1062CH12
AREAS RELATED TOCIRCLES
12.1 Introduction
You are already familiar with some methods of finding perimeters and areas of
simple
plane figures such as rectangles, squares, parallelograms, triangles and circles from
your earlier classes. Many objects that we come across in our daily life are related to
the circular shape in some form or the other. Cycle wheels, wheel barrow
(thela),
dartboard, round cake, papad, drain cover, various designs, bangles, brooches, circular
paths, washers, flower beds, etc. are some examples of such objects (see Fig. 12.1).
So, the problem of finding perimeters and areas related to circular figures is of great
practical importance. In this chapter, we shall begin our discussion with a review of
the concepts of perimeter (circumference) and area of a circle and
apply this knowledge
in finding the areas of two special 'parts' of a circular
region (or briefly of acircle)
known as sector and segment. We shall also see how to find the areas of some
Combinations of plane figures involving circles or their parts.
Design
Wheel
A barrow
Washer Cake
Wheel
Fig. 12.1
, MAwMe
12.2 Perimeter and Area ofa Circle A Review
around circle
Recall that the distance covered by travelling once a is ite
usually called its circumference. You also knoW Irom your earlier clase
Crcumference of a circle bears a constant ratio With its diameter, This conet classes, th
is denoted by the Greek letter t (read as 'pi'). In other words, tant Falio
circumference
diameter
circumference= 7 x diameter
Tx 2r (where r is the
radius of the circle
=27tr
The great Indian mathematician Aryabhata (C.E. 476-550) gave an approximat
62832
value of T. He stated that t= 000Which is nearly equal to 3.1416. It isalso
00
interesting to note that using an identity of the great mathematical genius Srinivas
Ramanujan (1887-1920) of India, mathematicians have been able to calculatethe
value of t correct to million places of decimals. As you know from Chapter 1of
Class IX, T is an irational number and its decimal
expansion is non-terminatingand
non-recurming (non-repeating). However, for practical purposes, we generally take
22
the value of T as o r 3.14, approximately.
You may also recall that area
of a circle is Tt, where r is the radius of the cir
Recall that you have verified it
in Class VII, by cutting a circle into a numoc
sectors and rearranging them as shown in Fig. 12.2.
14
13
/% WAAYAZN
(i)
Fig 12.2
, CIRCLES 225
eAs RELATED TO
2Tr
You can see that the shape in Fig. 12.2 (i) is nearly a rectangle of lengthx
r. This suggests that the area of the circle = x 2tr Tr. Let xr us
nd breadth
=
all the concepts learnt in earlier classes, through an example.
vample 1: The cost of fencing a circular field at the rate of T 24 per metre is
12
is to be ploughed at the rate of 0.50 per m. Find the cost of
5280. The field
loughing the field (Take T = 22
Total cost 5280 = 220
olution Length of the fence (in metres) =
Rate 24
So, circumference of the field 220 m =
Therefore, if r metres is the radius of the field, then
27tr = 220
2 x X r = 220
220x7
r= 35
2x 22
e., radius of the field is 35 m.
area of the field = trr= x x 35 x 35 m = 22 x 5 x 35 m
Therefore,
of ploughing 1 m2 of the field 7 0.50
Now, cost
=
tal cost of ploughing the field = R 22 x 5 x 35 x 0.50 = 7 1925
EXERCISE 12.1
Unless stated otherwise, use Tt=
The radii of two circles are 19 cm and 9 cm respectively.
Find the radius of the circle which has circumference equal WHITE
tO the sum of the circumferences of the two circles. BLACK
BLUE
8 and 6 cm respectively. Find
h e radii of two circles are cm
equal to the sum of the cOLD
tne radius of the circle having area
areas of the two circles.
with its five
8 depicts an archery target marked
12.3
as Gold, Red, Blue,
OTing regions from the centre outwards
olack and White. The diameter of the region representing
10.5
and each of the other bands is
cm
uold score is 21 cm Fig. 12.3
wide. Find the area of each of the five scoring regions.
