## Book: RS Aggarwal - Mathematics

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

#### Subject: Maths - Class 10th

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

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18
##### A chord of a circle of radius 30 cm makes an angle of 60° at the centre of the circle. Find the areas of the minor and major segments. [Take π = 3.14 and √3 = 1.732.]

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 = 471 cm2

Where a = side of the triangle

Area of triangle = √3/4×30×30

Area of triangle = √3/4×900

Area of triangle = (900×√(3 ))/4

(put √3 = 1.732)

Area of triangle = (1.732×900)/4

Area of triangle = 389.7 cm2

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

Area of minor segment = 471 – 389.7

Area of minor segment = 81.3 cm2

Area of major segment = πR2 – Area of minor segment

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

Area of major segment = π×(302) – 81.3 (put π = 3.14)

Area of major segment = 3.14×30×30 – 81.3

Area of major segment = 2826 – 81.3

= 2744.7 cm2

Area of major segment is 2744.7cm2 and of minor segment is 81.3 cm2.

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