Question

Four equal circles, each of radius a, touch each other. Show that the area between them is $\frac{6}{7}{a}^2$ (Take $\pi =\frac{22}{7}$)

Answer 1

The four circles can be arranged as:

Here, radius of each circle = a

∴ Each side of square = 2a

∴ Area of square = (2a)² = 4a²

Area of all the four sectors area equal,

∴ Area of 4 sectors = 4 × area of each sector

= 4 x $\frac{{90}^{\circ }}{{360}^{\circ }}$ x $\pi$ x $a^2$

= 4 x $\frac{1}{4}$ x $\pi$ x $a^2$

= $\frac{1}{4}$ x $\pi$ x $a^2$

Required area = Area of square – area of 4 sectors

=4 x $a^2$ - $\frac{22}{7}$ x $a^2$

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

= $a^2$ x $\frac{6}{7}$ )

= $\frac{6}{7}$ ) $a^2$

Hence, area between the circles is = $\frac{6}{7}$ $a^2$