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 =
