Let $[ABC]$ denote the area of figure $ABC$.
Clearly, $[BEDC]=[ABCD]-[ABE]$.  Using basic area formulas,

$[ABCD]=(BC)(BE)=80$
$[ABE]=(BE)(AE)/2 = 4(AE)$
Since $AE+ED=BC=10$ and $ED=6$, $AE=4$ and the area of $\triangle ABE$ is $4(4)=16$.
Finally, we have $[BEDC]=80-16=64\rightarrow \boxed{\text{D}}$
The [area](https://artofproblemsolving.com/wiki/index.php/Area) of the shaded region $\text{BEDC}$ in [parallelogram](https://artofproblemsolving.com/wiki/index.php/Parallelogram) $\text{ABCD}$ is

$\text{(A)}\ 24 \qquad \text{(B)}\ 48 \qquad \text{(C)}\ 60 \qquad \text{(D)}\ 64 \qquad \text{(E)}\ 80$