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Chapter 1
ALGEBRA
1.1 Simple expressions with two variables
We have learnt about rectangle. Its area is l # b in which the letters 'l' and 'b'
are variables.
Variables follow the rules of four fundamental operations of numbers.
Let us now translate a few verbal phrases into expressions using variables.
Operation
Verbal phrase
Algebraic Expression
Sum of x and y
x + y
Subtraction
Difference between
a
- b ^if a > b h ^ or h
a and b
b - a (if b > a)
Multiplication
product of x and y
x # y
(or) xy
Division
p divided by q
p
p ' q ( or )
q
The following table will help us to learn some of the words (phrases) that can
be used to indicate mathematical operations:
Subtraction
Multiplication
Division
The sum of
the difference of
the product of
the quotient of
increased by
decreased by
multiplied by
divided by
plus
minus
times
the ratio of
subtracted from
more than
less than
Example 1.1
Write the algebraic expressions for the following:
1)
Twice the sum of m and n.
2)
3)
b decreased by twice a.
Nu mbers x and y both squared and added.
4)
Product of
p
and
q

2

Algebra
5)
6)
7)
8)
Two times the product of a and b divided by 5.
x more than two-third of y.
Half a number x decreased by 3.
Su m of numbers m and n decreased by their product .
9)
4 ti mes
x
less than sum of
y
and 6.
10)
Double the sum of one third of a and m.
11)
Quotient of
y
by 5 added to
x.
Solution:
1)
2 ^ m + n h
2)
b - 2a
3)
2
2
x
+
y
4)
7 +
pq
2
ab
2
5)
6)
y + x
5
3
x
7)
- 3
8)
(m + n) - mn
2
1
9) ^ y + 6 h - 4x
10)
2
`
a
+
m
j
3
y
11)
+ x
5
Express each of the following as an algebraic expression
(i)
a
times
b .
(ii) 5 multiplied by the sum of a
and
b .
(iii) Twice m decreased by n .
(iv) Four times x divided by y .
(v)
Five times p multiplied by 3 times q.
Exercise 1.1
1.
Choose the correct answer:
(i)
The sum of 5 times x, 3 times
y
and 7
(A)
5 ^ x + 3y
+
7
h
(B)
5x +
3y + 7
(C)
5x + 3 ^ y + 7 h
(D)
5x + 3 ^ 7y h
(ii)
One half of the sum of numbers a
and
b
1
1
1
1
(A)
^
a + b
h
(B)
a + b
(C)
^
a - b
h
(D)
+ a + b
2
2
2
2
(iii)
Three times the difference of x and y
(A)
3x - y
(B) 3 - x - y
(C)
xy - 3
(D) 3 ^ y - x h

3

Chapter 1
(iv)
2 less than the product of
y
and
z
(A) 2 - yz
(B) 2 + yz
(C) yz - 2
(D) 2y - z
(v)
Half of
p
added to the product of 6 and
q
p
6
q
1
1
(A)
+ 6 q
(B) p
+
(C)
^
p
+ 6
q
h
(D)
^ 6 p + q
h
2
2
2
2
2.
Write the algebraic expressions for the following using variables, constants and
arithmetic operations:
(i)
(ii)
(iii)
(iv)
Sum of x and twice y.
Subtraction of z from y.
Product of x and y increased by 4
The difference between 3 times x and 4 times y.
(v)
(vi)
(vii)
(viii)
(ix)
The sum of 10, x and y.
Product of p and q decreased by 5.
Product of numbers m and n subtracted from 12.
Sum of numbers a and b subtracted from their product.
Number 6 added to 3 times the product of numbers c and d.
(x)
Four times the product of x and y divided by 3.
1.2 Simple Linear Equations
Malar’s uncle presented her a statue. She wants to
know the weight of that statue. She used a weighing balance to
measure its weight. She knows her weight is 40kg . She finds
that the weight of the statue and potatoes balance her weight.
i.e.,
Weight of
Weight of
Malar’s
Plus
Equal
statue
potatoes
weight
s
+
15
=
40
Table 1.1
Now we will think about a balance to find the value of s.

4

Algebra
Take away 15 from both sides.
Now the balance shows the weight of the statue.
s + 15
s + 15 – 15
s
=
=
40 (f rom Table 1.1)
40 – 15 (Taking away 15 from both the sides)
=
25
So the statue weighs 25 kg.
The statement s +
15 = 40
is
an
equation. i.e., a statement in which two
mathematical expressions are equal is called an equation.
In a balance, if we take away some weight from one side, to balance it we must
take away the same weight from the other side also.
If we add some weight to one side of the balance, to balance it we must add the
same weight on the other side also.
Similarly, an equation is like a weighing balance having equal weights on
each side. In an equation there is always an equality sign. This equality sign shows
that value of the expression on the left hand side (LHS) is equal to the value of the
expression on the right hand side (RHS).
¬
Consider the equation x + 7 = 15
Here LHS
is x + 7
RHS is 15
We shall subtract 7 from both sides of the equation
x + 7 - 7 =
15 - 7
x = 8
(Subtracting 7 reduces the LHS to x)
(variable x is separated)

5

Chapter 1
¬
Consider the equation n - 3 = 10
LHS is n - 3
RHS is
10
Adding 3 to both sides, we get
n - 3 + 3
= 10 + 3
n = 13
(variable n is separated)
¬
Consider the equation 4m = 28
Divide both sides by 4
4
m
28
=
4
4
m = 7
y
¬
Consider the equation
= 6
2
Multiply both sides by 2
y
# 2 =
6 # 2
2
y = 12
So, if we add (or subtract) any number on one side of an equation, we have
to add (or subtract) the same number the other side of the equation also to keep
the equation balanced. Similarly, if we multiply (or divide) both sides by the same
non-zero number, the equation is balanced. Hence to solve an equation, one has to
perform the arithmetical operations according to the given equations to separate the
variable from the equation.
Example 1.2
Solve
3p + 4 =
25
Solution:
3p
+ 4 - 4 = 25 - 4
(Subtracting 4 from both sides of the equation)
3p = 21
3
p
21
=
(Dividing both sides by 3)
3
3
p = 7
Example 1.3
Solve 7m - 5 = 30
Solution:
7m - 5 + 5 = 30 + 5
(adding 5 on both sides)

6

Algebra
7m = 35
7
m
35
=
(Dividing both sides by 7)
7
7
m = 5
While solving equations, the commonly used operation is adding or subtracting
the same number on both sides of the equation. Instead of adding or subtracting a
number on both sides of the equation, we can transpose the number.
Transposing a number (i.e., changing the side of the number) is the same as
adding or subtracting the number from both sides. While transposing a number we
should change its sign. Let us see some examples of transposing.
Example 1.4
Solve 2a - 12 = 14
Solution:
on both sides
Transposing
2a - 12 = 14
2a - 12 = 14
2a - 12 + 12 = 14 + 12 (adding 12 on
both sides)
Transpose (– 12) from LHS to RHS
2a = 14 + 12 (on transposing
2a = 26
-
12 becomes + 12 )
2
a
26
=
(dividing both sides by 2)
2a = 26
2
2
2
a
26
=
(Dividing both sides by 2)
a = 13
2
2
a = 13
Example 1.5
Solve 5x + 3 = 18
Solution: Transposing +3 from LHS to RHS
5x
= 18 - 3
(on Transposing +3 becomes - 3 )
5x = 15
5
x
15
=
(Dividing both sides by 5)
5
5
x = 3

7

Chapter 1
Example 1.6
Solve 2 ^ x + 4 h = 12
Solution: Divide both sides by 2 to remove the brackets in the LHS.
2
^
x +
4
h
12
=
2
2
x + 4
= 6
x =
6 - 4
(transposing +4 to RHS)
x = 2
Example 1.7
Solve
- 3 ^ m - 2 h
= 18
Solution: Divide both sides by ^- 3 h to remove the brackets in the LHS.
-
3
^
m
-
2
h
18
=
- 3
- 3
m - 2
= - 6
m = - 6 + 2
(t ransposing - 2 to RHS)
m
= - 4
Example 1.8
Solve
(3x + 1 ) – 7 = 12
Solution:
(3x + 1) – 7
3x + 1 – 7
3x – 6
=
12
=
12
=
12
3x
=
12 + 6
3
x
18
=
3
3
x
=
6
Example 1.9
Solve 5x + 3 = 17 - 2x
Solution:
5x + 3 = 17 - 2x

8

Algebra
5x + 2x
= 17 - 3
(transposing + 3 to RHS and - 2x to LHS)
7x = 14
7
x
14
=
7
7
x = 2
Example 1.10
Sum of three consecutive integers is 45. Find the integers.
Solution: Let the first integer be x .
& second
integer
=
x + 1
Third integer
=
x + 1 + 1 = x +
2
Their sum
=
x + ^ x + 1 h + ^ x + 2 h = 45
3x + 3 = 45
3
x =
42
x = 14
Hence, the integers are x = 14
x + 1 = 15
x + 2 = 16
Example 1.11
A number when added to 60 gives 75. What is the number?
Solution: Let the number be x .
The equation is
60 + x
=
75
x
= 75 - 60
x
= 15
Example 1.12
20 less than a number is 80. What is the number?
Solution: Let the number be x .
The equation is
x - 20 = 80
x
= 80 + 20
x
= 100

9

Chapter 1
Example 1.13
1
of a number is 63. What is the number?
10
Solution: Let the number be x .
1
The equation is
^
x h =
63
10
1
(x)
# 10
=
63 10
#
10
x = 630
Example 1.14
A number divided by 4 and increased by 6 gives 10. Find the number.
Solution: Let the number be x .
x
The equation is
+ 6 = 10
4
x
=
10 6
-
4
x
= 4
4
x
# 4 = 4 # 4
4
\ the number is 16.
Example 1.15
Thendral’s age is 3 less than that of Revathi. If Thendral’s age is 18, what is
Revathi’ s age?
Solution: Let Revathi’s age be x
& Thendral’s age = x - 3
Given, Thendral’s age is 18 years
&
x - 3 = 18
x
=
18 + 3
x = 21
Hence Revathi’s age is 21 years.

10

Algebra
Exercise 1.2
1.
Choose the correct answer.
(i)
If
p + 3 = 9 , then p
is
(A)
12
(B) 6
(C) 3
(D) 27
(ii)
If 12 - x = 8 ,
then x
is
(A)
4
(B)
20
(C) - 4
(D) - 20
q
(iii)
If
= 7 , then
q
is
6
1
7
(A)
13
(B) 42
(C) 42
(D)
6
(iv)
If
7 ^ x - 9 h = 35 , then x
is
(A)
5
(B) - 4
(C) 14
(D) 37
(v)
Three times a number is 60. Then the number is
(A)
63
(B) 57
(C) 180
(D) 20
2.
Solve :
(i)
x -
5 = 7
(ii) a + 3 = 10
(iii)
4 + y = - 2
(iv)
b -
3 = - 5
(v)
- x = 5
(vi)
- x = - 7
(vii)
3 - x = 8
(viii)
14 - n = 10
(ix)
7 - m = - 4
(x)
20 - y = - 7
3.
Solve :
(i)
2x = 100
(ii)
3l = 42
(iii)
36 = 9x
(iv)
51 = 17a
(v)
5x = - 45
(vi)
5t = - 20
(vii)
- 7x
= 42
(viii) - 10m = - 30
(ix) - 2x = 1
(x)
- 3x = - 18
4.
Solve :
1
a
n
(i)
x
= 7
(ii)
= 5
(iii)
= - 8
2
6
3
p
- x
- m
(iv)
= 8
(v)
= 2
(vi)
= - 4
- 7
5
3
5.
Solve :
(i)
3x + 1 = 10
(ii) 11 + 2x = - 19
(iii)
4z - 3 = 17
(iv)
4a - 5 = - 41
(v)
3 ^ x + 2 h = 15
(vi)
- 4 ^2 - x h = 12
y
+ 3
x
(vii)
= 14
(viii)
+ 5 = 7
(ix)
6y = 21 - y
5
3
(x)
11m = 42 +
4m
(xi)
- 3x = - 5x + 22 (xii)
6m - 1 = 2m + 1
(xiii) 3x - 14 = x - 8
(xiv)
5x - 2x + 7 = x + 1 (xv) 5t - 3 = 3t - 5

11

Chapter 1
6.
The sum of two numbers is 33. If one number is 18, what is the other number?
7.
A number increased by 12 gives 25. Find the number.
8.
If 60 is subtracted from a number, the result is 48. Find the number.
9.
5 times a number is 60. Find the number.
10.
3 times a number decreased by 6 gives 18. Find the number.
11.
The sum of 2 consecutive integers is 75. Find the numbers.
12.
Ram’s father gave him `70. Now he has `130. How much money did Ram have
in the beginning?
13.
8 years ago, I was 27 years old. How old am I now?
Solve:
(i)
y + 18 = - 70
(ii)
- 300 + x = 100
t
(iii)
3 - 5 = - 6
(iv)
2x + 9 = 19
(v)
3x + 4 = 2x + 11
Fun game
Ram asked his friends Arun, Saranya and Ravi to think of a number and told
them to add 50 to it. Then he asked them to double it. Next he asked them to add 48 to
the answer. Then he told them to divide it by 2 and subtract the number that they had
thought of. Ram said that the number could now be 74 for all of them. Check it out if
Arun had thought of 16, Saranya had thought of 20 and Ravi had thought of 7.
Arun
Saranya
Ravi
Think of a number
x
16 20
7
x+50
Double it
2x +
100
2x +
148
Divide by 2
x + 74
Take away the number
74
you thought of

12

Algebra
1.
Algebra is a branch of
Mathematics that involves alphabet, numbers and
mathematical operations.
2.
A variable or a literal is a quantity which can take various numerical values.
3.
A quantity which has a fixed numerical value is a constant.
4.
An algebraic expression is a combination of variables and constants connected
by the arithmetic operations.
5.
Expressions are made up of terms.
6.
Terms having the same variable or product of variables with same powers are
called Like terms. Terms having different variable or product of variables with
different powers are called Unlike terms.
7.
The degree of an expression of one variable is the highest value of the exponent
of the variable. The degree of an expression of more than one variable is the
highest value of the sum of the exponents of the variables in different terms
8.
A statement in which t wo expressions are equal is called an equation.
9.
An equation remains the same if the LHS and RHS are interchanged.
10. The value of the variable for which the equation is satisfied is called the
solution of the equation.

13

Chapter 2
LIFE MATHEMATICS
2.1 Percent
In the banners put up in the shops what do you understand by 25%, 20% ?
Ramu’s mother refers to his report card to analyze his performance in
Mathematics in standard VI.
His marks in Maths as given in his report card are
17 / 25 , 36 / 50 , 75 /100 , 80 /100 , 22 / 25, 45 / 50
She is unable to find his best mark and his least mark by just looking at the
marks.
So, she converts all the given marks for a maximum of 100 (equivalent fractions
with denominator 100) as given below:
Unit Test
Monthly
Quarterly
Half - yearly
Unit Test
Monthly
1
Test 1
Exam
Exam
2
Test 2
68
72
75
80
88
90
100
100
100
100
100
100

14

Life Mathematics
Now, all his marks are out of 100. So, she is able to compare his marks easily
and is happy that Ramu has improved consistently in Mathematics in standard V I.
Now let us learn about these special fractions.
Try and help the duck to trace the path through the maze from ‘Start’ to ‘End’.
Is there more than one path?
No, there is only one path that can be traced from ‘Start’ to ‘End’.
Total number of the smallest squares
=
10 0
Nu mber of shaded squares
=
41
Nu mber of unshaded squares
=
59
Nu mber of squares traced by the path
=
__
Now, look at the table below and fill in the blanks:
Ratio
Fraction
Percent
41 out of
41
41
: 100
41%
100
100
59 out of
59
59
: 100
59%
Portion
100
100
out
Portion traced
of 100
: 100
%
by the path
100
The fraction with its denominator 100 is called a Percent.
• The word ‘Percent’ is derived from the Latin word ‘Percentum’, which means
‘per hundred’ or ‘hundredth’ or ‘out of 100’.
Percentage also means ‘percent’.
Sy mbol used for percent is %
• Any ratio x : y, where y = 100 is called ‘Percent’.

15

Chapter 2
To Express Percent in Different Forms:
Pictorial
Representation
Shaded portion represented in the form of :
Ratio
5 : 10 0
17
: 100
43
:
100
5
17
43
Fraction
100
100
100
Percent
5%
17%
43%
Exercise 2.1
1)
Write the following as a percent:
93
1
100
(i)
20: 100 (ii) 100
(iii) 11 divided by 100
(iv) 100
(v) 100
2)
Write the following percent as a ratio:
(i)
43%
(ii) 75%
(iii) 5%
(iv) 17
1
%
(v) 33 %
1
2
3
3)
Write the following percent as a fraction:
(i)
25%
(ii) 12
1
%
(iii) 33%
(iv) 70%
(v) 82%
2
Think!
Find the selling price in percentage when 25% discount is given, in the first shop.
What is the reduction in percent given in the second shop?
Which shop offers better price?

16

Life Mathematics
2.2 To Express a Fraction and a Decimal as a Percent
5
1
. 2
175
We know that 100
= 5%,
= 1.2%, 100
= 175%.
100
5
To convert 10
to a percent
5
represented pictorially can be converted to a percent as shown below:
10
5
50
10
100
Multiply the numerator and denominator by 10 to make the denominator 100
5 # 10
50
=
= 0%
5
10 10
#
100
5
This can also be done by multiplying 10 by 100%
5
`
# 100 %
j
= 5
0%
10
50% of the circle is shaded. 25% of the circle is shaded.
Try drawing circles with (i) 50%, (ii) 25% portion shaded in
different ways.
Do you know?
Less than 1 and more than 100 can also be represented as a percent.
1
%
120%
2

17

Chapter 2
(i) Fractions with denominators that can be converted to 100
Example 2.1
3
Express 5 as a percent
Solution:
1) 2 ×
____
= 100
2) 5 × 20 = _____
5
multiplied by 20 gives 100
3) 4 × 25 = _____
3
# 20
60
=
= 60 %
5
# 20
100
4) 10 ×
= 100
3
= 60 %
5) 1 ×
= 100
5
Example 2.2
1
Express 6 4
as a percent
Solution:
1
25
=
6 4
4
4
multiplied by 25 gives 100
25 # 25
625
=
= 625 %
4
# 25
100
(ii) Fractions with denominators that cannot be converted to 100
Example 2.3
Express 4 as a percent
7
Solution: Multiply by 100%
4
400
`
# 100 %
j
=
%
7
7
1
=
57 %
=
57.14%
7
Example 2.4
1
Express 3 as a percent
Solution: Multiply by 100%
1
100
`
# 100 %
j
= `
j
%
3
3
= 33 % (or)33.33%
1
3
Example 2.5
There are 250 students in a school. 55 students like basketball, 75 students
like football, 63 students like throw ball, while the remaining like cricket. What is
the percent of students who like (a) basket ball? (b) throw ball?

18

Life Mathematics
Solution:
Total number of students = 250
(a)
Nu mber of students who like basket ball = 55
55
55
out of 250 like basket ball which can be represented as 250
55
Percentage of students who like basket ball =
`
# 100 %
j
250
= 22%
(b)
Number of students who like throw ball = 63
63
63
out of 250 like throw ball and that can be represented as 250
63
Percentage of students who like throw ball =
`
# 100 %
j
250
126
=
%
= 25.2%
5
22% like basket ball, 25.2% like throw ball.
(iii) To convert decimals to percents
Example 2.6
Express 0.07 as a percent
Solution:
Multiply by 100%
(0.07 # 100)% = 7%
Aliter:
7
0.07
=
= 7
%
100
Example 2.7
Express 0.567 as a percent
Solution:
Multiply by 100%
(0.567 # 100)% = 56.7%
567 =
567
Al iter: 0.567 =
1000
10 100
#
56
.
7
=
= 56 . 7 %
100
Note: To convert a fraction or a
decimal to a percent, multiply by 100%.

19

Chapter 2
Think!
9
1.
of your blood is water. What % of your blood is not water.
10
2
2.
of your body weight is muscle. What % of body is muscle?
5
2 of your body weight is water. Is muscle weight plus water weight
3
more or less than 100 %? What does that tell about your muscles?
Exercise 2.2
1.
Choose the correct answer:
(i)
6.25 =
(A)
62.5%
(B) 6250%
(C) 625%
(D) 6.25%
(ii)
0.0003 =
(A)
3%
(B) 0.3%
(C) 0.03%
(D) 0.0003%
(iii)
5 =
20
1
(A)
25%
(B)
(C) 0.25%
(D) 5%
4 %
(iv)
The percent of 20 minutes to 1 hour is
(A)
33 1 3
(B) 33
(C)
33 2 3
(D) none of these
(v)
The percent of 50 paise to Re. 1 is
1
(A)
500
(B)
(C) 50
(D) 20
2
2.
Convert the given fractions to percents
20
9
2
5
i)
ii) 50
iii) 5 1
iv)
20
4
3
v) 11
3.
Convert the given decimals to percents
i)
0.36
ii) 0.03
iii) 0.071
iv) 3.05
v) 0.75
4.
In a class of 35 students, 7 students were absent on a particular day. What
percentage of the students were absent?
5.
Ram bought 36 mangoes. 5 mangoes were rotten. What is the percentage of the
mangoes that were rotten?
6.
In a class of 50, 23 were girls and the rest were boys. What is the percentage of
girls and the percentage of boys?
7.
Ravi got 66 marks out of 75 in Mathematics and 72 out of 80 in Science. In
which subject did he score more?
8.
Shyam’s monthly income is `12,000. He saves `1,200 Find the percent of his
savings and his expenditure.

20

Life Mathematics
2.3 To Express a Percent as a Fraction (or) a Decimal
i) A percent is a fraction with its denominator 100. While expressing it as a
fraction, reduce the fraction to its lowest term.
Example 2.8
Express 12% as a fraction.
Solution:
12
12%
=
(reduce the fraction to its lowest terms)
100
3
=
25
Example 2.9
Percents that have easy
Express 233
1
% as a fraction.
3
fractions
Solution:
1
50%
=
2
700
233 %
1
=
%
1
3
3
25%
=
4
700
7
=
=
1
1
33 3 %
=
3
# 100
3
3
=
1
2 3
Find more of this kind
Example 2.10
1
Express 4 % as a fraction
Solution:
1
1
1
% =
=
4
4 # 100
400
(ii) A percent is a fraction with its denominator 100. To convert this fraction to
a decimal, take the numerator and move the decimal point to its left by 2 digits.
Example 2.11
Express 15% as a decimal.
Solution:
15
15%
=
= 0 . 15
100
Example 2.12
Express 25.7% as a decimal.
Solution:
25
. 7
25.7%
=
100
=
0.257

21

Chapter 2
Math game - To make a triplet (3 Matching cards)
This game can be played by 2 or 3 people.
Write an equivalent ratio and decimal for each of the given percent in different
cards as shown.
5%
1 : 20
0.05
1
1
: 3
0.33
33 3
%
Make a deck of 48 cards (16 such sets of cards) - 3 cards to represent one
particular value - in the form of %, ratio and decimal.
Shuffle the cards and deal the entire deck to all the players.
Players have to pick out the three cards that represent the same value of percent,
ratio and decimal and place them face up on the table.
The remaining cards are held by the players and the game begins.
One player chooses a single unknown card from the player on his left. If this
card completes a triple (3 matching cards) the 3 cards are placed face up on the table.
If triplet cannot be made, the card is added to the player’s hand. Play proceeds to the
left.
Players take turns to choose the cards until all triplets have been made.
The player with the most number of triplets is the winner.
TO FIND T HE VALUES OF PERCENTS
Colour 50% of the circle green and 25% of the circle red.
50
1
50% =
=
of the circle is to be coloured green.
100
2
25
25
1
Similarly, 25% =
=
=
100
100
4
1
of the circle is to be coloured red.
4

22

Life Mathematics
1
1
Now, try colouring 2
of the square, green and 4
of the
square, red.
Do you think that the green coloured regions are equal in
both the figures?
No, 50% of the circle is not equal to 50% of the square.
Similarly the red coloured regions, 25% of the circle is not equal to 25% of
the square.
Now, let’s find the value of 50% of `100 and 50% of `10.
What is 50% of `100?
What is 50% of `10?
50
1
50
1
50%
=
=
50% =
=
100
2
100
2
1
1
1
1
So, 2
of 100 =
# 100 = 50
of 10
=
#
10
=
5
2
2
2
50% of `100 = `50
50% of `10 = `5
Example 2.13
Find the value of 20% of 1000 kg.
Solution:
20
20% of 1000
=
of 1000
100
20
=
# 1000
100
20% of 1000 kg
=
20 0 kg.
Example 2.14
1
Find the value of
%
of 200.
2
Solution:
1
2
=
of 200
100
1
=
# 200
2
# 100
1
# 200 1
=
200
1
% of 200 1
=
2

23

Chapter 2
Example 2.15
Find the value of 0.75% of 40 kg.
Solution:
0
. 75
0.75%
=
100
0
.
75
0.75% of 40
=
# 40
100
3
=
= 0 . 3
10
0.75% of 40kg
=
0.3kg.
Example 2.16
In a class of 70, 60% are boys. Find the number of boys and girls.
Solution:
Total number of students
=
70
Nu mber of boys
=
60% of 70
60
=
# 70
100
=
42
Nu mber of boys
=
42
Nu mber of girls
=
Total students – Number of boys
=
70 – 42
=
28
Nu mber of girls
=
28
Example 2.17
In 2010, the population of a town is 1,50,000. If it is increased by 10% in the
next year, find the population in 2011.
Solution:
Population in 2010
=
1,50,000
10
Increase in population
=
# 1 , 50 , 000
100
=
15,000
Population in 2011
=
150000 + 15000
=
1,65,000

24

Life Mathematics
Exercise 2.3
1.
Choose the correct answer:
(i)
The common fraction of 30 % is
1
7
3
3
(A)
(B)
(C)
(D) 10
10
10
100
1
(ii)
The common fraction of
%
is
2
1
1
(A)
(B)
(C)
(D) 100
2
200
100 200
(iii)
The decimal equivalent of 25% is
(A)
0.25
(B) 25
(C) 0.0025
(D) 2.5
(iv)
10% of `300 is
(A)
`10
(B) `20
(C) `30
(D) `300
(v)
5% of `150 is
(A)
`7
(B) `7.50
(C) `5
(D) `100
2.
Convert the given percents to fractions:
1
i)
9%
ii) 75%
iii)
%
iv) 2.5%
v) 66 %
2
3
4
3.
Convert the given percents to decimals:
i)
7%
ii) 64%
iii) 375%
iv) 0.03%
v) 0.5%
4.
Find the value of:
i)
75% of 24
iv) 72% of 150
ii) 33 3 % of `72
v) 7.5% of 50kg
1
iii) 45% of 80m
5.
Ram spent 25% of his income on rent. Find the amount spent on rent, if his
income is `25,000.
6.
A team played 25 matches in a season and won 36% of them. Find the number of
matches won by the team.
7.
The population of a village is 32,000. 40% of them are men. 25% of them are
women and the rest are children. Find the number of men and children.
8.
The value of an old car is `45,000. If the price decreases by 15%, find its new
price.
9.
The percentage of literacy in a village is 47%. Find the number of illiterates in the
village, if the population is 7,500.

