A chord PQ of a circle with radius 15 cm subtends an angle of 60° with the centre of the circle. Find the area of the minor as well as the major segment. ( = 3.14, √3 = 1.73)

Given central angle = 60˚
Radius ,r = 15cm
Let chord PQ subtend ∠POQ = 60° at centre.
∴ θ = 60°
Area of minor segment = r²[(/360) – (sin/2)]
= 15²[3.14×(60/360) – sin60˚/2]
= 225[3.14×(1/6) – √3/4]
= 225[(3.14/6) – 1.73/4]
= 225[(6.28- 5.19)/12]
= 20.44
Hence area of minor segment is 20.44cm².
Area of circle = r²
= 3.14×15²
= 3.14×225
= 706.5cm²
Area of major segment = Area of circle – area of minor segment
= 706.5 – 20.44
= 686.06cm²
Hence area of major segment is 686.06cm².