Question

A chord PQ of a circle of radius 10 cm substends an angle of 60° at the centre of circle. Find the area of major and minor segments of the circle.

Answer 1

Radius of the circle, r = 10 cm

Area of sector OPRQ

`= 60^@/360^@ xx pir^2`

`= 1/6 xx 3.14 xx (10)^2`

`= 52.33 cm^3`

In ΔOPQ,

∠OPQ = ∠OQP (As OP = OQ)

∠OPQ + ∠OQP + ∠POQ = 180°

2∠OPQ = 120°

∠OPQ = 60°

ΔOPQ is an equilateral triangle.
So, area of ΔOPQ

`= sqrt3/4 xx ("Side")^2`

`= sqrt3/4 xx (10)^2`

`=( 100sqrt3)/4 cm^2`

= 43.30 cm²

Area of minor segment PRQ

= Area of sector OPRQ − Area of ΔOPQ

= 52.33 − 43.30

= 9.03 cm²

Area of major segment PSQ
= Area of circle − Area of minor segment PRQ

= π(10)² − 9.03

= 314 − 9.03

= 304.97 cm²