Suppose we have two circles of radius 2 each | vedclass

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(C) Let the centers of the two circles be $A$ and $B$. The radius of each circle is $r = 2$. The distance between the centers is $AB = 2 \sqrt{3}$.Let

Vedclass · Accepted Answer

$0.7$ and $0.75$

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(C) Let the centers of the two circles be $A$ and $B$. The radius of each circle is $r = 2$. The distance between the centers is $AB = 2 \sqrt{3}$.Let the circles intersect at points $P$ and $Q$. Let $C$ be the intersection of $AB$ and $PQ$. Since the circles are identical,$C$ is the midpoint of $AB$,so $AC = \frac{1}{2} AB = \sqrt{3}$.In $	riangle APC$,$\cos 	heta = \frac{AC}{AP} = \frac{\sqrt{3}}{2}$,which implies $	heta = 30^{\circ}$.The angle subtended by the chord $PQ$ at the center $A$ is $2	heta = 60^{\circ} = \frac{\pi}{3}$ radians.The area of the common region is twice the area of the circular segment of one circle cut by the chord $PQ$.Area of one segment = (Area of sector $APQ$ - Area of $	riangle APQ$)$= \frac{1}{2} r^2 (2	heta - \sin(2	heta)) = \frac{1}{2} (2)^2 (\frac{\pi}{3} - \sin 60^{\circ}) = 2 (\frac{\pi}{3} - \frac{\sqrt{3}}{2}) = \frac{2\pi}{3} - \sqrt{3}$.Total common area $= 2 	imes (\frac{2\pi}{3} - \sqrt{3}) = \frac{4\pi}{3} - 2\sqrt{3}$.Using $\pi \approx 3.14159$ and $\sqrt{3} \approx 1.732$:Area $\approx \frac{4 	imes 3.14159}{3} - 2 	imes 1.732 = 4.18879 - 3.464 = 0.72479$.Thus,the area lies between $0.7$ and $0.75$.

Suppose we have two circles of radius $2$ each in the plane such that the distance between their centers is $2 \sqrt{3}$. The area of the region common to both circles lies between

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