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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In the mentioned Figure,
O is the centre of the circle,
AB is a chord
AXB is a major arc,
OA = OB = radius = 15 cm
Arc AXB subtends an angle 60 degrees at O.
Area of sector AOB = 60/360 × π × r^2
= 60/360 × 3.14 × (15)^2
= 117.75 cm^2
Area of the minor segment (Area of Shaded region) = Area of sector AOB -Area of △ AOB
By trigonometry,
AC = 15 sin 30
OC = 15 cos 30
And, AB = 2AC
∴ AB = 2 × 15 sin 30 = 15 cm
∴ OC = 15 cos 30 = = 15 × 1.73/2 = 12.975 cm
∴ Area of △AOB = 0.5 × 15 × 12.975 = 97.3125 cm^2
∴ Area of minor segment (Area of Shaded region) :-
= 117.75 − 97.3125 = 20.4375 cm
Area of major segment = Area of circle − Area of minor segment
= (3.14 × 15 × 15) − 20.4375
= 686.0625 cm^2
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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
