If three circles of radius a each, are drawn such that each touches the 4 other two, prove that the area included between them is equal to [Take √3 = 1.73 and π = 3.14.]

Consider the figure shown below

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 = “a” units

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

Side of equilateral ∆ABC = 2×a units

∴ Area of all 3sectors = 3×Area of one sector

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

⟹ Area of enclosed region

(taking a^{2} common)

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