OpenSAT

A regular hexagon can be divided into 6 equilateral triangles. The area of an equilateral triangle with side length s is $\frac{\sqrt{3}}{4}s^2$, so the area of the hexagon is $6 \cdot \frac{\sqrt{3}}{4}s^2 = \frac{3\sqrt{3}}{2}s^2$. Substituting s = 6, we get $\frac{3\sqrt{3}}{2}(6)^2 = 54\sqrt{3}$.