# 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

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