jmc

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 $△BCX$ and $△EDY$ 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 $\frac{1}{2}(12+12+12\sqrt{2})(6\sqrt{2})=\boxed{72+72\sqrt{2}}$.