Answer :

Given: Radius of the circle = r = 6 cm

Thus, area of the circle = πr^{2}

= 3.14 × 36 = 113.04 cm^{2}

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)(r^{2}θ)

= 264/7 cm^{2}

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 (r^{2}sinθ)]

= 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 cm^{2}

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