154 MATHEMATICS
11
11.1 Areas of Sector and Segment of a Circle
You have already come across the terms sector and
segment of a circle in your earlier classes. Recall
that the portion (or part) of the circular region enclosed
by two radii and the corresponding arc is called a
sector of the circle and the portion (or part) of the
circular region enclosed between a chord and the
corresponding arc is called a segment of the circle.
Thus, in Fig. 11.1, shaded region OAPB is a sector
of the circle with centre O. ∠ AOB is called the
angle of the sector. Note that in this figure, unshaded region OAQB is also a sector of
the circle. For obvious reasons, OAPB is called the minor sector and
OAQB is called the major sector. You can also see that angle of the major sector is
360° – ∠ AOB.
Now, look at Fig. 11.2 in which AB is a chord
of the circle with centre O. So, shaded region APB is
a segment of the circle. You can also note that
unshaded region AQB is another segment of the circle
formed by the chord AB. For obvious reasons, APB
is called the minor segment and AQB is called the
major segment.
Remark : When we write ‘segment’ and ‘sector’
we will mean the ‘minor segment’ and the ‘minor
sector’ respectively, unless stated otherwise.
AREAS RELATED TO CIRCLES
Fig. 11.2
Fig. 11.1
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AREAS RELATED TO CIRCLES 155
Now with this knowledge, let us try to find some
relations (or formulae) to calculate their areas.
Let OAPB be a sector of a circle with centre
O and radius r (see Fig. 11.3). Let the degree
measure of Ð AOB be q .
You know that area of a circle (in fact of a
circular region or disc) is pr
2
.
In a way, we can consider this circular region to
be a sector forming an angle of 360° (i.e., of degree
measure 360) at the centre O. Now by applying the
Unitary Method, we can arrive at the area of the
sector OAPB as follows:
When degree measure of the angle at the centre is 360, area of the
sector = pr
2
So, when the degree measure of the angle at the centre is 1, area of the
sector =
Therefore, when the degree measure of the angle at the centre is q, area of the
sector =
=
.
Thus, we obtain the following relation (or formula) for area of a sector of a
circle:
Area of the sector of angle
qq
qq
q =
r
,
where r is the radius of the circle and q the angle of the sector in degrees.
Now, a natural question arises : Can we find
the length of the arc APB corresponding to this
sector? Yes. Again, by applying the Unitary
Method and taking the whole length of the circle
(of angle 360°) as 2pr, we can obtain the required
length of the arc APB as
.
So, length of an arc of a sector of angle
qq
qq
q =
r
.
Fig. 11.3
Fig. 11.4
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156 MATHEMATICS
Now let us take the case of the area of the
segment APB of a circle with centre O and radius r
(see Fig. 11.4). You can see that :
Area of the segment APB = Area of the sector OAPB – Area of OAB
=
2
– area of OAB
360
r
Note : From Fig. 11.3 and Fig. 11.4 respectively, you can observe that:
Area of the major sector OAQB = r
2
– Area of the minor sector OAPB
and Area of major segment AQB = r
2
– Area of the minor segment APB
Let us now take some examples to understand these concepts (or results).
Example 1 : Find the area of the sector of a circle
with radius 4 cm and of angle 30°. Also, find the area
of the corresponding major sector (Use = 3.14).
Solution : Given sector is OAPB (see Fig. 11.5).
Area of the sector =
2
360
r
=
2
30
3.1444cm
360
=
22
12.56
cm 4.19cm
3
(approx.)
Area of the corresponding major sector
= r
2
– area of sector OAPB
= (3.14 × 16 – 4.19) cm
2
= 46.05 cm
2
= 46.1 cm
2
(approx.)
Alternatively, area of the major sector =
2
(360 – )
360
r
=
2
360 30
3.14 16 cm
360
=
22
330
3.14 16cm 46.05 cm
360
= 46.1 cm
2
(approx.)