25

Chapter 2
Think!
1)
Is it true?
20% of 25 is same as 25% of 20.
2)
The tax in a restaurant is 1.5% of your total bill.
a)
Write the tax % as a decimal.
b)
A family of 6 members paid a bill of ` 750. What is the tax for
their bill amount?
c)
W hat is the total amount that they should pay at the restaurant?
2.4 Profit and Loss
Ram & Co. makes a profit of `1,50,000 in 2008.
Ram & Co. makes a loss of `25,000 in 2009.
Is it possible for Ram & Co. to make a profit in the first year and a loss in the
subsequent year?
Different stages of a leather product - bag are shown below:
Factory
Wholesale Dealer
Retailer
Where are the bags produced?
Do the manufactures sell the products directly?
Whom does the products reach finally?
Raja, the fruit stall owner buys fruits from the wholesale market and sells
them in his shop.
On a particular day, he buys apples, mangoes and bananas.

26

Life Mathematics
Each fruit has two prices, one at each shop, as shown in the price list.
The price at which Raja buys the fruit at the market is called the Cost Price
(C.P.). The price at which he sells the fruit in his stall is called the Selling Price (S.P.).
From the price list we can say that the selling price of the apples and the mangoes
in the shop are greater than their respective cost price in the whole sale market. (i.e.)
the shopkeeper gets some amount in addition to the cost price. This additional amount
is called the profit.
Selling Price of mango
=
Cost Price of mango + Profit
Selling price – Cost price
=
Profit
Profit
=
Selling Price – Cost Price
=
15 – 10
Profit
=
`5
i.e., Profit
=
Selling Price – Cost Price
In case of the apples,
Selling price of apple > Cost price of apple, there is a profit.
Profit
=
S.P. – C.P.
=
8 – 6
Profit
=
`2
As we know, bananas get rotten fast, the shop keeper wanted to sell them without
wasting them. So, he sells the bananas at a lower price (less than the cost price). The
amount by which the cost is reduced from the cost price is called Loss.
In case of bananas,
Cost price of banana > selling price of banana, there is a loss.
S.P. of the banana
=
C.P. of the banana – Reduced amount
S.P.
=
C.P. – Loss
Loss
=
C.P. – S.P.
Loss
=
3 – 2
Loss
=
`1

27

Chapter 2
So, we can say that
When the selling price of an article is greater than its cost price, then
there is a profit.
Profit = Selling Price - Cost Price
When the cost price of an article is greater than its selling price, then
there is a loss.
Loss = Cost Price – Selling Price
S.P = C.P + Profit
S.P = C.P - Loss.
To find Profit / Loss %
Rakesh buys articles for `10,000
and sells them for `11,000 and makes
a profit of `1,000, while Ramesh buys
articles for `1,00,000 and sells them
for `1,01,000 and makes a profit of
1)
Any fraction with its denominator
`1,000.
100 is called __________
2)
%
Both of them have made the
1 2 =
3)
35% =
( in fraction)
same amount of profit. Can you say
both of them are benefited equally?
4)
0.05 =
%
1
No.
5)
= __
%
4
To find who has gained more,
we need to compare their profit based on their investment.
We know that comparison becomes easier when numbers are expressed in
percent. So, let us find the profit %
Rakesh makes a profit of `1,000, when he invests `10,000.
Profit of `1,000 out of `10,000
1000
For each 1 rupee, he makes a profit of 10000
1000
Therefore for `100, profit =
# 100
10, 000
Profit =
10%

28

Life Mathematics
Ramesh makes a profit of `1000, when he invests `1,00,000.
1000
Profit of 1000 out of 1,00,000
=
100000
1000
Profit
=
# 100 = 1%
100000
So, from the above we can say that Rakesh is benefited more than Ramesh.
Profit #
So,
Profit Percentage
=
100
C.P
Loss % is also calculated in the same way.
Loss #
Loss Percentage
=
100
C.P.
Profit Percentage or Loss Percentage is always calculated
on the cost price of the article.
Example 2.18
A dealer bought a television set for `10,000 and sold it for `12,000. Find the
profit / loss made by him for 1 television set. If he had sold 5 television sets, find the
total profit/loss
Solution:
Selling Price of the television set
= `12,000
Cost Price of the television set
= `10,000
S.P.
>
C.P, there is a profit
Profit
=
S.P. – C. P.
=
12000 – 10000
Profit
= `2,000
Profit on 1 television set
= `2,000
Profit on 5 television sets
=
20 00 × 5
Profit on 5 television sets
= `10,000
Example 2.19
Sanjay bought a bicycle for `5,000.
He sold it for `600 less after two years.
Find the selling price and the loss percent.
Solution:
Cost Price of the bicycle
=
`50 00

29

Chapter 2
Loss
=
`60 0
Selling Price
=
Cost Price – Loss
=
50 00 – 600
Selling Price of the bicycle
=
`440 0
Loss #
Loss
=
100
C.P.
600
=
# 100
5000
=
12%
Loss
=
12%
Example 2.20
A man bought an old bicycle for `1,250. He spent `250 on its repairs. He then
sold it for `1400. Find his gain or loss %
Solution:
Cost Price of the bicycle
=
`1,250
Repair Charges
=
`250
Total Cost Price
=
1250 + 250 = `1,500
Selling Price
=
`1,400
C.P.
>
S.P., there is a Loss
Loss
=
Cost Price – Selling Price
=
1500 – 1400
=
10 0
Loss
=
`10 0
Loss #
Percentage of the loss
=
100
C.P.
100
=
# 100
1500
20
=
3
=
6 2 % (or) 6.67%
3
Loss
=
6.67%

30

Life Mathematics
Example 2.21
A fruit seller bought 8 boxes of grapes at `150 each. One box was damaged.
He sold the remaining boxes at `190 each. Find the profit / loss percent.
Solution:
Cost Price of 1 box of grapes
=
`150
Cost Price of 8 boxes of grapes
=
150 # 8
= `1200
Nu mber of boxes damaged
=
1
Nu mber of boxes sold
=
8 – 1
=
7
Selling Price of 1 box of grapes
=
`190
Selling Price of 7 boxes of grapes = 190 × 7
= `1330
S.P.
>
C.P, there is a Profit.
Profit
= Selling Price – Cost Price
=
1330 – 1200
=
130
Profit
= `130
Profit #
Percentage of the profit
=
100
C.P
130
=
# 100
1200
=
10.83
Profit
=
10.83 %
Example 2.22
Ram, the shopkeeper bought a pen for `50 and then sold it at a loss of `5. Find
his selling price.
Solution:
Cost price of the pen
=
`50
Loss
=
`5

31

Chapter 2
S.P.
=
C.P. – Loss
=
50 – 5
=
45
Selling price of the pen
=
`45.
Example 2.23
Sara baked cakes for the school festival. The cost of one cake was `55. She
sold 25 cakes and made a profit of `11 on each cake. Find the selling price of the
cakes and the profit percent.
Solution:
Cost price of 1 cake
=
`55
Nu mber of cakes sold
=
25
Cost price of 25 cakes
=
55 × 25 = `1375
Profit on 1 cake
=
`11
Profit on 25 cakes
=
11 × 25 = ` 275
S.P.
=
C.P. + Profit
=
1375 + 275
=
1,650
= `1,650
Profit #
Percentage of the profit
=
100
C. P
275
=
# 100
1375
=
20
Profit
=
20 %
Exercise 2.4
1.
Choose the correct answer:
i)
If the cost price of a bag is `575 and the selling price is `625, then
there is a profit of `
(A)
50
(B) 575
(C) 625
(D) none of these
ii)
If the cost price of the box is `155 and the selling price is `140, then there is a
loss of `
(A)
155
(B) 140
(C) 15
(D) none of these

32

Life Mathematics
iii)
If the selling price of a bag is `235 and the cost price is `200, then
there is a
(A)
profit of `235
(B) loss of `3
(C)
profit of `35
(D) loss of `200
iv)
Gain or loss percent is always calculated on
(A)
cost price
(B) selling price
(C) gain
(D) loss
v)
If a man makes a profit of `25 on a purchase of `250, then profit% is
(A)
25
(B) 10
(C) 250
(D) 225
2.
Complete the table by filling in the appropriate column:
C.P.
S.P.
Profit
Loss
`
`
`
`
144
168
59
38
600
635.45
26599
23237
107.50
100
3.
Find the selling price when cost price and profit / loss are given.
i)
Cost Price = `450
ii) Cost Price = `760
iii) Cost Price = `980
iv) Cost Price = `430
Profit = `80
Loss = `140
Profit = `47.50
Loss = `93.25
v)
Cost Price = `999.75
Loss = `56.25
4.
Vinoth purchased a house for `27, 50,000. He spent `2,50,000 on repairs and
painting. If he sells the house for `33,00,000 what is his profit or loss % ?
5.
A shop keeper bought 10 bananas for `100. 2 bananas were rotten. He sold the
remaining bananas at the rate of `11 per banana. Find his gain or loss %
6.
A shop keeper purchased 100 ball pens for `250. He sold each pen for
`4. Find the profit percent.
7.
A vegetable vendor bought 40 kg of onions for `360. He sold 36 kg at `11 per
kg. The rest were sold at `4.50 per kg as they were not very good. Find his
profit / loss percent.
Choose one product and find out the different stages it
crosses from the time it is produced in the factory to
the time it reaches the customer.

33

Chapter 2
Think!
Do you think direct selling by the manufacturer himself is more beneficial for
the costumers? Discuss.
Do it yourself
1.
A trader mixes two kinds of oil, one costing `100 per Kg. and the other costing
`80 per Kg. in the ratio 3: 2 and sells the mixture at `101.20 per Kg. Find his
profit or loss percent.
2.
Sathish sold a camera to Rajesh at a profit of 10 %. Rajesh sold it to John at a
loss of 12 %. If John paid `4,840, at what price did Sathish buy the camera?
3.
The profit earned by a book seller by selling a book at a profit of 5% is `15
more than when he sells it at a loss of 5%. Find the Cost Price of the book.
2.5 Simple Interest
Deposit `10,000 now. Get `20,000 at the end of 7 years.
Deposit `10,000 now. Get `20,000 at the end of 6 years.
Is it possible? What is the reason for these differences?
Lokesh received a prize amount of `5,000 which he deposited in a bank in
June 2008. After one year he got back `5,400.
Why does he get more money? How much more does he get?
If `5,000 is left with him in his purse, will he gain `400?
Lokesh deposited `5,000 for 1 year and received `5,400 at the end of the first
year.
When we borrow (or lend) money we pay (or receive) some additional amount
in addition to the original amount. This additional amount that we receive is termed
as Interest (I).

34

Life Mathematics
As we have seen in the above case, money can be borrowed deposited in
banks to get Interest.
In the above case, Lokesh received an interest of `400.
The amount borrowed / lent is called the Principal (P). In this case, the amount
deposited - `5,000 is termed as Principal (P).
The Principal added to the Interest is called the Amount (A).
In the above case, Amount =
Pr incipal +Interest
=
`5000 + `400 = `5,400.
Will this Interest remain the same always?
Definitely not. Now, look at the following cases
(i)
If the Principal deposited is increased from `5,000 to `10,000, then
will the interest increase?
(ii)
Si milarly, if `5,000 is deposited for more number of years, then will
the interest increase?
Yes in both the above said cases, interest will definitely increase.
From the above, we can say that interest depends on principal and duration of
time. But it also depends on one more factor called the rate of interest.
Rate of interest is the amount calculated annually for `100
(i.e.) if rate of interest is 10% per annum, then interest is `10 for `100 for 1 year.
So, Interest depends on:
Amount deposited or borrowed – Principal (P)
Period of time - mostly expressed in years (n)
Rate of Interest (r)
This Interest is termed as Simple Interest because it is always calculated on the
initial amount (ie) Principal.
Calculation of Interest
If ‘r’ is the rate of interest, principal is `100, then Interest
r
for 1 year
= 100 # 1 #
100
r
for 2 years
= 100 # 2 #
100
r
for 3 years
= 100 # 3 #
100
r
for n years
= 100 # n #
100

35

Chapter 2
Pnr
So,
I
=
100
A
=
P + I
Pnr
A
=
P
+
100
nr
A
=
P
`
1 +
j
100
Interest
=
Amount – Principal
I
=
A – P
The other formulae derived from
Pnr
I
=
are
100
100 I
r
=
P n
100 I
n
=
P r
100I
P
=
rn
Note: ‘n’ is always calculated in years. When ‘n’ is given in months \ days,
convert it into years.
Fill in the blanks
Principal
Interest
Amount
`
`
`
5,000
500
12,500
17,500
6,000
25,000
8,450
750
12,000
15,600
Example 2.24
Kamal invested `3,000 for 1 year at 7 % per annum. Find the simple interest
and the amount received by him at the end of one year.
Solution:
Pr incipal (P)
=
`3,000
Nu mber of years (n)
=
1
Rate of interest (r)
=
7 %

36

Life Mathematics
= Pnr
Interest (I)
100
3000
# 1 # 7
=
100
I
=
`210
A
=
P + I
=
30 00 + 210
A
=
`3,210
Example 2.25
Radhika invested `5,000 for 2 years at 11 % per annum. Find the simple interest
and the amount received by him at the end of 2 years.
Solution:
Pr incipal (P)
=
`5,000
Nu mber of years (n)
=
2 years
Rate of interest (r)
=
11 %
= Pnr
I
100
5000
# 2 # 11
=
100
=
1100
I
=
`1,100
Amount (A)
=
P + I
=
50 00 + 1100
A
=
`6,100
Example 2.26
Find the simple interest and the amount due on `7,500 at 8 % per annum for
1 year 6 months.
Know this
Solution:
12 months = 1 years
P
=
`7,500
6
6 months
=
year
n
=
1 yr 6 months
12
1
6
=
year
=
1
yrs
2
12
3
3 months
=
year
1
3
12
=
1
=
yrs
2
2
1
=
year
r
=
8 %
4

37

Chapter 2
= Pnr
I
100
3
7500 # # 8
2
=
100
7500 3 8
# #
=
2 # 100
=
900
I
=
`90 0
A
=
P + I
=
7500 + 900
=
`8,400
Interest
=
`90 0, Amount = `8,400
Aliter:
P
=
`7,500
3
n
=
years
2
r
=
8 %
nr
A
= P
`
1 +
j
100
3
# 8
e
2
o
=
7500 1 +
100
3 # 8
=
7500 1 +
c
m
2 # 100
28
= 7500
`
j
25
=
300 # 28
= 8400
A
=
`840 0
I
=
A – P
=
8400 – 7500
=
900
I
=
`90
0
Interest
=
`90
0
Amount
=
`8,400

38

Life Mathematics
Example 2.27
Find the simple interest and the amount due on `6,750 for 219 days at 10 %
per annum.
Solution:
P
=
`6,750
n
=
219 days
Know this
3
365 days
=
1 year
=
219 year =
year
365
5
219 days
219 year
r
=
10 %
= 365
3
= Pnr
=
year
I
5
100
73
73 days
=
year
6750 # 3 # 10
I
=
365
5 # 100
1
=
year
=
405
5
I
=
`405
A
=
P + I
=
6750 + 405
=
7,155
A
=
`7,155
Interest
=
`405, Amount = `7,155
Example 2.28
Rahul borrowed `4,000 on 7th of June 2006 and returned it on 19th August2006.
Find the amount he paid, if the interest is calculated at 5 % per annum.
Solution:
P
=
`4,000
r
=
5 %
Know this
Nu mber of days,
Ju ne
=
24 (30 – 6)
Thirty days
hath
September, April, June
July
=
31
and November. All the
rest have thirty
-
one
August
=
18
except February.
Total number of days
=
73
n
=
73 d ays

39

Chapter 2
73
=
year
365
1
=
year
5
nr
A
=
P
`
1 +
j
100
1
# 5
=
4000 1 +
c
m
5 # 100
1
=
4000 1 +
`
j
100
101
= 4000
`
j
100
=
4,040
Amount
=
`4,040
Example 2.29
Find the rate percent per annum when a principal of `7,000 earns a S.I. of
`1,680 in 16 months.
Solution:
P
=
`7,000
n
=
16 months
16
4
=
yr
=
yr
12
3
I
=
`1,680
r
=
?
r = 100 I
P n
100
#
1680
=
4
7000 #
3
100
#
1680 3
#
=
7000 # 4
=
18
r
=
18 %
Example 2.30
Vijay invested `10,000 at the rate of 5 % simple interest per annum.
He
received `11,000 after some years. Find the number of years.
Solution:
A
=
`11,000
P
=
`10,000

40

Life Mathematics
r
=
5 %
n
=
?
I
=
A – P
=
11,000 – 10,000
=
1,000
I
=
`1000
= 100 I
n
P r
100 1000
#
=
10000 # 5
n
=
2 years.
Aliter:
nr
A
=
P
`
1 +
j
100
n # 5
11000
=
10
000
`
1 +
j
100
11000
n
=
1
+
10000
20
11
= 20 + n
10
20
11
# 20 =
20 + n
10
22
=
20 + n
22 – 20
=
n
n
=
2 years
Example 2.31
A sum of money triples itself at 8 % per annum over a certain time. Find the
number of years.
Solution:
Let Principal be `P.
Amount
=
tr iple the principal
=
`3 P
r
=
8 %
n
=
?

41

Chapter 2
I
=
A – P
=
3P – P
=
2P
I
=
`2 P
100I
n
=
Pr
100 # 2 P
=
P # 8
n
=
25 years
Nu mber of years
=
25
A liter:
Let Principal be `100
Amount
=
3 # 100
=
`300
I
=
A – P
=
30 0 - 100
I
=
`200.
n = 100 I
100 200
#
=
P r
100 # 8
200
n
=
= 25
8
Nu mber of years
=
25.
Example 2.32
A certain sum of money amounts to `10,080 in 5 years at 8 % . Find the principal.
Solution:
A
=
`10,080
n
=
5 years
r
=
8 %
P
=
?
nr
A
=
P
`
1 +
j
100
5 # 8
10 080
=
P 1
`
+
j
100

42

Life Mathematics
7
10 080
=
P ` j
5
5
10080 #
=
P
7
7,200
=
P
Pr incipal
=
`7,200
Example 2.33
A certain sum of money amounts to `8,880 in 6 years and `7,920 in 4 years
respectively. Find the principal and rate percent.
Solution:
Amount at the end of 6 years
=
Pr incipal + interest for 6 years
=
P
+
I
=
8880
6
Amount at the end of 4 years
=
Pr incipal + Interest for 4 years
=
P
+
I
=
7920
4
=
8880 – 7920
I 2
= 960
Interest at the end of 2 years
=
`960
= 960
Interest at the end of 1st year
2
=
480
Interest at the end of 4 years
=
480 # 4
=
1,920
P +
=
7920
I 4
P + 1920 =
7920
P
=
7920 - 1920
P
=
6,000
Pr incipal
=
`6,000
100I
r
=
pn
100 1920
#
=
6000 # 4
r
=
8 %

43

Chapter 2
Exercise 2.5
1.
Choose the correct answer:
i)
Simple Interest on `1000 at 10 % per annum for 2 years is
(A)
`1000
(B) `200
(D) `2000
ii)
If Amount = `11,500,
(C) `100
Principal = `11,000, Interest is
(A)
`500
(B) `22,500
(C) `11,000
(D) `11,000
iii)
6 months =
1
1
3
(A)
yr
(B)
yr
(C)
yr
(D) 1 yr
2
4
4
iv)
292 days =
1
3
4
2
(A)
yr
(B) 5
yr
(C)
yr
(D)
yr
5
5
5
v)
If P = `14000
I = `1000, A is
(A)
`15000
(B) `13000
(C) `14000
(D) `1000
2.
Find the S.I. and the amount on `5,000 at 10 % per annum for 5 years.
3.
Find the S.I and the amount on `1,200 at 12
1
% per annum for 3 years.
2
4.
Lokesh invested `10,000 in a bank that pays an interest of 10 % per annum. He
withdraws the amount after 2 years and 3 months. Find the interest, he receives.
5.
Find the amount when `2,500 is invested for 146 days at 13 % per annum.
nd
6.
Fi nd the S.I and amount on `12,000 from May 21 st 1999 to August 2 1999 at
9 % per annum.
7.
Sathya deposited `6,000 in a bank and received `7500 at the end of 5 years.
Find the rate of interest.
1
8.
Fi nd the principal that earns `250 as S.I. in 2 2
years at 10 % per annum.
9.
In how many years will a sum of `5,000 amount to `5,800 at the rate of
8 % per annum.
10.
A su m of money doubles itself in 10 years. Find the rate of interest.

44

Life Mathematics
1
11.
A sum of money doubles itself at 12
%
per annum over a certain period of
2
time. Find the number of years.
12.
A certain sum of money amounts to `6,372 in 3 years at 6 % Find the principal.
1
13.
A certain sum of money amounts to `6,500 in 3 years and `5,750 in 1 2
years
respectively . Find the principal and the rate percent.
14.
Fi nd S.I. and amount on ` 3,600 at 15% p.a. for 3 years and 9 months.
1
15.
Fi nd the principal that earns ` 2,080 as S.I. in 3 4
years at 16% p.a.
Think!
9
1)
Fi nd the rate percent at which, a sum of money becomes 4
times
in 2 years.
2)
If Ram needs `6,00,000 after 10 years, how much should he
invest now in a bank if the bank pays 20 % interest p.a.
1.
A fraction whose denominator is 100 or a ratio whose second term is 100 is
termed as a percent.
2.
Percent means per hundred, denoted by %
3.
To convert a fraction or a decimal to a percent, multiply by 100.
4.
The price at which an article is bought is called the cost price of an article.
5.
The price at which an article is sold is called the selling price of an article.
6.
If the selling price of an article is more than the cost price, there is a profit.

45

Chapter 2
7.
If the cost price of an article is more than the selling price, there is a loss.
8.
Total cost price = Cost Price + Repair Charges / Transportation charges.
9.
Profit or loss is always calculated for the same number of ar ticles or same
units.
10.
Profit = Selling Price – Cost Price
11.
Loss = Cost Price – Selling Price
Profit #
12.
Profit% =
100
C.P.
Loss #
13.
Loss% =
100
C.P.
14.
Selling Price = Cost Price + Profit
15.
Selling Price = Cost Price - Loss
Pnr
16.
Simple interest is I =
100
17.
A
=
P + I
Pnr
=
P +
100
nr
=
P 1 +
`
j
100
18.
I
=
A – P
100I
19.
P
=
nr
100
I
20.
r
=
P n
100I
21.
n
=
Pr

46

Measurements
MEASUREMENTS
3.1 Trapezium
A trapezium is a quadrilateral with one pair of
opposite sides are parallel.
The distance between the parallel sides is the
height of the trapezium. Here the sides AD and BC
Fig. 3.1
are not parallel, but AB || DC.
If the non - parallel sides of a trapezium are equal ( AD = BC ), then it is known
as an isosceles trapezium.
Here
+ A
=
+ B ;
+ C =
+ D
AC
=
BD
+ A + + D
=
180° ;
+ B + + C = 180°
Fig. 3.2
Area of a trapezium
ABCD is a trapezium with parallel sides AB and DC measuring ‘a’ and ‘b’.
Let the distance between the two parallel sides be ‘h’. The diagonal BD divides the
trapezium into two triangles ABD and BCD.
Area of the trapezium
=
area of D ABD + area of D BCD
1
1
=
# AB # h +
# DC #
h
2
2
1
=
# h
6AB + DC @
Fig. 3.3
2
1
=
# h a
6
+
b
@ sq. units
2
1
` Area of a trapezium =
# height # (sum of the parallel sides) sq. units
2
Example 3.1
Find the area of the trapezium whose height is 10 cm and the parallel sides are
12 cm and 8 cm of length.

47

Chapter 3
Solution
Given: h = 10 cm, a = 12 cm, b = 8 cm
1
Area of a trapezium
=
# h a
^
+
b
h
2
1
=
# 10 # ^12 + 8 h = 5
# ^20 h
2
\ A rea of the trapezium
=
10 0 sq. cm 2
Example 3.2
The length of the two parallel sides of a trapezium are 15 cm and 10 cm. If its
area is 100 sq. cm. Find the distance between the parallel sides.
Solution
Given:
a = 15 cm, b = 10 cm, Area = 100 sq. cm.
Area of the trapezium
=
10 0
1
h a
^
+
b
h
=
100
2
1
# h # ^15 + 10 h
=
100
2
h # 25
=
200
200 = 8
h
=
25
` the distance between the parallel sides = 8 cm.
Example 3.3
The area of a trapezium is 102 sq. cm and its height is 12 cm. If one of its
parallel sides is 8 cm. Find the length of the other side.
Solution
Given:
Area = 102 cm 2 , h = 12 cm, a = 8 cm.
Area of a trapezium
=
102
1
h a
^
+
b
h
=
102
2
1
# 12 # ^8 + b
h
=
102
2
6 (8 + b)
=
102
8 + b
=
17
&
b = 17 – 8 = 9
` length of the other side =
9 cm
By paper folding method:
In a chart paper draw a trapezium ABCD of any measure. Cut and take the
trapezium separately. Fold the trapezium in such a way that DC lies on AB and crease
it on the middle to get EF.

