach the point P (4, 5). Then mark it. (As shown
in the adjoining figure)
Next, let us plot the point Q (5, 4). Here
the x-coordinate of Q is 5 and the y-coordinate
of Q is 4. And both are positive. Hence, this
point Q (5, 4) also lies in the quadrant I. To plot
this point Q (5, 4); start at the Origin. Move 5
units to the right along the x-axis . Then turn and
move 4 units up parallel to y-axis. You will reach the point Q (5, 4). Then mark it (As
shown in the above figure).
Conclusion: From the above figure, it is very clear that the points P(4,5) and
Q(5, 4) are two different points.
Example 7.2
Plot the following points on
a graph paper and find out in which
quadrant do they lie?
(i)A(3,5) (ii)B(-2,7)
(iii) C (-3,-5) (iv) D (2, - 7)
(V) O (0, 0)
Solution
Draw the x andy axes. Mark
the units along the x and y axes
with a suitable scale.
(i) To plot the point A (3 , 5)
Here, the x-coordinate of
A is 3 and the y-coordinate of A
is 5. Both are positive. Hence the
point A (3 , 5) lies in the quadrant
I. Start at the Origin. Move three
units to the right along the x-axis.
1
Scale:
I
.......
I
■\
r
X axis 1 cm = 1 unit
.±
^
/
\
Y axis " ' ' '
1
cm - 1 ur
1
III
P C
o
71
1
° \
' )
■7
-L
1
(
+ . +
)
A(
3,5
3)
-4
o.
"
J
_..
z
Y'
^
■4-
0,
0)
k \/
7-
6-
5-
4-
3-
2-
\
i
1 ^
■ J
i (
i
. •
•A
—+
n
„
-:i
c
/ *
i
'il
-4
I Vi
21
r
-&
R
n
il
7
D
2,
-7
-
:v
\
J
\
( 1
(-
9
-)
>
.\
u.
('
1
-)
Chapter 7
Then turn and move 5 units up parallel to Y-axis and mark the point A (3 , 5).
(ii) To plot the point B (-2 , 7)
Here, the x-coordinate of B is -2 which is negative and the y-coordinate of B
is 7 which is positive. Hence the point B (-2 , 7) lies in the quadrant II. Start at the
Origin. Move 2 units to the left along the x-axis. Then turn and move 7 units up
parallel to y-axis and mark the point B (-2 , 7).
(iii) To plot the point C (-3 , -5)
Here, the x-coordinate of C is -3 and the y-coordinate of C is -5. Both are
negative. Hence the point C (-3 , -5) lies in the quadrant III. Start at the Origin. Move
3 units to the left along the x-axis. Then turn and move 5 units down parallel toy-axis,
and mark the point C (-3 , -5).
(iv) To plot the point D (2 , -7)
Here, the x-coordinate of the point D is 2 which is positive and the y-coordinate
of D is -7 which is negative. Hence the point D (2 , -7) lies in the quadrant IV. Start
at the Origin. Move 2 units to the right along the x-axis. Then turn and move 7 units
down parallel to y-axis and mark the point D (2 , -7).
(v) To plot the point O (0, 0)
This is the origin. Both the x and y coordinates are zeros. It is the point of
intersection of the axes x and y. ,
Mark the point (0,0).
Example 7.3
Plot the following
points on a graph paper and
find out where do they lie?
Scale:
X axis 1 cm = 1 unit
Y axis 1 cm = 1 unit "
T
[T
^
'
—
t'
>
-)-
-7
"f
t
■■fr
......(
[.i.
;..i.
V-
■5
r^
0,
4)
.At
\j
1
(i)A(7,0) (ii)B(-5,0)
■rV
(iii) C (0,4) (iv)D(0,-3)
o
£.
i
H
Solution
v
B(
-5
.0)
A
(7,
0)
Draw the x and y axes.
A"
5
7-
6-
5-
4-
3-
2-
\
i
; <■
I
i (
) '
•
"X
4
r\
r\
O
Mark the units along the x and y
axes with a suitable scale.
-2-
-3-
U
(U,
- o
!
A
(i) To plot the point A (7, 0)
-fv
-&
[■V
Here the x-coordinate of
[TT
7
\
A is 7 which is positive and the
/
\
/
{-
5
-)
>
f
(
5
-;
v-coordinate of A is zero Hence
1
— .>
nnl."
,„l
T
1
T
T
1
n"
T
1
T
1
1
T
1, „
„J„„
T
1
Graphs
the point A (7, 0) hes on the x-axis. Start at the Origin. Move 7 units to the right along
the X-axis and mark it.
(ii) To plot the point B (-5 , 0)
Here, the x-coordinate of B is -5 which is negative and the y-coordinate is zero.
Hence the point B (-5 , 0) lies on the x-axis. Start at the Origin. Move 5 units to the
left along the x-axis and mark it.
(ill) To lot the point C (0 ,4)
Here, the x-coordinate of C is zero and the y-coordinate of C is 4 which is
positive. Hence the point C (0 , 4) lies on the y-axis. Start at the Origin. Move 4 units
up along the y-axis and mark it.
(iv) To plot the point D (0 , -3)
Here the x-coordinate of D is zero and the y-coordinate of D is -3 which is
negative. Hence the point D (0 , - 3) lies on they-axis. Start at the Origin. Move 3 units
down along the y-axis and mark it.
Do you know?)
Where do the points lie? How can we tell without actually plotting the points on
a graph sheet? To know this, observe the following table.
SI. No.
Examples
X coordinate of
the point
y coordinate of the
point
Location of the
point
(3,5)
Positive (+)
Positive (+)
Quadrant I
(-4,10)
Negative (-)
Positive (+)
Quadrant II
3.
(-5,-7)
Negative (-)
Negative (-)
Quadrant III
(2,-4)
Positive (+)
Negative (-)
Quadrant IV
(7,0)
Non zero
Zero
On the X axis
(0,-5)
Zero
Non-zero
On the Y axis
(0,0)
Zero
Zero
Origin
Can you tell, where do the following points lie without actually plotting them
on the graph paper?
(i) (2 , 7) (ii) (-2 , 7) (iii) (-2 , -7) (iv) (2 , -7)
(V) (2 , 0) (vi) (-2 , 0) (vii) (0 , 7) (viii) (0 , -7)
Chapter 7
1 7.4 Drawing straight lines and parallel lines to the coordinate axes
In this section first we are going to learn how to draw straight lines for the
I given two points and then how to draw lines parallel to coordinate axes. Also to find
I the area of plane figures.
7.4.1 Line joining two given points
Example 7.4
Draw the line joining the
following points.
(i) A (2,3) and B (5, 7),
(ii) P (-4,5) and Q (3,-4).
Solution
(i) To draw the line joining the
points A (2 , 3) and B(5 , 7):
First, plot the point (2 , 3) and
denote it by by A.
Next, plot the point (5 , 7) and
denote it by B.
Then, join the points A and B.
AB is the required line.
(ii) To draw the line joining
the points P (-4 , 5) and Q (3 , -4)
First, plot the point (-4 , 5)
and denote it by P.
Next, plot the point (3 , -4)
and denote it by Q.
Then, join the points P and Q.
PQ is the required line.
Scale:
X axis 1 cm = 1 unit
Y axis 1 cm = 1 unit
1
Iy,
1
•
1
/
-7
/
'B (5
7)
^
-
i /
/ i
1
!
;
;
V
1
1
(+,
+)
1 / 1
/
-f
/
/
/
{
A (2, 3)
'->
--
2
'
\ S 6
r ^^
>
f
''
.
T""""T"r-T"T"T MM
Scale:
-4 1-
.-.,.,,
\
X axis 1 cm = 1 unit
*
i
Y axis 1 cm = 1 unit "
1 1 1 1
[I
—
1::
- (-
,....+
I..
n 1
A
<^\
■■/■
-6
+,
+
)
......:^
P-
-3J
-5
\,
...
V
\
■■\
\
-^
T
v
■■■■2-
-f-
-|
\
4.
X'..^
7-
6-
5-
4-
3-
2-
I4.
X
*
i
»
> '
I
> f
»
^-x-
-a
\
V
P
/\
\
3,
-4)
...
r
\
n '
P
J '■
II]
_7.
Y
4...
- (-
~?
-)
>
f
(^
9
-)
~T ,
1
1
1
1
1
1
'
1
Graphs
7.4.2 Drawing straight parallel lines to axes
Example 7.5
(i) Draw the graph of x = 3.
(ii) Draw the graph of y = -5.
(iii) Draw the graph of x = 0.
Solution
(i) The equation x = 3 means:
Whatever may be y-coordinate,
x-coordinate is always 3. Thus, we have
X 3 3_
y 3 -4
Plot the points A (3, 3) and
B (3 , -4). Join these points and extend
this line on both sides to obtain the
graph of X = 3.
— 1
■ — T~i"n
Scale:
m
—n
rn
j.^
X axis 1 rm ~ 1 unit
\
Y axis 1 cm - 1 unit "
T
r
1-1 ■
■7-
........
i-
,J
)
(
■^.
...^.
^
/
-5
■ ro
\
^
■■■■4
-3
II
A
(3,
3)
1
■■■■■■f
Y'
^
,
7
6-
5-
4-
3-
2-
1.
•
;: ;i - ;i f
. .. y.
-^
-6^
B
3,
-4)
-4
........
-&
\'
m
7
IV
(-
?
-)
(t -
-)
>
f
(ii) The equation y = - 5 means:
Whatever may be the x-coordinate,
the y-coordinate is always - 5.
Thus we have.
X -2 6_
y -5 -5
Plot the points A (-2, -5) and
B (6 , -5). Join these points A and B and
extend this line on both sides to obtain
the graph of y = -5.
— 1
■ — r —
Scale:
X axis
. — .........
cm =
........
1 iir
1 —
lit"
/
i
Y axis 1 cm = 1 unit "
11
-7-
—
(-
,^
T
■5
.^..
"T
)
...4.
t
-±
f
X'-
'—
7
P
c;
A
*?
n
1
n •
1 ;
! '
I
i (
i
►X
'1
-2-
-^
A
-&
y
= -
-5
>-■■■■■
A (-2
,...-.
5)
R
3(6,-
5)
7
J\ T
Lil
':_!
\
(-
5
-)
\
/
)
Chapter 7
(iii) The equation x = means;
Whatever may be the y-coordinate,
x-coordinate is always 0.
Thus, we have
— 1
1
Scale:
1 r^»
m
■1 . ir
V
y ^
Y axis 1 cm - 1 unit "
1 1 1 1
....7.
T.T..
1
"*""
-
t-
11
5
.X...
■■■■fi
i+^
...^
)
)
....CL
5
--
^
K
,.
-3<
rt
en
''^
o
^'-'i
^^
--
V'
.4-
.4-.
f
7
P
c: A
3-
2-
s
•
;i :i ^. ;; (i ■
.-/v|
T
R en
•^^
-
■^
TT
-&
ill
r-:-
T
7
i
...1 V
)
-
-f-
-.,
>'
-
Plot the points A (0 , 3) and
B (0, - 3).
Join the points A and B and extend
this line to obtain the graph of x = 0.
