Find the side length of a hexagon In a convex hexagon $ABCDEF$, all six sides are congruent, $\angle A$ and $\angle D$ are right angles. And $\angle B$, $\angle C$, $\angle E$ and $\angle F$ are congruent. The area of the hexagonal region is $2116(\sqrt{2} +1)$. Find $AB$. It is pretty easy to set up a equation with variable $x$ for the side length. What is tricky is this. You have to find the answer without using a calculator.
 A: If we extend $AB$ and $DC$ to meet at $G$, and similarly, $AF$ and $DE$ to meet at $H$, then polygon $AGDH$ is a square of side length $x(1 + 
\frac{1}{\sqrt{2}})$, whose area is equal to the area of the hexagon plus the area of a square of side length $x/\sqrt{2}$.  Since we are given that the hexagon's area is $2116(\sqrt{2}+1)$, it follows that $$2116(\sqrt{2}+1) = |AGDH| - (x/\sqrt{2})^2 = x^2(1+\tfrac{1}{\sqrt{2}})^2 - \frac{1}{2}x^2.$$  The rest is simple algebra, easily done by hand.
A: Draw 2 lines, $FB$ and $EC$. 
We now have 2 triangles ($ABF$ & $CDE$) and 1 rectangle ($FBCE$). 
Let $AB$ be $x$ cm long. 
Area of triangles $ABF$ and $CDE$: $2*1/2*x*x = x^2$
Length of $FB$ = Length of $CE$ = $\sqrt{1+1}*x$ = $\sqrt2x$
Area of rectangle = $\sqrt2x*x$ = $\sqrt2x^2$
Area of 2 triangles plus rectangle: $x^2+\sqrt2x^2$ = $(\sqrt2+1)x^2$ = $2116(\sqrt2+1)$
I think I will leave you to answer the rest.
If anyone needs more help, a spoiler is provided.

! Rearranging the equation, we get:
  $x^2$ = 2116
  $x$ = $\sqrt{2116}$ = $\sqrt{2*2*529}$ = $2\sqrt{529}$ = $2*17$ = $34$

