Question

A chord of a circle of radius 15cmm subtends an angle of 60degree at the centre. find the area of the corresponding minor and major segments of the circle​

Answer 1

Answer:

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 × 152

= (225√3)/4 cm² = 97.3 cm²

Angle subtend at the centre by minor segment = 60°

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

= (1/6) × 152 π cm² = 225/6 π cm²

= (225/6) × 3.14 cm² = 117.75 cm²

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

= 117.75 cm² - 97.3 cm² = 20.4 cm²

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

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

= (5/6) × 152 π cm² = 1125/6 π cm²

= (1125/6) × 3.14 cm² = 588.75 cm²

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

= 588.75 cm² + 97.3 cm² = 686.05 cm²