7.4.3 Area of Plane Figures
Area of regions enclosed by plane figures like square, rectangle, parallelogram,
trapezium and triangle drawn in a graph sheet can be determined by actual counting
of unit squares in the graph sheet.
Example: 7.6
Plot the points A (5 , 3),
B (-3 , 3), C (-3 , -4), D (5 , -4) and
find the area of ABCD enclosed by
the figure.
Solution
Draw the x-axis and y-axis with
a suitable scale.
Plot the points A (5, 3),
B (- 3, 3), C (-3 , -4), D (5 , -4). Join
the points A and B, B and C, C and D
and D and A. We get a closed figure
ABCD. Clearly it is a rectangle. Count
the number of units squares enclosed
between the four sides. There are
altogether 56 unit squares. Hence the
area of the rectangle ABCD is 56 cm^.
^.
-T
"T-
~r
T"
T-
"T"
"T-
-r
"T"
~r
~n
Scale:
m
m
rn
vr
■^
X axis 1 cm = 1 unit
Y axis 1 cm = 1 unit "
[I
-7-
-
(-
_
-y
■■■■&
■■■■5
(
■^.
+
)
(-
■3,
3)
—
...4.
—
—
— '
A..
5,
3)
....4.
y
7-
6-
5-
4-
3-
2-
1 ,
■
;
»
1 '
■ !
(
i
, ^
— t-
A
'D-
;-57
~-4
)
C
^
y
= -
5
l-
i,-
4)
-&
/
II
^
-7-
(
w
>
(-
)
-)■
>
'
i
1
-)■
Graphs
Example 7.7
Plot the points A (2 , 8 ),
B (-3 , 3) , C (2 , 3) and
find the area of the region
enclosed by the figure ABC.
Solution
Draw the x-axis and
y-axis with a suitable scale.
Plot the points A (2 , 8),
B (- 3, 3), C (2 , 3). Join the
points A and B, B and C and
C and A. We get a closed
figure ABC. Clearly it is a
triangle. Count the number
of full squares. There are 10
full unit squares.
Count the number of half squares. There are 5 half unit squares.
Hence the area of a triangle is 10 + y = 10 + 2.5 = 12.5 cm^
' ■ I ■ I ■ 1 : ' Scale:
! ! '^ ! a. 10 a\ y =.vio ^ r-m - ^ unit
r ciAit> lull- 1 uiiii
A
II
jll 1
/ 1
(t,+) , ^
-^+' ^) ^
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;
! /^ *+
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^(-3,i)
; ! C (2, 3) i 1 ; \
'. \ : ; ! !
i! i i ! i
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— ! — ! — ! — ' — ! — ' — ! — ! — ! — ' — ! —
-is -12 -1
12 3 4 ^^^
(i \\ 1— i-J' i i i — K ■.(+ h 1— •—
U 5 Ji ! Y' ■ ■ ■ iV ' ? :/ !
EXERCISE 7.1
Plot the following points in the graph paper and find out where do they lie?
(i) A (2 , 3) (ii) B (- 3 , 2) (iii) C (-5 -5) (iv) D (5 , -8)
(V) E (6 , 0) (vi) F (-4 , 0) (vii) G (0 , 9) (viii) H (0 , -3)
(ix) J (7 , 8) (X) O (0 , 0)
State in which quadrant each of the following
points lie without actually plotting the points.
(i) (8 , 15) (ii) (-15 , 2)
(iii) (-20 , -10) (iv) (6 , -9)
(vi) (-17 , 0)
(viii) (-100 , -200)
(x) (-50 , 7500).
(V) (0 , 18)
(vii) (9 , 0)
(ix) (200 , 500)
Determine the quadrants and the coordinates
of the points A, B, C, D, E, F, G, H and O in
the given figure.
'
'
^
/
■fi
...4
A.
-3
H
.......
B
(
1
7-
6^
5-
4-
3-
2-
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1
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r
n
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Chapter 7
4. Plot the fohowing points and draw a hne through the points,
(i) (2,7), (-2 ,- 3) (ii) (5 , 4), (8 , -5)
(iii) (-3 , 4), (-7 , -2) (iv) (-5 , 3), (5 , -1)
(v) (2 , 0), (6 , 0) (vi) (0 , 7), (4 , -4)
5. Draw the graph of the following equations:
(i)y-O (ii)x-5 (iii)x = -7 (iv)y-4 (v)y = -3
6. Plot the following points and find out the area of enclosed figures,
(i) A(3 , 1), B (3 , 6), C (-5 , 6), D (-5 , 1)
(ii) A(-2 , -4), B (5 , -4), C (5 , 4), D (-2 , 4)
(in) A (3 , 3), B (-3 , 3), C (-3 , -3), D (3 , -3)
(iv) O (0 , 0), A (0 , 7), B (-7 , 7), C (-7 , 0)
(v)A(0,-2),B(-4,-6),C(4,-6)
(vi)A(l,2),B(9,2),C(7,4),D(3,4)
(vii) A (-4, 1), B (-4, 7), C (-7, 10), D (-7 , 4)
7. Find the perimeter of the rectangle and squares of the previous problems
6 (i), (ii), (iii) and (iv).
7.5 Linear Graphs
We have learnt to draw straight lines and the parallel lines in the graph sheet.
When we get a straight line by joining any two points, then the graph is called a linear
grapii.
7.5.1 Time and Distance Graph
Let us consider the following example to study the relationship between time
and distance.
Example 7.8
Amudha walks at a speed of 3 kilometers per hour. Draw a linear graph to show
the relationship between the time and distance.
Solution
Amudha walks at a speed of 3 kilometers per hour. It means she walks 3 Km in
1 hour, 6 Km in 2 hours, 9 km in 3 hours and so on.
Thus we have the table
Time in hours (x)
1
2
3
4
5
Distance in km (y)
3
6
9
12
15
Points: (0 , 0), (1 , 3), (2 , 6), (3 , 9), (4 , 12) and (5 , 15).
Graphs
> '
Scale:
Yaxis 1 cm = 3 units
...4...::...,
...4...
""U
/
4--
i
</
^_
j
4
/
(5, 15)
E
i
/
(4, 12
i
^ n
/
8
i5 ^
/
(3,9)
/
.b 1...
/
(2,6)
/
i
(1,3)
/lO.O)
-
1
Tim
!
) (
>
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f
.-t
' rr
■-f-
Plot the points (0 , 0), (1, 3),
(2, 6), (3, 9), (4, 12) and (5, 15). Join
all these points.
We get a straight line . Hence, it
is a linear graph.
Relationship between x andy:
We know that,
Distance = Speed x Time.
From the above table,
= 3x0
3=3x1
6 = 3x2
9 = 3x3
12 = 3x4
15 = 3x5
^y - '^X
[Here, y - Distance, x - Time in hour and 3 is the speed]
The linear equation of this problem is y = 3x.
7.5.2 Perimeter-side graph of a square
Example 7.9
Draw a linear graph to show the perimeter-side relationship of a square
Solution
We know that the perimeter
of a square is four times of its side,
(i.e) P = 4a.
(Here, P = Perimeter and a = side)
For different values of a, the values
of P are given in the following table.
a (in cm)
1
2
3
4
P = 4a (in cm)
4
8
12
16
Points: (1, 4), (2, 8), (3, 12), (4, 16).
Plot the above points. Join all the
points. We get the linear graph of P = 4a.
Scale:
■>
X axis 1 cm = 1 unit
y
Yaxis I crri - 4 units"
/
.^^"^y
A r-
y
/
(4,16
1
Ai-i
/
t
/
(3, 12
/
(2,8)
/
(1,4)
/
^
.___!___.
,
1
\
1
■
Chapter 7
7.5.3 Area as a function of side of a square
Example 7.10
Draw a graph to show the area-
side of a square.
Solution
We know that the area of a square is
the square of its side, (i.e) A = a^.
(liere, A = Area, a = side). For
different values of a, the values of A are
given in the following table.
a (in cm)
1
2
3
4
5
A = a^ (in cm^)
1
4
9
16
25
Points: (1 , 1), (2 , 4), (3 , 9), (4 ,16),
(5 , 25)
Plot the above points.
Join all the points. Observe the
graph. Is it a linear graph? No. It is a curve.
7.5.4 Plotting a graph of different multiples of numbers
Example 7.11
Draw a graph of multiples of 3.
Solution
Let us write the multiples of 3.
Multiples of 3 are 3, 6, 9, 12, 15... etc.
We can also write this as Multiples
of 3 = 3 X n, where n = l,2, 3, ...
m = 3 n. m is the multiple of 3
Thus, we have the following table.
'
1
J
1 Scale:
— t— ■
1 ,
k
ifi X axis 1 cm = 1 unit
T 't axis 1 cm = 2 units
/
(4,
lei
•1 A
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1
h
1
1
j
1
1
"*"
t
t
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/
/
/
A
/
/
(2,4)
1
/
1
r
1
o
/
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* 1 * f
>
r
>
j
!
n
1
2
3
4
5
m = 3n
3
6
9
12
15
Points: (1,3), (2, 6), (3, 9), (4, 12),
(5 , 15).
Plot all these points and join them.
We get the graph for multiples of 3.
'
Scale:
i
k
X axis 1 cm = 1 unit |
Y axis 1 cm = 2 units^
'^l
/(^
'
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V
An
/
/
(4, 12
An
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(3,9)
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(2,6)
1
A
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o
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(1. 3)
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i
Graphs
7.5.5. Simple Interest-Time graph
Example 7.12
Ashok deposited ? 10,000 in a bank at the rate of 8% per annum. Draw a linear
graph to show the simple interest-time relationship. Also, find the simple interest for
5 years.
Solution
We know that.
Simple interest, I =
Pnr
100
[where P ~ Principal, n = Time in years,
r = Rate of interest]
Principal, P = 10000
Time, n = ?
Rate, r = 8%
_ PXnXr
I
I
I =
100
_ 10000 X n X 8
100
800 n.
TTT
Scale:
m
>
■ T
X axis 1 cm = 1 unit
Y axis 1 cm = 800 units_
'
>
^
.^T'
f
O
0)
j
1
/
(5, 4000)
t
t
t
A
3
IT
^
(4, 32j)0)
/
(3, 2400)
/
yv
Q.
E
(2, 1600)
1
t
w
/
(1,80
0)
/
f
Time in yt
II li
^
1
ft
L
WnAmA
lm4mJ
It^-nli-mJ
Iffl4mtl
(Here, the simple interest, I depends upon N)
For different values of n, the values of I are given in the following table.
n (Time in Yrs)
1
2
3
4
5
I = 800 n (in X)
800
1600
2400
3200
4000
Points: (1, 800), (2, 1600), (3, 2400), (4, 3200), (5, 4000)
Plot all the points. Join them all. Draw the linear graph.
So, Ashok will get X 4000 as simple interest after 5 years. (In the graph, the
answer is shown by the dotted lines.)
7.6 Reading Linear Graphs
Money Exchange: The world has become very small today. It is inevitable to
do business with foreign countries. When we are doing business with other countries,
we have to transact our money (Indian currency) in terms of their currencies. Different
countries use different currencies under different names. Hence we should know the
concepts related to money exchange. Let us consider the following example.
