Answer :

Given: Radius of the circle = r = 6 cm



Thus, area of the circle = πr2


= 3.14 × 36 = 113.04 cm2


Join the vertices of the equilateral triangle to the center of the circle.


Thus, three isosceles triangles with equal sides of length equal to radius of the circle = 6 cm.


Also, BOC = AOB = AOC = θ = 120°


Area of shaded region = Area of the segment ABA + Area of the segment ACA + Area of the segment BCB


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


Thus, area of the sector AOBA = (1/2)(r2θ)



= 264/7 cm2


Area of the sector AOBA = Area of the sector AOCA = Area of the sector BOCB


Now, area of the shaded region = 3(Area of the minor segment)


= 3(Area of the one sector – Area of the one smaller triangle)


= 3[(264/7) – 1/2 (r2sinθ)]


= 3[(264/7) – 1/2 (36 × sin 120°)]


= 3[(264/7) – (18 × (√3/2))]


= 3[22.144]


= 66.432


Thus, area of the shaded region = 66.432 cm2


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