In [convex](https://artofproblemsolving.com/wiki/index.php/Convex_polygon) [hexagon](https://artofproblemsolving.com/wiki/index.php/Hexagon) $ABCDEF$, all six sides are congruent, $\angle A$ and $\angle D$ are [right angles](https://artofproblemsolving.com/wiki/index.php/Right_angle), and $\angle B, \angle C, \angle E,$ and $\angle F$ are [congruent](https://artofproblemsolving.com/wiki/index.php/Congruent). The area of the hexagonal region is $2116(\sqrt{2}+1).$ Find $AB$.
Let the side length be called $x$, so $x=AB=BC=CD=DE=EF=AF$. 
[2006 II AIME-1.png](https://artofproblemsolving.com/wiki/index.php/File:2006_II_AIME-1.png)
The diagonal $BF=\sqrt{AB^2+AF^2}=\sqrt{x^2+x^2}=x\sqrt{2}$. Then the areas of the triangles AFB and CDE in total are $\frac{x^2}{2}\cdot 2$,
and the area of the rectangle BCEF equals $x\cdot x\sqrt{2}=x^2\sqrt{2}$
Then we have to solve the equation 


$2116(\sqrt{2}+1)=x^2\sqrt{2}+x^2$.
$2116(\sqrt{2}+1)=x^2(\sqrt{2}+1)$
$2116=x^2$

$x=46$
Therefore, $AB$ is $\boxed{046}$.