48

Measurements
E
Fig. 3.4
EF divides the trapezium in to two parts as shown in the Fig. 4.40 (ii)
From D draw DG = EF. Cut the three parts separately.
Arrange three parts as shown in the Fig. 3.4 (iii)
The figure obtained is a rectangle whose length is AB + CD = a + b
1
1
and breadth is 2 ( height of trapezium ) =
h
2
` Area of trapezium
= area of rectangle as shown in Fig. 4.40 (iii)
= length # breadth
1
=
^
a + b
h
`
h
j
2
1
=
h a
^
+
b
h sq. units
2
Exercise 3.1
1.
Choose the correct answer.
i)
The area of trapezium is
____________
sq. units
1
1
(A)
h ^ a + b h
(B)
h (a + b)
(C)
h ^ a - b h
(D)
h (a – b)
2
2
ii)
In an isosceles trapezium
(A)
non parallel sides are equal
(B) parallel sides are equal
(C)
height = base
(D) parallel sides = non parallel sides
iii)
Th e sum of parallel sides of a trapezium is 18 cm and height is 15 cm. Then
its area is
(A)
105 cm 2
(B) 115 cm 2
(C) 125 cm 2
(D) 135 cm 2
iv)
The height of a trapezium whose sum of parallel sides is 20 cm and the area
8 0 cm 2 is
(A)
2 cm
(B) 4 cm
(C) 6 cm
(D) 8 cm

49

Chapter 3
2.
Find the area of a trapezium whose altitudes and parallel sides are given below:
i) altitude = 10 cm, parallel sides = 4 cm and 6 cm
ii) altitude = 11 cm, parallel sides = 7.5 cm and 4.5 cm
iii) altitude = 14 cm, parallel sides = 8 cm and 3.5 cm
3.
The area of a trapezium is 88 cm 2 and its height is 8 cm. If one of its parallel side
is 10 cm. Find the length of the other side.
4.
A garden is in the form of a trapezium. The parallel sides are 40 m and 30 m.
The perpendicular distance between the parallel side is 25 m. Find the area of the
garden.
5.
Area of a trapezium is 960 cm 2 . The parallel sides are 40 cm and 60 cm. Find the
distance between the parallel sides.
3.2 Circle
In our daily life, we come across a number of objects like wheels, coins, rings,
bangles,giant wheel, compact disc (C.D.)
What is the shape of the above said objects?
‘round’, ‘round’, ‘round’
Yes, it is round. In Mathematics it is called a circle.
Now, let us try to draw a circle.
Take a thread of any length and fix one end tightly at
a point O as shown in the figure. Tie a pencil (or a chalk) to
the other end and stretch the thread completely to a point A.
Holding the thread stretched tightly, move the pencil.
Stop it when the pencil again reaches the point A. Now see
Fig. 3.5
the path traced by the pencil.
Is the path traced by the pencil a circle or a straight line?
‘Circle’
Yes, the path traced by the point, which moves at a constant distance from a
fixed point on a given plane surface is called a circle.
Parts of a Circle
The fixed point is called the centre of the circle.
The constant distance between the fixed point and
the moving point is called the radius of the circle.
i.e. The radius is a line segment with one end point
at the centre and the other end on the circle. It is denoted
by ‘ r ’.
A line segment joining any two points on the circle
is called a chord.
Fig. 3.6

50

Measurements
Diameter is a chord passing through the centre of the circle. It is denoted by ‘ d ’.
The diameter is the longest chord. It is twice the radius.(i.e. d = 2r )
The diameter divides the circle into two equal parts. Each equal
part is a
semicircle.
Think it:
How many diameters can
a circle have ?
All the radii of a circle are equal.
Circumference of a circle:
Can you find the distance covered by an
athlete if he takes two rounds on a circular track.
Since it is a circular track, we cannot use
the ruler to find out the distance.
So, what can we do ?
Take a one rupee coin.Place it on a paper
Fig. 3.7
and draw its outline. Remove the coin. Mark a
point A on the outline as shown in the Fig. 3.7
Take a thread and fix one end at A. Now place the
thread in such a way that the thread coincides exactly with
the outline. Cut the other end of the thread when it reaches
the point A.
Length of the thread is nothing but the circumference
of the coin.
Fig. 3.8
So,
the distance around a circle is called the circumference of the circle, which is
denoted by ‘C’. i.e., The perimeter of a circle is known as its circumference.
Take a bottle cap or a bangle or any other circular objects and
find the circumference. If possible find the relation between
the circumference and the diameter of the circular objects.
Relation between diameter and circumference of the circle
Draw four circles with radii 3.5 cm, 7 cm, 5 cm, 10.5 cm in your note book.
Measure their circumferences using a thread and the diameter using a ruler as shown
in the Fig. 3.9 given below.

51

Chapter 3
Fig. 3.9
Fill in the missing values in table 3.1 and find the ratio of the circumference to
the diameter.
Diameter
Circumference
C
Circle
Ratio
`
j
( d )
(C)
d
1
3.5 cm
7 cm
22
cm
22
= 3.14
7
2
7
cm
14
cm
44
cm
44
22
=
= 3.14
14
7
3
5
cm
10
cm
----
----
4
10.5 cm
21
cm
----
----
Table 3.1
C
What do you infer from the above table?. Is this ratio
`
j approximately
d
the same?
Yes !
C
= 3.14
& C = ^3.14 h d
d
So, can you say that the circumference of a circle is always more than 3 times
its diameter ?
Yes !
In all the cases, the ratio C
is a constant and is denoted by the Greek letter r
d
(read as ‘pi’ ). Its approximate value is
22 or 3.14.
7
C
so,
= r
&
C = rd units
where d is the diameter of a circle.
d
We know that the diameter of a circle is twice the radius r . i.e., d = 2r .

52

Measurements
from the above formula, C = rd = r ^2r h &
C = 2rr units.
The value of r is calculated by many mathematicians.
22
Babylonians :
r = 3
Greeks
:
r =
or 3.14
7
1
10
62838
Archemides
:
Aryabhata :
r =
(or) 3.1416
3 7
< r < 3 71
2000
22
Now, we use
r =
or 3.14
7
Example 3.4
Find out the circumference of a circle whose diameter is 21 cm.
Solution
Ci rcumference of a circle =
rd
22
22
=
# 21
Here r =
7
7
=
66 cm.
Example 3.5
Find out the circumference of a circle whose radius is 3.5 m.
Solution
Ci rcumference of a circle =
2rr
22
=
2
#
# 3
.5
7
=
2 × 22 × 0.5
=
22 m
Example 3.6
A wire of length 88 cm is bent as a circle. What is the radius of the circle.
Solution
Length of the wire
=
88 cm
Ci rcumference of the circle
= Length of the wire
2rr
=
88
22
2
#
#
r
=
88
7
88 # 7
r
=
= 14 cm
2
# 22
` radius of a circle is 14 cm.
Example 3.7
The diameter of a bicycle wheel is 63 cm. How much distance will it cover in
20 revolutions?

53

Chapter 3
Solution
When a wheel makes one complete revolutions,
Distance covered in one rotation = Ci rcumference of wheel
` circumference of the wheel = rd units
22
=
# 63 cm
7
=
198 cm
For one revolution, the distance covered = 198 cm
` for 20 revolutions, the distance covered = 20 × 198 cm
= 3960 cm
=
39 m 60 cm
[100 cm = 1 m]
Example 3.8
A scooter wheel makes 50 revolutions to cover a distance of 8800 cm. Find
the radius of the wheel.
Solution
Distance travelled
= Nu mber of revolutions # Circumference
Distance travelled
Ci rcumference
=
Number of revolutions
8800
2rr =
50
i.e., 2rr
=
176
22
2
#
#
r
=
176
7
176 # 7
r =
2 # 22
r
=
28 cm
` radius of the wheel = 28 cm.
Example 3.9
The radius of a cart wheel is 70 cm. How many revolution does it make in
travelling a distance of 132 m.
Solution
Given: r = 70 cm, Distance travelled = 132 m.
` Circumference of a cart wheel = 2rr
22
=
2
#
# 70
7
=
440 cm

54

Measurements
Distance travelled
= Nu mber of revolutions # Circumference
Distance travelled
` Number of revolutions
=
Circumference
132 m
=
440 cm
13200 cm
cm
=
(1 m = 100 cm, 132 m = 13200 cm)
440
=
30
` Number of revolutions = 30.
Example 3.10
The circumference of a circular field is 44 m. A cow is tethered to a peg at the
centre of the field. If the cow can graze the entire field, find the length of the rope used
to tie the cow.
Solution
Length of the rope
= Radius of the circle
Ci rcumference
=
44 m
(given)
i.e., 2rr
=
44
22
2
#
#
r
=
44
7
44 # 7
` r
=
= 7 m
2 # 22
Fig. 3.10
` The length of the rope used to tie the cow is 7 m.
Example 3.11
The radius of a circular flower garden is 56 m. What is the cost of fencing it at
`10 a metre ?
Solution
Length to be fenced
=
Ci rcumference of the circular flower garden
Circumference of the flower garden
=
2rr
22
=
2
#
# 56 = 352 m
7
` Length of the fence =
352 m
Cost of fencing per metre =
`10
` cost of fencing 352 m =
`10 # 352
=
`3520
` Total cost of fencing is `3520.

55

Chapter 3
Example 3.12
The cost of fencing a circular park at the rate of `5 per metre is `1100. What
is the radius of the park.
Solution
Cost of fencing
= Ci rcumference # Rate
Cost of fencing
` Circumference
=
Rate
1100
i.e., 2rr
=
5
2rr
=
220
22
` 2
#
#
r
=
220
7
220 # 7
r
=
2 # 22
=
35 m
` Radius of the park
=
35 m.
Activity - Circular Geoboard
Take a square Board and draw a circle.
Fix nails on the circumference of the circle. ( See fig )
Using rubber band, form various diameters, chords, radii and
compare.
Exercise 3.2
1.
Choose the correct answer:
i)
The line segment that joins the centre of a circle to any point on the circle is
called
(A)
Diameter
(C) Chord
(D) None
ii)
A line segment joining any two points on the circle is called
(A)
Diameter
(C) Chord
(D) None
iii)
A chord passing through the centre is called
(A)
Diameter
(C) Chord
(D) None
iv)
The diameter of a circle is 1 m then its radius is
(A)
100 cm
(B) 50 cm
(C) 20 cm
(D) 10 cm
v)
The circumference of a circle whose radius is 14 cm is
(A)
22 cm
(B) 44 cm
(C) 66 cm
(D) 88 cm

56

Measurements
2.
Fill up the unknown in the following table:
radius ( r )
diameter ( d )
circumference (c)
(i)
35 cm
-----
-----
(ii)
-----
56 cm
-----
(iii)
-----
-----
30.8 cm
3.
Find the circumference of a circle whose diametre is given below:
(i)
35 cm
(ii) 84 cm
(iii) 119 cm
(iv) 147 cm
4.
Find the circumference of a circle whose radius is given below:
(i)
12.6 cm
(ii) 63 cm
(iii) 1.4 m
(iv) 4.2 m
5.
Find the radius of a circle whose circumference is given below:
(i)
110 cm
(ii) 132 cm
(iii) 4.4 m
(iv) 11 m
6.
The diameter of a cart wheel is 2.1 m. Find the distance travelled when it
completes 100 revolutions.
7.
The diameter of a circular park is 98 m. Find the cost of fencing it at `4 per
metre.
8.
A wheel makes 20 revolutions to cover a distance of 66 m. Find the diameter of
the wheel.
9.
The radius of a cycle wheel is 35 cm. How many revolutions does it make to
cover a distance of 81.40 m?
10.
The radius of a circular park is 63 m. Find the cost of fencing it at `12 per metre.
Area of a circle
Consider the following
A farmer levels a circular field of radius 70 m. What will be the cost of levelling?
What will be the cost of polishing a circular table-top of radius 1.5 m ?
How will you find the cost ?
To find the cost what do you need to find actually?
The region enclosed by
the circumference of a
Area or perimeter ?
circle is a circular region.
Area, area, area
Yes. In such cases we need to find the area of the circular region.
So far, you have learnt to find the area of triangles and quadrilaterals that made
up of straight lines. But, a cirlce is a plane figure made up of curved line different from
other plane figures.

57

Chapter 3
So, we have to find a new approach which will make the circle turn into a figure
with straight lines.
Take a chart paper and draw a circle. Cut the circle and take it separately. Shade
one half of the circular region. Now fold the entire circle into eight parts and cut along
the folds (see Fig. 3.11).
Fig. 3.11
Arrange the pieces as shown below.
Fig. 3.12
What is the figure obtained?
These eight pieces roughly form a parallelogram.
Similarly, if we divide the circle into 64 equal parts and arrange these, it gives
nearly a rectangle. (see Fig. 3.13)
Fig. 3.13
What is the breadth of this rectangle?
The breadth of the rectangle is the radius of the circle.
i.e.,
b = r
.....
(1)
What is the length of this rectangle ?
As the whole circle is divided into 64 equal parts and on each side we have 32
equal parts. Therefore, the length of the rectangle is the length of 32 equal parts, which
is half of the circumference of a circle.

58

Measurements
1
` length l
=
[circumference of the circle]
2
1
=
62r
r
@ = rr
2
` l
= rr
(2)
Area of the circle
= Area of the rectangle (from the Fig. 4.50)
=
l # b
=
(rr) # r
(from (1) and (2))
=
2
r
r
sq. units.
` Area of the circle
=
2
r
r
sq. units.
Example 3.13
Find the area of a circle whose diameter is 14 cm
Solution
Diameter d
=
14 cm
d
14
So,
=
=
= 7 cm
2
2
Area of circle
=
r r 2
22
=
# 7 # 7
7
=
154 sq. cm
` Area of circle
=
154 sq. cm
Example 3.14
A goat is tethered by a rope 3.5 m long. Find the maximum area that the goat
can graze.
Solution
Radius of the circle
= Length of the rope
7
=
3.5 m =
m
2
ma ximum area grazed by the goat =
r r 2
sq. units.
22
7
7
=
#
#
7
2
2
77
Fig. 3.14
=
= 38.5
sq. m
2
` maximum area grazed by the goat is 38.5 sq. m.

59

Chapter 3
Example 3.15
The circumference of a circular park is 176 m. Find the area of the park.
Solution
Ci rcumference
=
176 m
(given)
2rr
=
176
22
2
#
#
r
=
176
7
176 # 7
r
=
44
` r
=
28 m
Area of the park
=
r r 2
=
22 # 8 #
2
2 8
7
=
22 # 4 # 28
=
2464 sq. m.
Example 3.16
A silver wire when bent in the form of a square encloses an area of 121 sq. cm.
If the same wire is bent in the form of a circle. Find the area of the circle.
Solution
Let a be the side of the square
Area of the square
=
121 sq. cm.
(given)
a 2
=
121 & a = 11 cm
(11 # 11 = 121)
Perimeter of the square
=
4a units
=
4 # 11 cm
=
44 cm
Length of the wire
= Perimeter of the square
=
44 cm
The wire is bent in the form of a circle
The circumference of the circle = Length of the wire
` circumference of a circle = 44 cm
2rr
=
44
22
` 2
#
#
r
=
44
7
44 # 7
r
=
44
r
=
7 cm
` Area of the circle
=
r r 2
22
=
× 7 cm × 7 cm
7
Area of the circle
=
154 cm 2 .

60

Measurements
Example 3.17
When a man runs around circular plot of land 10 times, the distance covered by
him is 352 m. Find the area of the plot.
Solution
Distance covered in 10 times
= 352 m
352
Distance covered in one time
=
m = 35.2 m
10
The circumference of the circular plot
= Distance covered in one time
` circumference = 35.2 m
2rr
=
35.2
22
2
#
#
r
=
35.2
7
35 . 2 # 7
r
=
44
=
0.8 # 7
=
5.6 m
Area of the circular plot
=
r r 2
22
=
# 5 . 6 # 5 . 6
7
=
22 # 0.8 # 5.6
=
98.56 m 2
` Area of circular plot
=
98.56 m 2
Example 3.18
A wire in the shape of a rectangle of length 37 cm and width 29 cm is reshaped
in the form of a circle. Find the radius and area of the circle.
Solution
Length of the wire
= perimeter of the rectangle
=
2 [ length + breadth ]
=
2 [37 cm + 29 cm] = 2 # 66 cm
=
132 cm.
Since wire is bent in the form of a circle,
The circumference of the circle
= The length of the wire
` Circumference of a circle
=
132
2rr
=
132
22
2
#
#
r
=
132
7
132
r
=
# 7 =
21
44
` radius of the circle
=
21 cm
Area of the circle
=
r r 2
22
=
# 21 # 21 = 22 # 3 # 21
7
` Area of the circle
=
1386 sq. cm.

61

1.

2.

3.

4.

5.

6.

7.

### Chapter 3

Exercise 3.3

Find the area of the circles whose diameters are given below:

 (i) 7 cm (ii) 10.5 cm (iii) 4.9 m (iv) 6.3 m Find the area of the circles whose radii are given below: (i) 1.2 cm (ii) 14 cm (iii) 4.2 m (iv) 5.6 m

)

(take

(take

• 22 )

7

22

7

r =

r =

The diameter of a circular plot of ground is 28 m. Find the cost of levelling the

ground at the rate of `3 per sq. m.

A goat is tied to a peg on a grass land with a rope 7 m long. Find the maximum area of the land it can graze.

A circle and a square each have a perimeter of 88 cm. Which has a larger area?

A wheel goes a distance of 2200 m in 100 revolutions. Find the area of the wheel.

A wire is in the form of a circle of radius 28 cm. Find the area that will enclose, if it is bent in the form of a square having its perimeter equal to the circumference of the cirlce.

8.

The area of circular plot is 3850 m 2 . Find the radius of the plot. Find the cost of fencing the plot at `10 per metre.

9.

The radius of a circular ground is 70 m. Find the distance covered by a child walking along the boundary of the ground.

• 10. The area of a circular field is 154 m 2 . Find the time taken by an athlate to complete 2 rounds if she is jogging at the rate of 5 km/hr.

11 .

Ho w many circles of radius 7 cm can be cut from a paper of length 50 cm and width 32 cm.

### 3.3 Area of the path way

In our day - to - day life we go for a walk in a park, or in a play ground or even

around a swimming pool.

Can you represent the path way of a park diagrammatically ?

Have you ever wondered if it is possible to find the area of such paths ?

Can the path around the rectangular pool be related to the mount around the

photo in a photo frame ?

Can you think of some more examples?

In this section we will learn to find

• Area of rectangular pathway

• Area of circular pathway

62

Measurements
Area of rectangular pathway
(a) Area of uniform pathway outside the rectangle
Consider a rectangular building. A uniform flower garden is to be laid outside
the building. How do we find the area of the flower
garden?
The uniform flower garden including the
building is also a rectangle in shape. Let us call
it as outer rectangle. We call the building as inner
rectangle.
Fig. 3.15
Let l and b be the length and breadth of the
building.
` Area of the inner rectangle = l b sq.
units.
Fig. 3.16
Let w be the width of the flower garden.
What is the length and breadth of the outer rectangle ?
The length of the outer rectangle (L)
=
w +
l + w = ^l + 2w h units
The breadth of the outer rectangle (B)
=
w + b + w = ^ b + 2w h units
` area of the outer rectangle
=
L # B
=
^l + 2w h^ b + 2w h sq. units
Now, what is the area of the flower garden ?
Actually, the area of the flower garden is the pathway bounded between two
rectangles.
` Area of the flower garden =
(A rea of building and flower garden) –
(A rea of building)
Generally,
Area of the pathway = (Area of outer rectangle) – (Area of inner rectangle)
i.e. Area of the pathway = (l + 2w) (b + 2w) – lb.
Example 3.19
The area of outer rectangle is 360 m 2 . The area of inner rectangle is 280 m 2 . The
two rectangles have uniform pathway between them. What is the area of the pathway?
Solution
Area of the pathway = (A rea of outer rectangle) – (Area of inner rectangle)

63

Chapter 3
=
(360 – 280) m 2
= 80 m 2
` Area of the pathway =
80 m 2
Example 3.20
The length of a building is 20 m and its breadth is 10 m. A path of width 1 m is
made all around the building outside. Find the area of the path.
Solution
Inner rectangle (given)
Outer rectangle
l
= 20 m
width, w
= 1 m
b = 10 m
L
= l + 2w
Area
= l × b
= 20 + 2 = 22 m
Area
= 20 m × 10 m
B
= b + 2w
= 10 + 2 = 12 m
= 200 m 2
Area
= (l + 2w) (b + 2w)
Area
= 22 m #
12 m
= 264 m 2
Area of the path = (A rea of outer rectangle) – (Area of inner rectangle)
=
( 26 4 – 200 ) m 2 = 64 m 2
` Area of the path
= 64 m 2
Example 3.21
A school auditorium is 45 m long and 27 m
wide. This auditorium is surrounded by a varandha
of width 3 m on its outside. Find the area of the
varandha. Also, find the cost of laying the tiles in
the varandha at the rate of `100 per sq. m.
Solution
Fig. 3.17
Inner (given) rectangle
Outer rectangle
l
= 45 m
Width,
w
= 3 m
b
= 27 m
L
=
l + 2w
Area = 45m # 27 m
= 45 + 6 = 51 m
B
=
= 1215 m 2
b + 2w
= 27 + 6 = 33 m
Area
= 51m # 33 m
= 1683 m 2

64

Measurements
(i)
Area of the verandha
=
(A rea of outer rectangle) – (Area of inner rectangle)
=
(1683 – 1215) m 2
=
468 m 2
` Area of the verandha = 468 m 2 (or) 468 sq. m.
(ii)
Cost of laying tiles for 1 sq. m
= `100
Cost of laying tiles for 468 sq. m
= `100 # 468
= `46,800
` Cost of laying tiles in the verandha = `46,800
(b) Area of uniform pathway inside a rectangle
A swimming pool is built in the middle
of a rectangular ground leaving an uniform
width all around it to maintain the lawn.
If the pathway outside the pool is to be
grassed, how can you find its cost ?
If the area of the pathway and cost of
grassing per sq. unit is known, then the cost
Fig. 3.18
of grassing the pathway can be found.
Here, the rectangular ground is the outer rectangle where l and b are length and
` Area of the ground (outer rectangle) = l b sq. units
If w be the width of the pathway (lawn), what will be the length and breath of
the swimming pool ?
The length of the swimming pool
=
l -
w - w
=
l - 2w
The breadth of the swimming pool =
b - w - w
= b - 2w
` Area of the swimming pool (inner rectangle) = ^l - 2w h^ b - 2w h Sq. units
Area of the lawn
= Area of the ground – Area of the swimming pool.
Generally,
Area of the pathway
=
(A rea of outer rectangle) – (Area of inner rectangle)
=
lb – (l – 2w) (b – 2w)

65

Chapter 3
Example 3.22
The length and breadth
of
a
room are
8 m and 5 m respectively. A red colour border
of uniform width of 0.5 m has been painted all
around on its inside. Find the area of the border.
Solution
Fig. 3.19
Outer (given)rectangle
Inner rectangle
l
= 8 m
width, w
= 0.5 m
b
= 5 m
L
=
l - 2w
Area = 8m # 5 m
= (8 – 1) m
= 7 m
= 40 m 2
B
=
b - 2w
= (5 – 1) m
= 4 m
Area
= 7m × 4 m
= 28 m 2
Area of the path
= (A rea of outer rectangle) – (Area of inner rectangle)
=
(40 – 28) m 2
=
12 m 2
` Area of the border painted with red colour = 12 m 2
Example 3.23
A carpet measures 3 m # 2 m. A strip of 0.25 m wide is cut off from it on all
sides. Find the area of the remaining carpet and also find the area of strip cut out.
Solution
Outer rectangle
Inner rectangle
carpet before cutting the strip
carpet after cutting the strip
l
= 3 m
width, w
= 0.25 m
b
= 2 m
L
= l - 2w = (3 – 0.5) m
Area
= 3m # 2m
= 2.5 m
= 6 m 2
B
=
b - 2w = (2 – 0.5) m
= 1.5 m
Area
= 2.5m # 1.5m
= 3.75 m 2
The area of the carpet after cutting the strip = 3.75 m 2

66

Measurements
Area of the strip cut out = (A rea of the carpet) –
(Area of the remaining part)
=
(6 – 3.75) m 2
=
2.25 m 2
` Area of the strip cut out = 2.25 m 2
Note: If the length and breadth of the inner rectangle
is given, then the length and breadth of the outer rectangle
is l + 2w , b + 2w respectively where w is the width of the
path way.
Suppose the length and breadth of the outer rectangle
is given, then the length and breadth of the inner rectangle
is l - 2w , b - 2w respectively.
Fig. 3.20
Exercise 3.4
1.
A play ground 60 m # 40 m is extended on all sides by 3 m. What is the extended
area.
2.
A school play ground is rectangular in shape with length 80 m and breadth 60 m.
A cemented pathway running all around it on its outside of width 2 m is built.
Find the cost of cementing if the rate of cementing 1 sq. m is `20.
3.
A garden is in the form of a rectangle of dimension 30 m # 20 m. A path of width
1.5 m is laid all around the garden on the outside at the rate of `10 per sq. m.
What is the total expense.
4.
A picture is painted on a card board 50 cm long and 30 cm wide such that there is
a margin of 2.5 cm along each of its sides. Find the total area of the margin.
5.
A rectangular hall has 10 m long and 7 m broad. A carpet is spread in the centre
leaving a margin of 1 m near the walls. Find the area of the carpet. Also find the
area of the un covered floor.
6.
The outer length and breadth of a photo frame is 80 cm , 50 cm. If the width of
the frame is 3 cm all around the photo. What is the area of the picture that will be
visible?
Circular pathway
Concentric circles
Circles drawn in a plane with a common centre and different
radii are called concentric circles.
Circular pathway
Fig. 3.21
A track of uniform width is laid around a circular park for
walking purpose.
Can you find the area of this track ?