Example 7.13
On a particular day the exchange rate of 1 Euro was X 55. The following linear
Graph shows the relationship between the two currencies. Read the graph carefully,
and answer the questions given below:
Chapter 7
(i) Find the values of 4 Euros in terms of Rupees,
(ii) Find the values of 6 Euros in terms of Rupees,
(iii) Find the value of ? 275 in terms of Euros,
(iv) Find the value of ? 440 in terms of Euros.
Solution
(i) To find the value of 4
Euros. In this graph, draw a
dotted line at x = 4 parallel
to y-axis.
Locate the point of
intersection of this line
with the given line.
From this point draw
a dotted line parallel to
X-axis.
It cuts the y-axis at 220.
(See figure)
Hence the value of
4 Euros is ? 220.
ilctivity
■
1
Scale: |
X
axis 1 cm = 1 unit
<
1
>
k. "■"
i
!
^
/
/
/
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! >
V
A
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(4, 220)
/
/
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4
k
1
/
.
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....i
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/
, \ \ \ \ \ \ \ '
>
f
=uros
_».
.....
fi
1
Try and answer the remaining questions (ii), (iii) and (iv).
EXERCISE 7.2
^
1. Draw a linear graph for the following data,
(i)
X
5
5
5
5
5
5
(ii)
X
1
2
3
4
5
y
1
2
3
4
5
6
y
1
2
3
4
5
Draw the linear graph and find
the missing entries.
Draw the following graph of side-area
relationship of a square.
X
1
2
3
4
-
y
6
12
-
-
30
Side (in m)
2
3
4
5
6
Area (in m^)
4
9
16
25
36
4. Draw the graph of y = 7x.
5. If Akbar is driving a car at a uniform speed of 40 km/hr. Draw the distance time
graph. Also find the time taken by Akbar to cover a distance of 200 km.
6. Eliza deposited ? 20,000 in a bank at the rate of 10% per annum. Draw a linear
graph showing the time and simple interest relationship. Also, find the simple
interest for 4 years.
Data Handling
8.1 Introduction
8.2 Recalling the Formation of Frequency Table
8.3 Drawing Histogram and Frequency Polygon
for Grouped Data
8.4 Construction of Simple Pie chart
8.5 Measures of Central Tendency
8.1 Introduction
Everyday we come across different kinds of information in the
form of numbers through newspapers and other media of communication.
This information may be about food production in our country,
population of the world, import and export of different countries, drop-
outs of children from the schools in our state, the accidential deaths, etc.
In all these information, we use numbers. These numbers are
called data. The data help us in making decisions. They play a vital part
in almost all walks of life of every citizen. Hence, it is very important to
know how to get relevant and exact information from such data.
The calculated data may not be suitable for reading, understanding
and for analysing. The data should be carefully handled so that it can be
presented in various forms. A common man should be able to understand
and visualize and get more information on the data.
R.A. Fisher
[17th Feb., 1890 -
29th July, 1962]
Fisher was
interested in the
theory of errors
that eventually let
him to investigate
statistical
problems. He
became a teacher
in Mathematics
and Physics
between 1915
and 1919. He
studied the design
of experiments
by introducing
randomisation
and the analysis
of variance
procedures now
used throughout
the world. He is
known as
"Father of Modern
Statistics".
Chapter 8
8.2 Recalling the Formation of Frequency Table
We have learnt in seventh standard, how to form a frequency table. Let us
recall it.
8.2.1 Formation of frequency table for an ungrouped data
Example 8.1
Consider the following data:
15, 17, 17, 20, 15, 18, 16, 25, 16, 15,
16, 18, 20, 28, 30, 27, 18, 18, 20, 25,
16, 16, 20, 28, 15, 18, 20, 20, 20, 25. Form a frequency table.
Solution
The frequency table is given below:
Number
Tally Mark
Frequency
if)
15
4
16
N
5
17
2
18
M
5
20
N
>
7
25
3
27
1
28
2
30
1
Total
30
8.2.2 Formation of frequency table for a grouped data
Example: 8.2
The marks obtained by 50 students in a Mathematics test with maximum marks
of 100 are given as follows:
Data Handling
43, 88, 25, 93, 68, 81, 29, 41, 45, 87, 34, 50, 61, 75, 51, 96, 20, 13, 18, 35,
25, 77, 62, 98, 47, 36, 15, 40, 9, 25, 39, 60, 37, 50, 19, 86, 42, 29, 32, 61,
45, 68, 41, 87, 61, 44, 67, 30, 54, 8.
Prepare a frequency table for the above data using class interval.
Solution
Total number of values = 50
Range = Highest value -Lowest value
= 98 - 8 = 90
Let us divide the given data into 10 classes.
Length of the Class interval =
Range
Number of class interval
90 ^Q
10
The frequency table of the marks obtained by 50 students in a mathematics test
is as follows:
Class
Interval (C.I)
Tally Mark
Frequency (/)
0-10
2
10-20
4
20-30
Is
s
s|
6
30-40
N
s
si
7
40-50
N
s
s|
9
50-60
4
60-70
Is
s
8
70-80
2
80-90
Is
s
s|
5
90 - 100
3
Total
50
Chapter 8
Thus the given data can be grouped and tabulated as folloM^s:
Class
Interval
(C.I)
0-10
10-20
20-30
30-40
40-50
50-60
60-70
70-80
80-90
90-100
Frequency
if)
2
4
6
7
9
4
8
2
5
3
8.3 Drawing Histogram and Frequency Polygon for Grouped Data
The statistical data can be represented by means of geometrical figures or
diagrams which are normally called "graphs". The graphical representation of data
makes itself interesting for reading, consuming less time and easily understandable.
There are many ways of representing numerical data graphically. In this chapter, we
will study the following two types of diagrams:
(1) Histogram
(ii) Frequency Polygon
8.3.1 Histogram
A two dimensional graphical representation of a continuous frequency
distribution is called a histogram.
In histogram, the bars are placed continuously side by side with no gap between
adjacent bars. That is, in histogram rectangles are erected on the class intervals of
the distribution. The areas of rectangle are proportional to the frequencies.
8.3.1 (a) Drawing a histogram for continuous frequency distribution
Procedure:
Step 1 : Represent the data in the continuous (exclusive) form if it is in the
discontinuous (inclusive) form.
Step 2 : Mark the class intervals along the X-axis on a uniform scale.
Step 3 : Mark the frequencies along the Y-axis on a uniform scale.
Step 4 : Construct rectangles with class intervals as bases and corresponding
frequencies as heights.
The method of drawing a histogram is explained in the following example.
Example 8.3
Draw a histogram for the following table which represent the marks obtained by
100 students in an examination:
Data Handling
Marks
0-10
10-20
20-30
30-40
40-50
50-60
60-70
70-80
Number of
students
5
10
15
20
25
12
8
5
Solution
The class intervals are all equal with length of 10 marks. Let us denote these
class intervals along the X-axis. Denote the number of students along the Y-axis, with
appropriate scale. The histogram is given below.
;;;;;;|; ;;;;;;;;;;;
llllllllll
::::::::!: :::::: :: :::
Scale
X axi s 1 cm = 1 Marks IIIIII
: : :
V pvio 1 ^*^ — ^ QUiA
iiiiiii
i!ii
i i i
; ; ;
:>::::
... ^
3;;:; ;;;; ;;;;;; ;; ;;;
3;
5
... ,
::: -
M
5;;;;;;;;;;;;;;;;;;
i
a::::
3
|i i iiii iiiiii -^^^
:::
j
►
^f ;:::
t 10
70 10 ^n ^0 ^a 70 sn on
11 II II 1 1 1 II II 1 II III II III
II 11 III 1 1 III 1 1 11
^
^
'-'
^ w
—
:;^pIi^1s&:
::i
Fig. 8.1
Note: In the above diagram, the bars are drawn continuously. The rectangles are of
lengths (heights) proportional to the respective frequencies. Since the class intervals
are equal, the areas of the bars are proportional to the respective frequencies.
8.3.1 (b) Drawing a histogram when class intervals are not continuous
Example 8.4:
The heights of trees in a forest are given as follows. Draw a histogram to
represent the data.
Heights in metre
16-20
21-25
26-30
31-35
36-40
41-45
46-50
51-55
Number of trees
10
15
25
30
45
50
35
20
Chapter 8
Solution
In this problem, the given class intervals are discontinuous (inclusive) form.
If we draw a histogram as it is, we will get gaps between the class intervals. But in
a histogram the bars should be continuously placed without any gap. Therefore we
should make the class intervals continuous. For this we need an adjustment factor.
Adjustment Factor = ~ [(lower limit of a class interval) -
(upper limit of the preceding class interval)]
_ 1
(21-20)= 0.5
In the above class interval, we subtract 0.5 from each lower limit and add 0.5 in
each upper limit. Therefore we rewrite the given table into the following table.
Heights in
metre
15.5-20.5
20.5-25.5
25.5-30.5
30.5-35.5
35.5-40.5
40.5-45.5
45.5-50.5
50.5-55.5
Number of
Trees
10
15
25
30
45
50
35
20
Now the above table becomes continuous frequency distribution. The histogram
is given below
m
'||I||||||||||||||||||||||||||||||||||||||||H^
::|||||||||||||||'scaiel#W-
; X axis 1 cm = 5 Metres ; ;
; : Y axis 1 cm = 5 Trees
iiiiiiiii i
iiiiiiiii i
! 1 1 ! II 1 1 1
■ u
35llllllllllllllllllllllllllllllllllllllllllllllim
■ P ,
■tlSB
:;Ss;
:; !
■■fS-
■ " aS'
jjigrr
25=iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
■ ■ a A
5.5 20.5 25.5 30.5 35.5 4C
.5 45.5 50.5 55.5 60.^^ 1 1
Wu
lllllllllllllllllllll
Hfflwr
ifflffl+r
mM
lllllllllllllllllllllllllllllllllllllllll :;;:;:;;:;:
Fig. 8.2
Data Handling
Note: In the histogram (Fig. 8.2) along the X-axis the
first value starts from 15.5, therefore a break (kink)
is indicated near the origin to signify that the graph is
drawn beginning at 15.5 and not at the origin.
^Do you knowj^
The break is indicated
by a Zig - Zag curve.
8.3.2 Frequency polygon
Frequency Polygon is another method of representing frequency distribution
graphically.
Draw a histogram for the given continuous data. Mark the middle points of the
tops of adjacent rectangles. If we join these middle points successively by line segment,
we get a polygon. This polygon is called the frequency polygon. It is customary to
bring the ends of the polygon down to base level by assuming a lower class of a
frequency and highest class of a frequency.
Frequency Polygon can be constructed in two ways:
(i) Using histogram
(ii) Without using histogram.
8.3.2 (a) To draw frequency polygon using histogram
Procedure:
Step 1 : Obtain the frequency distribution from the given data and draw a
histogram.
Step 2 : Join the mid points of the tops of adjacent rectangles of the histogram
by means of line segments.
Step 3 : Obtain the mid points of two assumed class intervals of zero
frequency, one adjacent to the first bar on its left and another
adjacent to the last bar on its right. These class intervals are known
as imagined class interval.
