jmc

Since triangle $DGF$ is similar to triangle $AHF$, we have $$\frac{DG}{GF}=\frac{AH}{HF}.$$ Substituting we find $$\frac{DG}{3}=\frac{10}{10+3},$$ which we solve to get $DG=30/13$ inches. The area of triangle $DGF$ is $\frac{1}{2}\left(\frac{30}{13}\right)(3)=\frac{45}{13}$ square inches. The area of the $3\text{ in.}\times 3\text{ in.}$ square is $9$ square inches, so the area of the shaded region is $9-\frac{45}{13}=\boxed{\frac{72}{13}}$ square inches.