A chord AB of a circle of radius 3.5 subtends an angle of 60degree at the centre o of the circle. Find the area of the corresponding minor segment of the circle.
Answers
Answer:
Ans:- 1.11 cm2
Step-by-step explanation:
Radius of the circle = 3.5 cm
OA=OB= 3.5
ΔAOB is isosceles as two sides are equal.
∴ ∠A = ∠B
Sum of all angles of triangle = 180°
∠A + ∠B + ∠O = 180°
⇒ 2 ∠A = 180° - 60°
⇒ ∠A = 120°/2
⇒ ∠A = 60°
Triangle is equilateral as ∠A = ∠B = ∠O = 60°
∴ OA = OB = AB = 15 cm
Area of equilateral ΔAOB = 1/2 r² sinθ
1/2 r² √3/2= √3/4 r²
√3/4 × (OA)2 = √3/4 × 12.25
= (12.25√3)/4 = 5.30 cm2
Angle subtend at the centre by minor segment = 60°
Area of Minor sector making angle 60° = (60°/360°) × π r2 cm2
= (1/6) × 12.25 π = 12.25/6 π
= (12.25/6) × 3.14 = 6.41 cm2
Area of the minor segment = Area of Minor sector - Area of equilateral ΔAOB
= 6.41 - 5.30 = 1.11 cm2