Question

A chord of a circle of raduis 15 CM subtend an angle of 60 degree at the centre. Find the areas of the corresponding minor and major segments of the circle (use π =3.14 and √3= 1.73)

Answer 1

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

Answer 2

area of a sector =
$\frac{1}{2} \times {r}^{2} \times \alpha$
angle = 60° = 60*pi/180 = pi/3

r= 15cm

area of minor segment = (1/2)*(15^2)*(pi/3)

= 117.75 cm2

area of circle =
$\pi \times {r}^{2}$
706.5 cm2

area of major segment = area of circle - area of minor segment

= 706.5 - 117.75

= 588.75 cm2