Question

Four equal circles, each of radius a units, touch each other. Show that the area between them is $\left(\frac{6}{7}{a}^2\right)$ sq units.

Answer 1

When four circles touch each other, their centres form the vertices of a square. The sides of the square are 2a units.

Area of the square = ${\left(2a\right)}^2=4a^2\mathrm{sq}.\mathrm{units}$

Area occupied by the four sectors
$$\begin{gathered} =4\times \frac{90}{360}\times \mathrm{\pi }\times a^2 \\ =\pi a^2\mathrm{sq}.\mathrm{units} \end{gathered}$$

Area between the circles = Area of the square $-$ Area of the four sectors
$$\begin{gathered} =\left(4-\frac{22}{7}\right)a^2 \\ =\frac{6}{7}{a}^2\mathrm{sq}.\mathrm{units} \end{gathered}$$