In the diagram below, $ABCD$ is a rectangle with side lengths $AB=3$ and $BC=11$, and $AECF$ is a rectangle with side lengths $AF=7$ and $FC=9,$ as shown. The area of the shaded region common to the interiors of both rectangles is $\frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
Let $G$ be the intersection of $AD$ and $FC$.
From vertical angles, we know that $\angle FGA= \angle DGC$. Also, because we are given that $ABCD$ and $AFCE$ are rectangles, we know that $\angle AFG= \angle CDG=90 ^{\circ}$.
Therefore, by AA similarity, we know that $\triangle AFG\sim\triangle CDG$.
Let $AG=x$. Then, we have $DG=11-x$. By similar triangles, we know that $FG=\frac{7}{3}(11-x)$ and $CG=\frac{3}{7}x$. We have $\frac{7}{3}(11-x)+\frac{3}{7}x=FC=9$.
Solving for $x$, we have $x=\frac{35}{4}$.
The area of the shaded region is just $3\cdot \frac{35}{4}=\frac{105}{4}$.
Thus, the answer is $105+4=\framebox{109}$. 
~yuanyuanC