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}