Step 4 : Complete the polygon by joining the mid points of first and the
last class intervals to the mid point of the imagined class intervals
adjacent to them.
Example 8.5
Draw a frequency polygon imposed on the histogram for the following
distribution
Class interval
10-20
20-30
30-40
40-50
50-60
60-70
70-80
80-90
Frequency
4
6
8
10
12
14
7
5
Chapter 8
Solution
Take the class-intervals along the X-axis and frequencies along the Y-axis with
appropriate scale as shown in the Fig 8.3.
Draw a histogram for the given data. Now mark the mid points of the upper
sides of the consecutive rectangles. We also mark the midpoints of the assumed class
intervals 0-10 and 90-100. The mid points are joined with the help of a ruler. The ends
of the polygon are joined with the mid points of 0-10 and 90-100. Now, we get the
frequency polygon. Refer Fig 8.3.
; :
;
; ;
: :
:
: :
1
iSca e
: :
:
: :
: :
::: : :::
: :
X axis
cm =
lOUnitsl
: :
: :
Y axis 1 cm =
2 Units
ii
\
::
... . /
;:
m
: :
/
..
::
: /
ij,im
:
: ^
3:
u
3
T
';
SffiTTW
!
■
/
fMfl
::
\
/
■■
::: /
..
^^ /
' '
MM
M
H
1^
>
X
nrq'^
mi
;:!:
^Cr
^^^itr
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!!:
!!:
;;!(!:::::::
Id
■
;Cla
ISS 1
nt(
rrtn
jrvals
i
ffl
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;;;
Fig. 8.3
Example 8.6
Draw a frequency polygon of the following data using histogram
Class interval
0-10
10-20
20-30
30-40
40-50
50-60
Frequency
5
10
25
16
12
8
Solution
Mark the class intervals along the X-axis and the frequencies along the Y-axis
with appropriate scale shown in Fig 8.4.
Data Handling
Draw a histogram for the given data. Now, mark the mid points of the upper sides
of the consecutive rectangles. Also we take the imagined class interval (-10) - and
60 - 70. The mid points are joined with the help of a ruler. The ends of the polygon are
joined with the mid points of the imagined class intervals (-IO)-O and 60 - 70. Now we
get the frequency polygon. (Refer Fig 8.4).
ililillllllll
: :
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= 5 Units
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:::::::: Y axis 1 Cn
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:::::::::::::::«.::::
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llllll|iiiiiiiiiiiiiiiiTf 1 1 I]
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Fig. 8A
Note: Sometimes imagined class intervals do not exist. For example, in case of
marks obtained by the students in a test, we cannot go below zero and beyond
maximum marks on the two sides. In such cases, the extreme line segments are
only partly drawn and are brought down vertically so that they meet at the mid
points of the vertical left and right sides of first and last rectangles respectively.
Using this note, we will draw a frequency polygon for the following example:
Example 8.7
Draw a frequency polygon for the following data using histogram
Marks
0-10
10-20
20-30
30-40
40-50
50-60
60-70
70-80
80-90
90-100
1
Number of
students
5
4
6
8
5
7
4
9
5
7
1
Chapter 8
Solution
Mark the class intervals along the X-axis and the number of students along the
Y-axis. Draw a histogram for the given data. Now mark the mid points of the upper
sides of the consecutive rectangles. The mid points are joined with the help of a ruler.
Note that, the first and last edges of the frequency polygon meet at the mid point of
the vertical edges of the first and last rectangles.
Fig. 8.5
8.3.2 (b) To draw a frequency polygon without using histogram
Procedure:
Step 1 : Obtain the frequency distribution and compute the mid points of
each class interval.
Step 2 : Represent the mid points along the X-axis and the frequencies along
the Y-axis.
Step 3 : Plot the points corresponding to the frequency at each mid point.
Step 4 : Join these points, by straight lines in order.
Step 5 : To complete the polygon join the point at each end immediately to the
lower or higher class marks (as the case may be at zero frequency) on
the X-axis.
Data Handling
Example 8.8
Draw a frequency polygon for the following data without using histogram.
Class interval
10-20
20-30
30-40
40-50
50-60
60-70
70-80
80-90
Frequency
4
6
8
10
12
14
7
5
Solution:
Mark the class intervals along the
X-axis and the frequency along the Y-axis.
We take the imagined classes 0-10 at the
beginning and 90-100 at the end, each
with frequency zero. We have tabulated
the data as shown.
Using the adjacent table, plot the
points A (5, 0), B (15, 4), C (25, 6),
D (35, 8), E (45, 10), F (55, 12),
G (65, 14), H (75, 7), I (85, 5)
and J (95, 0).
We draw the line segments AB, BC, CD, DE, EF, FG, GH, HI, IJ to obtain the
required frequency polygon ABCDEFGHIJ, which is shown in Fig 8.6.
Class interval Midpoints
Frequency
0-10 5
10-20
15
4
20-30
25
6
30-40
35
8
40-50
45
10
50-60
55
12
60-70
65
14
70-80
75
7
80-90
85
5
90-100 ^^^
1^^
:::
::
1-1*
:
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: :
::::::::::::::.
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Xaxis 1 cm= 10 Units
Y axis 1 cm = 2 Units
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Fig. 8.6
Chapter 8
EXERCISES 8.1
1. Draw a histogram to represent the foUowing data
Class intervals
0-10
10-20
20-30
30-40
40-50
50-60
Frequency
8
12
6
14
10
5
2. Draw a histogram with the help of the following table
Yield per acre (Quintal)
11-15
16-20
21-25
26-30
31-35
36-40
Number of rice field
3
5
18
15
6
4
3. Draw a histogram to represent the following data of the spectators in a cricket
match
Age in years
10-19
20-29
30-39
40-49
50-59
60-69
Number of spectators
4
6
12
10
8
2
4. In a study of diabetic patients in a village, the following observations were noted
Age (in years)
10-20
20-30
30-40
40-50
50-60
60-70
Number of patients
3
6
13
20
10
5
Represent the above data by a frequency polygon using histogram.
5. Construct a histogram and frequency polygon for the following data
Class interval
0-10
10-20
20-30
30-40
40-50
50-60
Frequency
7
10
23
11
8
5
6. The following table shows the performance of 150 candidates in an Intelligence
test. Draw a histogram and frequency polygon for this distribution
Intelligent Quotient
55-70
70-85
85-100
100-115
115-130
130-145
Number of candidates
20
40
30
35
10
15
7. Construct a frequency polygon from the following data using histogram.
Marks
0-10
10-20
20-30
30-40
40-50
50-60
60-70
70-80
80-90
90-100
No of
students
9
3
4
6
2
3
4
5
7
8
Data Handling
8. Draw a frequency polygon for the following data without using histogram
Age (in years)
0-10
10-20
20-30
30-40
40-50
50-60
60-70
Number of persons
6
11
25
35
18
12
6
9. Construct a frequency polygon for the following data without using histogram
Class interval
30-34
35-39
40-44
45-49
50-54
55-59
Frequency
12
16
20
8
10
4
10. The following are the marks obtained by 40 students in an English examination
(out of 50 marks). Draw a histogram and frequency polygon for the data
29, 45, 23, 40, 31, 11, 48, 1, 30, 24, 25, 29, 25, 32, 31, 22, 9, 49, 19, 13,
32, 39, 25, 3, 27, 41, 12, 13, 2, 44, 7, 43, 15, 35, 40, 3, 12, 48, 49, 18.
8.4 Construction of Simple Pie Chart
Have you ever come across the data represented in circular form as shown
in Figure 8.7 and Figure 8.8 ?
The time spent by a school student Viewers watching different types of
during a day (24 hours). channels on TV.
Fig. 8.7
The figures similar to the above are called circle
graphs. A circle graph shows the relationship between
a whole and its parts. Here, the whole circle is divided
into sectors. The size of each sector is proportional
to the activity or information it represents. Since,
the sectors resemble the slices of a pie, it is called a
pie chart.
Fig. 8.8
Pie is an American
food item
Chapter 8
For example, in the pie chart (Fig 8.7).
1 The proportion of the sector
1 for hours spent in sleeping
_ number of sleeping hours ^1
whole day
_ 8 hours _ 1 J
^^ 24 hours 3 ^^^^^
So, this sector is drawn ^rd part of the circle.
The proportion of the sector
_ number of school hours
for hours spent in school
Whole day
_ 6 hours _ 1
24 hours 4
So, this sector is drawn 4^th of the circle.
4
The proportion of the sector
_ number of play hours
for hours spent in play
Whole day
_ 3 hours _ 1
i
24 hours 8
So, this sector is drawn -i- th of the circle.
o
The proportion of the sector
for hours spent in homework
number of home work hours
whole day
3 hours _J_
24hours 8
So, this sector is drawn ^ th of the circle.
8
[The proportion of the sector
24hours
So, this sector is drawn ^ th of the circle.
6
number of others hours
whole day
4 hours _ 1^
^
Adding the above fractions for all activities
We get the total =
_ 8+6+3+3+4
34886
24
= 1.
24 24
The sum of all fractions is equal to one. Here the time spent by a school
student during a day is represented using a circle and the whole area of the circle is
taken as one. The different activities of the school student are represented in various
sectors by calculating their proportion. This proportional part can also be calculated
using the measure of angle. Since, the sum of the measures of all angles at the central
point is 360°, we can represent each sector by using the measure of angle.
Data Handling
In the following example, we are going to illustrate how a pie chart can be
constructed by using the measure of angle.
Example 8.9
The number of hours spent by a school student on various activities on a working
day, is given below. Construct a pie chart using the angle measurment.
Activity
Sleep
School
Play
Homework
Others
Number of hours
8
6
3
3
4
Solution
Number of hours spent in different activities in a day of 24 hours are converted
into component parts of 360°. Since the duration of sleep is 8 hours, it should be
represented by ^ X 360° =120°.
Therefore, the sector of the circle representing sleep hours should have a central
angle of 120°.
Similarly, the sector representing other activities such as school, play, homework,
and others are calculated in the same manner in terms of degree, which is given in the
following table :
Activity
Duration in hours
Central angle
Sleep
8
^, X 360° - 120°
24
School
6
6 X 360° - 90°
Play
3
^X 360° = 45°
Homework
3
1^ X 360° - 45°
Others
4
^^ X 360° - 60°
Total
24
360°
Drawing pie chart
Now we draw a circle of any convenient radius. In that circle, start with any
radius and measure 120° at the centre of the circle and mark the second arm of this
angle. This sector represents the hours spent in sleeping.
Chapter 8
From this arm now mark
off a second sector, by measuring
an angle of 90° in the same sense
as before. This sector represents
the school hours. Proceeding
like this, we construct the sectors
for play and home work. The
remaining sector will represent
the last class (i.e. others).
The sectors may be
shaded or coloured differently
to distinguish one from the
other. The completed pie chart is
represented in the figure as shown above
The time spent by a school student
during a day (24 hours)
I I Sleep
I I School
□ Play
^1 Homework
I I Others
Fig. 8.9
Note: In a pie chart, the various observations or components are represented by
the sectors of a circle and the whole circle represents the sum of the value of all the
components .Clearly, the total angle of 360° at the centre of the circle is divided
according to the values of the components .
