Q15 of 45 Page 12

A chord of a circle of radius 14 cm subtends an angle of 120° at the center. Find the area of the corresponding minor segment of the circle. Use

π = 22/7 and √3 = 1.73. (CBSE 2011)

Radius of the circle = r = 14 cm


Central angle of the sector AOBA = 120°


Central angle of the sector AOBA (in radians) = θ = 120 × (π/180) = 2π/3


Now, area of the minor sector AOBA = 1/2 × (r2θ)


= 1/2 × (14)2 × (2π/3)


= 205.33 cm2



Area of the triangle AOB = 1/2 (r2 × sin120°)


= 1/2 [(14)2 × (√3/2)]


= 49 × √3


= 49 × 1.73


= 84.77 cm2


Area of the minor segment = Area of minor sector – Area of the triangle


= 205.33 – 74.77 = 120.56 cm2

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