In the figure, equilateral hexagon $ABCDEF$ has three nonadjacent acute interior angles that each measure $30^\circ$. The enclosed area of the hexagon is $6\sqrt{3}$. What is the perimeter of the hexagon?

$\textbf{(A)} \: 4 \qquad \textbf{(B)} \: 4\sqrt3 \qquad \textbf{(C)} \: 12 \qquad \textbf{(D)} \: 18 \qquad \textbf{(E)} \: 12\sqrt3$
Divide the equilateral hexagon $ABCDEF$ into isosceles triangles $ABF$, $CBD$, and $EDF$ and triangle $BDF$. The three isosceles triangles are congruent by SAS congruence. By CPCTC, $BF=BD=DF$, so triangle $BDF$ is equilateral. 
Let the side length of the hexagon be $s$. The area of each isosceles triangle is \[\frac{1}{2} a b \sin\angle C = \frac{1}{2} \cdot s \cdot s \cdot \sin{30^{\circ}} = \frac{1}{4}s^2.\]
By the [Law of Cosines](https://artofproblemsolving.com/wiki/index.php/Law_of_Cosines) on triangle $ABF$, \[BF^2=s^2+s^2-2s^2\cos{30^{\circ}}=2s^2-\sqrt{3}s^2.\]
Hence, the [area of the equilateral triangle](https://artofproblemsolving.com/wiki/index.php/Area_of_an_equilateral_triangle) $BDF$ is \[\frac{\sqrt{3}}{4} BF^2 = \frac{\sqrt{3}}{4}\left(2s^2-\sqrt{3}s^2\right)=\frac{\sqrt{3}}{2}s^2-\frac{3}{4}s^2.\]
The total area of the hexagon is thrice the area of each isosceles triangle plus the area of the equilateral triangle, or \[3\left(\frac{1}{4}s^2\right)+ \left( \frac{\sqrt{3}}{2}s^2-\frac{3}{4}s^2 \right)=\frac{\sqrt{3}}{2}s^2=6\sqrt{3}.\] Hence, $s=2\sqrt{3}$, and the perimeter of the hexagon is $6s=\boxed{\textbf{(E)} \: 12\sqrt3}$.