Value of the component
The central angle of a component
Total value
X 360'
Sometimes, the value of the components are expressed in percentages. In such
cases, the central angle of a component =
Percentage value of the component
100
X360"
Steps for construction of pie chart of a given data:
1. Calculate the central angle for each component , using the above formula.
2. Draw a circle of convenient radius.
3. Within this circle, draw a horizontal radius.
4. Draw radius making central angle of first component with horizontal
radius; this sector represents the first component. From this radius, draw
next radius with central angle of second component; this sector represents
second component and so on, until we exhaust all components.
5. Shade each sector differently and mark the component it represents.
6. Give a key.
7. Give the heading.
Thus, we obtain the required pie chart for the given data.
Data Handling
Example 8.10
The following table shows the monthly budget of a family
Particulars
Food
House
Rent
Clothing
Education
Savings
Miscella-
neous
Expenses
(in?)
4800
2400
1600
800
1000
1400
Draw a pie chart using the angle measurement.
Solution
The central angle for various components may be calculated as follows:
Particulars
Expenses (in Rs)
Central angle
Food
4800
^^^^ X 360° - 144°
12000
House rent
2400
2400 y -3^0°- 7^°
12000 ^ ^^^ '^
Clothing
1600
1600 y -5^0°- 4R°
12000 ^ ^^^ ^^
Education
800
12000^^^^ ^^
Savings
1000
1000 y '\f.r\° - rip.o
12000 >^^^" ^^
Miscellaneous
1400
^'^^^ X360° -42°
12000
Total
12000
360°
Clearly, we obtain the required pie chart as shown below.
Monthly Budget of a Family
1cEY
r
Food
House Rent
Clothing
Education
Savings
Miscellaneous
Fig. 8.10
Chapter 8
Example 8.11
The S.S.L.C Pubhc Examination resuh of a school is as follows:
Result
Passed in
first class
Passed in
second class
Passed in
third class
Failed
Percentage of
students
25%
35%
30%
10%
Draw a pie chart to represent the above information.
Solution
„ ^ 1 , r , Percentage value of the component ^^„o
Central angle for a component = - ^no X 360
We may calculate the central angles for various components as follows:
Result
Percentage of students
Central angle
Passed in first class
25%
^X360° = 90°
Passed in second class
35%
^^^ X 360° - 126°
Passed in third class
30%
^X 360° = 108°
Failed
10%
^^^ X 360° - 36°
Total
100%
360°
Clearly, we obtain the required pie chart as shown below:
SSLC Public Examination Results
r I Passed in First Class
I I Passed in Second Class
^1 Passed in Third Class
I I Failed
Fig. 8.11
Data Handling
EXERCISE 8.2
1. Yugendran's progress report card shows his marks as follows:
Subject
Tamil
English
Mathematics
Science
Social science
Marks
72
60
84
70
74
Draw a pie chart exhibiting his mark in various subjects.
2. There are 36 students in class VIII. They are members of different clubs:
Clubs
Mathematics
N.C.C
J.R.C
Scout
Number of students
12
6
10
8
Represent the data by means of a pie chart.
3. The number of students in a hostel speaking different languages is given below:
Languages
Tamil
Telugu
Malayalam
Kannada
English
Others
Number of students
36
12
9
6
5
4
Represent the data in a pie chart.
4. In a school, the number of students interested in taking part in various hobbies
from class VIII is given below. Draw a pie chart.
Hobby
Music
Pottery
Dance
Drama
Social service
Number of students
20
25
27
28
20
5. A metal alloy contains the following metals. Represent the data by a pie chart.
Metal
Gold
Lead
Silver
Copper
Zinc
Weights (gm)
60
100
80
150
60
6. On a particular day, the sales (in ? ) of different items of a baker's shop are given
below. Draw a pie chart for this data.
Item
Ordinary Bread
Fruit Bread
Cakes
Biscuits
Others
Cost (?)
320
80
160
120
40
7. The money spent on a book by a publisher is given below:
Item
Paper
Printing
Binding
Publicity
Royalty
Money spent (?)
25
12
6
9
8
Represent the above data by a pie chart.
8. Expenditure of a farmer for cultivation is given as follows:
Item
Ploughing
Fertilizer
Seeds
Pesticides
Irrigation
Amount (?)
2000
1600
1500
1000
1100
Represent the data in a pie chart.
Chapter 8
9. There are 900 creatures in a zoo as per list below:
Creatures
Wild animal
Birds
Other land animals
Water animals
Reptiles
Number of
Creatures
400
120
135
170
75
Represent the above data by a pie chart.
10. In a factory, five varieties of vehicles were manufactured in a year, whose break
up is given below. Represent this data with the help of a pie chart.
Vehicle
Scooter
Motorbike
Car
Jeep
Van
Number
3000
4000
1500
1000
500
11. A food contains the following nutrients. Draw a pie chart representing the data.
Nutrients
Protein
Fat
Carbohydrates
Vitamins
Minerals
Percentage
30%
10%
40%
15%
5%
12. The favorite flavours of ice cream for students of a school is given in
percentages as follows
Flavours
Chocolate
Vanilla
Strawberry
Other flavours
Percentage of Students
preferring the flavours
40%
30%
20%
10%
Represent this data by a pie chart.
13. The data on modes of transport used by students to come to school are given
below:
Mode of transport
Bus
Cycle
Walking
Scooter
Car
Percentage of students
40%
30%
15%
10%
5%
Represent the data with the help of a pie chart.
14. Mr. Rajan Babu spends 20% of his income on house rent, 30% on food and 10%
for his children's education. He saves 25%, while the remaining is used for other
expenses. Make a pie chart exhibiting the above information.
15. The percentage of various categories of workers in a state are given in following
table
Category of
workers
Cultivators
Agricultural
labours
Industrial
Workers
Commercial
Workers
Others
Percentage
40%
25%
12.5%
10%
12.5%
Represent the information given above in the form of a pie chart.
Data Handling
8.5 Measures of Central Tendency
Even after tabulating the collected mass of data, we get only a hazy general
picture of the distribution. To obtain a more clear picture, it would be ideal if we can
describe the whole mass of data by a single number or representative number. To get
more information about the tendency of the data to deviate about a particular value,
there are certain measures which characterise the entire data. These measures are
called the Measures of Central Tendency. Such measures are
(i) Arithmetic Mean, (ii) Median and (iii) Mode
8.5.1 Arithmetic Mean (A.M)
The arithmetic mean is the ratio of the sum of all the observations to the total
number of observations.
8.5.1. (a) Arithmetic mean for ungrouped data
If there are n observations x^, x^, x^,
mean is denoted by x and it is given by x
If there are n observations x, x, x^,---,x for the variable x then their arithmetic
V 2' 3' ' II
■Aj I ^-j I .At I • • • I ^
n
In Mathematics, the symbol
in Greek letter 2, is called Sigma.
This notation is used to represent the
summation. With this symbol, the sum of
Xj, X-,, X3, • ■ ■ ,x„ is denoted as^ x. or simply
1=1
as Zxi. Then we have x = ^^^.
n
Note: Arithmetic mean is also known
as Average or Mean.
More about Notation : Z
3
1+2+3=6
k= 1
6
3 + 4 + 5 + 6 = 18
« = 3
4
Y2n =
2x2 + 2x3 + 2x4 = 18
n = l
3
k= 1
X5xk°
k= 1
=
5 X 1" + 5 X 2° + 5 X 3°
=
5 + 5 + 5 = 15
Z(^-i) =
K = 2
(2-l) + (3-l) + (4-l) = 6
Example 8.12
The marks obtained by 10 students in a test are 15, 75, 33, 67, 76, 54, 39, 12, 78,
11. Find the arithmetic mean.
Solution
Here, the number of observations, n = 10
ji^^M = X = 15 + 75 + 33 + 67 + 76 + 54 + 39+12 + 78 + 11
^ = 460 =46 ^°
X ^^ 4b.
Chapter 8
Example 8.13
If the average of the values 9, 6, 7, 8, 5 and x is 8. Find the value of x.
Solution
Here, the given values are 9, 6, 7, 8, 5 and x, also n = 6.
By formula, A.M. = x
9 + 6 + 7 + 8 + 5+x _ 35+;c
By data, x
SO, g
i.e. 35 + X
6 6
8
8
48
X = 48 - 35 = 13.
Example 8.14
The average height of 10 students in a class was calculated as 166 cm. On
verification it was found that one reading was wrongly recorded as 160 cm instead of
150 cm. Find the correct mean height.
Solution
Here, x = 166 cm and n= 10
We have x ^ ^^ - Zx
n
Hx
10
10
i.e. 166 = ^ or Ex = 1660
The incorrect Ex = 1660
The correct Ex = incorrect Ex - the wrong value + correct value
= 1660 - 160 + 150 = 1650
Hence, the correct A.M.
1650
10
165 cm.
8.5.1 (b) Arithmetic mean for grouped data
Arithmetic mean for grouped data can be obtained in two methods which are
(i) Direct Method and (ii) Assumed Mean Method
(i) To calculate Arithmetic Mean (Direct Method)
Suppose we have the following frequency distribution
Variable
^,
^2
^3
X
n
Frequency
h
U
u
fn
Then this table is to be interpreted in the following manner:
The value : xi occurs / times
x^ occurs f^ times
Xg occurs f^ times
X occurs f times.
Data Handling
Here x,, x^, x„ • • • ,x„ are the distinct values of the variable x.
In this case, the total number of observations is usually denoted by N.
(i.e.,)/+^+^+-+X=N (or) ± f, = N
i= 1
Then the total values observed
= (x, + Xj + Xj + • • ■/'times) + (x^ + x^ + x^ + • • -/times) + • • • + (x_ + x, + x^^+ ■• -/times)!
= f,Xx,+f,Xx,+ ---+f„Xx„ = Zfx,
Total values observed _ ^fi^i
Hence, x
Usually, it is written as x
Total number of observations
■> - ■> where N = 2/
n
Z/ N '
Example 8.15
Calculate the Arithmetic mean of the following data by direct method
Solution
X
5
10
15
20
25
30
f
4
5
7
4
3
2
X
/
f^
5
4
20
10
5
50
15
7
105
20
4
80
25
3
75
30
2
60
Total
N = 25
Zfx = 390
Arithmetic Mean, x =
N
390
25
= 15.6 .
(11) To calculate Arithmetic Mean (Assumed Mean Method)
In the above example multiplication looks very simple, since the numbers
are small . When the numbers are huge, their multiplications are always tedious or
uninteresting and leads to errors.
Chapter 8
To overcome this difficulty another simpler method is devised. In this method
we assume one of the values as mean (A). This assumed value A is known as assumed
mean. Then we calculate the deviation d^,d^,d^,---,d^^ of each of the variables
x^, x,_, Xj, • • • ,x„ from the assumed mean A.
Now, multiply d^,d^ ,d^,---,d^^ respectively by f^J^J^'^Jn ^^^d add all these
values to get Z/J. Now,
Arithmetic mean
X
X
= A +
Zfd
A + -^— (Here A is assumed mean and N = Z /)
Now, we can calculate the A.M. for the above problem (example 8.15) by
assumed mean method.
