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 $\frac{4}{25}{a}^2.$

Answer 1

When three circles touch each other, their centres form an equilateral triangle, with each side being 2a.

Area of the triangle = $\frac{\sqrt{3}}{4}\times 2a\times 2a=\sqrt{3}{a}^2$

Total area of the three sectors of circles = $3\times \frac{60}{360}\times \frac{22}{7}\times a^2=\frac{1}{2}\times \frac{22}{7}\times a^2=\frac{11}{7}{a}^2$

Area of the region between the circles = $\mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{triangle}-\mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{three}\mathrm{sectors}$
$$\begin{gathered} =\left(\sqrt{3}-\frac{11}{7}\right)a^2 \\ =\left(1.73-1.57\right)a^2 \\ =0.16a^2 \\ =\frac{4}{25}{a}^2 \end{gathered}$$