Question

Four equal circles, each of radius $a$, touch each other. Show that the area between then is $\frac{6}{7}{a}^2$

(Take $\pi$$=\frac{22}{7}$).

Answer 1

$\Rightarrow$ Here, radius of each circle $=a$

$\therefore$ Each side of square $=2a$

$\therefore$ Area of square $=(2a{)}^2=4a^2$

$\Rightarrow$ Area of $4$ sectors $=4\times \frac{\theta }{{360}^o}\pi r^2$

$\Rightarrow$ Area of $4$ sectors $=4\times \frac{{90}^o}{{360}^o}\times \pi \left(a^2\right)$

$\Rightarrow$ Area of $4$ sectors $=4\times \frac{1}{4}\times \frac{22}{7}{a}^2$

$\therefore$ Area of $4$ sectors $=\frac{22}{7}{a}^2$

$\Rightarrow$ Required area = Area of square - Area of $4$ sectors.

$\Rightarrow$ Required area $=4a^2-\frac{22}{7}{a}^2$

$\Rightarrow$ Required area $=a^2(4-\frac{22}{7})$

$\Rightarrow$ Required area $=a^2\times \frac{6}{7}=\frac{6}{7}{a}^2$

$\therefore$ Area between the circle is $\frac{6}{7}{a}^2$

Hence Proved