67

Chapter 3
Yes. Area of the track is the area bounded between two
concentric circles. In Fig. 3.22, O is the common centre of the two
circles.Let the radius of the outer circle be R and inner circle be r.
The shaded portion is known as the circular ring or the
circular pathway. i.e. a circular pathway is the portion bounded
between two concentric circles.
Fig. 3.22
width of the pathway, w
=
R –
r units
i.e., w
=
R – r & R
= w + r units
r
= R – w units.
The area of the circular path
= (a rea of the outer circle) – (area of the inner circle)
=
2
2
r
R
- r
r
=
2
2
r ^ R
- r
h sq. units
` The area of the circular path =
2
2
r ^ R - r
h sq. units
=
r ^ R + r h^ R - r h sq. units
Example 3.24
The adjoining figure shows two concentric circles. The radius
of the larger circle is 14 cm and the smaller circle is 7 cm. Find
(i)
The area of the larger circle.
(ii)
The area of the smaller circle.
(iii)
The area of the shaded region between two circles.
Solution
Fig. 3.23
i) Larger circle
ii) Smaller circle
R
=
14
r
=
7
area
=
r R 2
area =
r r 2
22
=
# 14 # 14
= 22 # 7 # 7
7
7
=
22 # 28
=
22 # 7
=
616 cm 2
=
154 cm 2
iii) The area of the shaded region
= (A rea of larger circle) – (Area of smaller circle)
=
(616 – 154) cm 2
= 462 cm 2
Example 3.25
From a circular sheet of radius 5 cm, a concentric circle of radius 3 cm is
removed. Find the area of the remaining sheet ? (Take r = 3.14 )

68

Measurements
Solution
Given: R = 5 cm,
r = 3 cm
Area of the remaining sheet
=
2
2
r ^ R - r
h
=
3.14 (5 2 – 3 2 )
=
3.14 (25 – 9)
=
3.14 × 16
=
50.24 cm 2
Aliter:
Outer circle
Inner circle
R
= 5 cm
r
= 3 cm
Area
=
r R 2
sq. units
Area =
r r 2
sq. units
=
3.14 # 5 # 5
=
3.14 # 3 # 3
= 3.14
# 25
= 3.14 # 9
= 78.5 cm 2
= 28.26 cm 2
Area of the remaining sheet = (A rea of outer circle) – (Area of inner circle)
= (78.5 – 28.26) cm 2
= 50.24 cm 2
` Area of the remaining sheet = 50.24 cm 2
Example 3.26
A circular flower garden has an area 500 m 2 . A sprinkler at the centre of the
garden can cover an area that has a radius of 12 m. Will the sprinkler water the entire
garden (Take r = 3.14)
Solution
Given,
area of the garden
=
50 0 m 2
Area covered by a sprinkler
=
r r 2
=
3.14 # 12 # 12
=
3.14 # 144
Fig. 3.24
=
452 .16 m 2
Since, the area covered by a sprinkler is less than the area of the circular flower
garden, the sprinkler cannot water the entire garden.
Example 3.27
A uniform circular path of width 2 m is laid out side a circular park of radius
50 m. Find the cost of levelling the path at the rate of `5 per m 2 (Take r = 3.14)

69

Chapter 3
Solution
Given: r = 50 m, w = 2 m, R = r + w = 50 + 2 = 52 m
Area of the circular path
=
r ^ R + r h^ R - r h
=
3.14 # ^52 + 50 h^52 - 50 h
=
3.14 # 102 # 2
=
3.14 # 204
Fig. 3.25
= 640.56 m 2
The cost of levelling the path of area 1 sq m = `5
The cost of levelling the path of 640.56 m 2 = `5 # 640.56
= ` 3202.80
` the cost of levelling the path = ` 3202.80
Exercise 3.5
1.
A circus tent has a base radius of 50 m. The ring at the centre for the performance
by an artists is 20 m in radius. Find the area left for the audience.
(Take r = 3.14)
2.
A circular field of radius 30 m has a circular path of width 3 m inside its boundary.
Find the area of the path (Take r = 3.14)
3.
A ring shape metal plate has an internal radius of 7 cm and an external radius of
10.5 cm. If the cost of material is `5 per sq. cm, find the cost of 25 rings.
4.
A circular well has radius 3 m. If a platform of uniform width of 1.5 m is laid
around it, find the area of the platform . (Take r = 3.14)
5.
A uniform circular path of width 2.5 m is laid outside a circular park of radius
56m. Find the cost of levelling the path at the rate of `5 per m 2 (Take r = 3.14)
6.
The radii of 2 concentric circles are 56 cm and 49 cm.
pathway.
Find the area of the
7.
The area of the circular pathway is 88 m 2 .
m, find the radius of the inner circle.
If the radius of the outer circle is 8
8.
The cost of levelling the area of the circular pathway is ` 12,012 at the rate of
` 6 per m 2 . Find the area of the pathway.

70

Measurements
Figure
Area
Forumula
1
1
× height × sum of
× h × (a + b) sq.
2
2
parallel sides
units
Perimeter of the circle =
2 r r units
2 × r × radius
Area of the circle =
r r 2 sq. units
Area of the pathway
Area of outer
rectangle – Area of
i) area of the rectangular
inner rectangle
pathway
Area of outer circle –
Area of inner circle
ii) area of the circular
= r (R 2 – r 2 ) sq. units
pathway
= r (R + r) (R – r)
sq. units

71

Chapter 4
GEOMETRY
4.1 Triangle: Revision
A
triangle is a closed plane
figure made of three line
segments.
In
Fig.
4.1
the
line
segments
AB,
BC
and
CA
form
a
closed
figure.
This
is
a triangle
and
is denoted by
D ABC. This triangle may be named as
D ABC or
D BCA or
Fig. 4.1
D CAB.
The line segments forming a triangle are the three sides of the triangle. In
Fig.4.1 AB , BC and CA are the three sides of the triangle.
The point where any two of the three line segments of a triangle intersect is
called the vertex of the triangle. In Fig. 4.1 A,B and C are the three vertices of the
D ABC.
When two line segments intersect, they form an angle at that point. In the
triangle in Fig. 4.1 AB
and
BC intersect at B and form an angle at that vertex.
This angle at B is read as angle B or + B or + ABC.
Thus a triangle has three
angles + A, + B and + C.
In Fig. 4.1 D ABC has
Sides
: AB,BC,CA
Angles
:
+ CAB, + ABC, + BCA
Vertices :
A, B, C
The side opposite to the vertices A, B, C are BC, AC and AB respectively. The
angle opposite to the side BC, CA and AB is + A, + B and + C respectively.
A triangle is a closed figure made of three line segments. It has three vertices,
three sides and three angles.

72

Geometry
4.2
Types of Triangles
Based on sides
A triangle is said to be
Equilateral, when all its sides are equal.
Isosceles, when two of its sides are equal.
Scalene, when its sides are unequal.
Based on angles
A triangle is said to be
Right angled, when one of its angle is a right angle and the other two angles are
acute.
Obtuse - angled, when one of its angle is obtuse and the other two angles are
acute.
Acute - angled, when all the three of its angles are acute.
The sum of the lengths of any two sides of a triangle is always greater than the
length of the third side.
4.3
Angle sum property of a triangle:
Activity 1
Draw any triangle ABC on a sheet of paper
Is it possible to form a
triangle whose sides are
and mark the angles 1, 2 and 3 on both sides of the
7cm, 5cm and 13cm?
paper as shown in Fig. 4.2 (i).
Fig. 4.2
Cut a triangle ABC in a paper. Fold the vertex A to touch the side BC as shown
in the Fig. 4.2 (ii) Fold the vertices B and C to get a rectangle as shown in the Fig. 4.2
(iii) Now you see that + 1, + 2 and + 3 make a straight line.

73

Chapter 4
From this you observe that
+ 1 + + 2
+ + 3
=
180°
+ A + + B + + C
=
180°
T he sum of the three angles of a triangle is 180°
Activity 2
Draw a triangle. Cut on the three angles.
Re arrange them as shown in Fig. 4.2 (ii). You
observe that the three angles now constitute
one angle. This angle is a straight angle and so
has measure 180°
Fig. 4.3
The sum of the three angles of a triangle is 180°
Think it.
1. Can you have a triangle with the three angles less than 60°?
2. Can you have a triangle with two right angles?
4.4 Exterior angle of a triangle and its property
Activity 3
Fig. 4.4
Draw a triangle ABC and
produce
one
of
its
sides, say
BC as shown
in
Fig. 4.4 (i) observe the angles ACD formed at the point C. This angle lies in the
exterior of T ABC formed at vertex C.
+ BCA is an adjacent angle to + ACD.
The remaining two angles of the
triangle namely + A nd + B are called the two interior opposite angles.
Now cut out (or make trace copies of) + A and + B and place them adjacent to
each other as shown in Fig. 4.4 (ii)
You observe that these two pieces together entirely cover + ACD.

74

Geometry
From this we conclude that the exterior angle of a triangle is equal to the sum of
the two interior opposite angles.
The relation between an exterior angle and its two interior angles is referred to
as the exterior angle property of a triangle.
Draw a triangle ABC and produce one of its sides BC as shown in
Fig. 4.4 (i) + ACD formed at the point C. Now take a protractor
and measure + ACD, + A and + B.
Find the sum + A + + B. and compare it with the measure of
+ ACD. Do you observe that + ACD = + A + B?
Example 4.1
In the given figure find the value of x.
Solution
+ CAB + + ABC + + BCA
= 180°
(Since sum of the three
40° + x + x
= 180°
angles of a triangle is 180°)
40° + 2 x
= 180°
2 x
= 180° – 40°
2
x
= 140°
140
°
x
=
= 70°
2
The value of x
= 70°.
Example 4.2
Two angles of a triangle are 40° and 60°.
Find the
third angle.
Solution
+ RPQ + + PQR + + QRP =
180°
(Since sum of the three angles of a
x + 40° + 60°
=
180°
tr iangle is 180°)
x + 100°
=
180°
x
=
180° – 100°
=
80°
`
The third angle x
=
80°
Example 4.3
In the given figure, find the measure of + A.
Solution
+ CAB + + ABC + + BCA
=
180°
(Since sum of the three angles of a
2 x + 120° + x =
180°
tr iangle is 180°)

75

Chapter 4
3x + 120°
=
180°
3x
=
180° – 120°
3x =
60°
60
°
x =
3
=
20°
`
+ A =
2 x = 2 × 20° = 40°
Example 4.4
In the given figure. Find the value of x.
Solution
In the figure exterior angle = + ABD = 110°.
Sum of the two interior opposite angle = + BCA + + CAB
= x + 50°
x + 50°
=
110°
(Since the sum of the two interior opposite
x
=
110° – 50°
angle is equal to the exterior angle)
=
60°
` The value of x
is 60°.
Example 4.5
In the given figure find the values of x and y.
Solution
In the give figure,
Exterior angle
=
+ DCA = 130°
50° + x
=
130°
(Since sum of the two interior opposite
x
= 130° – 50° angle is equal to the exterior angle)
=
80°
In
D ABC,
+ A + + B + + C
=
180° (Since sum of three angles of a triangle is 180°)
50° + x + y
=
180°
50° + 80° + y
=
18 0°
130° + y
=
18 0°
y
=
18 0° – 130° = 50°
` The values of x = 80° and y = 50°.

76

Geometry
Aliter:
+ ACB + + DCA
=
180° (Since sum of the adjacent angles on a line is 180°)
y + 130°
=
180°
y
=
180° – 130°
=
50°
In D ABC,
+ A
+ + B + + C
=
180° (Since sum of the three angles of a triangle is 180°)
50° + x + y
=
180°
50° + x + 50°
=
180°
10 0° + x
=
180°
x
=
180° – 100°
=
80°
Example 4.6
Three angles of a triangle are 3x + 5°, x + 20°, x + 25°. Find the measure of each
angle.
Solution
Su m of the three angles of a triangle
=
180°
3x + 5° + x + 20° + x + 25°
=
180°
5x + 50°
=
180°
5x
=
180° – 50°
5x
=
130°
130
°
x
=
5
=
26°
3x + 5°
=
(3 × 26°) + 5° = 78° + 5° = 83°
x
+ 20°
=
26° + 20° = 46°
x
+ 25°
=
26° + 25° = 51°
` The three angles of a triangle are 83°, 46° and 51°.

77

Chapter 4
Exercise 4.1
1.
Choose the correct answer.
i)
The sum of the three angles of a triangle is
(A)
90°
(B) 180°
(C) 270°
(D) 360°
ii)
In a triangle, all the three angles are equal, then the measure of each angle is
(A)
30 0
(B) 45 0
(C) 60 0
(D) 90 0
iii)
Which of the following can be angles of a triangle?
0
0
0
0
0
0
0
0
0
(A)
50 , 30 , 105 0 (B) 36 , 44
, 90 0
(C) 70 , 30 , 80 0
(D)
45 , 45 80
iv)
Two angles of a triangle are 40° and 60°, then the third angle is
(A)
20 0
(B) 40 0
(C) 60 0
(D) 80 0
v)
In T ABC, BC is produced to D and + ABC = 50 0 ,
+ ACD = 105 0 , then + BAC will be equal to
(A)
75°
(B) 15°
(C)
40°
(D) 55°
2.
State which of the following are triangles.
(i)
+ A = 25 0
+ B
=
35 0
+ C = 120 0
(ii) + P = 90 0
+ Q
= 30 0
+
R = 50 0
(iii) + X = 40 0 + Y = 70 0
+ Z = 80 0
3.
Two angles of a triangle is given, find the third angle.
0
0
0
0
0
0
0
(i)
75 , 45 0 (ii) 80 , 30
(iii) 40 , 90
(iv) 45
, 85
4.
Find the value of the unknown x in the following diagrams:

78

Geometry
5.
Find the values of the unknown x and y in the following diagrams:
6.
Three angles of a triangle are x + 5°, x + 10° and x + 15° find x.

79

Chapter 4
1.
The sum of the three angles of a triangle is 180°.
2.
In a triangle an exterior angle is equal to the sum of the two interior opposite
angles.

80

Practical Geometry
P RACTICAL G EO METRY
5.1 Construction of triangles
In the previous class, we have learnt the various types of triangles on the
basis of their sides and angles. Now let us recall the different types of triangles
and some properties of triangle.
Classification of triangles
No.
Name of Triangle
Figure
Note
1
Equilateral triangle
Three sides are equal
Isosceles triangle
2
Any two sides are equal
3
Scalene triangle
Sides are unequal
All the three angles are
4
Acute angled triangle
acute (less than 90°)
Any one of the angles is
5
Obtuse angled triangle
obtuse (more than 90°)
Any one of the angles is
6
Right angled triangle
right angle (90°)
CLASSIFICATION OF TRIANGLES
on the basis ANGLES
on the basis SIDES

81

Chapter 5
Some properties of triangle
1.
The sum of the lengths of any two sides of a triangle is greater
than the third side.
2.
The sum of all the three angles of a triangle is 180°.
To construct a triangle we need three measurements in which at least the
length of one side must be given. Let us construct the following types of triangles
with the given measurements.
(i)
Th ree sides (SSS).
(ii)
Two sides and included angle between them (SAS).
(iii)
Two angles and included side between them (ASA).
(i) To construct a triangle when three sides are given (SSS Criterion)
Example 5.1
Construct a triangle ABC given that AB = 4cm, BC = 6 cm and AC = 5 cm.
Solution
Rough Diagram
Given measurements
AB = 4cm
BC = 6 cm
AC = 5 cm.
Steps for construction
Step 1
:
Draw a line segment BC = 6cm
Step 2
:
With ‘B’ as centre, draw an arc of radius 4 cm above the line BC.
Step 3
:
With ‘C’ as centre, draw an arc of 5 cm to intersect the previous arc at ‘A’
Step 4
:
Join AB and AC.
Now ABC is the required triangle.

82

3 cm
Practical Geometry
By using protector measure all the angles of a triangle. Find
the sum of all the three angles of a triangle.
1.
A student attempted to draw a triangle with given
measurements PQ = 2cm, QR = 6cm, PR = 3 cm. (as in
Rough Figure
P
the rough figure). First he drew QR = 6cm. Then he drew
an arc of 2cm with Q as centre and he drew another arc
of radius 3 cm with R as centre. They could not intersect
each to get P.
Q
R
6 cm
(i) What is the reason?
(ii) What is the triangle property in connection with this?
The sum of any two sides of a triangle is always greater than the third side.
Draw the bisectors of the three angles of
a triangle. Check whether all of them pass
through a same point. This point is incentre.
(ii) To construct a triangle when Two sides and an angle included between
them are given. (SAS Criterion)
Example 5.2
Construct a triangle PQR given that PQ = 4 cm, QR = 6.5 cm and + PQR = 60°.
Solution
Given measurements
PQ
=
4 cm
QR
=
6.5 cm
+ PQR
=
60°
2 cm

83

Chapter 5
Steps for construction
Step 1
:
Draw the line segment QR = 6.5 cm.
Step 2
:
At Q, draw a line QX making an angle of 60 0 with QR.
Step 3
:
With Q as centre, draw an arc of radius 4 cm to cut the line (QX)
at P.
Step 4
:
Join PR.
PQR is the required triangle.
Construct a
triangle
with
the given measurements
XY = 6cm, YZ = 6cm and + XYZ = 70°. Measure the angles
of
the triangle opposite to the equal sides. What do
you
observe?
(iii) To construct a triangle when two of its angles and a side
included between them are given. (ASA criterion)
Example 5.3
Construct a triangle XYZ given that XY = 6 cm,
+ ZXY
=
30°
and
+ XYZ = 100°. Examine whether the third angle
measures 50°.
Solution
Given measurements
XY = 6 cm
+ ZXY = 30°
+ XYZ = 100°

84

Practical Geometry
Step 1
:
Draw the line segment XY = 6cm.
Step 2
:
At X, draw a ray XP making an angle of 30° with XY.
Step 3
:
At Y, draw another ray YQ making an angle of 100° with XY. The
rays XP and YQ intersect at Z.
Step 4
:
The third angle measures 50° i.e + Z = 50°.
Construct a triangle PQR given that PQ = 7 cm,
+ Q = 70°, + R = 40°.
Hint: Use the Angle Sum Property of a triangle.
Exercise : 5.1
I.
Construct the triangles for the following given measurements.
1.
Construct 3 PQR, given that PQ = 6cm, QR = 7cm, PR = 5cm.
2.
Construct an equilateral triangle with the side 7cm. Using protector measure each
angle of the triangle. Are they equal?
3.
Draw a triangle DEF such that DE = 4.5cm, EF = 5.5cm and DF = 4.5cm. Can
you indentify the type of the triangle? Write the name of it.
II.
Construct the triangles for the following given measurements.
4.
Construct 3 XYZ, given that YZ = 7cm, ZX = 5cm, + Z = 50°.
5.
Construct 3 PQR when PQ = 6cm, PR = 9cm and + P = 100°.
6.
Construct 3 ABC given that AB = 6 cm, BC = 8 cm and + B = 90° measure
length of AC.
III.
Construct the triangles for the following given measurements.
7.
Construct 3 XYZ, when + X = 50°, + Y = 70° and XY = 5cm.
8.
Construct 3 ABC when + A = 120°, + B = 30° and AB = 7cm.
9.
Construct 3 LMN, given that + L = 40°, + M = 40° and LM = 6cm. Measure and
write the length of sides opposite to the + L and + M. Are they equal? What type
of Triangle is this?

85

Chapter 6
DATA HANDLING
6.1 Mean, Median and Mode of ungrouped data
Arithmetic mean
We use the word ‘average’ in our day to day life.
Poovini spends on an average of about 5 hours daily for her studies.
In the month of May, the average temperature at Chennai is 40 degree celsius.
What do the above statements tell us?
Poovini usually studies for 5 hours. On some days, she may study for less
number of hours and on other days she may study longer.
The average temperature of 40 degree celsius means that, the temperature in
the month of May at chennai is 40 degree celsius. Some times it may be less than 40
degree celsius and at other times it may be more than 40 degree celsius.
Average lies between the highest and the lowest value of the given data.
Rohit gets the following marks in different subjects in an examination.
62, 84, 92, 98, 74
In order to get the average marks scored by him in the examination, we first
add up all the marks obtained by him in different subjects.
62 + 84 + 92 + 98 + 74 = 410.
and then divide the sum by the total number of subjects. (i.e. 5)
410 = 82
The average marks scored by Rohit =
.
5
This number helps
us
to understand the general level of his academic
achievement and is referred to as mean.
` The average or arithmetic mean or mean is defined as follows.
Sum of all observations
Mean =
Total number of observations

86

Data Handling
Example 6.1
Gayathri studies for 4 hours, 5 hours and 3 hours respectively on 3 consecutive
days. How many hours did she study daily on an average?
Solution:
Total number of study hours
Average study time =
Number of days for which she studied.
4
+ +
5
3
=
hours
3
12
=
3
=
4 hours per day.
Thus we can say that Gayathri studies for 4 hours daily on an average.
Example 6.2
The monthly income of 6 families are ` 3500, ` 2700, ` 3000, ` 2800, ` 3900 and
` 2100. Find the mean income.
Solution:
Total income of 6 familes
Average monthly income
=
Number of families
= ` 3500 2700 3000 2800 3900 2100
+
+
+
+
+
6
18000
=
`
6
=
` 3,000.
Example 6.3
The mean price of 5 pens is ` 75. What is the total cost of 5 pens?
Solution:
Total cost of 5 pens
Mean
=
Number of pens
Total cost of 5 pens
= Mean # Number of pens
=
`
75 # 5
=
` 375
Median
Consider a group of 11 students with the following height (in cm)
106, 110, 123, 125, 115, 120, 112, 115, 110, 120, 115.
The Physical EducationTeacher Mr. Gowtham wants to divide the students into
two groups so that each group has equal number of students. One group has height
lesser than a particular height and the other group has student with height greater than
the particular height.

87

Chapter 6
Now, Mr. Gowtham arranged the students according to their height in ascending
order.
106, 110, 110, 112, 115, 115,115, 120, 120, 123, 125
The middle value in the data is 115 because this value divides the students into
two equal groups of 5 students each. This value is called as median. Median refers
to the value 115 which lies in the middle of the data.Mr. Gowtham decides to keep
the middle student as a referee in the game.
Median is defined as the middle value of the data when the data is arranged
in ascending or descending order.
Find
the
actual
Find the median of the following:
distance
between
40, 50, 30, 60, 80, 70
your
school
and
Arrange the given data in ascending order.
house. Find
the
30, 40, 50, 60, 70, 80.
median of the place.
Here the number of terms is 6 which is even. So the third and fourth terms are
middle terms. The average value of the these terms is the median.
+
60
110
(i.e)
Median = 50
=
= 55 .
2
2
(i) When the number of observations is odd, the middle number is the
median.
(ii)
When the number of observations is even, the median is the average of
the two middle numbers.
Example 6.4
Find the median of the following data.
3, 4, 5, 3, 6, 7, 2.
Solution:
Arrange the data in ascending order.
In highways, the yellow
2, 3, 3, 4, 5, 6, 7
line represents the median.
The number of observation is 7 which is odd.
` The middle value 4 is the median.
Example 6.5
Find the median of the data
12, 14, 25, 23, 18, 17, 24, 20.
Solution:
Arrange the data in ascending order

88

Data Handling
12, 14, 17, 18, 20, 23, 24, 25.
T he number of observation is 8 which is even.
` Median is the average of the two middle terms 18 and 20.
18 20
+
38
Median =
=
= 19
2
2
Example 6.6
Find the median of the first 5 prime numbers.
Solution:
The first five prime numbers are 2, 3, 5, 7, 11.
The number of observation is 5 which is odd.
` The middle value 5 is the median.
Mode
Look at the following example,
Mr. Raghavan, the owner of a ready made dress shop says that the most popular
size of shirts he sells is of size 40 cm.
Observe that here also, the owner is concerned about the number of shirts of
different sizes sold. He is looking at the shirt size that is sold, the most. The highest
occurring event is the sale of size 40 cm. This value is called the mode of the data.
Mode is the variable which occurs most frequently in the given data.
Mode of Large data
Putting the same observation together and counting them is not easy if the
number of observation is large. In such cases we tabulate the data.
Example 6.7
Following are the margin of victory in the foot ball matches of a league.
1, 3, 2, 5, 1, 4, 6, 2, 5, 2, 2, 2, 4, 1, 2, 3, 2, 3, 2, 3,
1, 1, 2, 3, 2, 6, 4, 3, 2, 1, 1, 4, 2, 1,
5, 3, 4, 2, 1, 2. Find the mode of this data.
Solution:
Margin of victory
Tally Marks
Number of Matches
1
||||
||||
9
2
||||
||||
||||
14
3
||||
||
7
4
||||
5
5
||
3
6
||
2
Total
40
Table 6.1

89

Chapter 6
Now we quickly say that ‘2’ is the mode. Since 2 has occurred the more number
of times, then the most of the matches have been won with a victory margin of 2 goals.
Example 6.8
Find the mode of the following data.
3, 4, 5, 3, 6, 7
Find
the
mode of
the transport in your
Solution:
place.
3
occurs the most number of times.
` Mode of the data is 3.
Example 6.9
Find the mode of the following data.
2, 2, 2, 3, 3, 4, 5,5, 5, 6,6, 8
Solution:
2
and 5 occur 3 times.
` Mode of the data is 2 and 5.
Example 6.10
Find the mode of the following data
Find the mode
90, 40, 68, 94, 50, 60.
of the flower.
Solution:
Here there are no frequently occurring values. Hence this data has no mode.
Example 6.11
The number of children in 20 families are 1, 2, 2, 1, 2, 1, 3, 1, 1, 3
1, 3, 1, 1, 1, 2, 1, 1, 2, 1. Find the mode.
Solution:
Number of
Number of
Tally Marks
Children
Families
1
||||
||||
||
12
2
||||
5
3
|||
3
Total
20
Table 6.2
12 families have 1 child only, so the mode of the data is 1.

