Question

Two circles, each of radius $7cms$, intersect each other. The distance between their centres is $7\sqrt{2}cms$. Find the area common to both the circles.

Answer 1

$△ABC$ is isasceles right angle triangle.

Draw a perpendicular $AO$ from $A$ on side $BC$

In $△ABO$:

$\sin {45}^0=\frac{AO}{AB}$

$\Rightarrow \frac{1}{\sqrt{2}}=\frac{AD}{7}$

$\Rightarrow AO=\frac{7}{\sqrt{2}}cm$

$\therefore AO$ is the perpendicular bisector of $BC$ and also , $BC=7\sqrt{2}$ (given)

$\therefore △ABO\cong △AOC\Rightarrow BO=\frac{1}{2}BC\Rightarrow BO=\frac{7\sqrt{2}}{2}=\frac{7}{\sqrt{2}}$

Also, $BO=CO$ (cpct)

Area of $△AOB=$ Area of $△ACO=\frac{1}{2}\times CO\times AO$

$=\frac{1}{2}\times \frac{7}{\sqrt{2}}\times 7\sqrt{2}$

$=\frac{49}{4}$

For area of sector $ACD$:

$2\pi$ angle has area $\to \pi r^2$

$\Rightarrow \frac{\pi }{4}$ angle has area =Area $ACD=\frac{\pi r^2}{2\pi }\times \frac{\pi }{4}$

$=\frac{\pi r^2}{8}$

$=\frac{\pi \left(49\right)}{8}=\frac{49\pi }{8}c{m}^2$

Area of portion $AOD=(\frac{49\pi }{8}-\frac{49}{4})c{m}^2$

$=\frac{49}{4}(\frac{\pi }{2}-1)c{m}^2$

Total common area $=4\times$ Area of $AOD$

(Due to symmetricity)

$=4\times \frac{49}{4}(\frac{\pi }{2}-1)$

$=49(\frac{\pi }{2}-1)$

$=\frac{49}{2}(\pi -2)c{m}^2$