jmc

Diagonals $\overline{AC}$, $\overline{CE}$, $\overline{EA}$, $\overline{AD}$, $\overline{CF}$, and $\overline{EB}$ divide the hexagon into twelve congruent 30-60-90 triangles, six of which make up equilateral $△ACE$.



Because $AC=\sqrt{7^2+1^2}=\sqrt{50}$, the area of $△ACE$ is $\frac{\sqrt{3}}{4}{\left(\sqrt{50}\right)}^2=\frac{25}{2}\sqrt{3}$. The area of hexagon $ABCDEF$ is $2\left(\frac{25}{2}\sqrt{3}\right)=\boxed{25\sqrt{3}}$.

An alternate way to start: let $O$ be the center of the hexagon. Then triangles $ABC,CDE,$ and $EFA$ are congruent to triangles $AOC,COE,$ and $EOA$, respectively. Thus the area of the hexagon is twice the area of equilateral $△ACE$. Then proceed as in the first solution.