90

### Data Handling

Exercise: 6.1
1.
Choose the correct answer:
i)
The arithmetic mean of 1, 3, 5, 7 and 9 is
(A)
5
(B) 7
(C) 3
(D) 9
ii)
The average marks of 5 children is 40 then their total mark is
(A)
20
(B) 200
(C) 8
(D) 4
iii)
The median of 30,50, 40, 10, 20 is
(A)
40
(B) 20
(C) 30
(D) 10
iv)
The median of 2, 4, 6, 8, 10, 12 is
(A)
6
(B) 8
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
You are on page 1of 101
Chapter 1
ALGEBRA
1.1 Simple expressions with two variables
We have learnt about rectangle. Its area is l # b in which the letters 'l' and 'b'
are variables.
Variables follow the rules of four fundamental operations of numbers.
Let us now translate a few verbal phrases into expressions using variables.
Operation
Verbal phrase
Algebraic Expression
Sum of x and y
x + y
Subtraction
Difference between
a
- b ^if a > b h ^ or h
a and b
b - a (if b > a)
Multiplication
product of x and y
x # y
(or) xy
Division
p divided by q
p
p ' q ( or )
q
The following table will help us to learn some of the words (phrases) that can
be used to indicate mathematical operations:
Subtraction
Multiplication
Division
The sum of
the difference of
the product of
the quotient of
increased by
decreased by
multiplied by
divided by
plus
minus
times
the ratio of
subtracted from
more than
less than
Example 1.1
Write the algebraic expressions for the following:
1)
Twice the sum of m and n.
2)
3)
b decreased by twice a.
Nu mbers x and y both squared and added.
4)
Product of
p
and
q

2

Algebra
5)
6)
7)
8)
Two times the product of a and b divided by 5.
x more than two-third of y.
Half a number x decreased by 3.
Su m of numbers m and n decreased by their product .
9)
4 ti mes
x
less than sum of
y
and 6.
10)
Double the sum of one third of a and m.
11)
Quotient of
y
by 5 added to
x.
Solution:
1)
2 ^ m + n h
2)
b - 2a
3)
2
2
x
+
y
4)
7 +
pq
2
ab
2
5)
6)
y + x
5
3
x
7)
- 3
8)
(m + n) - mn
2
1
9) ^ y + 6 h - 4x
10)
2
`
a
+
m
j
3
y
11)
+ x
5
Express each of the following as an algebraic expression
(i)
a
times
b .
(ii) 5 multiplied by the sum of a
and
b .
(iii) Twice m decreased by n .
(iv) Four times x divided by y .
(v)
Five times p multiplied by 3 times q.
Exercise 1.1
1.
Choose the correct answer:
(i)
The sum of 5 times x, 3 times
y
and 7
(A)
5 ^ x + 3y
+
7
h
(B)
5x +
3y + 7
(C)
5x + 3 ^ y + 7 h
(D)
5x + 3 ^ 7y h
(ii)
One half of the sum of numbers a
and
b
1
1
1
1
(A)
^
a + b
h
(B)
a + b
(C)
^
a - b
h
(D)
+ a + b
2
2
2
2
(iii)
Three times the difference of x and y
(A)
3x - y
(B) 3 - x - y
(C)
xy - 3
(D) 3 ^ y - x h

3

Chapter 1
(iv)
2 less than the product of
y
and
z
(A) 2 - yz
(B) 2 + yz
(C) yz - 2
(D) 2y - z
(v)
Half of
p
added to the product of 6 and
q
p
6
q
1
1
(A)
+ 6 q
(B) p
+
(C)
^
p
+ 6
q
h
(D)
^ 6 p + q
h
2
2
2
2
2.
Write the algebraic expressions for the following using variables, constants and
arithmetic operations:
(i)
(ii)
(iii)
(iv)
Sum of x and twice y.
Subtraction of z from y.
Product of x and y increased by 4
The difference between 3 times x and 4 times y.
(v)
(vi)
(vii)
(viii)
(ix)
The sum of 10, x and y.
Product of p and q decreased by 5.
Product of numbers m and n subtracted from 12.
Sum of numbers a and b subtracted from their product.
Number 6 added to 3 times the product of numbers c and d.
(x)
Four times the product of x and y divided by 3.
1.2 Simple Linear Equations
Malar’s uncle presented her a statue. She wants to
know the weight of that statue. She used a weighing balance to
measure its weight. She knows her weight is 40kg . She finds
that the weight of the statue and potatoes balance her weight.
i.e.,
Weight of
Weight of
Malar’s
Plus
Equal
statue
potatoes
weight
s
+
15
=
40
Table 1.1
Now we will think about a balance to find the value of s.

4

Algebra
Take away 15 from both sides.
Now the balance shows the weight of the statue.
s + 15
s + 15 – 15
s
=
=
40 (f rom Table 1.1)
40 – 15 (Taking away 15 from both the sides)
=
25
So the statue weighs 25 kg.
The statement s +
15 = 40
is
an
equation. i.e., a statement in which two
mathematical expressions are equal is called an equation.
In a balance, if we take away some weight from one side, to balance it we must
take away the same weight from the other side also.
If we add some weight to one side of the balance, to balance it we must add the
same weight on the other side also.
Similarly, an equation is like a weighing balance having equal weights on
each side. In an equation there is always an equality sign. This equality sign shows
that value of the expression on the left hand side (LHS) is equal to the value of the
expression on the right hand side (RHS).
¬
Consider the equation x + 7 = 15
Here LHS
is x + 7
RHS is 15
We shall subtract 7 from both sides of the equation
x + 7 - 7 =
15 - 7
x = 8
(Subtracting 7 reduces the LHS to x)
(variable x is separated)

5

Chapter 1
¬
Consider the equation n - 3 = 10
LHS is n - 3
RHS is
10
Adding 3 to both sides, we get
n - 3 + 3
= 10 + 3
n = 13
(variable n is separated)
¬
Consider the equation 4m = 28
Divide both sides by 4
4
m
28
=
4
4
m = 7
y
¬
Consider the equation
= 6
2
Multiply both sides by 2
y
# 2 =
6 # 2
2
y = 12
So, if we add (or subtract) any number on one side of an equation, we have
to add (or subtract) the same number the other side of the equation also to keep
the equation balanced. Similarly, if we multiply (or divide) both sides by the same
non-zero number, the equation is balanced. Hence to solve an equation, one has to
perform the arithmetical operations according to the given equations to separate the
variable from the equation.
Example 1.2
Solve
3p + 4 =
25
Solution:
3p
+ 4 - 4 = 25 - 4
(Subtracting 4 from both sides of the equation)
3p = 21
3
p
21
=
(Dividing both sides by 3)
3
3
p = 7
Example 1.3
Solve 7m - 5 = 30
Solution:
7m - 5 + 5 = 30 + 5
(adding 5 on both sides)

6

Algebra
7m = 35
7
m
35
=
(Dividing both sides by 7)
7
7
m = 5
While solving equations, the commonly used operation is adding or subtracting
the same number on both sides of the equation. Instead of adding or subtracting a
number on both sides of the equation, we can transpose the number.
Transposing a number (i.e., changing the side of the number) is the same as
adding or subtracting the number from both sides. While transposing a number we
should change its sign. Let us see some examples of transposing.
Example 1.4
Solve 2a - 12 = 14
Solution:
on both sides
Transposing
2a - 12 = 14
2a - 12 = 14
2a - 12 + 12 = 14 + 12 (adding 12 on
both sides)
Transpose (– 12) from LHS to RHS
2a = 14 + 12 (on transposing
2a = 26
-
12 becomes + 12 )
2
a
26
=
(dividing both sides by 2)
2a = 26
2
2
2
a
26
=
(Dividing both sides by 2)
a = 13
2
2
a = 13
Example 1.5
Solve 5x + 3 = 18
Solution: Transposing +3 from LHS to RHS
5x
= 18 - 3
(on Transposing +3 becomes - 3 )
5x = 15
5
x
15
=
(Dividing both sides by 5)
5
5
x = 3

7

Chapter 1
Example 1.6
Solve 2 ^ x + 4 h = 12
Solution: Divide both sides by 2 to remove the brackets in the LHS.
2
^
x +
4
h
12
=
2
2
x + 4
= 6
x =
6 - 4
(transposing +4 to RHS)
x = 2
Example 1.7
Solve
- 3 ^ m - 2 h
= 18
Solution: Divide both sides by ^- 3 h to remove the brackets in the LHS.
-
3
^
m
-
2
h
18
=
- 3
- 3
m - 2
= - 6
m = - 6 + 2
(t ransposing - 2 to RHS)
m
= - 4
Example 1.8
Solve
(3x + 1 ) – 7 = 12
Solution:
(3x + 1) – 7
3x + 1 – 7
3x – 6
=
12
=
12
=
12
3x
=
12 + 6
3
x
18
=
3
3
x
=
6
Example 1.9
Solve 5x + 3 = 17 - 2x
Solution:
5x + 3 = 17 - 2x

8

Algebra
5x + 2x
= 17 - 3
(transposing + 3 to RHS and - 2x to LHS)
7x = 14
7
x
14
=
7
7
x = 2
Example 1.10
Sum of three consecutive integers is 45. Find the integers.
Solution: Let the first integer be x .
& second
integer
=
x + 1
Third integer
=
x + 1 + 1 = x +
2
Their sum
=
x + ^ x + 1 h + ^ x + 2 h = 45
3x + 3 = 45
3
x =
42
x = 14
Hence, the integers are x = 14
x + 1 = 15
x + 2 = 16
Example 1.11
A number when added to 60 gives 75. What is the number?
Solution: Let the number be x .
The equation is
60 + x
=
75
x
= 75 - 60
x
= 15
Example 1.12
20 less than a number is 80. What is the number?
Solution: Let the number be x .
The equation is
x - 20 = 80
x
= 80 + 20
x
= 100

9

Chapter 1
Example 1.13
1
of a number is 63. What is the number?
10
Solution: Let the number be x .
1
The equation is
^
x h =
63
10
1
(x)
# 10
=
63 10
#
10
x = 630
Example 1.14
A number divided by 4 and increased by 6 gives 10. Find the number.
Solution: Let the number be x .
x
The equation is
+ 6 = 10
4
x
=
10 6
-
4
x
= 4
4
x
# 4 = 4 # 4
4
\ the number is 16.
Example 1.15
Thendral’s age is 3 less than that of Revathi. If Thendral’s age is 18, what is
Revathi’ s age?
Solution: Let Revathi’s age be x
& Thendral’s age = x - 3
Given, Thendral’s age is 18 years
&
x - 3 = 18
x
=
18 + 3
x = 21
Hence Revathi’s age is 21 years.

10

Algebra
Exercise 1.2
1.
Choose the correct answer.
(i)
If
p + 3 = 9 , then p
is
(A)
12
(B) 6
(C) 3
(D) 27
(ii)
If 12 - x = 8 ,
then x
is
(A)
4
(B)
20
(C) - 4
(D) - 20
q
(iii)
If
= 7 , then
q
is
6
1
7
(A)
13
(B) 42
(C) 42
(D)
6
(iv)
If
7 ^ x - 9 h = 35 , then x
is
(A)
5
(B) - 4
(C) 14
(D) 37
(v)
Three times a number is 60. Then the number is
(A)
63
(B) 57
(C) 180
(D) 20
2.
Solve :
(i)
x -
5 = 7
(ii) a + 3 = 10
(iii)
4 + y = - 2
(iv)
b -
3 = - 5
(v)
- x = 5
(vi)
- x = - 7
(vii)
3 - x = 8
(viii)
14 - n = 10
(ix)
7 - m = - 4
(x)
20 - y = - 7
3.
Solve :
(i)
2x = 100
(ii)
3l = 42
(iii)
36 = 9x
(iv)
51 = 17a
(v)
5x = - 45
(vi)
5t = - 20
(vii)
- 7x
= 42
(viii) - 10m = - 30
(ix) - 2x = 1
(x)
- 3x = - 18
4.
Solve :
1
a
n
(i)
x
= 7
(ii)
= 5
(iii)
= - 8
2
6
3
p
- x
- m
(iv)
= 8
(v)
= 2
(vi)
= - 4
- 7
5
3
5.
Solve :
(i)
3x + 1 = 10
(ii) 11 + 2x = - 19
(iii)
4z - 3 = 17
(iv)
4a - 5 = - 41
(v)
3 ^ x + 2 h = 15
(vi)
- 4 ^2 - x h = 12
y
+ 3
x
(vii)
= 14
(viii)
+ 5 = 7
(ix)
6y = 21 - y
5
3
(x)
11m = 42 +
4m
(xi)
- 3x = - 5x + 22 (xii)
6m - 1 = 2m + 1
(xiii) 3x - 14 = x - 8
(xiv)
5x - 2x + 7 = x + 1 (xv) 5t - 3 = 3t - 5

11

Chapter 1
6.
The sum of two numbers is 33. If one number is 18, what is the other number?
7.
A number increased by 12 gives 25. Find the number.
8.
If 60 is subtracted from a number, the result is 48. Find the number.
9.
5 times a number is 60. Find the number.
10.
3 times a number decreased by 6 gives 18. Find the number.
11.
The sum of 2 consecutive integers is 75. Find the numbers.
12.
Ram’s father gave him `70. Now he has `130. How much money did Ram have
in the beginning?
13.
8 years ago, I was 27 years old. How old am I now?
Solve:
(i)
y + 18 = - 70
(ii)
- 300 + x = 100
t
(iii)
3 - 5 = - 6
(iv)
2x + 9 = 19
(v)
3x + 4 = 2x + 11
Fun game
Ram asked his friends Arun, Saranya and Ravi to think of a number and told
them to add 50 to it. Then he asked them to double it. Next he asked them to add 48 to
the answer. Then he told them to divide it by 2 and subtract the number that they had
thought of. Ram said that the number could now be 74 for all of them. Check it out if
Arun had thought of 16, Saranya had thought of 20 and Ravi had thought of 7.
Arun
Saranya
Ravi
Think of a number
x
16 20
7
x+50
Double it
2x +
100
2x +
148
Divide by 2
x + 74
Take away the number
74
you thought of

12

Algebra
1.
Algebra is a branch of
Mathematics that involves alphabet, numbers and
mathematical operations.
2.
A variable or a literal is a quantity which can take various numerical values.
3.
A quantity which has a fixed numerical value is a constant.
4.
An algebraic expression is a combination of variables and constants connected
by the arithmetic operations.
5.
Expressions are made up of terms.
6.
Terms having the same variable or product of variables with same powers are
called Like terms. Terms having different variable or product of variables with
different powers are called Unlike terms.
7.
The degree of an expression of one variable is the highest value of the exponent
of the variable. The degree of an expression of more than one variable is the
highest value of the sum of the exponents of the variables in different terms
8.
A statement in which t wo expressions are equal is called an equation.
9.
An equation remains the same if the LHS and RHS are interchanged.
10. The value of the variable for which the equation is satisfied is called the
solution of the equation.

13

Chapter 2
LIFE MATHEMATICS
2.1 Percent
In the banners put up in the shops what do you understand by 25%, 20% ?
Ramu’s mother refers to his report card to analyze his performance in
Mathematics in standard VI.
His marks in Maths as given in his report card are
17 / 25 , 36 / 50 , 75 /100 , 80 /100 , 22 / 25, 45 / 50
She is unable to find his best mark and his least mark by just looking at the
marks.
So, she converts all the given marks for a maximum of 100 (equivalent fractions
with denominator 100) as given below:
Unit Test
Monthly
Quarterly
Half - yearly
Unit Test
Monthly
1
Test 1
Exam
Exam
2
Test 2
68
72
75
80
88
90
100
100
100
100
100
100

14

Life Mathematics
Now, all his marks are out of 100. So, she is able to compare his marks easily
and is happy that Ramu has improved consistently in Mathematics in standard V I.
Now let us learn about these special fractions.
Try and help the duck to trace the path through the maze from ‘Start’ to ‘End’.
Is there more than one path?
No, there is only one path that can be traced from ‘Start’ to ‘End’.
Total number of the smallest squares
=
10 0
Nu mber of shaded squares
=
41
Nu mber of unshaded squares
=
59
Nu mber of squares traced by the path
=
__
Now, look at the table below and fill in the blanks:
Ratio
Fraction
Percent
41 out of
41
41
: 100
41%
100
100
59 out of
59
59
: 100
59%
Portion
100
100
out
Portion traced
of 100
: 100
%
by the path
100
The fraction with its denominator 100 is called a Percent.
• The word ‘Percent’ is derived from the Latin word ‘Percentum’, which means
‘per hundred’ or ‘hundredth’ or ‘out of 100’.
Percentage also means ‘percent’.
Sy mbol used for percent is %
• Any ratio x : y, where y = 100 is called ‘Percent’.

15

Chapter 2
To Express Percent in Different Forms:
Pictorial
Representation
Shaded portion represented in the form of :
Ratio
5 : 10 0
17
: 100
43
:
100
5
17
43
Fraction
100
100
100
Percent
5%
17%
43%
Exercise 2.1
1)
Write the following as a percent:
93
1
100
(i)
20: 100 (ii) 100
(iii) 11 divided by 100
(iv) 100
(v) 100
2)
Write the following percent as a ratio:
(i)
43%
(ii) 75%
(iii) 5%
(iv) 17
1
%
(v) 33 %
1
2
3
3)
Write the following percent as a fraction:
(i)
25%
(ii) 12
1
%
(iii) 33%
(iv) 70%
(v) 82%
2
Think!
Find the selling price in percentage when 25% discount is given, in the first shop.
What is the reduction in percent given in the second shop?
Which shop offers better price?

16

Life Mathematics
2.2 To Express a Fraction and a Decimal as a Percent
5
1
. 2
175
We know that 100
= 5%,
= 1.2%, 100
= 175%.
100
5
To convert 10
to a percent
5
represented pictorially can be converted to a percent as shown below:
10
5
50
10
100
Multiply the numerator and denominator by 10 to make the denominator 100
5 # 10
50
=
= 0%
5
10 10
#
100
5
This can also be done by multiplying 10 by 100%
5
`
# 100 %
j
= 5
0%
10
50% of the circle is shaded. 25% of the circle is shaded.
Try drawing circles with (i) 50%, (ii) 25% portion shaded in
different ways.
Do you know?
Less than 1 and more than 100 can also be represented as a percent.
1
%
120%
2

17

Chapter 2
(i) Fractions with denominators that can be converted to 100
Example 2.1
3
Express 5 as a percent
Solution:
1) 2 ×
____
= 100
2) 5 × 20 = _____
5
multiplied by 20 gives 100
3) 4 × 25 = _____
3
# 20
60
=
= 60 %
5
# 20
100
4) 10 ×
= 100
3
= 60 %
5) 1 ×
= 100
5
Example 2.2
1
Express 6 4
as a percent
Solution:
1
25
=
6 4
4
4
multiplied by 25 gives 100
25 # 25
625
=
= 625 %
4
# 25
100
(ii) Fractions with denominators that cannot be converted to 100
Example 2.3
Express 4 as a percent
7
Solution: Multiply by 100%
4
400
`
# 100 %
j
=
%
7
7
1
=
57 %
=
57.14%
7
Example 2.4
1
Express 3 as a percent
Solution: Multiply by 100%
1
100
`
# 100 %
j
= `
j
%
3
3
= 33 % (or)33.33%
1
3
Example 2.5
There are 250 students in a school. 55 students like basketball, 75 students
like football, 63 students like throw ball, while the remaining like cricket. What is
the percent of students who like (a) basket ball? (b) throw ball?

18

Life Mathematics
Solution:
Total number of students = 250
(a)
Nu mber of students who like basket ball = 55
55
55
out of 250 like basket ball which can be represented as 250
55
Percentage of students who like basket ball =
`
# 100 %
j
250
= 22%
(b)
Number of students who like throw ball = 63
63
63
out of 250 like throw ball and that can be represented as 250
63
Percentage of students who like throw ball =
`
# 100 %
j
250
126
=
%
= 25.2%
5
22% like basket ball, 25.2% like throw ball.
(iii) To convert decimals to percents
Example 2.6
Express 0.07 as a percent
Solution:
Multiply by 100%
(0.07 # 100)% = 7%
Aliter:
7
0.07
=
= 7
%
100
Example 2.7
Express 0.567 as a percent
Solution:
Multiply by 100%
(0.567 # 100)% = 56.7%
567 =
567
Al iter: 0.567 =
1000
10 100
#
56
.
7
=
= 56 . 7 %
100
Note: To convert a fraction or a
decimal to a percent, multiply by 100%.

19

Chapter 2
Think!
9
1.
of your blood is water. What % of your blood is not water.
10
2
2.
of your body weight is muscle. What % of body is muscle?
5
2 of your body weight is water. Is muscle weight plus water weight
3
more or less than 100 %? What does that tell about your muscles?
Exercise 2.2
1.
Choose the correct answer:
(i)
6.25 =
(A)
62.5%
(B) 6250%
(C) 625%
(D) 6.25%
(ii)
0.0003 =
(A)
3%
(B) 0.3%
(C) 0.03%
(D) 0.0003%
(iii)
5 =
20
1
(A)
25%
(B)
(C) 0.25%
(D) 5%
4 %
(iv)
The percent of 20 minutes to 1 hour is
(A)
33 1 3
(B) 33
(C)
33 2 3
(D) none of these
(v)
The percent of 50 paise to Re. 1 is
1
(A)
500
(B)
(C) 50
(D) 20
2
2.
Convert the given fractions to percents
20
9
2
5
i)
ii) 50
iii) 5 1
iv)
20
4
3
v) 11
3.
Convert the given decimals to percents
i)
0.36
ii) 0.03
iii) 0.071
iv) 3.05
v) 0.75
4.
In a class of 35 students, 7 students were absent on a particular day. What
percentage of the students were absent?
5.
Ram bought 36 mangoes. 5 mangoes were rotten. What is the percentage of the
mangoes that were rotten?
6.
In a class of 50, 23 were girls and the rest were boys. What is the percentage of
girls and the percentage of boys?
7.
Ravi got 66 marks out of 75 in Mathematics and 72 out of 80 in Science. In
which subject did he score more?
8.
Shyam’s monthly income is `12,000. He saves `1,200 Find the percent of his
savings and his expenditure.

20

Life Mathematics
2.3 To Express a Percent as a Fraction (or) a Decimal
i) A percent is a fraction with its denominator 100. While expressing it as a
fraction, reduce the fraction to its lowest term.
Example 2.8
Express 12% as a fraction.
Solution:
12
12%
=
(reduce the fraction to its lowest terms)
100
3
=
25
Example 2.9
Percents that have easy
Express 233
1
% as a fraction.
3
fractions
Solution:
1
50%
=
2
700
233 %
1
=
%
1
3
3
25%
=
4
700
7
=
=
1
1
33 3 %
=
3
# 100
3
3
=
1
2 3
Find more of this kind
Example 2.10
1
Express 4 % as a fraction
Solution:
1
1
1
% =
=
4
4 # 100
400
(ii) A percent is a fraction with its denominator 100. To convert this fraction to
a decimal, take the numerator and move the decimal point to its left by 2 digits.
Example 2.11
Express 15% as a decimal.
Solution:
15
15%
=
= 0 . 15
100
Example 2.12
Express 25.7% as a decimal.
Solution:
25
. 7
25.7%
=
100
=
0.257

21

Chapter 2
Math game - To make a triplet (3 Matching cards)
This game can be played by 2 or 3 people.
Write an equivalent ratio and decimal for each of the given percent in different
cards as shown.
5%
1 : 20
0.05
1
1
: 3
0.33
33 3
%
Make a deck of 48 cards (16 such sets of cards) - 3 cards to represent one
particular value - in the form of %, ratio and decimal.
Shuffle the cards and deal the entire deck to all the players.
Players have to pick out the three cards that represent the same value of percent,
ratio and decimal and place them face up on the table.
The remaining cards are held by the players and the game begins.
One player chooses a single unknown card from the player on his left. If this
card completes a triple (3 matching cards) the 3 cards are placed face up on the table.
If triplet cannot be made, the card is added to the player’s hand. Play proceeds to the
left.
Players take turns to choose the cards until all triplets have been made.
The player with the most number of triplets is the winner.
TO FIND T HE VALUES OF PERCENTS
Colour 50% of the circle green and 25% of the circle red.
50
1
50% =
=
of the circle is to be coloured green.
100
2
25
25
1
Similarly, 25% =
=
=
100
100
4
1
of the circle is to be coloured red.
4

22

Life Mathematics
1
1
Now, try colouring 2
of the square, green and 4
of the
square, red.
Do you think that the green coloured regions are equal in
both the figures?
No, 50% of the circle is not equal to 50% of the square.
Similarly the red coloured regions, 25% of the circle is not equal to 25% of
the square.
Now, let’s find the value of 50% of `100 and 50% of `10.
What is 50% of `100?
What is 50% of `10?
50
1
50
1
50%
=
=
50% =
=
100
2
100
2
1
1
1
1
So, 2
of 100 =
# 100 = 50
of 10
=
#
10
=
5
2
2
2
50% of `100 = `50
50% of `10 = `5
Example 2.13
Find the value of 20% of 1000 kg.
Solution:
20
20% of 1000
=
of 1000
100
20
=
# 1000
100
20% of 1000 kg
=
20 0 kg.
Example 2.14
1
Find the value of
%
of 200.
2
Solution:
1
2
=
of 200
100
1
=
# 200
2
# 100
1
# 200 1
=
200
1
% of 200 1
=
2

23

Chapter 2
Example 2.15
Find the value of 0.75% of 40 kg.
Solution:
0
. 75
0.75%
=
100
0
.
75
0.75% of 40
=
# 40
100
3
=
= 0 . 3
10
0.75% of 40kg
=
0.3kg.
Example 2.16
In a class of 70, 60% are boys. Find the number of boys and girls.
Solution:
Total number of students
=
70
Nu mber of boys
=
60% of 70
60
=
# 70
100
=
42
Nu mber of boys
=
42
Nu mber of girls
=
Total students – Number of boys
=
70 – 42
=
28
Nu mber of girls
=
28
Example 2.17
In 2010, the population of a town is 1,50,000. If it is increased by 10% in the
next year, find the population in 2011.
Solution:
Population in 2010
=
1,50,000
10
Increase in population
=
# 1 , 50 , 000
100
=
15,000
Population in 2011
=
150000 + 15000
=
1,65,000

24

Life Mathematics
Exercise 2.3
1.
Choose the correct answer:
(i)
The common fraction of 30 % is
1
7
3
3
(A)
(B)
(C)
(D) 10
10
10
100
1
(ii)
The common fraction of
%
is
2
1
1
(A)
(B)
(C)
(D) 100
2
200
100 200
(iii)
The decimal equivalent of 25% is
(A)
0.25
(B) 25
(C) 0.0025
(D) 2.5
(iv)
10% of `300 is
(A)
`10
(B) `20
(C) `30
(D) `300
(v)
5% of `150 is
(A)
`7
(B) `7.50
(C) `5
(D) `100
2.
Convert the given percents to fractions:
1
i)
9%
ii) 75%
iii)
%
iv) 2.5%
v) 66 %
2
3
4
3.
Convert the given percents to decimals:
i)
7%
ii) 64%
iii) 375%
iv) 0.03%
v) 0.5%
4.
Find the value of:
i)
75% of 24
iv) 72% of 150
ii) 33 3 % of `72
v) 7.5% of 50kg
1
iii) 45% of 80m
5.
Ram spent 25% of his income on rent. Find the amount spent on rent, if his
income is `25,000.
6.
A team played 25 matches in a season and won 36% of them. Find the number of
matches won by the team.
7.
The population of a village is 32,000. 40% of them are men. 25% of them are
women and the rest are children. Find the number of men and children.
8.
The value of an old car is `45,000. If the price decreases by 15%, find its new
price.
9.
The percentage of literacy in a village is 47%. Find the number of illiterates in the
village, if the population is 7,500.

