jmc

Draw $DH$.



The overlap of the two squares is quadrilateral $CDEH$. The area of each square is 16, so the side length of each square is $\sqrt{16}=4$.

Then $DE=4$ and $HE=EF/2=4/2=2$, so the area of triangle $DEH$ is $DE\cdot EH/2=4\cdot 2/2=4$. By symmetry, the area of triangle $CDH$ is also 4, so the area of quadrilateral $CDEH$ is $4+4=8$.

Then the area of pentagon $ADEHB$ is $16-8=8$, and the area of pentagon $CDGFH$ is also $16-8=8$. Hence, the area of polygon $ABHFGD$ is $8+8+8=\boxed{24}$.