$ABCDEFGH$ is a regular octagon of side 12cm. Find the area in square centimeters of trapezoid $BCDE$. Express your answer in simplest radical form.


[asy] real x = 22.5; draw(dir(0+x)--dir(45+x)--dir(90+x)--dir(90+45+x)-- dir(90+2*45+x)--dir(90+3*45+x)-- dir(90+4*45+x)-- dir(90+5*45+x)--dir(90+6*45+x));

label("$A$", dir(90+45+x), W); label("$B$", dir(90+x), NW);label("$C$", dir(45+x), NE); label("$D$", dir(x), E);label("$E$", dir(90+5*45+x), E);label("$F$", dir(90+4*45+x), SE); label("$G$", dir(90+3*45+x), SW);label("$H$", dir(90+2*45+x), W);
draw( dir(90+x)--dir(90+5*45+x) );
[/asy]
Let the perpendiculars from $C$ and $D$ to $BE$ intersect $BE$ at $X$ and $Y$, respectively.  These perpendiculars split trapezoid $BCDE$ into two isosceles right triangles $\triangle BCX$ and $\triangle EDY$ and one rectangle $CDYX$.

In isosceles right triangles (which have angles 45-45-90), the ratio of the leg length to the hypotenuse length is $1:\sqrt{2}$; hence, we have $BX=CX=DY=YE=12/\sqrt{2}=6\sqrt{2}$.  We also have $XY=CD=12$, as opposite sides of a rectangle are equal.

Thus, trapezoid $BCDE$ has bases of length $CD=12$ and $BE=6\sqrt{2}+12+6\sqrt{2}=12+12\sqrt{2}$, and height of length $6\sqrt{2}$.  Hence its area is $\frac{1}{2}(12 + 12 + 12\sqrt{2}) (6\sqrt{2}) = \boxed{72 + 72\sqrt{2}}$.