## Book: RS Aggarwal - Mathematics

### Chapter: 18. Area of Circle, Sector and Segment

#### Subject: Maths - Class 10th

##### Q. No. 17 of Exercise 18B

Listen NCERT Audio Books to boost your productivity and retention power by 2X.

17
##### Find the area of both the segments of a circle of radius 42 cm with central angle 120°. [Given, sin 120° = √3/2 and √3 = 1.73.]

Given R = 42 cm and central angle of sector = 120°

Area of minor segment = Area of sector – Area of triangle eqn1

Where R = radius of the circle and θ = central angle of the sector

Area of sector = 1848 cm2

Area of right angle triangle = 1/2×base×height×sin θ

Where θ = central angle of the sector

Area of triangle = 1/2×42×42×√3/2

Area of triangle = (42×42×√3)/4

(put √3 = 1.73)

= 762.93 cm2

Put the values of area of triangle and area of sector in equation 1

Area of minor segment = 1848 – 762.93

= 1085.07 cm2

Area of major segment = πR2 – Area of minor segment

Put the value of area of minor segment and R in above equation

= π(422) – 1085.07

Area of major segment = 22/7×42×42-1085.07

(put π = 22/7)

Area of major segment = 5544 – 1085.07

Area of major segment = 4458.93 cm2

Area of major segment is 4458.93 cm2 and of minor segment is 1085.07 cm2.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60