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

Let r be the radius of the circle = 10 cm

The figure is given below:

Angle subtended at the center of the sector = θ = 60°

OP = OQ = r = radius

In triangle POQ, ∠POQ = 60°, ∠OPQ = ∠OQP = ϕ

Since, sum of all interior angles of a triangle is 180°

∴θ+ ϕ + ϕ = 18o°

⇒ 60° + 2 ϕ = 180°

⇒ 2 ϕ = 180° - 60° = 120°

⇒ ϕ = 60°

∴ Each angle is of 60° and hence the triangle POQ is an equilateral triangle.

Now, Area of the minor segment = Area of the sector POQR – Area of triangle POQ

Angle subtended at the center of the sector =60°

Angle subtended at the center (in radians) = θ = 60π/180 = π/3

∴ Area of sector POQR of a circle = (1/2) × r² × θ

= (1/2) × 100 × (π/3)

= 50π/3

= (50 × 3.14)/3

= 52.33 cm²

Area of the equilateral triangle POQ = (√3/4) × (Side)²

= (√3/4) × (10)²

= 25√3 cm²

= 43.30 cm²

∴ Area of minor segment = Area of the sector POQR – Area of triangle POQ = (52.33 – 43.30) cm²

= 9.03 cm²

Now, area of the circle = πr² = 3.14 × 100 = 314 cm²

Area of the major segment = Area of circle – Area of the minor segment = (314 – 9.03) cm² = 304.97 cm²