Question

A chord of a circle of radius 10 cm subtends an angle of 60° at the centre O. Find the area of the shaded region. `("Use"  sqrt(3) = 1.73, sqrt(2) = 1.41 and π = 3.14)`

Answer 1

Radius of circle (r) = 10 cm

Sector angle (θ) = 60°

Area of sector OACB = `θ/360 xx πr^2`

= `60/360 xx 3.14 xx 10 xx 10`

= `314/6`

= `157/3`

= 52.33 cm²

In the ΔAОВ, 

OA = OB   ...(∵ OA and OB are radius of circle.)

∠AOB = 60°

So, ∠OAB = ∠OBA   ...(∵ OA = OB)

⇒ ∠OAB + ∠OBA + ∠AOB = 180°   ...(By angle sum property)

⇒ ∠OAB + ∠OBA + 60° = 180°

⇒ ∠OAB + ∠OBA = 180° – 60°

⇒ 2∠OAB = 120°   ...(∵ ∠OAB = ∠OBA)

⇒ ∠OAB = `120^circ/2` 

⇒ ∠OAB = 60°

∴ ∠OAB = ∠OBA = 60°

ΔAOB is an equilateral triangle.

So, Area of ΔΑΟΒ = `sqrt(3)/4 xx (a)^2`

= `sqrt(3)/4 xx 10 xx 10`

= `25sqrt(3)`

= 25 × 1.73

= 43.25 cm²

Area of the shaded region = Area of sector ОАСВ – Area of ΔАОВ

= 52.33 – 43.25

= 9.08 cm²