Find the area of the minor segment of a circle of radius 14 cm, when the angle of the corresponding sector is 60°.

#### Solution

Given that, radius of circle (r) = 14 cm

And angle of the corresponding sector i.e., central angle (θ) = 60°

Since, in ΔAOB, OA = OB = Radius of circle i.e., ΔAOB is isosceles.

⇒ ∠AOB = ∠OBA = θ

Now, In ΔAOB,

∠AOB + ∠OAB = ∠OBA = 180° ......[Since, sum of interior angles of any triangle is 180°]

⇒ 60° + θ + θ = 180°

⇒ 2θ = 120°

⇒ θ = 60°

i.e. ∠OAB = ∠OBA = 60° = ∠AOB

Since, all angles of ΔAOB are equal to 60° i.e., ΔAOB ia an equilateral triangle.

Also, OA = OB = AB = 14 cm

So, Area of ΔOAB = `sqrt(3)/4` (side)^{2}

= `sqrt(3)/4 xx (14)^2` ......[∵ Area of an equilateral triangle = `sqrt(3)/4` (sides)^{2}]

= `sqrt(3)/4 xx 196`

= `49sqrt(3)` cm^{2}