Find the area of the circle

Consider my diagram. Let $A$ be the center of the circle, and it is also the centroid of the triangle (since it is equilateral). The area of an equilateral triangle is given by $A=\dfrac{\sqrt{3}}{4} x^2$ where $x$ is the side length. We have

$9\sqrt{3}=\dfrac{\sqrt{3}}{4}x^2$ $\implies$ $x=6$

It follows that side length of the square is also $6$ . By pythagorean theorem, we have $h=\sqrt{6^2-3^2}=3\sqrt{3}$ . It follows that $a=\dfrac{1}{3}h=\dfrac{1}{3}(3\sqrt{3})=\sqrt{3}$ . So $AC=a+6=\sqrt{3}+6$ . By pythagorean theorem on triangle $ACB$ , we have

$r^2=3^2+(6+\sqrt{3})^2=9+(36+12\sqrt{3}+3)=48+12\sqrt{3}$

Thus, the area of the circle is

$A=\pi r^2=\dfrac{22}{7}(48+12\sqrt{3})=\boxed{\dfrac{1056}{7}+\dfrac{264}{7}\sqrt{3}}$