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

Let r be the radius of the circle = 14 cm

Angle subtended at the center of the sector = θ = 60°

In triangle AOB, ∠AOB = 60°, ∠OAB = ∠OBA = θ

Since, sum of all interior angles of a triangle is 180°

∴ θ + θ + 60 = 180

⇒ 2 θ = 120

⇒ θ = 60

∴ Each angle is of 60° and hence the triangle AOB is an equilateral triangle.

Now, Area of the minor segment = Area of the sector AOBC – Area of triangle AOB

Angle subtended at the center of the sector = 60°

Angle subtended at the center (in radians) = θ = 60π/100 = π/3

∴ Area of a sector of a circle = r²θ/2

=

= 308/3 cm²

Area of the equilateral triangle =

∴ Area of minor segment = = 17.796 cm²