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$
Solution
Assume $AB=x$. Then $BF^2=x^2+x^2-2x^2\cdot \frac{\sqrt3}2=(2-\sqrt3)x^2$.
The total area = $3\cdot \frac12 x^2\cdot \frac12+\frac{\sqrt3}4\cdot BF^2=\frac34x^2+\frac{\sqrt3}4\cdot(2-\sqrt3)x^2=\frac{\sqrt3}2x^2$.
$\frac{\sqrt3}2x^2=6\sqrt3$ gives $x^2=12$, $x=2\sqrt3$, and the perimeter is $6x=12\sqrt3$.
In the figure, equilateral hexagon $ABCDEF$ has three nonadjacent acute interior angles that each measure $45^\circ$. The enclosed area of the hexagon is $3\sqrt2+2\sqrt3-\sqrt6$. 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$ 
