jmc

Noticing that $ADE$ is a 3-4-5 right triangle scaled by a factor of 2, we have $AD=2\cdot 5=10$. Thus, the area of square $ABCD$ is $10\cdot 10=100$. The area of triangle $ADE$ is $\frac{1}{2}\cdot 6\cdot 8=3\cdot 8=24$. Finally, we calculate the area of pentagon $ABCDE$ to be the difference of the two: $100-24=\boxed{76}\text{ sq units}$.