Let $A B C D$ be a square, and let $\ell$ be a line passing through the midpoint of segment $\overline{A B}$ that intersects segment $\overline{B C}$. Given that the distances from $A$ and $C$ to $\ell$ are 4 and 7 , respectively, compute the area of $A B C D$.
Solution: Consider the line $\ell^{\prime}$ through $B$ parallel to $\ell$, and drop perpendiculars from $A$ to $\ell^{\prime}$ and $C$ to $\ell^{\prime}$. Note that because $\ell$ passes through the midpoint of segment $A B$, the distance from $B$ to $\ell$ is 4 . Thus, the distances from $A$ to $\ell^{\prime}$ and from $C$ to $\ell^{\prime}$ are $4+4=8$ and $4+7=11$, respectively. Let $P$ be the foot from $A$ to $\ell^{\prime}$. Rotating the square $90^{\circ}$ from $B$ to $A$ sends the altitude from $C$ to $\ell^{\prime}$ to the segment along $\ell^{\prime}$ between $B$ and the foot from $A$ to $\ell^{\prime}$; hence $B P=11$. So the side length of the square is $\sqrt{A P^{2}+B P^{2}}=\sqrt{8^{2}+11^{2}}$, which means the area of the square is $8^{2}+11^{2}=185$. $\fbox{185}$.