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: $3\pi (2+ \sqrt3)^2$

Question

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: $3\pi (2+ \sqrt3)^2$

Team Careers360 · Accepted Answer

Correct Answer: $\frac{\pi}{3} (2+ \sqrt3)^2$

Solution :

Since radius of each circle = 1 cm,
AB = BC = AC = 2 cm
⇒ $ riangle$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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Question : Three equal circles of unit radius touch one another. Then the area of the circle circumscribing the three circles is:Option 1: 6π(2+3)2Option 2: π6(2+3)2Option 3: π3(2+3)2Option 4: 3π(2+3)2

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Question : Three equal circles of unit radius touch one another. Then the area of the circle circumscribing the three circles is:
Option 1: $6 π (2 + \sqrt{3})^{2}$
Option 2: $\frac{π}{6} (2 + \sqrt{3})^{2}$
Option 3: $\frac{π}{3} (2 + \sqrt{3})^{2}$
Option 4: $3 π (2 + \sqrt{3})^{2}$