a chord of a circle of radius 15 CM substance and angle 60 degree at the centre find the area of the corresponding minor and major segment of circle used by is equal to 3.14 and root 3 is equal to 1.73
Answers
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
Hope this helps :)
Answer:
hope it's help you...
Step-by-step explanation:
pls mark as brainliest pls pls...