25

Chapter 2
Think!
1)
Is it true?
20% of 25 is same as 25% of 20.
2)
The tax in a restaurant is 1.5% of your total bill.
a)
Write the tax % as a decimal.
b)
A family of 6 members paid a bill of ` 750. What is the tax for
their bill amount?
c)
W hat is the total amount that they should pay at the restaurant?
2.4 Profit and Loss
Ram & Co. makes a profit of `1,50,000 in 2008.
Ram & Co. makes a loss of `25,000 in 2009.
Is it possible for Ram & Co. to make a profit in the first year and a loss in the
subsequent year?
Different stages of a leather product - bag are shown below:
Factory
Wholesale Dealer
Retailer
Where are the bags produced?
Do the manufactures sell the products directly?
Whom does the products reach finally?
Raja, the fruit stall owner buys fruits from the wholesale market and sells
them in his shop.
On a particular day, he buys apples, mangoes and bananas.

26

Life Mathematics
Each fruit has two prices, one at each shop, as shown in the price list.
The price at which Raja buys the fruit at the market is called the Cost Price
(C.P.). The price at which he sells the fruit in his stall is called the Selling Price (S.P.).
From the price list we can say that the selling price of the apples and the mangoes
in the shop are greater than their respective cost price in the whole sale market. (i.e.)
the shopkeeper gets some amount in addition to the cost price. This additional amount
is called the profit.
Selling Price of mango
=
Cost Price of mango + Profit
Selling price – Cost price
=
Profit
Profit
=
Selling Price – Cost Price
=
15 – 10
Profit
=
`5
i.e., Profit
=
Selling Price – Cost Price
In case of the apples,
Selling price of apple > Cost price of apple, there is a profit.
Profit
=
S.P. – C.P.
=
8 – 6
Profit
=
`2
As we know, bananas get rotten fast, the shop keeper wanted to sell them without
wasting them. So, he sells the bananas at a lower price (less than the cost price). The
amount by which the cost is reduced from the cost price is called Loss.
In case of bananas,
Cost price of banana > selling price of banana, there is a loss.
S.P. of the banana
=
C.P. of the banana – Reduced amount
S.P.
=
C.P. – Loss
Loss
=
C.P. – S.P.
Loss
=
3 – 2
Loss
=
`1

27

Chapter 2
So, we can say that
When the selling price of an article is greater than its cost price, then
there is a profit.
Profit = Selling Price - Cost Price
When the cost price of an article is greater than its selling price, then
there is a loss.
Loss = Cost Price – Selling Price
S.P = C.P + Profit
S.P = C.P - Loss.
To find Profit / Loss %
Rakesh buys articles for `10,000
and sells them for `11,000 and makes
a profit of `1,000, while Ramesh buys
articles for `1,00,000 and sells them
for `1,01,000 and makes a profit of
1)
Any fraction with its denominator
`1,000.
100 is called __________
2)
%
Both of them have made the
1 2 =
3)
35% =
( in fraction)
same amount of profit. Can you say
both of them are benefited equally?
4)
0.05 =
%
1
No.
5)
= __
%
4
To find who has gained more,
we need to compare their profit based on their investment.
We know that comparison becomes easier when numbers are expressed in
percent. So, let us find the profit %
Rakesh makes a profit of `1,000, when he invests `10,000.
Profit of `1,000 out of `10,000
1000
For each 1 rupee, he makes a profit of 10000
1000
Therefore for `100, profit =
# 100
10, 000
Profit =
10%

28

Life Mathematics
Ramesh makes a profit of `1000, when he invests `1,00,000.
1000
Profit of 1000 out of 1,00,000
=
100000
1000
Profit
=
# 100 = 1%
100000
So, from the above we can say that Rakesh is benefited more than Ramesh.
Profit #
So,
Profit Percentage
=
100
C.P
Loss % is also calculated in the same way.
Loss #
Loss Percentage
=
100
C.P.
Profit Percentage or Loss Percentage is always calculated
on the cost price of the article.
Example 2.18
A dealer bought a television set for `10,000 and sold it for `12,000. Find the
profit / loss made by him for 1 television set. If he had sold 5 television sets, find the
total profit/loss
Solution:
Selling Price of the television set
= `12,000
Cost Price of the television set
= `10,000
S.P.
>
C.P, there is a profit
Profit
=
S.P. – C. P.
=
12000 – 10000
Profit
= `2,000
Profit on 1 television set
= `2,000
Profit on 5 television sets
=
20 00 × 5
Profit on 5 television sets
= `10,000
Example 2.19
Sanjay bought a bicycle for `5,000.
He sold it for `600 less after two years.
Find the selling price and the loss percent.
Solution:
Cost Price of the bicycle
=
`50 00

29

Chapter 2
Loss
=
`60 0
Selling Price
=
Cost Price – Loss
=
50 00 – 600
Selling Price of the bicycle
=
`440 0
Loss #
Loss
=
100
C.P.
600
=
# 100
5000
=
12%
Loss
=
12%
Example 2.20
A man bought an old bicycle for `1,250. He spent `250 on its repairs. He then
sold it for `1400. Find his gain or loss %
Solution:
Cost Price of the bicycle
=
`1,250
Repair Charges
=
`250
Total Cost Price
=
1250 + 250 = `1,500
Selling Price
=
`1,400
C.P.
>
S.P., there is a Loss
Loss
=
Cost Price – Selling Price
=
1500 – 1400
=
10 0
Loss
=
`10 0
Loss #
Percentage of the loss
=
100
C.P.
100
=
# 100
1500
20
=
3
=
6 2 % (or) 6.67%
3
Loss
=
6.67%

30

Life Mathematics
Example 2.21
A fruit seller bought 8 boxes of grapes at `150 each. One box was damaged.
He sold the remaining boxes at `190 each. Find the profit / loss percent.
Solution:
Cost Price of 1 box of grapes
=
`150
Cost Price of 8 boxes of grapes
=
150 # 8
= `1200
Nu mber of boxes damaged
=
1
Nu mber of boxes sold
=
8 – 1
=
7
Selling Price of 1 box of grapes
=
`190
Selling Price of 7 boxes of grapes = 190 × 7
= `1330
S.P.
>
C.P, there is a Profit.
Profit
= Selling Price – Cost Price
=
1330 – 1200
=
130
Profit
= `130
Profit #
Percentage of the profit
=
100
C.P
130
=
# 100
1200
=
10.83
Profit
=
10.83 %
Example 2.22
Ram, the shopkeeper bought a pen for `50 and then sold it at a loss of `5. Find
his selling price.
Solution:
Cost price of the pen
=
`50
Loss
=
`5

31

Chapter 2
S.P.
=
C.P. – Loss
=
50 – 5
=
45
Selling price of the pen
=
`45.
Example 2.23
Sara baked cakes for the school festival. The cost of one cake was `55. She
sold 25 cakes and made a profit of `11 on each cake. Find the selling price of the
cakes and the profit percent.
Solution:
Cost price of 1 cake
=
`55
Nu mber of cakes sold
=
25
Cost price of 25 cakes
=
55 × 25 = `1375
Profit on 1 cake
=
`11
Profit on 25 cakes
=
11 × 25 = ` 275
S.P.
=
C.P. + Profit
=
1375 + 275
=
1,650
= `1,650
Profit #
Percentage of the profit
=
100
C. P
275
=
# 100
1375
=
20
Profit
=
20 %
Exercise 2.4
1.
Choose the correct answer:
i)
If the cost price of a bag is `575 and the selling price is `625, then
there is a profit of `
(A)
50
(B) 575
(C) 625
(D) none of these
ii)
If the cost price of the box is `155 and the selling price is `140, then there is a
loss of `
(A)
155
(B) 140
(C) 15
(D) none of these

32

Life Mathematics
iii)
If the selling price of a bag is `235 and the cost price is `200, then
there is a
(A)
profit of `235
(B) loss of `3
(C)
profit of `35
(D) loss of `200
iv)
Gain or loss percent is always calculated on
(A)
cost price
(B) selling price
(C) gain
(D) loss
v)
If a man makes a profit of `25 on a purchase of `250, then profit% is
(A)
25
(B) 10
(C) 250
(D) 225
2.
Complete the table by filling in the appropriate column:
C.P.
S.P.
Profit
Loss
`
`
`
`
144
168
59
38
600
635.45
26599
23237
107.50
100
3.
Find the selling price when cost price and profit / loss are given.
i)
Cost Price = `450
ii) Cost Price = `760
iii) Cost Price = `980
iv) Cost Price = `430
Profit = `80
Loss = `140
Profit = `47.50
Loss = `93.25
v)
Cost Price = `999.75
Loss = `56.25
4.
Vinoth purchased a house for `27, 50,000. He spent `2,50,000 on repairs and
painting. If he sells the house for `33,00,000 what is his profit or loss % ?
5.
A shop keeper bought 10 bananas for `100. 2 bananas were rotten. He sold the
remaining bananas at the rate of `11 per banana. Find his gain or loss %
6.
A shop keeper purchased 100 ball pens for `250. He sold each pen for
`4. Find the profit percent.
7.
A vegetable vendor bought 40 kg of onions for `360. He sold 36 kg at `11 per
kg. The rest were sold at `4.50 per kg as they were not very good. Find his
profit / loss percent.
Choose one product and find out the different stages it
crosses from the time it is produced in the factory to
the time it reaches the customer.

33

Chapter 2
Think!
Do you think direct selling by the manufacturer himself is more beneficial for
the costumers? Discuss.
Do it yourself
1.
A trader mixes two kinds of oil, one costing `100 per Kg. and the other costing
`80 per Kg. in the ratio 3: 2 and sells the mixture at `101.20 per Kg. Find his
profit or loss percent.
2.
Sathish sold a camera to Rajesh at a profit of 10 %. Rajesh sold it to John at a
loss of 12 %. If John paid `4,840, at what price did Sathish buy the camera?
3.
The profit earned by a book seller by selling a book at a profit of 5% is `15
more than when he sells it at a loss of 5%. Find the Cost Price of the book.
2.5 Simple Interest
Deposit `10,000 now. Get `20,000 at the end of 7 years.
Deposit `10,000 now. Get `20,000 at the end of 6 years.
Is it possible? What is the reason for these differences?
Lokesh received a prize amount of `5,000 which he deposited in a bank in
June 2008. After one year he got back `5,400.
Why does he get more money? How much more does he get?
If `5,000 is left with him in his purse, will he gain `400?
Lokesh deposited `5,000 for 1 year and received `5,400 at the end of the first
year.
When we borrow (or lend) money we pay (or receive) some additional amount
in addition to the original amount. This additional amount that we receive is termed
as Interest (I).

34

Life Mathematics
As we have seen in the above case, money can be borrowed deposited in
banks to get Interest.
In the above case, Lokesh received an interest of `400.
The amount borrowed / lent is called the Principal (P). In this case, the amount
deposited - `5,000 is termed as Principal (P).
The Principal added to the Interest is called the Amount (A).
In the above case, Amount =
Pr incipal +Interest
=
`5000 + `400 = `5,400.
Will this Interest remain the same always?
Definitely not. Now, look at the following cases
(i)
If the Principal deposited is increased from `5,000 to `10,000, then
will the interest increase?
(ii)
Si milarly, if `5,000 is deposited for more number of years, then will
the interest increase?
Yes in both the above said cases, interest will definitely increase.
From the above, we can say that interest depends on principal and duration of
time. But it also depends on one more factor called the rate of interest.
Rate of interest is the amount calculated annually for `100
(i.e.) if rate of interest is 10% per annum, then interest is `10 for `100 for 1 year.
So, Interest depends on:
Amount deposited or borrowed – Principal (P)
Period of time - mostly expressed in years (n)
Rate of Interest (r)
This Interest is termed as Simple Interest because it is always calculated on the
initial amount (ie) Principal.
Calculation of Interest
If ‘r’ is the rate of interest, principal is `100, then Interest
r
for 1 year
= 100 # 1 #
100
r
for 2 years
= 100 # 2 #
100
r
for 3 years
= 100 # 3 #
100
r
for n years
= 100 # n #
100

35

Chapter 2
Pnr
So,
I
=
100
A
=
P + I
Pnr
A
=
P
+
100
nr
A
=
P
`
1 +
j
100
Interest
=
Amount – Principal
I
=
A – P
The other formulae derived from
Pnr
I
=
are
100
100 I
r
=
P n
100 I
n
=
P r
100I
P
=
rn
Note: ‘n’ is always calculated in years. When ‘n’ is given in months \ days,
convert it into years.
Fill in the blanks
Principal
Interest
Amount
`
`
`
5,000
500
12,500
17,500
6,000
25,000
8,450
750
12,000
15,600
Example 2.24
Kamal invested `3,000 for 1 year at 7 % per annum. Find the simple interest
and the amount received by him at the end of one year.
Solution:
Pr incipal (P)
=
`3,000
Nu mber of years (n)
=
1
Rate of interest (r)
=
7 %

36

Life Mathematics
= Pnr
Interest (I)
100
3000
# 1 # 7
=
100
I
=
`210
A
=
P + I
=
30 00 + 210
A
=
`3,210
Example 2.25
Radhika invested `5,000 for 2 years at 11 % per annum. Find the simple interest
and the amount received by him at the end of 2 years.
Solution:
Pr incipal (P)
=
`5,000
Nu mber of years (n)
=
2 years
Rate of interest (r)
=
11 %
= Pnr
I
100
5000
# 2 # 11
=
100
=
1100
I
=
`1,100
Amount (A)
=
P + I
=
50 00 + 1100
A
=
`6,100
Example 2.26
Find the simple interest and the amount due on `7,500 at 8 % per annum for
1 year 6 months.
Know this
Solution:
12 months = 1 years
P
=
`7,500
6
6 months
=
year
n
=
1 yr 6 months
12
1
6
=
year
=
1
yrs
2
12
3
3 months
=
year
1
3
12
=
1
=
yrs
2
2
1
=
year
r
=
8 %
4

37

Chapter 2
= Pnr
I
100
3
7500 # # 8
2
=
100
7500 3 8
# #
=
2 # 100
=
900
I
=
`90 0
A
=
P + I
=
7500 + 900
=
`8,400
Interest
=
`90 0, Amount = `8,400
Aliter:
P
=
`7,500
3
n
=
years
2
r
=
8 %
nr
A
= P
`
1 +
j
100
3
# 8
e
2
o
=
7500 1 +
100
3 # 8
=
7500 1 +
c
m
2 # 100
28
= 7500
`
j
25
=
300 # 28
= 8400
A
=
`840 0
I
=
A – P
=
8400 – 7500
=
900
I
=
`90
0
Interest
=
`90
0
Amount
=
`8,400

38

Life Mathematics
Example 2.27
Find the simple interest and the amount due on `6,750 for 219 days at 10 %
per annum.
Solution:
P
=
`6,750
n
=
219 days
Know this
3
365 days
=
1 year
=
219 year =
year
365
5
219 days
219 year
r
=
10 %
= 365
3
= Pnr
=
year
I
5
100
73
73 days
=
year
6750 # 3 # 10
I
=
365
5 # 100
1
=
year
=
405
5
I
=
`405
A
=
P + I
=
6750 + 405
=
7,155
A
=
`7,155
Interest
=
`405, Amount = `7,155
Example 2.28
Rahul borrowed `4,000 on 7th of June 2006 and returned it on 19th August2006.
Find the amount he paid, if the interest is calculated at 5 % per annum.
Solution:
P
=
`4,000
r
=
5 %
Know this
Nu mber of days,
Ju ne
=
24 (30 – 6)
Thirty days
hath
September, April, June
July
=
31
and November. All the
rest have thirty
-
one
August
=
18
except February.
Total number of days
=
73
n
=
73 d ays

39

Chapter 2
73
=
year
365
1
=
year
5
nr
A
=
P
`
1 +
j
100
1
# 5
=
4000 1 +
c
m
5 # 100
1
=
4000 1 +
`
j
100
101
= 4000
`
j
100
=
4,040
Amount
=
`4,040
Example 2.29
Find the rate percent per annum when a principal of `7,000 earns a S.I. of
`1,680 in 16 months.
Solution:
P
=
`7,000
n
=
16 months
16
4
=
yr
=
yr
12
3
I
=
`1,680
r
=
?
r = 100 I
P n
100
#
1680
=
4
7000 #
3
100
#
1680 3
#
=
7000 # 4
=
18
r
=
18 %
Example 2.30
Vijay invested `10,000 at the rate of 5 % simple interest per annum.
He
received `11,000 after some years. Find the number of years.
Solution:
A
=
`11,000
P
=
`10,000

40

Life Mathematics
r
=
5 %
n
=
?
I
=
A – P
=
11,000 – 10,000
=
1,000
I
=
`1000
= 100 I
n
P r
100 1000
#
=
10000 # 5
n
=
2 years.
Aliter:
nr
A
=
P
`
1 +
j
100
n # 5
11000
=
10
000
`
1 +
j
100
11000
n
=
1
+
10000
20
11
= 20 + n
10
20
11
# 20 =
20 + n
10
22
=
20 + n
22 – 20
=
n
n
=
2 years
Example 2.31
A sum of money triples itself at 8 % per annum over a certain time. Find the
number of years.
Solution:
Let Principal be `P.
Amount
=
tr iple the principal
=
`3 P
r
=
8 %
n
=
?

41

Chapter 2
I
=
A – P
=
3P – P
=
2P
I
=
`2 P
100I
n
=
Pr
100 # 2 P
=
P # 8
n
=
25 years
Nu mber of years
=
25
A liter:
Let Principal be `100
Amount
=
3 # 100
=
`300
I
=
A – P
=
30 0 - 100
I
=
`200.
n = 100 I
100 200
#
=
P r
100 # 8
200
n
=
= 25
8
Nu mber of years
=
25.
Example 2.32
A certain sum of money amounts to `10,080 in 5 years at 8 % . Find the principal.
Solution:
A
=
`10,080
n
=
5 years
r
=
8 %
P
=
?
nr
A
=
P
`
1 +
j
100
5 # 8
10 080
=
P 1
`
+
j
100

42

Life Mathematics
7
10 080
=
P ` j
5
5
10080 #
=
P
7
7,200
=
P
Pr incipal
=
`7,200
Example 2.33
A certain sum of money amounts to `8,880 in 6 years and `7,920 in 4 years
respectively. Find the principal and rate percent.
Solution:
Amount at the end of 6 years
=
Pr incipal + interest for 6 years
=
P
+
I
=
8880
6
Amount at the end of 4 years
=
Pr incipal + Interest for 4 years
=
P
+
I
=
7920
4
=
8880 – 7920
I 2
= 960
Interest at the end of 2 years
=
`960
= 960
Interest at the end of 1st year
2
=
480
Interest at the end of 4 years
=
480 # 4
=
1,920
P +
=
7920
I 4
P + 1920 =
7920
P
=
7920 - 1920
P
=
6,000
Pr incipal
=
`6,000
100I
r
=
pn
100 1920
#
=
6000 # 4
r
=
8 %

43

Chapter 2
Exercise 2.5
1.
Choose the correct answer:
i)
Simple Interest on `1000 at 10 % per annum for 2 years is
(A)
`1000
(B) `200
(D) `2000
ii)
If Amount = `11,500,
(C) `100
Principal = `11,000, Interest is
(A)
`500
(B) `22,500
(C) `11,000
(D) `11,000
iii)
6 months =
1
1
3
(A)
yr
(B)
yr
(C)
yr
(D) 1 yr
2
4
4
iv)
292 days =
1
3
4
2
(A)
yr
(B) 5
yr
(C)
yr
(D)
yr
5
5
5
v)
If P = `14000
I = `1000, A is
(A)
`15000
(B) `13000
(C) `14000
(D) `1000
2.
Find the S.I. and the amount on `5,000 at 10 % per annum for 5 years.
3.
Find the S.I and the amount on `1,200 at 12
1
% per annum for 3 years.
2
4.
Lokesh invested `10,000 in a bank that pays an interest of 10 % per annum. He
withdraws the amount after 2 years and 3 months. Find the interest, he receives.
5.
Find the amount when `2,500 is invested for 146 days at 13 % per annum.
nd
6.
Fi nd the S.I and amount on `12,000 from May 21 st 1999 to August 2 1999 at
9 % per annum.
7.
Sathya deposited `6,000 in a bank and received `7500 at the end of 5 years.
Find the rate of interest.
1
8.
Fi nd the principal that earns `250 as S.I. in 2 2
years at 10 % per annum.
9.
In how many years will a sum of `5,000 amount to `5,800 at the rate of
8 % per annum.
10.
A su m of money doubles itself in 10 years. Find the rate of interest.

44

Life Mathematics
1
11.
A sum of money doubles itself at 12
%
per annum over a certain period of
2
time. Find the number of years.
12.
A certain sum of money amounts to `6,372 in 3 years at 6 % Find the principal.
1
13.
A certain sum of money amounts to `6,500 in 3 years and `5,750 in 1 2
years
respectively . Find the principal and the rate percent.
14.
Fi nd S.I. and amount on ` 3,600 at 15% p.a. for 3 years and 9 months.
1
15.
Fi nd the principal that earns ` 2,080 as S.I. in 3 4
years at 16% p.a.
Think!
9
1)
Fi nd the rate percent at which, a sum of money becomes 4
times
in 2 years.
2)
If Ram needs `6,00,000 after 10 years, how much should he
invest now in a bank if the bank pays 20 % interest p.a.
1.
A fraction whose denominator is 100 or a ratio whose second term is 100 is
termed as a percent.
2.
Percent means per hundred, denoted by %
3.
To convert a fraction or a decimal to a percent, multiply by 100.
4.
The price at which an article is bought is called the cost price of an article.
5.
The price at which an article is sold is called the selling price of an article.
6.
If the selling price of an article is more than the cost price, there is a profit.

45

Chapter 2
7.
If the cost price of an article is more than the selling price, there is a loss.
8.
Total cost price = Cost Price + Repair Charges / Transportation charges.
9.
Profit or loss is always calculated for the same number of ar ticles or same
units.
10.
Profit = Selling Price – Cost Price
11.
Loss = Cost Price – Selling Price
Profit #
12.
Profit% =
100
C.P.
Loss #
13.
Loss% =
100
C.P.
14.
Selling Price = Cost Price + Profit
15.
Selling Price = Cost Price - Loss
Pnr
16.
Simple interest is I =
100
17.
A
=
P + I
Pnr
=
P +
100
nr
=
P 1 +
`
j
100
18.
I
=
A – P
100I
19.
P
=
nr
100
I
20.
r
=
P n
100I
21.
n
=
Pr

46

Measurements
MEASUREMENTS
3.1 Trapezium
A trapezium is a quadrilateral with one pair of
opposite sides are parallel.
The distance between the parallel sides is the
height of the trapezium. Here the sides AD and BC
Fig. 3.1
are not parallel, but AB || DC.
If the non - parallel sides of a trapezium are equal ( AD = BC ), then it is known
as an isosceles trapezium.
Here
+ A
=
+ B ;
+ C =
+ D
AC
=
BD
+ A + + D
=
180° ;
+ B + + C = 180°
Fig. 3.2
Area of a trapezium
ABCD is a trapezium with parallel sides AB and DC measuring ‘a’ and ‘b’.
Let the distance between the two parallel sides be ‘h’. The diagonal BD divides the
trapezium into two triangles ABD and BCD.
Area of the trapezium
=
area of D ABD + area of D BCD
1
1
=
# AB # h +
# DC #
h
2
2
1
=
# h
6AB + DC @
Fig. 3.3
2
1
=
# h a
6
+
b
@ sq. units
2
1
` Area of a trapezium =
# height # (sum of the parallel sides) sq. units
2
Example 3.1
Find the area of the trapezium whose height is 10 cm and the parallel sides are
12 cm and 8 cm of length.

47

Chapter 3
Solution
Given: h = 10 cm, a = 12 cm, b = 8 cm
1
Area of a trapezium
=
# h a
^
+
b
h
2
1
=
# 10 # ^12 + 8 h = 5
# ^20 h
2
\ A rea of the trapezium
=
10 0 sq. cm 2
Example 3.2
The length of the two parallel sides of a trapezium are 15 cm and 10 cm. If its
area is 100 sq. cm. Find the distance between the parallel sides.
Solution
Given:
a = 15 cm, b = 10 cm, Area = 100 sq. cm.
Area of the trapezium
=
10 0
1
h a
^
+
b
h
=
100
2
1
# h # ^15 + 10 h
=
100
2
h # 25
=
200
200 = 8
h
=
25
` the distance between the parallel sides = 8 cm.
Example 3.3
The area of a trapezium is 102 sq. cm and its height is 12 cm. If one of its
parallel sides is 8 cm. Find the length of the other side.
Solution
Given:
Area = 102 cm 2 , h = 12 cm, a = 8 cm.
Area of a trapezium
=
102
1
h a
^
+
b
h
=
102
2
1
# 12 # ^8 + b
h
=
102
2
6 (8 + b)
=
102
8 + b
=
17
&
b = 17 – 8 = 9
` length of the other side =
9 cm
By paper folding method:
In a chart paper draw a trapezium ABCD of any measure. Cut and take the
trapezium separately. Fold the trapezium in such a way that DC lies on AB and crease
it on the middle to get EF.

