11. A circle is drawn inside a regular hexagon so that it touches all six sides of the hexagon. The area of the circle is $\pi \times 64 \sqrt{3}$. What is the area of the hexagon?
Solution
384

Let $O$ be the centre of the hexagon.
Let $A B$ be an edge of the hexagon with midpoint $M$.
For the circle we have $\pi r^{2}=\pi \times 64 \sqrt{3}$.
Therefore $r^{2}=64 \sqrt{3}$.
In $\triangle O M B, O B=\frac{O M}{\cos 30^{\circ}}=\frac{2 r}{\sqrt{3}}$.
The area of the equilateral $\triangle O A B$ is $\frac{1}{2} \times A B \times O M=\frac{1}{2} \times O B \times O M$, since $A B=O B$.
Therefore, area $\triangle O A B=\frac{1}{2} \times \frac{2 r}{\sqrt{3}} \times r=\frac{r^{2}}{\sqrt{3}}$
The area of the hexagon is $6 \times \frac{r^{2}}{\sqrt{3}}=6 \times \frac{64 \sqrt{3}}{\sqrt{3}}=6 \times 64=384$.