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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)
$$=4\times \dfrac{49}{4}(\dfrac{\pi}{2}-1)$$
$$=49(\dfrac{\pi}{2}-1)$$
$$=\dfrac{49}{2}(\pi-2) cm^{2}$$





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