Take the assumed mean A = 15
X
f
d = x-A
fd
5
4
-10
-40
10
5
-5
-25
15
7
20
4
5
20
25
3
10
30
30
2
15
30
Total
N = 25
Zfd = 15
Arithmetic Mean =
X
A +
N
= 15.6.
78
5
8.5.2 Weighted Arithmetic Mean (W.A.M.)
Sometimes the variables are associated with various weights and in
those cases the A.M. can be calculated, such an arithmetic mean is known as
Weighted Arithmetic Mean (W.A.M.).
For example, let us assume that the variable xi is associated with the weight
w, , X2 is associated with the weight Wi etc. and finally, x„ is associated with the weight
w„ then
W.A.M. =
W^X^ + VV^X, + W^X^ + ••• + W,,X„
w + W^ + w + ■ • ■ + w
Zwx
.w
Data Handling
Example 8.16
Find tlie weighted A. M of the price for the following data:
Food stuff
Quantity (in kg) w.
Price per kg (in ^ ) x.
Rice
25
30
Sugar
12
30
Oil
8
70
Solution
Here the x-values are the price of the given food stuff and the weights associated
are the quantities (in Kg)
Then, the W.A.M =
WjTj + W^X^ + W3X3 +
+ W X„
w, + w, + w, + • • • + w„
= 25x30 + 12x30 + 8x70 = 1670
25+12 + 8 45
= ?37.U.
8.5.3 Median
Another measure of central tendency is the Median.
8.5.3 (a) To find Median for ungrouped data
The median is calculated as follows:
(i) Suppose there are an odd number of observations, write them in ascending
or descending order. Then the middle term is the Median.
For example: Consider the five observations 33, 35, 39, 40, 43. The middle
most value of these observation is 39. It is the Median of these observation,
(ii) Suppose there are an even number of observations, write them in
ascending or descending order. Then the average of the two middle terms
is the Median.
For example, the median of 33, 35, 39, 40, 43, 48 is 39±40 = 39 5
Note: The Median is that value of the variable which is such that there are as many |
observations above and below it.
Example 8.17
Find the median of 17, 15, 9, 13, 21, 7, 32.
Solution
Arrange the values in the ascending order as 7, 9, 13, 15, 17, 21,32,
Here, n = 7 (odd number)
Therefore, Median = Middle value
_ / n + 1 \""
Hence, the median is 15.
( " + M " value = ( ^2^ )'" value = 4th value.
Chapter 8
Example 8.18
A cricket player has taken the runs 13, 28, 61, 70, 4, 11, 33, 0, 71, 92. Find the
median.
Solution
Arrange the runs in ascending order as 0, 4, 11, 13, 28, 33, 61, 70, 71, 92.
Here n = 10 (even number).
There are tw^o middle values 28 and 33.
.•. Median = Average of the two middle values
_ 28 + 33 _ 61 _
30.5
8.5.3 (b) To find Median for grouped data
Cumulative frequency
Cumulative frequency of a class is nothing but the total frequency upto that
class.
Example 8.19
Find the median for marks of 50 students
Marks
20
27
34
43
58
65
89
Number of students
2
4
6
11
12
8
7
Solution
Marks (x)
Number of students (f)
Cumulative
frequency
20
2
2
27
4
(2 + 4 = ) 6
34
6
(6 + 6 = ) 12
43
11
(11 + 12 = ) 23
58 12 (23 + 12 = ) 35
65
8
(35 + 8 = ) 43
89
7
(43 + 7 =) 50
Here, the total frequency, N = Z/ = 50
N _ 50 _
25.
The median is (^) value = 25th value.
Now, 25th value occurs in the cummulative frequency 35, whose
corresponding marks is 58.
Hence, the median = 58.
Data Handling
8.5.4 Mode
Mode is also a measure of central tendency.
The Mode can be calculated as follows:
8.5.4 (a) To find Mode for ungrouped data (Discrete data)
If a set of individual observations are given, then the Mode is the value which
occurs most often.
Example 8.20
Find the mode of 2, 4, 5, 2, 1, 2, 3, 4, 4, 6, 2.
Solution
In the above example the number 2 occurs maximum number of times.
ie, 4 times. Hence mode = 2.
Example 8.21
Find the mode of 22, 25, 21, 22, 29, 25, 34, 37, 30, 22, 29, 25.
Solution
Here 22 occurs 3 times and 25 also occurs 3 times
.-. Both 22 and 25 are the modes for this data. We observe that there are two
modes for the given data.
Example 8.22
Find the mode of 15, 25, 35, 45, 55, 65,
Solution
Each value occurs exactly one time in the series. Hence there is no mode for
this data.
8.5.4 (b) To find Mode for grouped data (Frequency distribution)
If the data are arranged in the form of a frequency table, the class corresponding
to the maximum frequency is called the modal class. The value of the variate of the
modal class is the mode.
Example: 8.23
Find the mode for the following frequency table
Wages (Rs)
250
300
350
400
450
500
Number of workers
10
15
16
12
11
13
Chapter 8
Solution
Wages (?)
Number of workers
250
10
300
15
^^^^■^^H^^^B
400
12
450
11
500
13
We observe from the above table that the maximum frequency is 16. The value
of the variate (wage) corresponding to the maximum frequency 16 is 350. This is
the mode of the given data.
p
r
Unimodal
Bimodal
Trimodal
Muhimodal
If there is
If there are
If there are
If there are
only one mode in
two modes in a
three modes in a
more than three
a given series, then
given series, then it
given series, then
modes in the
it is called
is called
it is called
series it is called
Unimodal.
Bimodal.
Trimodal.
Multimodal.
Example :
Example:
Example:
Example:
10, 15, 20, 25, 15,
20, 25, 30, 30,
60, 40, 85, 30, 85,
1, 2, 3, 8, 5, 4, 5,
18, 12, 15.
15, 10, 25.
45, 80, 80, 55, 50,
J, 4, Z, J, 1, J, D,
Here, Mode is 15.
Here 25, 30 are
60. Here 60, 80,
2, 7, 4, 1. Here
Bimodal.
85 are Trimodal.
1, 2, 3, 4, 5 are
Multimodal.
EXERCISE 8.3
I. Prohlems on Arithmetic Mean
1. Find the mean of 2, 4, 6, 8, 10 , 12, 14, 16.
2. If the average of the values 18, 41, x , 36, 31, 24, 37, 35, 27, 36, is 31. Find the
value of X.
3. If in a class of 20 students, 5 students have scored 76 marks, 7 students have
scored 77 marks and 8 students have scored 78 marks, then compute the mean
of the class.
4. The average height of 20 students in a class was calculated as 160 cm. On
verification it was found that one reading was wrongly recorded as 132 cm
instead of 152 cm. Find the correct mean height.
Data Handling
5. Calculate the Arithmetic mean of the following data:
II.
1.
X
15
25
35
45
55
65
75
85
f
12
20
15
14
16
11
7
8
The following data give the number of boys of a particular age in a class of 40
students. Calculate the mean age of the students
Age (in years)
13
14
15
16
17
18
Number of students
3
8
9
11
6
3
7. Obtain the A.M of the foUowing data:
Marks
65
70
75
80
85
90
95
100
Number of students
6
11
3
5
4
7
10
4
8. The following table shows the weights of 12 workers in a factory
Weight (in Kg)
60
64
68
70
72
Number of workers
3
4
2
2
1
Find the mean weight of the workers.
For a month, a family requires the commodities listed in the table below. The
weights to each commodity is given. Find the Weighted Arithmetic Mean.
Commodity
Weights (in kg)
Price per kg (in ?)
Rice
25
30
Wheet
5
20
Pulses
4
60
Vegetables
8
25
Oil
3
65
10. Find the Weighted Arithmetic Mean for the following data:
Item
Number of Item
Cost of Item
Powder
2
?45
Soap
4
f 12
Pen
5
?15
Instruments box
4
? 25.50
Problems on Median
Find the median of the following set of values:
(i) 83, 66, 86, 30, 81.
(ii) 45, 49, 46, 44, 38, 37, 55, 51.
(iii) 70, 71, 70, 68, 67, 69, 70.
(iv) 51, 55, 46, 47, 53, 55, 51, 46.
Chapter 8
2. Find the median for the following data:
X
1
2
3
4
5
6
7
8
f
9
11
5
6
8
1
3
7
3. The height ( in cm) of 50 students in a particular class are given below:
Height (in cm)
156
155
154
153
152
151
150
Number of smdents
8
4
6
10
12
3
7
Find the median.
4. The hearts of 60 patients were examined through X-ray and the observations
obtained are given below:
Diameter of heart (in mm)
130
131
132
133
134
135
Number of patients
7
9
15
12
6
11
Find the median.
5. The salary of 43 employees are given in the following table. Find the median.
Salary (in ^)
4000
5500
6000
8250
10,000
17,000
25,000
Number of employees
7
5
4
3
13
8
3
III. Problems on Mode
1. Find the mode of the following data:
i) 74, 81, 62, 58, 77, 74. iii) 43, 36, 27, 25, 36, 66, 20, 25.
ii) 55, 51, 62, 71, 50, 32. vi) 24, 20, 27, 32, 20, 28, 20.
2. Find the mode for the following frequency table:
X
5
10
15
20
25
30
f
14
25
37
16
8
5
3. Find the mode for the following table:
Temperature in °c
29
32.4
34.6
36.9
38.7
40
Number of days
7
2
6
4
8
3
4. The demand of different shirt sizes, as obtained by a survey, is given below.
Size
38
39
40
41
42
43
44
Number of persons (wearing it)
27
40
51
16
14
8
6
Calculate the Mode.
IV. Problems on Mean, Median and Mode
1. Find the mean, median and mode for the following frequency table:
X
10
20
25
30
37
55
f
5
12
14
15
10
4
Data Handling
2. The age of the employees of a company are given below.
Age (in years)
Number of persons
19
13
21
15
23
20
25
18
27
16
29
17
31
13
Find the mean, median and mode.
The following table shows the weights of 20 students.
Weight (in kg)
47
50
53
56
60
Number of students
4
3
7
2
4
Calculate the mean, median and mode.
Concept Summary
Histogram and frequency polygon are the two types of graphical
representations of a frequency distribution.
In the graphical representation of histogram and frequency polygon the
class-intervals are taken along the X-axis and the corresponding frequencies
are taken along the Y-axis.
In a histogram the rectangles are placed continuously side by side with no
gap between the adjacent rectangles.
The polygon obtained by joining the mid points of the top of the adjacent
rectangles of a histogram and also the midpoints of the preceding and
succeeding class intervals is known as a frequency polygon.
Value of the component
X360°
The central angle of a component
Total value
Arthmetic mean is the ratio of the sum of all the observations to the total
number of observations.
Formula for finding A.M.
(hi)
X -
,...-_ Z fx
X
= A
2/
when A is the assumed mean and d — x — A.
^ The weighted Arithmetic mean (W.A.M.)
2 w-
'V The median is that value of the variable which is such that there are as
many observations above and below it.
^ Mode is that value which occurs most frequently in a distribution.
