Question

Three equal circle of unit radius touch each other. then the area of the circle circumscribing the three circle is:

• A. $6\pi (2+\sqrt{3}{)}^2$
• B. $\frac{\pi }{6}(2+\sqrt{3}{)}^2$
• C. $\frac{\pi }{3}(2+\sqrt{3}{)}^2$
• D. $3\pi 2+\sqrt{3}{)}^2$

Answer 1

The correct option is C $\frac{\pi }{3}(2+\sqrt{3}{)}^2$

Let the $O,A,B,C$ be the center of the circumscribing circle, and three touching circles respectively.

Area of $\Delta ABC=\frac{\sqrt{3}}{4}\times a^2$

$\frac{1}{2}\times 2\times BD=\sqrt{3}$

$BD=\sqrt{3}$

$BO=\frac{2}{3}\times BD$

$BO=\frac{2}{3}\times \sqrt{3}$

$BO=\frac{2\sqrt{3}}{3}$

So, radius of circumscribing circle is $=1+OB$

$=1+\frac{2\sqrt{3}}{3}$

$=\frac{3+2\sqrt{3}}{3}$

Therefore, Area of circumscribing circle $=\pi r^2$

$=\pi {\left(\frac{3+2\sqrt{3}}{3}\right)}^2$

$=\frac{\pi }{3}{\left(2+\sqrt{3}\right)}^2$