48

Measurements
E
Fig. 3.4
EF divides the trapezium in to two parts as shown in the Fig. 4.40 (ii)
From D draw DG = EF. Cut the three parts separately.
Arrange three parts as shown in the Fig. 3.4 (iii)
The figure obtained is a rectangle whose length is AB + CD = a + b
1
1
and breadth is 2 ( height of trapezium ) =
h
2
` Area of trapezium
= area of rectangle as shown in Fig. 4.40 (iii)
= length # breadth
1
=
^
a + b
h
`
h
j
2
1
=
h a
^
+
b
h sq. units
2
Exercise 3.1
1.
Choose the correct answer.
i)
The area of trapezium is
____________
sq. units
1
1
(A)
h ^ a + b h
(B)
h (a + b)
(C)
h ^ a - b h
(D)
h (a – b)
2
2
ii)
In an isosceles trapezium
(A)
non parallel sides are equal
(B) parallel sides are equal
(C)
height = base
(D) parallel sides = non parallel sides
iii)
Th e sum of parallel sides of a trapezium is 18 cm and height is 15 cm. Then
its area is
(A)
105 cm 2
(B) 115 cm 2
(C) 125 cm 2
(D) 135 cm 2
iv)
The height of a trapezium whose sum of parallel sides is 20 cm and the area
8 0 cm 2 is
(A)
2 cm
(B) 4 cm
(C) 6 cm
(D) 8 cm

49

Chapter 3
2.
Find the area of a trapezium whose altitudes and parallel sides are given below:
i) altitude = 10 cm, parallel sides = 4 cm and 6 cm
ii) altitude = 11 cm, parallel sides = 7.5 cm and 4.5 cm
iii) altitude = 14 cm, parallel sides = 8 cm and 3.5 cm
3.
The area of a trapezium is 88 cm 2 and its height is 8 cm. If one of its parallel side
is 10 cm. Find the length of the other side.
4.
A garden is in the form of a trapezium. The parallel sides are 40 m and 30 m.
The perpendicular distance between the parallel side is 25 m. Find the area of the
garden.
5.
Area of a trapezium is 960 cm 2 . The parallel sides are 40 cm and 60 cm. Find the
distance between the parallel sides.
3.2 Circle
In our daily life, we come across a number of objects like wheels, coins, rings,
bangles,giant wheel, compact disc (C.D.)
What is the shape of the above said objects?
‘round’, ‘round’, ‘round’
Yes, it is round. In Mathematics it is called a circle.
Now, let us try to draw a circle.
Take a thread of any length and fix one end tightly at
a point O as shown in the figure. Tie a pencil (or a chalk) to
the other end and stretch the thread completely to a point A.
Holding the thread stretched tightly, move the pencil.
Stop it when the pencil again reaches the point A. Now see
Fig. 3.5
the path traced by the pencil.
Is the path traced by the pencil a circle or a straight line?
‘Circle’
Yes, the path traced by the point, which moves at a constant distance from a
fixed point on a given plane surface is called a circle.
Parts of a Circle
The fixed point is called the centre of the circle.
The constant distance between the fixed point and
the moving point is called the radius of the circle.
i.e. The radius is a line segment with one end point
at the centre and the other end on the circle. It is denoted
by ‘ r ’.
A line segment joining any two points on the circle
is called a chord.
Fig. 3.6

50

Measurements
Diameter is a chord passing through the centre of the circle. It is denoted by ‘ d ’.
The diameter is the longest chord. It is twice the radius.(i.e. d = 2r )
The diameter divides the circle into two equal parts. Each equal
part is a
semicircle.
Think it:
How many diameters can
a circle have ?
All the radii of a circle are equal.
Circumference of a circle:
Can you find the distance covered by an
athlete if he takes two rounds on a circular track.
Since it is a circular track, we cannot use
the ruler to find out the distance.
So, what can we do ?
Take a one rupee coin.Place it on a paper
Fig. 3.7
and draw its outline. Remove the coin. Mark a
point A on the outline as shown in the Fig. 3.7
Take a thread and fix one end at A. Now place the
thread in such a way that the thread coincides exactly with
the outline. Cut the other end of the thread when it reaches
the point A.
Length of the thread is nothing but the circumference
of the coin.
Fig. 3.8
So,
the distance around a circle is called the circumference of the circle, which is
denoted by ‘C’. i.e., The perimeter of a circle is known as its circumference.
Take a bottle cap or a bangle or any other circular objects and
find the circumference. If possible find the relation between
the circumference and the diameter of the circular objects.
Relation between diameter and circumference of the circle
Draw four circles with radii 3.5 cm, 7 cm, 5 cm, 10.5 cm in your note book.
Measure their circumferences using a thread and the diameter using a ruler as shown
in the Fig. 3.9 given below.

51

Chapter 3
Fig. 3.9
Fill in the missing values in table 3.1 and find the ratio of the circumference to
the diameter.
Diameter
Circumference
C
Circle
Ratio
`
j
( d )
(C)
d
1
3.5 cm
7 cm
22
cm
22
= 3.14
7
2
7
cm
14
cm
44
cm
44
22
=
= 3.14
14
7
3
5
cm
10
cm
----
----
4
10.5 cm
21
cm
----
----
Table 3.1
C
What do you infer from the above table?. Is this ratio
`
j approximately
d
the same?
Yes !
C
= 3.14
& C = ^3.14 h d
d
So, can you say that the circumference of a circle is always more than 3 times
its diameter ?
Yes !
In all the cases, the ratio C
is a constant and is denoted by the Greek letter r
d
(read as ‘pi’ ). Its approximate value is
22 or 3.14.
7
C
so,
= r
&
C = rd units
where d is the diameter of a circle.
d
We know that the diameter of a circle is twice the radius r . i.e., d = 2r .

52

Measurements
from the above formula, C = rd = r ^2r h &
C = 2rr units.
The value of r is calculated by many mathematicians.
22
Babylonians :
r = 3
Greeks
:
r =
or 3.14
7
1
10
62838
Archemides
:
Aryabhata :
r =
(or) 3.1416
3 7
< r < 3 71
2000
22
Now, we use
r =
or 3.14
7
Example 3.4
Find out the circumference of a circle whose diameter is 21 cm.
Solution
Ci rcumference of a circle =
rd
22
22
=
# 21
Here r =
7
7
=
66 cm.
Example 3.5
Find out the circumference of a circle whose radius is 3.5 m.
Solution
Ci rcumference of a circle =
2rr
22
=
2
#
# 3
.5
7
=
2 × 22 × 0.5
=
22 m
Example 3.6
A wire of length 88 cm is bent as a circle. What is the radius of the circle.
Solution
Length of the wire
=
88 cm
Ci rcumference of the circle
= Length of the wire
2rr
=
88
22
2
#
#
r
=
88
7
88 # 7
r
=
= 14 cm
2
# 22
` radius of a circle is 14 cm.
Example 3.7
The diameter of a bicycle wheel is 63 cm. How much distance will it cover in
20 revolutions?

53

Chapter 3
Solution
When a wheel makes one complete revolutions,
Distance covered in one rotation = Ci rcumference of wheel
` circumference of the wheel = rd units
22
=
# 63 cm
7
=
198 cm
For one revolution, the distance covered = 198 cm
` for 20 revolutions, the distance covered = 20 × 198 cm
= 3960 cm
=
39 m 60 cm
[100 cm = 1 m]
Example 3.8
A scooter wheel makes 50 revolutions to cover a distance of 8800 cm. Find
the radius of the wheel.
Solution
Distance travelled
= Nu mber of revolutions # Circumference
Distance travelled
Ci rcumference
=
Number of revolutions
8800
2rr =
50
i.e., 2rr
=
176
22
2
#
#
r
=
176
7
176 # 7
r =
2 # 22
r
=
28 cm
` radius of the wheel = 28 cm.
Example 3.9
The radius of a cart wheel is 70 cm. How many revolution does it make in
travelling a distance of 132 m.
Solution
Given: r = 70 cm, Distance travelled = 132 m.
` Circumference of a cart wheel = 2rr
22
=
2
#
# 70
7
=
440 cm

54

Measurements
Distance travelled
= Nu mber of revolutions # Circumference
Distance travelled
` Number of revolutions
=
Circumference
132 m
=
440 cm
13200 cm
cm
=
(1 m = 100 cm, 132 m = 13200 cm)
440
=
30
` Number of revolutions = 30.
Example 3.10
The circumference of a circular field is 44 m. A cow is tethered to a peg at the
centre of the field. If the cow can graze the entire field, find the length of the rope used
to tie the cow.
Solution
Length of the rope
= Radius of the circle
Ci rcumference
=
44 m
(given)
i.e., 2rr
=
44
22
2
#
#
r
=
44
7
44 # 7
` r
=
= 7 m
2 # 22
Fig. 3.10
` The length of the rope used to tie the cow is 7 m.
Example 3.11
The radius of a circular flower garden is 56 m. What is the cost of fencing it at
`10 a metre ?
Solution
Length to be fenced
=
Ci rcumference of the circular flower garden
Circumference of the flower garden
=
2rr
22
=
2
#
# 56 = 352 m
7
` Length of the fence =
352 m
Cost of fencing per metre =
`10
` cost of fencing 352 m =
`10 # 352
=
`3520
` Total cost of fencing is `3520.

55

Chapter 3
Example 3.12
The cost of fencing a circular park at the rate of `5 per metre is `1100. What
is the radius of the park.
Solution
Cost of fencing
= Ci rcumference # Rate
Cost of fencing
` Circumference
=
Rate
1100
i.e., 2rr
=
5
2rr
=
220
22
` 2
#
#
r
=
220
7
220 # 7
r
=
2 # 22
=
35 m
` Radius of the park
=
35 m.
Activity - Circular Geoboard
Take a square Board and draw a circle.
Fix nails on the circumference of the circle. ( See fig )
Using rubber band, form various diameters, chords, radii and
compare.
Exercise 3.2
1.
Choose the correct answer:
i)
The line segment that joins the centre of a circle to any point on the circle is
called
(A)
Diameter
(C) Chord
(D) None
ii)
A line segment joining any two points on the circle is called
(A)
Diameter
(C) Chord
(D) None
iii)
A chord passing through the centre is called
(A)
Diameter
(C) Chord
(D) None
iv)
The diameter of a circle is 1 m then its radius is
(A)
100 cm
(B) 50 cm
(C) 20 cm
(D) 10 cm
v)
The circumference of a circle whose radius is 14 cm is
(A)
22 cm
(B) 44 cm
(C) 66 cm
(D) 88 cm

56

Measurements
2.
Fill up the unknown in the following table:
radius ( r )
diameter ( d )
circumference (c)
(i)
35 cm
-----
-----
(ii)
-----
56 cm
-----
(iii)
-----
-----
30.8 cm
3.
Find the circumference of a circle whose diametre is given below:
(i)
35 cm
(ii) 84 cm
(iii) 119 cm
(iv) 147 cm
4.
Find the circumference of a circle whose radius is given below:
(i)
12.6 cm
(ii) 63 cm
(iii) 1.4 m
(iv) 4.2 m
5.
Find the radius of a circle whose circumference is given below:
(i)
110 cm
(ii) 132 cm
(iii) 4.4 m
(iv) 11 m
6.
The diameter of a cart wheel is 2.1 m. Find the distance travelled when it
completes 100 revolutions.
7.
The diameter of a circular park is 98 m. Find the cost of fencing it at `4 per
metre.
8.
A wheel makes 20 revolutions to cover a distance of 66 m. Find the diameter of
the wheel.
9.
The radius of a cycle wheel is 35 cm. How many revolutions does it make to
cover a distance of 81.40 m?
10.
The radius of a circular park is 63 m. Find the cost of fencing it at `12 per metre.
Area of a circle
Consider the following
A farmer levels a circular field of radius 70 m. What will be the cost of levelling?
What will be the cost of polishing a circular table-top of radius 1.5 m ?
How will you find the cost ?
To find the cost what do you need to find actually?
The region enclosed by
the circumference of a
Area or perimeter ?
circle is a circular region.
Area, area, area
Yes. In such cases we need to find the area of the circular region.
So far, you have learnt to find the area of triangles and quadrilaterals that made
up of straight lines. But, a cirlce is a plane figure made up of curved line different from
other plane figures.

57

Chapter 3
So, we have to find a new approach which will make the circle turn into a figure
with straight lines.
Take a chart paper and draw a circle. Cut the circle and take it separately. Shade
one half of the circular region. Now fold the entire circle into eight parts and cut along
the folds (see Fig. 3.11).
Fig. 3.11
Arrange the pieces as shown below.
Fig. 3.12
What is the figure obtained?
These eight pieces roughly form a parallelogram.
Similarly, if we divide the circle into 64 equal parts and arrange these, it gives
nearly a rectangle. (see Fig. 3.13)
Fig. 3.13
What is the breadth of this rectangle?
The breadth of the rectangle is the radius of the circle.
i.e.,
b = r
.....
(1)
What is the length of this rectangle ?
As the whole circle is divided into 64 equal parts and on each side we have 32
equal parts. Therefore, the length of the rectangle is the length of 32 equal parts, which
is half of the circumference of a circle.

58

Measurements
1
` length l
=
[circumference of the circle]
2
1
=
62r
r
@ = rr
2
` l
= rr
(2)
Area of the circle
= Area of the rectangle (from the Fig. 4.50)
=
l # b
=
(rr) # r
(from (1) and (2))
=
2
r
r
sq. units.
` Area of the circle
=
2
r
r
sq. units.
Example 3.13
Find the area of a circle whose diameter is 14 cm
Solution
Diameter d
=
14 cm
d
14
So,
=
=
= 7 cm
2
2
Area of circle
=
r r 2
22
=
# 7 # 7
7
=
154 sq. cm
` Area of circle
=
154 sq. cm
Example 3.14
A goat is tethered by a rope 3.5 m long. Find the maximum area that the goat
can graze.
Solution
Radius of the circle
= Length of the rope
7
=
3.5 m =
m
2
ma ximum area grazed by the goat =
r r 2
sq. units.
22
7
7
=
#
#
7
2
2
77
Fig. 3.14
=
= 38.5
sq. m
2
` maximum area grazed by the goat is 38.5 sq. m.

59

Chapter 3
Example 3.15
The circumference of a circular park is 176 m. Find the area of the park.
Solution
Ci rcumference
=
176 m
(given)
2rr
=
176
22
2
#
#
r
=
176
7
176 # 7
r
=
44
` r
=
28 m
Area of the park
=
r r 2
=
22 # 8 #
2
2 8
7
=
22 # 4 # 28
=
2464 sq. m.
Example 3.16
A silver wire when bent in the form of a square encloses an area of 121 sq. cm.
If the same wire is bent in the form of a circle. Find the area of the circle.
Solution
Let a be the side of the square
Area of the square
=
121 sq. cm.
(given)
a 2
=
121 & a = 11 cm
(11 # 11 = 121)
Perimeter of the square
=
4a units
=
4 # 11 cm
=
44 cm
Length of the wire
= Perimeter of the square
=
44 cm
The wire is bent in the form of a circle
The circumference of the circle = Length of the wire
` circumference of a circle = 44 cm
2rr
=
44
22
` 2
#
#
r
=
44
7
44 # 7
r
=
44
r
=
7 cm
` Area of the circle
=
r r 2
22
=
× 7 cm × 7 cm
7
Area of the circle
=
154 cm 2 .

60

Measurements
Example 3.17
When a man runs around circular plot of land 10 times, the distance covered by
him is 352 m. Find the area of the plot.
Solution
Distance covered in 10 times
= 352 m
352
Distance covered in one time
=
m = 35.2 m
10
The circumference of the circular plot
= Distance covered in one time
` circumference = 35.2 m
2rr
=
35.2
22
2
#
#
r
=
35.2
7
35 . 2 # 7
r
=
44
=
0.8 # 7
=
5.6 m
Area of the circular plot
=
r r 2
22
=
# 5 . 6 # 5 . 6
7
=
22 # 0.8 # 5.6
=
98.56 m 2
` Area of circular plot
=
98.56 m 2
Example 3.18
A wire in the shape of a rectangle of length 37 cm and width 29 cm is reshaped
in the form of a circle. Find the radius and area of the circle.
Solution
Length of the wire
= perimeter of the rectangle
=
2 [ length + breadth ]
=
2 [37 cm + 29 cm] = 2 # 66 cm
=
132 cm.
Since wire is bent in the form of a circle,
The circumference of the circle
= The length of the wire
` Circumference of a circle
=
132
2rr
=
132
22
2
#
#
r
=
132
7
132
r
=
# 7 =
21
44
` radius of the circle
=
21 cm
Area of the circle
=
r r 2
22
=
# 21 # 21 = 22 # 3 # 21
7
` Area of the circle
=
1386 sq. cm.

61

1.

2.

3.

4.

5.

6.

7.

### Chapter 3

Exercise 3.3

Find the area of the circles whose diameters are given below:

 (i) 7 cm (ii) 10.5 cm (iii) 4.9 m (iv) 6.3 m Find the area of the circles whose radii are given below: (i) 1.2 cm (ii) 14 cm (iii) 4.2 m (iv) 5.6 m

)

(take

(take

• 22 )

7

22

7

r =

r =

The diameter of a circular plot of ground is 28 m. Find the cost of levelling the

ground at the rate of `3 per sq. m.

A goat is tied to a peg on a grass land with a rope 7 m long. Find the maximum area of the land it can graze.

A circle and a square each have a perimeter of 88 cm. Which has a larger area?

A wheel goes a distance of 2200 m in 100 revolutions. Find the area of the wheel.

A wire is in the form of a circle of radius 28 cm. Find the area that will enclose, if it is bent in the form of a square having its perimeter equal to the circumference of the cirlce.

8.

The area of circular plot is 3850 m 2 . Find the radius of the plot. Find the cost of fencing the plot at `10 per metre.

9.

The radius of a circular ground is 70 m. Find the distance covered by a child walking along the boundary of the ground.

• 10. The area of a circular field is 154 m 2 . Find the time taken by an athlate to complete 2 rounds if she is jogging at the rate of 5 km/hr.

11 .

Ho w many circles of radius 7 cm can be cut from a paper of length 50 cm and width 32 cm.

### 3.3 Area of the path way

In our day - to - day life we go for a walk in a park, or in a play ground or even

around a swimming pool.

Can you represent the path way of a park diagrammatically ?

Have you ever wondered if it is possible to find the area of such paths ?

Can the path around the rectangular pool be related to the mount around the

photo in a photo frame ?

Can you think of some more examples?

In this section we will learn to find

• Area of rectangular pathway

• Area of circular pathway

62

Measurements
Area of rectangular pathway
(a) Area of uniform pathway outside the rectangle
Consider a rectangular building. A uniform flower garden is to be laid outside
the building. How do we find the area of the flower
garden?
The uniform flower garden including the
building is also a rectangle in shape. Let us call
it as outer rectangle. We call the building as inner
rectangle.
Fig. 3.15
Let l and b be the length and breadth of the
building.
` Area of the inner rectangle = l b sq.
units.
Fig. 3.16
Let w be the width of the flower garden.
What is the length and breadth of the outer rectangle ?
The length of the outer rectangle (L)
=
w +
l + w = ^l + 2w h units
The breadth of the outer rectangle (B)
=
w + b + w = ^ b + 2w h units
` area of the outer rectangle
=
L # B
=
^l + 2w h^ b + 2w h sq. units
Now, what is the area of the flower garden ?
Actually, the area of the flower garden is the pathway bounded between two
rectangles.
` Area of the flower garden =
(A rea of building and flower garden) –
(A rea of building)
Generally,
Area of the pathway = (Area of outer rectangle) – (Area of inner rectangle)
i.e. Area of the pathway = (l + 2w) (b + 2w) – lb.
Example 3.19
The area of outer rectangle is 360 m 2 . The area of inner rectangle is 280 m 2 . The
two rectangles have uniform pathway between them. What is the area of the pathway?
Solution
Area of the pathway = (A rea of outer rectangle) – (Area of inner rectangle)

63

Chapter 3
=
(360 – 280) m 2
= 80 m 2
` Area of the pathway =
80 m 2
Example 3.20
The length of a building is 20 m and its breadth is 10 m. A path of width 1 m is
made all around the building outside. Find the area of the path.
Solution
Inner rectangle (given)
Outer rectangle
l
= 20 m
width, w
= 1 m
b = 10 m
L
= l + 2w
Area
= l × b
= 20 + 2 = 22 m
Area
= 20 m × 10 m
B
= b + 2w
= 10 + 2 = 12 m
= 200 m 2
Area
= (l + 2w) (b + 2w)
Area
= 22 m #
12 m
= 264 m 2
Area of the path = (A rea of outer rectangle) – (Area of inner rectangle)
=
( 26 4 – 200 ) m 2 = 64 m 2
` Area of the path
= 64 m 2
Example 3.21
A school auditorium is 45 m long and 27 m
wide. This auditorium is surrounded by a varandha
of width 3 m on its outside. Find the area of the
varandha. Also, find the cost of laying the tiles in
the varandha at the rate of `100 per sq. m.
Solution
Fig. 3.17
Inner (given) rectangle
Outer rectangle
l
= 45 m
Width,
w
= 3 m
b
= 27 m
L
=
l + 2w
Area = 45m # 27 m
= 45 + 6 = 51 m
B
=
= 1215 m 2
b + 2w
= 27 + 6 = 33 m
Area
= 51m # 33 m
= 1683 m 2

64

Measurements
(i)
Area of the verandha
=
(A rea of outer rectangle) – (Area of inner rectangle)
=
(1683 – 1215) m 2
=
468 m 2
` Area of the verandha = 468 m 2 (or) 468 sq. m.
(ii)
Cost of laying tiles for 1 sq. m
= `100
Cost of laying tiles for 468 sq. m
= `100 # 468
= `46,800
` Cost of laying tiles in the verandha = `46,800
(b) Area of uniform pathway inside a rectangle
A swimming pool is built in the middle
of a rectangular ground leaving an uniform
width all around it to maintain the lawn.
If the pathway outside the pool is to be
grassed, how can you find its cost ?
If the area of the pathway and cost of
grassing per sq. unit is known, then the cost
Fig. 3.18
of grassing the pathway can be found.
Here, the rectangular ground is the outer rectangle where l and b are length and
` Area of the ground (outer rectangle) = l b sq. units
If w be the width of the pathway (lawn), what will be the length and breath of
the swimming pool ?
The length of the swimming pool
=
l -
w - w
=
l - 2w
The breadth of the swimming pool =
b - w - w
= b - 2w
` Area of the swimming pool (inner rectangle) = ^l - 2w h^ b - 2w h Sq. units
Area of the lawn
= Area of the ground – Area of the swimming pool.
Generally,
Area of the pathway
=
(A rea of outer rectangle) – (Area of inner rectangle)
=
lb – (l – 2w) (b – 2w)

65

Chapter 3
Example 3.22
The length and breadth
of
a
room are
8 m and 5 m respectively. A red colour border
of uniform width of 0.5 m has been painted all
around on its inside. Find the area of the border.
Solution
Fig. 3.19
Outer (given)rectangle
Inner rectangle
l
= 8 m
width, w
= 0.5 m
b
= 5 m
L
=
l - 2w
Area = 8m # 5 m
= (8 – 1) m
= 7 m
= 40 m 2
B
=
b - 2w
= (5 – 1) m
= 4 m
Area
= 7m × 4 m
= 28 m 2
Area of the path
= (A rea of outer rectangle) – (Area of inner rectangle)
=
(40 – 28) m 2
=
12 m 2
` Area of the border painted with red colour = 12 m 2
Example 3.23
A carpet measures 3 m # 2 m. A strip of 0.25 m wide is cut off from it on all
sides. Find the area of the remaining carpet and also find the area of strip cut out.
Solution
Outer rectangle
Inner rectangle
carpet before cutting the strip
carpet after cutting the strip
l
= 3 m
width, w
= 0.25 m
b
= 2 m
L
= l - 2w = (3 – 0.5) m
Area
= 3m # 2m
= 2.5 m
= 6 m 2
B
=
b - 2w = (2 – 0.5) m
= 1.5 m
Area
= 2.5m # 1.5m
= 3.75 m 2
The area of the carpet after cutting the strip = 3.75 m 2

66

Measurements
Area of the strip cut out = (A rea of the carpet) –
(Area of the remaining part)
=
(6 – 3.75) m 2
=
2.25 m 2
` Area of the strip cut out = 2.25 m 2
Note: If the length and breadth of the inner rectangle
is given, then the length and breadth of the outer rectangle
is l + 2w , b + 2w respectively where w is the width of the
path way.
Suppose the length and breadth of the outer rectangle
is given, then the length and breadth of the inner rectangle
is l - 2w , b - 2w respectively.
Fig. 3.20
Exercise 3.4
1.
A play ground 60 m # 40 m is extended on all sides by 3 m. What is the extended
area.
2.
A school play ground is rectangular in shape with length 80 m and breadth 60 m.
A cemented pathway running all around it on its outside of width 2 m is built.
Find the cost of cementing if the rate of cementing 1 sq. m is `20.
3.
A garden is in the form of a rectangle of dimension 30 m # 20 m. A path of width
1.5 m is laid all around the garden on the outside at the rate of `10 per sq. m.
What is the total expense.
4.
A picture is painted on a card board 50 cm long and 30 cm wide such that there is
a margin of 2.5 cm along each of its sides. Find the total area of the margin.
5.
A rectangular hall has 10 m long and 7 m broad. A carpet is spread in the centre
leaving a margin of 1 m near the walls. Find the area of the carpet. Also find the
area of the un covered floor.
6.
The outer length and breadth of a photo frame is 80 cm , 50 cm. If the width of
the frame is 3 cm all around the photo. What is the area of the picture that will be
visible?
Circular pathway
Concentric circles
Circles drawn in a plane with a common centre and different
radii are called concentric circles.
Circular pathway
Fig. 3.21
A track of uniform width is laid around a circular park for
walking purpose.
Can you find the area of this track ?

