Radius of the circle = 15 cm

ΔAOB is isosceles as two sides are equal.

∴ ∠A = ∠B

Sum of all angles of triangle = 180°

∠A + ∠B + ∠C = 180°

⇒ 2 ∠A = 180° - 60°

⇒ ∠A = 120°/2

⇒ ∠A = 60°

Triangle is equilateral as ∠A = ∠B = ∠C = 60°

∴ OA = OB = AB = 15 cm

Area of equilateral ΔAOB = √3/4 × (OA)^{2 }= √3/4 × 15^{2 }

= (225√3)/4 cm^{2 }= 97.3 cm^{2}

Angle subtend at the centre by minor segment = 60°

Area of Minor sector making angle 60° = (60°/360°) × π r^{2 }cm^{2}

= (1/6) × 15^{2 }π^{ }cm^{2} =^{ }225/6 π^{ }cm^{2}

^{ } = ^{ }(225/6) × 3.14 cm^{2 }= 117.75 ^{ }cm^{2}

Area of the minor segment = Area of Minor sector - Area of equilateral ΔAOB

= 117.75 ^{ }cm^{2} - 97.3 cm^{2 }= 20.4 cm^{2}

Angle made by Major sector = 360° - 60° = 300°

Area of the sector making angle 300° = (300°/360°) × π r^{2 }cm^{2}

^{ }= (5/6) × 15^{2 }π^{ }cm^{2} =^{ }1125/6 π^{ }cm^{2}

^{ } = ^{ }(1125/6) × 3.14 cm^{2 }= 588.75 ^{ }cm^{2}

Area of major segment = Area of Minor sector + Area of equilateral ΔAOB

= 588.75 ^{ }cm^{2} + 97.3 cm^{2 }= 686.05 cm^{2}