Three equal circles, each of radius 6 cm, touch one another as shown in the figure. Find the area enclosed between them. [Take π = 3.14 and √3 = 1.732.]

Consider the above figure,

Here, first we join the center of all adjacent circles then the distance between the center of circles touching each other is equal to the side of an equilateral triangle formed by joining the center of adjacent circles. Therefore, we can say that the side of the equilateral triangle is equal to the twice of the radius of circle. Now by simply calculating the area of the 3 sectors and then subtracting it from the area of the equilateral triangle we can easily calculate the area of the enclosed region.

Given radius of each circle = r = 6 cm

Central angle of each sector = θ = 60° (∵ ∆ABC is equilateral)

Side of equilateral ∆ABC = a = 2×r = 2×6

∴ Side of equilateral ∆ABC = a = 12 cm

⇒ Area of one sector = 6π cm^{2}→ eqn1

Area of all the 3 sector = 3×Area of one sector

= 3×6π (from eqn1)

= 18π cm^{2}→ eqn2

Area of enclosed region = Area of equilateral ∆ABC – Area of all 3 sectors

⟹ Area of enclosed region = 36√3-18π (from eqn 3 and eqn 2)

⟹ Area of enclosed region = (36×1.732)-(18×3.14)

(put π = 3.14 &√3 = 1.732)

= 62.352 - 56.52

= 5.832 cm^{2}

The area of enclosed region is 5.832 cm^{2}.

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