A chord of circle radius 15cm subtends an angle of 60 degree find the area of corresponding major and minor segment of the circle (use pie = 3.14)
Answers
Answer:
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
Area of major segment = Area of Minor sector + Area of equilateral ΔAOB
= 588.75 cm2 + 97.3 cm2 = 686.05 cm2
#answerwithquality #BAL
Solution : Radius (r) of the circle = 15 cm
Area of sector OPRQ =
=
=
= 117.75 cm²
In ∆OPQ
⇒ ∠OPQ = ∠OQP (As OP = OQ)
⇒ ∠OPQ + ∠OQP + ∠POQ = 180°
⇒ 2 ∠OPQ = 120°
⇒ ∠OPQ = 120°/2
⇒ ∠OPQ = 60°
∆OPQ is an equilateral triangle.
⇒ ∴ Area of ∆OPQ =
⇒
⇒56.25√3
⇒97.312 cm²
⇒ Now, Area of Minor segment PRQ = Area of sector OPRQ – Area of ∆OPQ
⇒ 117.75 ― 97.3125
⇒ 20.437 cm²
Area of Major segment PSQ = Area of circle ― Area of segment PRQ
⇒ 3.14 × 225 ― 20.437
⇒ 706.5 ― 20.437
⇒ 686.06 cm²