Answers
(
ANSWERS )
Chapter 1
Exercise 1.1
1. i) A
ii)
C
iii)
B iv) D
V)
A
2. i) Commutative
ii)
Associative
iii)
Commutative
iv) Additive
identity
V)
Additive inverse
3. i) Commutative
ii)
Multiplicative identity
iii) Multiplicative Inverse
iv)
Associative
v) Distributive property
of multiplication over addition
fi i^ -505
^- '^ 252
ii)
-1
14
Exercise 1.2
1. i)§
ii)
23
84
-)|^
iv) 53
^ 24
2 i) 31 51
^ 70' 14C
)
11)111
MIO
243
220
iii) 11
9
20
iv)-l 1
''24' 12
3 i) 3 5
9
32
ii)
41 83 167
60' 120' 240
m) i'^2' 8'
-5
48
^^^48'
11
96'
23
192
Note: In the above
probelms 1, 2 and 3; the given answes are one of the possibilities.
Exercise 1.3
1. i) A
B
iii)
C iv) A
V)
B
2. i)2j^
16
17
iii)
11 iv) 1 7
32 ^ 18
V)
-8
19
vi) 4 23
vii)
4
viii)
5^1
60
Exercise 1.4
1. i) C
B
iii)
A iv) D
V)
C
vi) A
vii)
B
viii)
B ix) B
X)
D
2- ^) 'el
1
64
111)625 iv) ^
V)
1
322
vi)54
vii)
1
viii) 256 p" ix) 231
x)5^ 1
Answers
3. i)
5
ii)
1
2
HI) 29
iv)
1
V)
5 16 ^'^ /21
4. i)
m = 2
ii)
m = 3
HI) m = 3
iv)
m = 3
V)
m = - 6 vl) m = —
^ 4
5. a)
1)4
ii)
4
HI) 256
iv)
64
V)
1
4
5. b)
1)4
ii)
2187
HI) 9
iv)
6561
V)
1
9
Exercise 1.5
1.
(11), (ill
), (v) are not perfect squares.
2. i)
4
ii)
9
111) 1
iv)
5
V)
4
3. i)
64
ii)
16
111) 81
4. i)
1 + 3 +
5 + 7-
f 9+11
+ 13 H) 1 + 3
+5+7+9+
11 + 13 + 15 + 17
ill)
1 + 3 +
5 + 7-
f 9
Iv) 1 + 3
+5+7+9+
11
+ 13 + 15 + 17 + 19 + 21
5. i)
9
64
ii)
49
100
"» 25
iv)
4
9
V)
961
1600
6. i)
9
ii)
49
111) 0.09
iv)
4
9
V)
16 "' °-36
7. a)
42 + 52
+ 202 =
= 212
b) 10000200001
52 + 52
+ 302 =
= 312
100000020000001
52 + 72
+ 422 =
= 432
Exercise 1.6
1. i)
12
ii)
10
111) 27
iv)
385
2. i)
3
8
ii)
1
4
111) 7
iv)
4
3. i)
48
ii)
67
111) 59
iv)
23
V)
57
vi)
37
vU)
76
vlll) 89
Ix)
24
X)
56
4. i)
27
ii)
20
111) 42
iv)
64
V)
88
vi)
98
vi)
77
vlll) 96
ix)
23
X)
90
5. i)
1.6
ii)
2.7
111) 7.2
iv)
6.5
V)
5.6
vi)
0.54
vU)
3.4
vlll) 0.043
6. i)
2
ii)
53
111) 1
iv)
41
V)
31
7. i)
4
ii)
14
111) 4
iv)
24
V)
149
Answers
8. i)
1.41
ii)
2.24
iii)
0.13
iv)
0.94
V)
1.04
9.
21m
10.
i)15
^ 56
ii)
46
59
iii)
23
42
iv)
ll3
76
Exercise 1.7
1. i)
A
C
iii)
B
iv)
A
V)
B
vi)
D
vii)
A
viii)
A
ix)
A
X)
D
2. ii)
216
iii)
729
V)
1000
3. i)
128
100
V)
72
vi)
625
4. i)
3
2
iii)
5
iv)
3
V)
11 vi) 5
5. i)
3
2
iii)
3
iv)
5
V)
10
6. i)
9
7
iii)
8
iv)
0.4
V)
0.6
vi) 1.75
7. 2.7 cm
1.1 viii) -30
Exercise 1.8
1. i) 12.57
iv) 56.60 m
2. i) 0.052 m
iv) 0.133 gm
3. i) 250 ii) 150
ii) 25.42 kg
v) 41.06 m
ii) 3.533 km
v) 365.301
iii) 39.93 m
vi) 729.94 km
iii) 58.294/
vi) 100.123
iii) 6800 iv) 10,000
v) 36 lakhs vi) 104 crores
4. i) 22 ii) 777 iii) 402 iv) 306 v) 300 vi) 10,000
Exercise 1.9
1. i) 25, 20, 15
iv) 7.7, 8.8, 9.9
vu) 125, 216, 343
ii) 6, 8, 10
v) 15,21,28
iii) 63, 56, 49
vi) 34, 55, 89
Answers
2. a) 11 jumps b) 5 jumps
3. a) 10 rows of apples = 55 apples
b) 210 apples
Rows
1
2
3
4
5
6
7
8
9
I
Total
apples
1
3
6
10
15
21
28
36
45
1
Chapter
2
xercise 2.1
1. i) C
ii) B
vi) D
vii) C
2.
iii) A iv) A v) D
SI. No.
Terms
Coeffficients of variables
i)
3 abc
-5 ca
3
-5
ii)
1, X, /
constant term, 1, 1
iii)
3x2/
-3xyz
3
3
1
iv)
-7
2pq
^-
qp
constant term
2
-5
7
1
V)
X
2-
-y
2
- 0.3 xy
1
2
-1
2
-0.3
3. Monomials : 3x^
Binomials : 3x + 2, x^ - 7, a^b + b^c , 2/ + 2m.
Trinomials : x^ - 4x + 2, x^ + 3xy + /, s^ + 3sr - 2f
4. i) 5x2-x-2 ii) 2x^ + ^-2 iii) -3f-2t-3
iv) v) 2 (a^ + fa2 + c^ + ab + be + ca)
Answers
5. i) a ii) - 4x - 18y iii) Sab - 7bc + 13ca
iv) - x2 + 5x2 + 3x + 1 ^) 5^2^ _ g^^ _ 7^ + i2y + 25
6. i) 7, 5 ii) 13, - 1 iii) 7,-1 iv) 8, 1 v) 8, - 2
Exercise 2.2
1. i) 21 X
vii) xy^
2.
ii) - 21 xy iii) - 15a^b iv) - 20a-^ v) ^x^ vi) x^
viii) a'^b^d ix) x^y^z^ x) a^fa^c
First Monomial -*
Second Monomial i
2x
-3y
4x2
- 5xy
7x2y
- 6x2y2
2x
4x2
- 6xy
8x3
- 10x2y
14x3y
- 12xy
-3y
-6 xy
9/
- 12 x2y
15xy2
-21x2/
18x3/
4x2
Sx^
- 12 x^y
16 x^
- 20 xy
28xV
- 24 xy
- 5 xy
- 10 x^y
15 x/
- 20 xy
25 x2/
35 xy
30x3/
7xy
Mx^y
-21xy
28 xV
-35x3/
49 xV'
- 42 xy
- 6xY
- 12xy
18 xy
-24x^2
30 xy
-42X''/
36 xV'
3. i) 30a^ ii) 72 xyz iii) a^bV iv) - 72m^ v) xyz^
vi) F m^ n^ vii) - 30 p^q
4. i) 8a23 ii) - 2x3 _ 3;^ + 20 iii) 3x2 + 3^^^ _ 3^2 i^-, l2x2-x-6
iv) ^a^b^
' 4
5. i) 2a3 - 3a2fa - lab"- + 35^
iii) x^ + 2xy + y'^ -i}
6. i) 2 (x^ - 2xy + yz-xz-y^)
ii) 2x3 + ^2^ _ ^2 + 3^3
iv) a^ + 3a2fa + 3afa2 + £,3 ^-j ^3 _ ^3
ii) 17a2 + 14a5 - 21ac
Exercise 2.3
1. i) C ii) D
2. i) x^ + 6x + 9
iv) a^ _ 2 + J-
a
vii) 4/2 - 9m2
iii) B iv) D
ii) 4m2 + 12m + 9
v) 9x2-4
Vlll) T7^ - X2
io
v) A vi) B
iii) 4x2-20x + 25
vi) 25a2 - 30 ab + 952
ix) \-\ x)9991
X y
Answers
3. i)
x^ + llx + 28
ii)
25x2 + 35^ + 12
iii)
49x2 _ gy
iv)
64x2 _ 5g^ + iQ
V)
4m2 + 14 mn + 12n
Wi)
x2y2 _ Sxy + 6
vii)
a^^(' + y]a^ 1 viii)
\ xy 1 xy
A + 2x-2y -xy
4. i)
p^-2pq + q^
ii)
a2 - 10a + 25
iii)
9x2 + 30;^ + 25
iv)
25x2 - 40x + 16
V)
49x2 + A2xy + 3y^
vi)
100m2 - 180mn + 81n2
vii)
0.16a2-0.4ab + 0.25b^
viii)
x2 2+1
X
ix)
x2 "^y + y^
3 9
X)
0.08
5. i)
10609 ii) 2304
iii)
2916 iv) 8464
V)
996004 vi) 2491
vii)
9984 viii) 896
ix)
6399 x) 7.84
xi)
84 xii) 95.06
7.
ab = ^,a^ + b^ = 4^
8.
i) 80, 16, ii) 196, 196
9.
625
10.
x^ + (a + b + c)x^ + (ab
+ be + ca)x + abc.
Exercise 2.4
1. i)
C ii) D
iii)
A iv) C
V)
B
2. i)
3 (x - 15)
ii)
7{x-2y)
iii)
5 a (a + 7)
iv)
4y(5/-3)
V)
Sab (3a + 7)
vi)
pq(l-r)
vii)
9m (2m2 - 5n^) viii)
17 (/2 + 5m2)
ix)
3x2 ^2xy -4y + 5x^)
X)
2a^b (aW -7b + 2a)
3. i)
a (2fa + 3) + 2b (or) 25 (
a +
l) + 3a
vi)
(a + fa) (ax + by + c)
ii)
(3x-2)(2y-3)
vii)
(ax - fa) (x2 + 1)
iii)
(x+y)(3y + 2)
viii)
(x-y)(m-n)
iv)
(5b - x2) (3b - 1)
ix)
(2m2 + 3) (m - 1)
V)
(ax + y) {ax + fa)
X)
(a + Ufa) (a + 1)
4. i)
(a + 7)2 ii) (x-6)2
iii)
(2p + 5q)(2p-5q)
iv)
(5x-2y)2
V)
(13m + 25n)(13m-25n)
vi)
(-if
Answers
vii) (11a + 7by
viii)
3x (x + 5) (x - 5)
ix) (6 + 7x)(6-7x)
x) (1 - 3xy
5. i) (x + 3) (x + 4)
ii)
(p-2)(p-4)
iii) (m-7)(m + 3)
iv) (x - 9) (x - 5)
V)
(x - 18) (x - 6)
vi) (a + 12)(a + l)
vii) (x - 2) (x - 3)
viii)
(x-2y)ix-12y)
ix) (m - 24) (m +
3) X)
(x-22)(x-6)
Exercise 2.5
1. i) y- ii)
- 6y iii)
^a'b'd
iv) 7m - 6
v) yxy vi)
9P m' n^
2. i) 5/ -4y + 3
ii)
3x3 _ 5;^2 _ 7
iii) |-x2-2x+|
iv) X + y - 7
V)
8x3 _ 4y + 3^3_
3. i) (x + 5) ii)
(a + 12) iii)
(x-2) iv) (5m-
2n)
v) (2a + 3b) vi)
(a2 + b^) (a +
b)
Exercise 2.6
1. i) X = 6 ii)
y = -7 iii)
y = 4 iv) X = 12
V) y = -77
vi) X = - 6 vii)
X = 2 viii)
X = 12 ix) X = 6
X) m = |
2. i) 18 ii)
29, 30, 31
iii) /=19,
fa = 11 iv) 12,48
v) 12, 9 vi)
45, 27 vii)
4000 viii) ^
ix) Nandhini's present age is 15
years and Mary's present age is 45 years.
x) Rs.3,00,000
Chapter 3
Exercise 3.1
1. i) D ii)
C iii)
B iv) B
v) A
2. i) 200 litres ii)
20,000 km
iii) ? 1,550
iv) 50 minutes
V)
?50
3. ^ 40,000
4.
