A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use π = 3.14 and √3 = 1.73).
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Answers
Step-by-step explanation:
refer to the attached image for solutions
Step-by-step explanation:
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