Fig. 11.5
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AREAS RELATED TO CIRCLES 157
Example 2 : Find the area of the segment AYB
shown in Fig. 11.6, if radius of the circle is 21 cm and
AOB = 120°. (Use =
22
7
)
Solution : Area of the segment AYB
= Area of sector OAYB –
Area of OAB (1)
Now, area of the sector OAYB =
120 22
21 21
360 7
cm
2
= 462 cm
2
(2)
For finding the area of OAB, draw OM AB as shown in Fig. 11.7.
Note that OA = OB. Therefore, by RHS congruence, AMO BMO.
So, M is the mid-point of AB and AOM = BOM =
1
120 60
2
.
Let OM = x cm
So, from OMA,
OM
OA
= cos 60°
or,
21
x
=
11
cos 60° =
22
or, x =
21
2
So, OM =
21
2
cm
Also,
AM
OA
= sin 60° =
3
2
So, AM =
21 3
2
cm
Therefore, AB = 2 AM =
2213
cm = 21 3cm
2
Fig. 11.6
Fig. 11.7
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158 MATHEMATICS
So, area of OAB =
1
AB × OM
2
=
2
121
21 3 cm
22
=
2
441
3cm
4
(3)
Therefore, area of the segment AYB =
2
441
462 3 cm
4
[From (1), (2) and (3)]
=
2
21
(88 – 21 3)cm
4
EXERCISE 11.1
Unless stated otherwise, use =
22
7
.
1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.
2. Find the area of a quadrant of a circle whose circumference is 22 cm.
3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute
hand in 5 minutes.
4. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of
the corresponding : (i) minor segment (ii) major sector. (Use = 3.14)
5. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:
(i) the length of the arc (ii) area of the sector formed by the arc
(iii) area of the segment formed by the corresponding chord
6. A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find
the areas of the corresponding minor and major
segments of the circle.
(Use = 3.14 and
3
= 1.73)
7. A chord of a circle of radius 12 cm subtends an
angle of 120° at the centre. Find the area of the
corresponding segment of the circle.
(Use = 3.14 and
3
= 1.73)
8. A horse is tied to a peg at one corner of a square
shaped grass field of side 15 m by means of a 5 m
long rope (see Fig. 11.8). Find
Fig. 11.8
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AREAS RELATED TO CIRCLES 159
(i) the area of that part of the field in which the
horse can graze.
(ii) the increase in the grazing area if the rope
were 10 m long instead of 5 m. (Use = 3.14)
9. A brooch is made with silver wire in the form
of a circle with diameter 35 mm. The wire is also
used in making 5 diameters which divide the
circle into 10 equal sectors as shown in
Fig. 11.9. Find :
(i) the total length of the silver wire required.
(ii) the area of each sector of the brooch.
10. An umbrella has 8 ribs which are equally spaced
(see Fig. 11.10). Assuming umbrella to be a flat
circle of radius 45 cm, find the area between the
two consecutive ribs of the umbrella.
11. A car has two wipers which do not overlap. Each
wiper has a blade of length 25 cm sweeping
through an angle of 115°. Find the total area
cleaned at each sweep of the blades.
12. To warn ships for underwater rocks, a lighthouse
spreads a red coloured light over a sector of
angle 80° to a distance of 16.5 km. Find the area
of the sea over which the ships are warned.
(Use = 3.14)
13. A round table cover has six equal designs as
shown in Fig. 11.11. If the radius of the cover is
28 cm, find the cost of making the designs at the
rate of ` 0.35 per cm
2
. (Use
3
= 1.7)
14. Tick the correct answer in the following :
Area of a sector of angle p (in degrees) of a circle
with radius R is
(A)
2R
180
p
(B)
2
R
180
p
(C)
2R
360
p
(D)
2
2R
720
p
Fig. 11.9
Fig. 11.10
Fig. 11.11
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160 MATHEMATICS
11.2 Summary
In this chapter, you have studied the following points :
1. Length of an arc of a sector of a circle with radius r and angle with degree measure is
2
360
r
2. Area of a sector of a circle with radius r and angle with degree measure is
2
360
r
3. Area of segment of a circle
= Area of the corresponding sector – Area of the corresponding triangle.
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