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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