Question

Find the area of the minor segment of a circle of radius 14 cm, when its central angle is
60˚. Also find the area of the corresponding major segment.
22

Answer 1

Radius of the circle = 14 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 = 14 cm

Area of equilateral ΔAOB = √3/4 × (OA)2 = √3/4 × 142

= (196√3)/4 cm2 = 84.87 cm2

Angle subtend at the centre by minor segment = 60°

Area of Minor sector making angle 60° = (60°/360°) × π r2 cm2

= (1/6) × 142 π cm2 = 196/6 π cm2

= (196/6) × 3.14 cm2 = 102.57 cm2

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

= 102.75 cm2 - 84.87 cm2 = 17.87 cm2

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

Area of the sector making angle 300° = (300°/360°) × π r2 cm2

= (5/6) × 142 π cm2 = 980/6 π cm2

= (980/6) × 3.14 cm2 = 512.86 cm2

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

= 512.86 cm2 + 84.87 cm2 = 597.73 cm2