Question

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 `sqrt3 = 1.73`]

Answer 1

Radius (r) of circle = 15 cm

Area of sector OPRQ = `60^@/360^@xxpir^2`

= `1/6 xx 3.14 xx (15)^2`

= 117.75 cm²

In ΔOPQ,

∠OPQ = ∠OQP (As OP = OQ)

∠OPQ + ∠OQP + ∠POQ = 180°

2 ∠OPQ = 120°

∠OPQ = 60°

ΔOPQ is an equilateral triangle.

Area of ΔOPQ = `(sqrt3)/4 xx (r)^2`

`= sqrt3/4 xx (15)^2`

= `(225sqrt3)/4  "cm"^2`

`= 56.25sqrt3`

97.3125 cm²

Area of segment PRQ = Area of sector OPRQ − Area of ΔOPQ

= 117.75 − 97.3125

= 20.4375 cm²

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

= πr² − 20.4375

= π × (15)² − 20.4375

= 3.14 × 225 − 20.4375

= 706.5 − 20.4375

= 686.0625 cm²