An arc of circle radius 14 cm makes an angle of 60degree at the centre find the area of the minor and the major segment
Answers
Answer Is:
Radius of the circle = 14 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 = 14 cm
Area of equilateral ΔAOB = √3/4 × (OA)2 = √3/4 × 142
= (196√3)/4 cm2 = 84.87 cm2
Angle subtend at the centre by minor segment = 60°
Area of Minor sector making angle 60° = (60°/360°) × π r2 cm2
= (1/6) × 142 π cm2 = 196/6 π cm2
= (196/6) × 3.14 cm2 = 102.57 cm2
Area of the minor segment = Area of Minor sector - Area of equilateral ΔAOB
= 102.75 cm2 - 84.87 cm2 = 17.87 cm2
Angle made by Major sector = 360° - 60° = 300°
Area of the sector making angle 300° = (300°/360°) × π r2 cm2
= (5/6) × 142 π cm2 = 980/6 π cm2
= (980/6) × 3.14 cm2 = 512.86 cm2
Area of major segment = Area of Minor sector + Area of equilateral ΔAOB
= 512.86 cm2 + 84.87 cm2 = 597.73 cm2
Radius of the circle = 14 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 = 14 cm
Area of equilateral ΔAOB = √3/4 × (OA)2 = √3/4 × 142
= (196√3)/4 cm2 = 84.87 cm2
Angle subtend at the centre by minor segment = 60°
Area of Minor sector making angle 60° = (60°/360°) × π r2 cm2
= (1/6) × 142 π cm2 = 196/6 π cm2
= (196/6) × 3.14 cm2 = 102.57 cm2
Area of the minor segment = Area of Minor sector - Area of equilateral ΔAOB
= 102.75 cm2 - 84.87 cm2 = 17.87 cm2
Angle made by Major sector = 360° - 60° = 300°
Area of the sector making angle 300° = (300°/360°) × π r2 cm2
= (5/6) × 142 π cm2 = 980/6 π cm2
= (980/6) × 3.14 cm2 = 512.86 cm2
Area of major segment = Area of Minor sector + Area of equilateral ΔAOB
= 512.86 cm2 + 84.87 cm2 = 597.73cm^2