3750
Answers
5. i) 90% ii) 94%
iii) 98% iv) 88%
V)
95% vi) 93%
6. 5
7. ^9,000
8.
? 1,020
9. 180, 1320
10. 6kgs.
11. i) 26,100 ii) 5,220
12. 25%,? 8,600
13. She scored better in maths by 20% 14. ^6,250
15. 20%
Exercise 3.2
1. i) ? 7490 ii) ? 500
iii) ? 9,000 iv)? 2,246
v) ? 6,57,500
2. i) Profit ? 64, Profit % =
= 20%
ii) Profit ? 200, Profit %
= 8%
iii) Loss ? 19, Loss % =
5%
iv) S.P. = ^ 38, Loss % =
5%
v) S.P. = ? 5,500, Profit % = 10%.
3. i) ? 787.50
ii) ? 1,260
iii)
? 2,835
4. ? 1,200
5. 33^%
6.
25%
7. ^ 22,80,000
8. ? 34,40,000
9. 11 ^%
10. Overall gain ? 113.68
Exercise 3.3
1. i) A ii) D
iii) B iv) B
V)
C
2. ? 360
3. ? 8,000
4.
? 49,220
5. ? 18,433.40
6. ? 4,950
7.
? 13,000
8. 33%
9. ? 9,832.50
10.
20%
11. ? 1,310.40 per shirt
12. i) Amount of discount =
= ? 460; S.P. = ? 1,840
ii) Amount of discount =
= ? 35; Rate of discount = 25%
iii) M.P. = ? 20,000; Amount of discount = ? 4,000
iv) Rate of discount = 5%; Amount of discount = ^ 725
v) Amount of discount =
= ? 403; S.P. = ? 2,821
Answers
Exercise 3.4
1. i) A = ? 1,157.63, Interest = ? 157.63
ii) A = ? 4,840, Interest = ^ 840
iii) A = ^ 22,869, Interest = ? 4,869
2. ? 2,125
3. i) ? 88,200 ii) ? 4,410
4. A = ?27,783, C.I. =? 3,783
5. ? 9,826
6. C.I. =? 1,951 7. ? 20,000
8. ? 36,659.70
9. i) ? 92,400 ii) ? 92,610, Difference = ^ 210
10. ^6 11. ^25 12. ^2,000
13. Suja pays more interest of ? 924.10
14. P = ? 1,25,000
15. 2 years 16. 10% 17. 10%
Exercise 3.5
1. 2,205 2. ? 2,55,150
3. ? 46,000
4. 5,31,616.25 5. 5,415
6. ? 20,000
7. 10,000
Exercise 3.6
1. ^ 27,000 2. ? 86,250
3. ? 10,800
4. ? 200 5. 9%
6. ? 1,250
7. ? 19,404 8. E.M.I. = ? 875, Total Amount = ? 8,750
Exercise 3.7
1. 24 days 2. 10; 1250 3. 36 compositors
4. 15 Workers
5. 24 days 6. ? 192
Exercise 3.8
1. 3 days 2. 30 days 3. 2 days
4. 12 minutes
5. A = ? 360, B = ^ 240 6. 6 days
7. 1 hour
Answers
Chapter 4
Exercise 4.1
1. i) C ii) B
iii)
A iv) D v) A
vi) D vii) B
^iii)
C ix) A x) C
2. i) 180 cm, 1925 cm^
ii)
54 cm, 173.25 cm^
iii) 32.4 m, 62.37 m^
iv)
25.2 m, 37.73 m^
3. i) 7.2 cm, 3.08 cm^
ii)
144 cm, 1232 cm^
iii) 216 cm, 2772 cm^
iv)
288m, 4928 m^
4. i) 350 cm, 7546 cm^
ii)
250 cm, 3850 cm^
iii) 150 m, 1386 m^
iv)
100 m, 616 m^
5. 77 cm^, 38.5 cm^
6.
Rs.540
Exercise 4.2
1. i) 32 cm ii) 40 cm
iii)
32.6 cm
iv) 40 cm v) 98 cm
2. i) 124 cm^ ii) 25 m^
iii)
273 cm^ iv) 49.14 cm^ v) 10.40m2
3. i) 24 m^ ii) 284 cm^
iii)
308 cm^
iv) 10.5 cm^ v) 135.625
cm^
vi) 6.125 cm^
4. 770 cm2
5.
1286 m^ 6. 9384 m^
7. 9.71 cm^
8.
203 cm^ 9. 378 cm^
10. i) 15,100 m2,ii)55000C
)m2
Chapter 5
Revision Exercise
1. y° = 52° 2. x° = 40°
3.
ZA = 110° 4. x° = 40°
5. x° = 105° 6. i) Correspond
ing angle, ii) Alternate angle, iii) Coresponding ang
Answers
Exercise 5.1
^^
^^
^^
1. i) B
ii) A
iii)
A
iv) B
V)
A
2. x° = 65°
3. x°
= 42
° 5.
i)x°
= 58°,y° =
108° ii)
x° =
30°, y°
= 30°
iii) x° = 42°
, y° = 40°
6.
x° =
= 153°, y° =
132°, z°
= 53°
■
Exercise 5.2
l.i)C ii) C
iii)C
iv)
C
v)B
vi) A
vii)
B
2. x° = 66°, y° --
= 132° 3.
x°
= 70°
4.x° = 15°, y° =
55° 7.
x° =
30°,
y° = (
50°, z° = 60
3
Exercise 5.3
l.i)D ii) C
iii) A
iv)
B
2.The shortest side is BC
3.
QR
= 26 cm
4. It forms a right an
gled triangle
5.QR = 5 cm
6.
x =
9 m 7.Altitude x
= 5.
r^ cm
8.Yes
9. 2V
'51
ft
Exercise 5.4
1. i) D
ii) D
iii) C
2. radius = 5 cm.
Chapter 7
Exercise 7.1
2. i) Quadrant I
iv) Quadrant IV
vii) on the x-axis
x) Quadrant II
3.
ii) Quadrant II
v) on the y-axis
viii) Quadrant III
iii) Quadrant III
vi) on the x-axis
ix) Quadrant I
Point
Quadrants/Axes
Coordinates
A
On the y - axis
(0,4)
B
Quadrant II
(-3,2)
C
On the X-axis
(-5,0)
D
Quadrant III
(-4,-6)
E
On the y-axis
(0.-3)
F
Quadrant IV
(7,-1)
G
On the X-axis
(4,0)
H
Quadrant I
(6,3)
O
The origin
(0,0)
Answers
6. i) 40 cm^ ii) 56 cm^
iii) 36 cm^
iv)
49 cm^
v) 16 cm^ vi) 12 cm^
vii) 18 cm^
Exercise 7.2
5. 5 hours 6. ? 8,000
7.i) 26 cm ii) 30 cm iii) 24 cm
iv)
28 cm
Chapter 8
Exercise 8.3
I, Problems on Arithmetic Mean
1. 9 2. x = 25 3. 77.15
4.
161cm
5.
45
6. 15.45 7. 82.1 8. 65.33
9.
?33
10.
?21
II, Problems on Median
1. i) 81 ii) 45.5 iii) 70
iv)
51
2. 3 3. 153 4. 132
5.
? 10,000
III, Problems on Mode
1. i) 74 ii) No mode
iii)
25 and 36
iv)
20
2. 15 3. 38.7°C 4. 40
IV. 1. Mean 28; Median 25; Mode 30
2. Mean 25; Median 25; Mode 23
3. Mean 53.05; Median 53; Mode 53
Sequential Inputs of numbers with 8
1x8 + 1
=
9
12 X 8 + 2
=
98
123 X 8 + 3
=
987
1234 X 8 + 4
=
9876
12345 X 8 + 5
=
98765
123456 X 8 + 6
=
987654
1234567 X 8 + 7
=
9876543
12345678 x 8 + 8
=
98765432
123456789 x 8 + 9
=
987654321
Sequential 8's with 9
9x9+7 =
98 X 9 + 6 =
987 X 9 + 5 =
9876 X 9 + 4 =
98765 X 9 + 3 =
987654 X 9 + 2 =
9876543 x 9 + 1 =
98765432 x 9 + =
88
888
8888
88888
888888
8888888
88888888
888888888
Without 8
12345679 x 9
=
111111111
12345679 x 18
=
222222222
12345679 x 27
=
333333333
12345679 x 36
=
444444444
12345679 x 45
=
555555555
12345679 x 54
=
666666666
12345679 x 63
=
777777777
12345679 x 72
=
888888888
12345679 x 81
=
999999999
Numeric Palindrome with I's 1
1x1
=
1
11x11
=
121
111 X 111
=
12321
1111 X 1111
=
1234321
mil X mil
=
123454321
mm X mm
=
12345654321
1111111 X 1111111
=
1234567654321
11111111 X 11111111
=
123456787654321
111111111 X 111111111
=
12345678987654321
What sum will become 4913 in 1.5 years if the rate of interest is 12.5% compounded half yearly?Hence the sum will be Rs. 4096. Was this answer helpful?
What sum will amount to Rs 4913 in 3 years if rate of interest is 6.25% pa compounded annually?Given A = 4913 rupees, R = 6.25%, n = 3 years. 4096 = P. Therefore the sum = 4096 rupees. Hope this helps!
What sum will become Rs 9826 in 18 months if the rate of interest is 2 ½ per annum and interest is compounded half yearly?Hence, required sum = Rs 9466.54. Q.
What sum will amount to Rs 4590 at 12% pa simple interest in 3 years?Hence, ₨ 3375 will amount to ₨ 4590 in 3 years at 12 % interest per year. Q.
|
What sum will amount to Rs 4913 in 18 months if the rate of interest is 12.5% per annum compounded half yearly *?
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