Question

If three circles of radius $a$ each, are drawn such that each touches the other two, prove that the area included between them is equal to $\frac{4}{25}$ $a^2$.

[ Take $\sqrt{3}=1.73and\pi =\mathrm{3.14.}$]

Answer 1

Given, three circles touch each other having radius a.

Construction: Join the centres of the circles.

Note: recheck your proof part.

Proof:
On joining the centres of the circles, we get an equilateral triangle ABC with side 2a.
Thus,
Area formed between the circles= Area of shaded region
= Area of equilateral triangle ABC- $3\times$( Area of sector)

=$\frac{\sqrt{3}}{4}\times (2a{)}^2-3(\frac{60°}{360°}\pi \times a^2)$

=$\sqrt{3}{a}^2-3(\frac{1}{6}\times \pi \times a^2)$


=$(\sqrt{3}-\frac{\pi }{2})a^2$

putting values of $\pi$ and $\sqrt{3}$ as 3.14 and 1.73 respectively , we get

Required area = $0.16a^2=\frac{4}{25}{a}^2$

Hence proved.