## Book: RS Aggarwal - Mathematics

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

#### Subject: Maths - Class 10th

##### Q. No. 16 of Exercise 18B

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

16
##### A chord 10 cm long is drawn in a circle whose radius is 5√2 cm. Find the areas of both the segments. [Take π = 3.14.]

Consider the figure shown above.

In this, the triangle AOB is an isosceles triangle. So here we will construct a perpendicular bisector from O on AB and as this triangle is isosceles therefore this perpendicular will also act as median and angle bisector.

Therefore,

Draw a perpendicular bisector from O which meets AB at D and bisects AB, as ABO is an isosceles triangle therefore OD acts as a median.

So, consider right angle triangle AOD right angled at D

Let AOD = θ Perpendicular = AD and Hypotenuse = AO = R

Given Radius of circle = R = 5√2 cm

Length of chord AB = 10 cm, AD = 5 cm

θ = 45°

AOD = 45°

Area of minor segment = Area of sector – Area of right angle triangle

eqn1

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

Area of sector = 39.25 cm2

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

As this is isosceles right-angle triangle

height = base = 5√2 cm

Area of right angle triangle = 1/2 ×5√2×5√2 = = 25 cm2

Put the value of area of sector and area of right angle triangle in equation 1,

Area of minor segment = 39.25 -25

= 14.25 cm2

Area of major segment = πR2 – area of minor segment

Area of major segment = 157 – 14.25 = 142.75 cm2

Area of major segment is 142.75 cm2 and of minor segment is 14.25 cm 2.

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