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

Given that, three circles are in such a way that each of them touches the other two.

Now, join AB, BC and CA.

Since, radius of each circle is 3.5 cm.

So, AB = 2 × radius of circle

= 2 × 3.5 cm

= 7 cm

∴ AC = BC = AB = 7 cm

So, ΔABC is an equilateral triangle with side 7 cm.

We know that, each angle between two adjacent sides of an equilateral triangle is 60°.

∴ Area of sector with ∠A = 60°

= `(∠"A")/360^circ xx pi"r"^2`

= `60^circ/360^circ xx pi xx (3.5)^2 "cm"^2`

So, area of each sector = 3 × area of sector with angle A

= `3 xx 60^circ/360^circ xx pi xx (3.5)^2 "cm"^2`

= `1/2 xx 22/7 xx 3.5 xx 3.5  "cm"^2`

= `11 xx 5/10 xx 35/10 "cm"^2`

= `77/4 "cm"^2`

= 19.25 cm²

And area of ΔABC = `sqrt(3)/4 xx (7)^2 "cm"^2`   ...[∵ Area of an equilateral triangle = `sqrt(3)/4 xx ("side")^2`]

= `(49sqrt(3))/4 "cm"^2`

∴ Area of shaded region enclosed between these circles

= Area of ΔABC – Area of each sector

= `(49sqrt(3))/4 - 19.25`

= `12.25 xx sqrt(3) - 19.25`

= 21.2176 – 19.25

= 1.9676 cm²

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