AREAS RELATED TOCIRCLES
12.1 Introduction
You are already familiar with some methods of finding perimeters and areas of
simple
plane figures such as rectangles, squares, parallelograms, triangles and circles from
your earlier classes. Many objects that we come across in our daily life are related to
the circular shape in some form or the other. Cycle wheels, wheel barrow
(thela),
dartboard, round cake, papad, drain cover, various designs, bangles, brooches, circular
paths, washers, flower beds, etc. are some examples of such objects (see Fig. 12.1).
So, the problem of finding perimeters and areas related to circular figures is of great
practical importance. In this chapter, we shall begin our discussion with a review of
the concepts of perimeter (circumference) and area of a circle and
apply this knowledge
in finding the areas of two special 'parts' of a circular
region (or briefly of acircle)
known as sector and segment. We shall also see how to find the areas of some
Combinations of plane figures involving circles or their parts.
Design
Wheel
A barrow
Washer Cake
Wheel
Fig. 12.1
, MAwMe
12.2 Perimeter and Area ofa Circle A Review
around circle
Recall that the distance covered by travelling once a is ite
usually called its circumference. You also knoW Irom your earlier clase
Crcumference of a circle bears a constant ratio With its diameter, This conet classes, th
is denoted by the Greek letter t (read as 'pi'). In other words, tant Falio
circumference
diameter
circumference= 7 x diameter
Tx 2r (where r is the
radius of the circle
=27tr
The great Indian mathematician Aryabhata (C.E. 476-550) gave an approximat
62832
value of T. He stated that t= 000Which is nearly equal to 3.1416. It isalso
00
interesting to note that using an identity of the great mathematical genius Srinivas
Ramanujan (1887-1920) of India, mathematicians have been able to calculatethe
value of t correct to million places of decimals. As you know from Chapter 1of
Class IX, T is an irational number and its decimal
expansion is non-terminatingand
non-recurming (non-repeating). However, for practical purposes, we generally take
22
the value of T as o r 3.14, approximately.
You may also recall that area
of a circle is Tt, where r is the radius of the cir
Recall that you have verified it
in Class VII, by cutting a circle into a numoc
sectors and rearranging them as shown in Fig. 12.2.
14
13
/% WAAYAZN
(i)
Fig 12.2
, CIRCLES 225
eAs RELATED TO
2Tr
You can see that the shape in Fig. 12.2 (i) is nearly a rectangle of lengthx
r. This suggests that the area of the circle = x 2tr Tr. Let xr us
nd breadth
=
all the concepts learnt in earlier classes, through an example.
vample 1: The cost of fencing a circular field at the rate of T 24 per metre is
12
is to be ploughed at the rate of 0.50 per m. Find the cost of
5280. The field
loughing the field (Take T = 22
Total cost 5280 = 220
olution Length of the fence (in metres) =
Rate 24
So, circumference of the field 220 m =
Therefore, if r metres is the radius of the field, then
27tr = 220
2 x X r = 220
220x7
r= 35
2x 22
e., radius of the field is 35 m.
area of the field = trr= x x 35 x 35 m = 22 x 5 x 35 m
Therefore,
of ploughing 1 m2 of the field 7 0.50
Now, cost
=
tal cost of ploughing the field = R 22 x 5 x 35 x 0.50 = 7 1925
EXERCISE 12.1
Unless stated otherwise, use Tt=
The radii of two circles are 19 cm and 9 cm respectively.
Find the radius of the circle which has circumference equal WHITE
tO the sum of the circumferences of the two circles. BLACK
BLUE
8 and 6 cm respectively. Find
h e radii of two circles are cm
equal to the sum of the cOLD
tne radius of the circle having area
areas of the two circles.
with its five
8 depicts an archery target marked
12.3
as Gold, Red, Blue,
OTing regions from the centre outwards
olack and White. The diameter of the region representing
10.5
and each of the other bands is
cm
uold score is 21 cm Fig. 12.3
wide. Find the area of each of the five scoring regions.
