Work Outwards!

There are multiple methods to solve this problem, feel free to upload your solutions!

The side of the triangle (i.e. $AC$ and the other 2 sides) is $8 \sqrt {3}$ cm (This can be found very easily, for example using $\frac {1}{2} ab \sin c$ or other methods).

Now let's focus on Triangle $ABC$ , where point $B$ is the center of the circle. It is isosceles since $AB$ and $BC$ are radii of the circle. If we find the length of $AB$ or $BC$ , then we can just find the area of the circle.

Since $\angle ABC = 120 ^ \circ$ , we have a side and an opposite angle, therefore by using the sine rule in a triangle we get:

$\dfrac {8 \sqrt {3}} {\sin 120} = \dfrac {r} {\sin 30}$ where $r$ is the radius of the circle.

We get $r = 8$ . Therefore the area of the circle is $64π$ .

The answer is $64$ .

Just divide the figure into $3$ equal parts thru' the centre.

The angle at centre of each sector is $120\degree$ and the pink area $= 16\sqrt{3}$

$\cfrac{1}{2} r^2 \sin(120\degree) = 16\sqrt{3} \implies r^2 = 64$

Divide the triangle into 6 identical right triangles. The radius $r$ bisects the $60^{\circ}$ angles of the equilateral triangle.

The area of one of these right triangles is $\frac{48\sqrt{3}}{6} = 8\sqrt{3}$ . We can solve for the radius of the circle using this quantity. $\frac{1}{2} \cdot r\cos(30) \cdot r\sin(30) = \frac{1}{2} \cdot \frac{\sqrt{3}r}{2} \cdot \frac{r}{2} = \frac{\sqrt{3}r^2}{8} = 8\sqrt{3}$ $r^2 = 64$ Therefore, the area of the circle is $A = \pi r^2 = \boxed{64\pi}$