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?

1. Let the side length be $x$. The hexagon has a $120^\circ$ rotational symmetry, so $\triangle ABF$, $\triangle CBD$ and $\triangle EDF$ are congruent isosceles triangles with legs $x$ and apex angle $45^\circ$, and $\triangle BDF$ is equilateral. By the law of cosines, $BF^2=x^2+x^2-2x^2\cos45^\circ=(2-\sqrt2)x^2$.
2. The enclosed area is the three isosceles triangles plus the equilateral triangle: \begin{align*}3\cdot\frac12x^2\sin45^\circ+\frac{\sqrt3}4BF^2&=\frac{3\sqrt2}4x^2+\frac{\sqrt3}4(2-\sqrt2)x^2\\&=\frac{3\sqrt2+2\sqrt3-\sqrt6}4\,x^2.\end{align*}
3. Setting this equal to $3\sqrt2+2\sqrt3-\sqrt6$ gives $x^2=4$, so $x=2$.
4. The perimeter is $6x=\boxed{12}$.