Question

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.​

Answer 1

$\huge\bf\fbox\pink{Answer:-}$

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 × 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.

Answer 2

Answer:

The correct answer of your question is 686.05cm^2

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 × 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.