Question

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

Solution

## $$\triangle ABC$$ is isasceles right angle triangle.Draw a perpendicular $$AO$$ from $$A$$ on side $$BC$$In $$\triangle ABO$$:$$\sin 45^{0}=\dfrac{AO}{AB}$$$$\Rightarrow \dfrac{1}{\sqrt{2}}=\dfrac{AD}{7}$$$$\Rightarrow AO=\dfrac{7}{\sqrt{2}} cm$$$$\therefore AO$$ is the perpendicular bisector of $$BC$$ and also , $$BC=7\sqrt{2}$$ (given)$$\therefore \triangle ABO\cong \triangle AOC \Rightarrow BO=\dfrac{1}{2}BC \Rightarrow BO=\dfrac{7\sqrt{2}}{2}=\dfrac{7}{\sqrt{2}}$$Also, $$BO=CO$$ (cpct)Area of $$\triangle AOB=$$ Area of $$\triangle ACO=\dfrac{1}{2}\times CO \times AO$$$$=\dfrac{1}{2}\times \dfrac{7}{\sqrt{2}}\times {7}{\sqrt{2}}$$$$=\dfrac{49}{4}$$For area of sector $$ACD$$:$$2\pi$$  angle has area $$\rightarrow \pi r^{2}$$$$\Rightarrow \dfrac{\pi}{4}$$ angle has area =Area $$ACD=\dfrac{\pi r^{2}}{2\pi}\times \dfrac{\pi}{4}$$$$=\dfrac{\pi r^{2}}{8}$$$$=\dfrac{\pi(49)}{8}=\dfrac{49\pi}{8} cm^{2}$$Area of portion $$AOD=(\dfrac{49\pi}{8}-\dfrac{49}{4})cm^{2}$$$$=\dfrac{49}{4}(\dfrac{\pi}{2}-1) cm^{2}$$Total common area $$=4\times$$ Area of $$AOD$$(Due to symmetricity)$$=4\times \dfrac{49}{4}(\dfrac{\pi}{2}-1)$$$$=49(\dfrac{\pi}{2}-1)$$$$=\dfrac{49}{2}(\pi-2) cm^{2}$$Maths

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