jmc

Since $AE=2$, $EB=8$, but since $EFGH$ is a square, $EH=EF$, and $AH=EB$ by ASA congruence of right triangles $AHE$ and $BEF$. By the Pythagorean theorem, $(EH{)}^2=(AE{)}^2+(AH{)}^2=2^2+8^2=\boxed{68}$, which is also the area of square $EFGH$ as the square of one of its sides.