jmc

The y-coordinate of $F$ must be $4$. All other cases yield non-convex and/or degenerate hexagons, which violate the problem statement. Letting $F=(f,4)$, and knowing that $\angle FAB=120^{\circ }$, we can use rewrite $F$ using complex numbers: $f+4i=(b+2i)\left(e^{i(2\pi /3)}\right)=(b+2i)\left(-1/2+\frac{\sqrt{3}}{2}i\right)=-\frac{b}{2}-\sqrt{3}+\left(\frac{b\sqrt{3}}{2}-1\right)i$. We solve for $b$ and $f$ and find that $F=\left(-\frac{8}{\sqrt{3}},4\right)$ and that $B=\left(\frac{10}{\sqrt{3}},2\right)$. The area of the hexagon can then be found as the sum of the areas of two congruent triangles ($EFA$ and $BCD$, with height $8$ and base $\frac{8}{\sqrt{3}}$) and a parallelogram ($ABDE$, with height $8$ and base $\frac{10}{\sqrt{3}}$). $A=2\times \frac{1}{2}\times 8\times \frac{8}{\sqrt{3}}+8\times \frac{10}{\sqrt{3}}=\frac{144}{\sqrt{3}}=48\sqrt{3}$. Thus, $m+n=\boxed{51}$.