jmc

Because $\angle B$, $\angle C$, $\angle E$, and $\angle F$ are congruent, the degree-measure of each of them is $\frac{720-2\cdot 90}{4}=135$. Lines $BF$ and $CE$ divide the hexagonal region into two right triangles and a rectangle. Let $AB=x$. Then $BF=x\sqrt{2}$. Thus $$\begin{array}{rl}2116(\sqrt{2}+1)& =[ABCDEF]\\ & =2\cdot \frac{1}{2}{x}^2+x\cdot x\sqrt{2}=x^2(1+\sqrt{2}),\end{array}$$so $x^2=2116$, and $x=\boxed{46}$.