Question

Three circles each of radius 3.5 cm are drawn in such a way that each of them touches the other two. Find the area enclosed between these circles.

Answer 1

The three circles are drawn such that each of them touches the other two.

By joining the centers of the three circles,

We get,

AB = BC = CA = 2(radius) = 7 cm

Therefore, triangle ABC is an equilateral triangle with each side 7 cm.

∴ Area of the triangle = (√3 /4) × a², where a is the side of the triangle.

= (√3 /4) × 7²

= (49/4) √3 cm²

= 21.2176 cm²

Now, Central angle of each sector = ∅ = 60° (60π/180)

= π/3 radians

Thus, area of each sector = (1/2) r²θ

= (1/2) × (3.5)² × (π/3)

= 12.25 × (22/ (7×6))

= 6.4167 cm²

Total area of three sectors = 3 × 6.4167 = 19.25 cm²

∴ Area enclosed between three circles = Area of triangle ABC – Area of the three sectors

= 21.2176 – 19.25

= 1.9676 cm²

Hence, the required area enclosed between these circles is 1.967 cm² (approx.).