Question : Three equal circles of unit radius touch one another. Then the area of the circle circumscribing the three circles is:
Option 1: $6 \pi (2+ \sqrt3)^2$
Option 2: $\frac{\pi}{6} (2+ \sqrt3)^2$
Option 3: $\frac{\pi}{3} (2+ \sqrt3)^2$
Option 4: $3\pi (2+ \sqrt3)^2$
Correct Answer: $\frac{\pi}{3} (2+ \sqrt3)^2$
Solution :
Since radius of each circle = 1 cm,
AB = BC = AC = 2 cm
⇒ $\triangle$ABC is an equilateral triangle
⇒ AP = $\frac{\sqrt3 a}{2}$ = $\frac{\sqrt3}{2}×2$ = $\sqrt3$ cm
Point O is the centroid.
OA = $\frac{2}{3}×\sqrt3$ = $\frac{2}{\sqrt3}$
Radius of larger circle, $R$ = OM = $\frac{2}{\sqrt3}+1$ = $\frac{2+\sqrt3}{\sqrt3}$ cm
Required area = $ \pi R^2$ = $ \pi (\frac{2+\sqrt3}{\sqrt3})^2$ = $\frac{\pi}{3} (2+ \sqrt3)^2$
Hence, the correct answer is $\frac{\pi}{3} (2+ \sqrt3)^2$.
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