jmc

In square $ABCD$, diagonal $\overline{AC}$ divides the square into 2 equal areas. Thus, the area of $△ACD$ is one-half of the area of square $ABCD$, and therefore is $\frac{1}{4}$ of the total area of the two squares.

Since $\overline{AC}$ is the diagonal of square $ABCD$, we have $\angle ACB=45^{\circ }$, so $\angle HCJ=45^{\circ }$, which means $△CHJ$ is an isosceles right triangle. Since $HJ=\frac{HG}{2}$, the area of $△CHJ$ is $\frac{1}{2}(CH)(HJ)=\frac{1}{2}\cdot \frac{HG}{2}\cdot \frac{HG}{2}=\frac{1}{8}H{G}^2$, which means that the area of $△CHJ$ is $\frac{1}{8}$ of the area of one of the squares, or $\frac{1}{16}$ of the total area of the two squares. Combining the two shaded regions, $\frac{1}{4}+\frac{1}{16}=\boxed{\frac{5}{16}}$ of the two squares is shaded.