67

Chapter 3
Yes. Area of the track is the area bounded between two
concentric circles. In Fig. 3.22, O is the common centre of the two
circles.Let the radius of the outer circle be R and inner circle be r.
The shaded portion is known as the circular ring or the
circular pathway. i.e. a circular pathway is the portion bounded
between two concentric circles.
Fig. 3.22
width of the pathway, w
=
R –
r units
i.e., w
=
R – r & R
= w + r units
r
= R – w units.
The area of the circular path
= (a rea of the outer circle) – (area of the inner circle)
=
2
2
r
R
- r
r
=
2
2
r ^ R
- r
h sq. units
` The area of the circular path =
2
2
r ^ R - r
h sq. units
=
r ^ R + r h^ R - r h sq. units
Example 3.24
The adjoining figure shows two concentric circles. The radius
of the larger circle is 14 cm and the smaller circle is 7 cm. Find
(i)
The area of the larger circle.
(ii)
The area of the smaller circle.
(iii)
The area of the shaded region between two circles.
Solution
Fig. 3.23
i) Larger circle
ii) Smaller circle
R
=
14
r
=
7
area
=
r R 2
area =
r r 2
22
=
# 14 # 14
= 22 # 7 # 7
7
7
=
22 # 28
=
22 # 7
=
616 cm 2
=
154 cm 2
iii) The area of the shaded region
= (A rea of larger circle) – (Area of smaller circle)
=
(616 – 154) cm 2
= 462 cm 2
Example 3.25
From a circular sheet of radius 5 cm, a concentric circle of radius 3 cm is
removed. Find the area of the remaining sheet ? (Take r = 3.14 )

68

Measurements
Solution
Given: R = 5 cm,
r = 3 cm
Area of the remaining sheet
=
2
2
r ^ R - r
h
=
3.14 (5 2 – 3 2 )
=
3.14 (25 – 9)
=
3.14 × 16
=
50.24 cm 2
Aliter:
Outer circle
Inner circle
R
= 5 cm
r
= 3 cm
Area
=
r R 2
sq. units
Area =
r r 2
sq. units
=
3.14 # 5 # 5
=
3.14 # 3 # 3
= 3.14
# 25
= 3.14 # 9
= 78.5 cm 2
= 28.26 cm 2
Area of the remaining sheet = (A rea of outer circle) – (Area of inner circle)
= (78.5 – 28.26) cm 2
= 50.24 cm 2
` Area of the remaining sheet = 50.24 cm 2
Example 3.26
A circular flower garden has an area 500 m 2 . A sprinkler at the centre of the
garden can cover an area that has a radius of 12 m. Will the sprinkler water the entire
garden (Take r = 3.14)
Solution
Given,
area of the garden
=
50 0 m 2
Area covered by a sprinkler
=
r r 2
=
3.14 # 12 # 12
=
3.14 # 144
Fig. 3.24
=
452 .16 m 2
Since, the area covered by a sprinkler is less than the area of the circular flower
garden, the sprinkler cannot water the entire garden.
Example 3.27
A uniform circular path of width 2 m is laid out side a circular park of radius
50 m. Find the cost of levelling the path at the rate of `5 per m 2 (Take r = 3.14)

69

Chapter 3
Solution
Given: r = 50 m, w = 2 m, R = r + w = 50 + 2 = 52 m
Area of the circular path
=
r ^ R + r h^ R - r h
=
3.14 # ^52 + 50 h^52 - 50 h
=
3.14 # 102 # 2
=
3.14 # 204
Fig. 3.25
= 640.56 m 2
The cost of levelling the path of area 1 sq m = `5
The cost of levelling the path of 640.56 m 2 = `5 # 640.56
= ` 3202.80
` the cost of levelling the path = ` 3202.80
Exercise 3.5
1.
A circus tent has a base radius of 50 m. The ring at the centre for the performance
by an artists is 20 m in radius. Find the area left for the audience.
(Take r = 3.14)
2.
A circular field of radius 30 m has a circular path of width 3 m inside its boundary.
Find the area of the path (Take r = 3.14)
3.
A ring shape metal plate has an internal radius of 7 cm and an external radius of
10.5 cm. If the cost of material is `5 per sq. cm, find the cost of 25 rings.
4.
A circular well has radius 3 m. If a platform of uniform width of 1.5 m is laid
around it, find the area of the platform . (Take r = 3.14)
5.
A uniform circular path of width 2.5 m is laid outside a circular park of radius
56m. Find the cost of levelling the path at the rate of `5 per m 2 (Take r = 3.14)
6.
The radii of 2 concentric circles are 56 cm and 49 cm.
pathway.
Find the area of the
7.
The area of the circular pathway is 88 m 2 .
m, find the radius of the inner circle.
If the radius of the outer circle is 8
8.
The cost of levelling the area of the circular pathway is ` 12,012 at the rate of
` 6 per m 2 . Find the area of the pathway.

70

Measurements
Figure
Area
Forumula
1
1
× height × sum of
× h × (a + b) sq.
2
2
parallel sides
units
Perimeter of the circle =
2 r r units
2 × r × radius
Area of the circle =
r r 2 sq. units
Area of the pathway
Area of outer
rectangle – Area of
i) area of the rectangular
inner rectangle
pathway
Area of outer circle –
Area of inner circle
ii) area of the circular
= r (R 2 – r 2 ) sq. units
pathway
= r (R + r) (R – r)
sq. units

71

Chapter 4
GEOMETRY
4.1 Triangle: Revision
A
triangle is a closed plane
figure made of three line
segments.
In
Fig.
4.1
the
line
segments
AB,
BC
and
CA
form
a
closed
figure.
This
is
a triangle
and
is denoted by
D ABC. This triangle may be named as
D ABC or
D BCA or
Fig. 4.1
D CAB.
The line segments forming a triangle are the three sides of the triangle. In
Fig.4.1 AB , BC and CA are the three sides of the triangle.
The point where any two of the three line segments of a triangle intersect is
called the vertex of the triangle. In Fig. 4.1 A,B and C are the three vertices of the
D ABC.
When two line segments intersect, they form an angle at that point. In the
triangle in Fig. 4.1 AB
and
BC intersect at B and form an angle at that vertex.
This angle at B is read as angle B or + B or + ABC.
Thus a triangle has three
angles + A, + B and + C.
In Fig. 4.1 D ABC has
Sides
: AB,BC,CA
Angles
:
+ CAB, + ABC, + BCA
Vertices :
A, B, C
The side opposite to the vertices A, B, C are BC, AC and AB respectively. The
angle opposite to the side BC, CA and AB is + A, + B and + C respectively.
A triangle is a closed figure made of three line segments. It has three vertices,
three sides and three angles.

72

Geometry
4.2
Types of Triangles
Based on sides
A triangle is said to be
Equilateral, when all its sides are equal.
Isosceles, when two of its sides are equal.
Scalene, when its sides are unequal.
Based on angles
A triangle is said to be
Right angled, when one of its angle is a right angle and the other two angles are
acute.
Obtuse - angled, when one of its angle is obtuse and the other two angles are
acute.
Acute - angled, when all the three of its angles are acute.
The sum of the lengths of any two sides of a triangle is always greater than the
length of the third side.
4.3
Angle sum property of a triangle:
Activity 1
Draw any triangle ABC on a sheet of paper
Is it possible to form a
triangle whose sides are
and mark the angles 1, 2 and 3 on both sides of the
7cm, 5cm and 13cm?
paper as shown in Fig. 4.2 (i).
Fig. 4.2
Cut a triangle ABC in a paper. Fold the vertex A to touch the side BC as shown
in the Fig. 4.2 (ii) Fold the vertices B and C to get a rectangle as shown in the Fig. 4.2
(iii) Now you see that + 1, + 2 and + 3 make a straight line.

73

Chapter 4
From this you observe that
+ 1 + + 2
+ + 3
=
180°
+ A + + B + + C
=
180°
T he sum of the three angles of a triangle is 180°
Activity 2
Draw a triangle. Cut on the three angles.
Re arrange them as shown in Fig. 4.2 (ii). You
observe that the three angles now constitute
one angle. This angle is a straight angle and so
has measure 180°
Fig. 4.3
The sum of the three angles of a triangle is 180°
Think it.
1. Can you have a triangle with the three angles less than 60°?
2. Can you have a triangle with two right angles?
4.4 Exterior angle of a triangle and its property
Activity 3
Fig. 4.4
Draw a triangle ABC and
produce
one
of
its
sides, say
BC as shown
in
Fig. 4.4 (i) observe the angles ACD formed at the point C. This angle lies in the
exterior of T ABC formed at vertex C.
+ BCA is an adjacent angle to + ACD.
The remaining two angles of the
triangle namely + A nd + B are called the two interior opposite angles.
Now cut out (or make trace copies of) + A and + B and place them adjacent to
each other as shown in Fig. 4.4 (ii)
You observe that these two pieces together entirely cover + ACD.

74

Geometry
From this we conclude that the exterior angle of a triangle is equal to the sum of
the two interior opposite angles.
The relation between an exterior angle and its two interior angles is referred to
as the exterior angle property of a triangle.
Draw a triangle ABC and produce one of its sides BC as shown in
Fig. 4.4 (i) + ACD formed at the point C. Now take a protractor
and measure + ACD, + A and + B.
Find the sum + A + + B. and compare it with the measure of
+ ACD. Do you observe that + ACD = + A + B?
Example 4.1
In the given figure find the value of x.
Solution
+ CAB + + ABC + + BCA
= 180°
(Since sum of the three
40° + x + x
= 180°
angles of a triangle is 180°)
40° + 2 x
= 180°
2 x
= 180° – 40°
2
x
= 140°
140
°
x
=
= 70°
2
The value of x
= 70°.
Example 4.2
Two angles of a triangle are 40° and 60°.
Find the
third angle.
Solution
+ RPQ + + PQR + + QRP =
180°
(Since sum of the three angles of a
x + 40° + 60°
=
180°
tr iangle is 180°)
x + 100°
=
180°
x
=
180° – 100°
=
80°
`
The third angle x
=
80°
Example 4.3
In the given figure, find the measure of + A.
Solution
+ CAB + + ABC + + BCA
=
180°
(Since sum of the three angles of a
2 x + 120° + x =
180°
tr iangle is 180°)

75

Chapter 4
3x + 120°
=
180°
3x
=
180° – 120°
3x =
60°
60
°
x =
3
=
20°
`
+ A =
2 x = 2 × 20° = 40°
Example 4.4
In the given figure. Find the value of x.
Solution
In the figure exterior angle = + ABD = 110°.
Sum of the two interior opposite angle = + BCA + + CAB
= x + 50°
x + 50°
=
110°
(Since the sum of the two interior opposite
x
=
110° – 50°
angle is equal to the exterior angle)
=
60°
` The value of x
is 60°.
Example 4.5
In the given figure find the values of x and y.
Solution
In the give figure,
Exterior angle
=
+ DCA = 130°
50° + x
=
130°
(Since sum of the two interior opposite
x
= 130° – 50° angle is equal to the exterior angle)
=
80°
In
D ABC,
+ A + + B + + C
=
180° (Since sum of three angles of a triangle is 180°)
50° + x + y
=
180°
50° + 80° + y
=
18 0°
130° + y
=
18 0°
y
=
18 0° – 130° = 50°
` The values of x = 80° and y = 50°.

76

Geometry
Aliter:
+ ACB + + DCA
=
180° (Since sum of the adjacent angles on a line is 180°)
y + 130°
=
180°
y
=
180° – 130°
=
50°
In D ABC,
+ A
+ + B + + C
=
180° (Since sum of the three angles of a triangle is 180°)
50° + x + y
=
180°
50° + x + 50°
=
180°
10 0° + x
=
180°
x
=
180° – 100°
=
80°
Example 4.6
Three angles of a triangle are 3x + 5°, x + 20°, x + 25°. Find the measure of each
angle.
Solution
Su m of the three angles of a triangle
=
180°
3x + 5° + x + 20° + x + 25°
=
180°
5x + 50°
=
180°
5x
=
180° – 50°
5x
=
130°
130
°
x
=
5
=
26°
3x + 5°
=
(3 × 26°) + 5° = 78° + 5° = 83°
x
+ 20°
=
26° + 20° = 46°
x
+ 25°
=
26° + 25° = 51°
` The three angles of a triangle are 83°, 46° and 51°.

77

Chapter 4
Exercise 4.1
1.
Choose the correct answer.
i)
The sum of the three angles of a triangle is
(A)
90°
(B) 180°
(C) 270°
(D) 360°
ii)
In a triangle, all the three angles are equal, then the measure of each angle is
(A)
30 0
(B) 45 0
(C) 60 0
(D) 90 0
iii)
Which of the following can be angles of a triangle?
0
0
0
0
0
0
0
0
0
(A)
50 , 30 , 105 0 (B) 36 , 44
, 90 0
(C) 70 , 30 , 80 0
(D)
45 , 45 80
iv)
Two angles of a triangle are 40° and 60°, then the third angle is
(A)
20 0
(B) 40 0
(C) 60 0
(D) 80 0
v)
In T ABC, BC is produced to D and + ABC = 50 0 ,
+ ACD = 105 0 , then + BAC will be equal to
(A)
75°
(B) 15°
(C)
40°
(D) 55°
2.
State which of the following are triangles.
(i)
+ A = 25 0
+ B
=
35 0
+ C = 120 0
(ii) + P = 90 0
+ Q
= 30 0
+
R = 50 0
(iii) + X = 40 0 + Y = 70 0
+ Z = 80 0
3.
Two angles of a triangle is given, find the third angle.
0
0
0
0
0
0
0
(i)
75 , 45 0 (ii) 80 , 30
(iii) 40 , 90
(iv) 45
, 85
4.
Find the value of the unknown x in the following diagrams:

78

Geometry
5.
Find the values of the unknown x and y in the following diagrams:
6.
Three angles of a triangle are x + 5°, x + 10° and x + 15° find x.

79

Chapter 4
1.
The sum of the three angles of a triangle is 180°.
2.
In a triangle an exterior angle is equal to the sum of the two interior opposite
angles.

80

Practical Geometry
P RACTICAL G EO METRY
5.1 Construction of triangles
In the previous class, we have learnt the various types of triangles on the
basis of their sides and angles. Now let us recall the different types of triangles
and some properties of triangle.
Classification of triangles
No.
Name of Triangle
Figure
Note
1
Equilateral triangle
Three sides are equal
Isosceles triangle
2
Any two sides are equal
3
Scalene triangle
Sides are unequal
All the three angles are
4
Acute angled triangle
acute (less than 90°)
Any one of the angles is
5
Obtuse angled triangle
obtuse (more than 90°)
Any one of the angles is
6
Right angled triangle
right angle (90°)
CLASSIFICATION OF TRIANGLES
on the basis ANGLES
on the basis SIDES

81

Chapter 5
Some properties of triangle
1.
The sum of the lengths of any two sides of a triangle is greater
than the third side.
2.
The sum of all the three angles of a triangle is 180°.
To construct a triangle we need three measurements in which at least the
length of one side must be given. Let us construct the following types of triangles
with the given measurements.
(i)
Th ree sides (SSS).
(ii)
Two sides and included angle between them (SAS).
(iii)
Two angles and included side between them (ASA).
(i) To construct a triangle when three sides are given (SSS Criterion)
Example 5.1
Construct a triangle ABC given that AB = 4cm, BC = 6 cm and AC = 5 cm.
Solution
Rough Diagram
Given measurements
AB = 4cm
BC = 6 cm
AC = 5 cm.
Steps for construction
Step 1
:
Draw a line segment BC = 6cm
Step 2
:
With ‘B’ as centre, draw an arc of radius 4 cm above the line BC.
Step 3
:
With ‘C’ as centre, draw an arc of 5 cm to intersect the previous arc at ‘A’
Step 4
:
Join AB and AC.
Now ABC is the required triangle.

82

3 cm
Practical Geometry
By using protector measure all the angles of a triangle. Find
the sum of all the three angles of a triangle.
1.
A student attempted to draw a triangle with given
measurements PQ = 2cm, QR = 6cm, PR = 3 cm. (as in
Rough Figure
P
the rough figure). First he drew QR = 6cm. Then he drew
an arc of 2cm with Q as centre and he drew another arc
of radius 3 cm with R as centre. They could not intersect
each to get P.
Q
R
6 cm
(i) What is the reason?
(ii) What is the triangle property in connection with this?
The sum of any two sides of a triangle is always greater than the third side.
Draw the bisectors of the three angles of
a triangle. Check whether all of them pass
through a same point. This point is incentre.
(ii) To construct a triangle when Two sides and an angle included between
them are given. (SAS Criterion)
Example 5.2
Construct a triangle PQR given that PQ = 4 cm, QR = 6.5 cm and + PQR = 60°.
Solution
Given measurements
PQ
=
4 cm
QR
=
6.5 cm
+ PQR
=
60°
2 cm

83

Chapter 5
Steps for construction
Step 1
:
Draw the line segment QR = 6.5 cm.
Step 2
:
At Q, draw a line QX making an angle of 60 0 with QR.
Step 3
:
With Q as centre, draw an arc of radius 4 cm to cut the line (QX)
at P.
Step 4
:
Join PR.
PQR is the required triangle.
Construct a
triangle
with
the given measurements
XY = 6cm, YZ = 6cm and + XYZ = 70°. Measure the angles
of
the triangle opposite to the equal sides. What do
you
observe?
(iii) To construct a triangle when two of its angles and a side
included between them are given. (ASA criterion)
Example 5.3
Construct a triangle XYZ given that XY = 6 cm,
+ ZXY
=
30°
and
+ XYZ = 100°. Examine whether the third angle
measures 50°.
Solution
Given measurements
XY = 6 cm
+ ZXY = 30°
+ XYZ = 100°

84

Practical Geometry
Step 1
:
Draw the line segment XY = 6cm.
Step 2
:
At X, draw a ray XP making an angle of 30° with XY.
Step 3
:
At Y, draw another ray YQ making an angle of 100° with XY. The
rays XP and YQ intersect at Z.
Step 4
:
The third angle measures 50° i.e + Z = 50°.
Construct a triangle PQR given that PQ = 7 cm,
+ Q = 70°, + R = 40°.
Hint: Use the Angle Sum Property of a triangle.
Exercise : 5.1
I.
Construct the triangles for the following given measurements.
1.
Construct 3 PQR, given that PQ = 6cm, QR = 7cm, PR = 5cm.
2.
Construct an equilateral triangle with the side 7cm. Using protector measure each
angle of the triangle. Are they equal?
3.
Draw a triangle DEF such that DE = 4.5cm, EF = 5.5cm and DF = 4.5cm. Can
you indentify the type of the triangle? Write the name of it.
II.
Construct the triangles for the following given measurements.
4.
Construct 3 XYZ, given that YZ = 7cm, ZX = 5cm, + Z = 50°.
5.
Construct 3 PQR when PQ = 6cm, PR = 9cm and + P = 100°.
6.
Construct 3 ABC given that AB = 6 cm, BC = 8 cm and + B = 90° measure
length of AC.
III.
Construct the triangles for the following given measurements.
7.
Construct 3 XYZ, when + X = 50°, + Y = 70° and XY = 5cm.
8.
Construct 3 ABC when + A = 120°, + B = 30° and AB = 7cm.
9.
Construct 3 LMN, given that + L = 40°, + M = 40° and LM = 6cm. Measure and
write the length of sides opposite to the + L and + M. Are they equal? What type
of Triangle is this?

85

Chapter 6
DATA HANDLING
6.1 Mean, Median and Mode of ungrouped data
Arithmetic mean
We use the word ‘average’ in our day to day life.
Poovini spends on an average of about 5 hours daily for her studies.
In the month of May, the average temperature at Chennai is 40 degree celsius.
What do the above statements tell us?
Poovini usually studies for 5 hours. On some days, she may study for less
number of hours and on other days she may study longer.
The average temperature of 40 degree celsius means that, the temperature in
the month of May at chennai is 40 degree celsius. Some times it may be less than 40
degree celsius and at other times it may be more than 40 degree celsius.
Average lies between the highest and the lowest value of the given data.
Rohit gets the following marks in different subjects in an examination.
62, 84, 92, 98, 74
In order to get the average marks scored by him in the examination, we first
add up all the marks obtained by him in different subjects.
62 + 84 + 92 + 98 + 74 = 410.
and then divide the sum by the total number of subjects. (i.e. 5)
410 = 82
The average marks scored by Rohit =
.
5
This number helps
us
to understand the general level of his academic
achievement and is referred to as mean.
` The average or arithmetic mean or mean is defined as follows.
Sum of all observations
Mean =
Total number of observations

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Data Handling
Example 6.1
Gayathri studies for 4 hours, 5 hours and 3 hours respectively on 3 consecutive
days. How many hours did she study daily on an average?
Solution:
Total number of study hours
Average study time =
Number of days for which she studied.
4
+ +
5
3
=
hours
3
12
=
3
=
4 hours per day.
Thus we can say that Gayathri studies for 4 hours daily on an average.
Example 6.2
The monthly income of 6 families are ` 3500, ` 2700, ` 3000, ` 2800, ` 3900 and
` 2100. Find the mean income.
Solution:
Total income of 6 familes
Average monthly income
=
Number of families
= ` 3500 2700 3000 2800 3900 2100
+
+
+
+
+
6
18000
=
`
6
=
` 3,000.
Example 6.3
The mean price of 5 pens is ` 75. What is the total cost of 5 pens?
Solution:
Total cost of 5 pens
Mean
=
Number of pens
Total cost of 5 pens
= Mean # Number of pens
=
`
75 # 5
=
` 375
Median
Consider a group of 11 students with the following height (in cm)
106, 110, 123, 125, 115, 120, 112, 115, 110, 120, 115.
The Physical EducationTeacher Mr. Gowtham wants to divide the students into
two groups so that each group has equal number of students. One group has height
lesser than a particular height and the other group has student with height greater than
the particular height.

87

Chapter 6
Now, Mr. Gowtham arranged the students according to their height in ascending
order.
106, 110, 110, 112, 115, 115,115, 120, 120, 123, 125
The middle value in the data is 115 because this value divides the students into
two equal groups of 5 students each. This value is called as median. Median refers
to the value 115 which lies in the middle of the data.Mr. Gowtham decides to keep
the middle student as a referee in the game.
Median is defined as the middle value of the data when the data is arranged
in ascending or descending order.
Find
the
actual
Find the median of the following:
distance
between
40, 50, 30, 60, 80, 70
your
school
and
Arrange the given data in ascending order.
house. Find
the
30, 40, 50, 60, 70, 80.
median of the place.
Here the number of terms is 6 which is even. So the third and fourth terms are
middle terms. The average value of the these terms is the median.
+
60
110
(i.e)
Median = 50
=
= 55 .
2
2
(i) When the number of observations is odd, the middle number is the
median.
(ii)
When the number of observations is even, the median is the average of
the two middle numbers.
Example 6.4
Find the median of the following data.
3, 4, 5, 3, 6, 7, 2.
Solution:
Arrange the data in ascending order.
In highways, the yellow
2, 3, 3, 4, 5, 6, 7
line represents the median.
The number of observation is 7 which is odd.
` The middle value 4 is the median.
Example 6.5
Find the median of the data
12, 14, 25, 23, 18, 17, 24, 20.
Solution:
Arrange the data in ascending order

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Data Handling
12, 14, 17, 18, 20, 23, 24, 25.
T he number of observation is 8 which is even.
` Median is the average of the two middle terms 18 and 20.
18 20
+
38
Median =
=
= 19
2
2
Example 6.6
Find the median of the first 5 prime numbers.
Solution:
The first five prime numbers are 2, 3, 5, 7, 11.
The number of observation is 5 which is odd.
` The middle value 5 is the median.
Mode
Look at the following example,
Mr. Raghavan, the owner of a ready made dress shop says that the most popular
size of shirts he sells is of size 40 cm.
Observe that here also, the owner is concerned about the number of shirts of
different sizes sold. He is looking at the shirt size that is sold, the most. The highest
occurring event is the sale of size 40 cm. This value is called the mode of the data.
Mode is the variable which occurs most frequently in the given data.
Mode of Large data
Putting the same observation together and counting them is not easy if the
number of observation is large. In such cases we tabulate the data.
Example 6.7
Following are the margin of victory in the foot ball matches of a league.
1, 3, 2, 5, 1, 4, 6, 2, 5, 2, 2, 2, 4, 1, 2, 3, 2, 3, 2, 3,
1, 1, 2, 3, 2, 6, 4, 3, 2, 1, 1, 4, 2, 1,
5, 3, 4, 2, 1, 2. Find the mode of this data.
Solution:
Margin of victory
Tally Marks
Number of Matches
1
||||
||||
9
2
||||
||||
||||
14
3
||||
||
7
4
||||
5
5
||
3
6
||
2
Total
40
Table 6.1

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Chapter 6
Now we quickly say that ‘2’ is the mode. Since 2 has occurred the more number
of times, then the most of the matches have been won with a victory margin of 2 goals.
Example 6.8
Find the mode of the following data.
3, 4, 5, 3, 6, 7
Find
the
mode of
the transport in your
Solution:
place.
3
occurs the most number of times.
` Mode of the data is 3.
Example 6.9
Find the mode of the following data.
2, 2, 2, 3, 3, 4, 5,5, 5, 6,6, 8
Solution:
2
and 5 occur 3 times.
` Mode of the data is 2 and 5.
Example 6.10
Find the mode of the following data
Find the mode
90, 40, 68, 94, 50, 60.
of the flower.
Solution:
Here there are no frequently occurring values. Hence this data has no mode.
Example 6.11
The number of children in 20 families are 1, 2, 2, 1, 2, 1, 3, 1, 1, 3
1, 3, 1, 1, 1, 2, 1, 1, 2, 1. Find the mode.
Solution:
Number of
Number of
Tally Marks
Children
Families
1
||||
||||
||
12
2
||||
5
3
|||
3
Total
20
Table 6.2
12 families have 1 child only, so the mode of the data is 1.

90

### Data Handling

Exercise: 6.1
1.
Choose the correct answer:
i)
The arithmetic mean of 1, 3, 5, 7 and 9 is
(A)
5
(B) 7
(C) 3
(D) 9
ii)
The average marks of 5 children is 40 then their total mark is
(A)
20
(B) 200
(C) 8
(D) 4
iii)
The median of 30,50, 40, 10, 20 is
(A)
40
(B) 20
(C) 30
(D) 10
iv)
The median of 2, 4, 6, 8, 10, 12 is
(A)
6
(B) 8
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