Question

Three equal circles, each of radius 6 cm, touch one another as shown in the figure. Find the area of enclosed between them.

Answer 1

Join ABC. All sides are equal, so it is an equilateral triangle.
Now,
Area of the equilateral triangle =$\frac{\sqrt{3}}{4}\times {\mathrm{Side}}^2$

$$\begin{gathered} =\frac{1.73}{4}\times 12\times 12 \\ =62.28{\mathrm{cm}}^2 \end{gathered}$$


$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{arc}\mathrm{of}\mathrm{the}\mathrm{circle}=\frac{60}{360}{\mathrm{\pi r}}^2 \\ =\frac{1}{6}{\mathrm{\pi r}}^2 \\ =\frac{1}{6}\times \frac{22}{7}\times 6\times 6 \\ =18.86{\mathrm{cm}}^2 \\ \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{three}\mathrm{sectors}=3\times 18.86=56.57{\mathrm{cm}}^2 \end{gathered}$$


Area of the shaded portion = Area of the triangle $-$ Area of the three quadrants

$$\begin{gathered} =62.28-56.57 \\ =5.71{\mathrm{cm}}^2 \end{gathered}$$