A chord of a circle of radius 15 cm subtends an angle of 60° at the Centre. Find the areas of the corresponding A chord of a circle of radius 15 cm subtends an angle of 60° at the Centre. Find the areas of the corresponding minor and major segments of the circle. (Use π= 3.14 and √3 = 1.73)minor and major segments of the circle. (Use π= 3.14 and √3 = 1.73)

Radius of circle 15 cm
Area of sector OPRQ
= 60°/360° × πr² = 1/6 × π(15)² = 1/6 × 3.14 × 15 × 15 = 117.75 cm²
Area of minor segment
= Area of minor sector OPRQ – Area of △0PQ
= 117.75 cm² – √3/4(15) cm² [As triangle OPQ is an equilateral triangle]
= 117.75 cm² – 225√3/4 cm²
= 117.75 cm² – 56.25 x 1.73 cm²
= 231 cm² – 97.3125 cm²
= 20.4375 cm²
Area of major segment = Area of circle – Area of minor segment
= πr² – 20.4375 cm²
= π(15)² – 20.4375] cm²
= [3.14 x 15 x 15 – 20.4375] cm²
= [706.5 – 20.4375] cm²
= 686.0625 cm²

For better understanding to the above question see here😎👇