The diagram below shows a rectangle with side lengths $12$ and $18$ and a square with side length $13$. Three vertices of the square lie on three different sides of the rectangle, as shown. What is the area of the region inside both the square and the rectangle?

Place the rectangle with corners $(0,0),(18,0),(18,12),(0,12)$, and let the square have vertex $A=(a,0)$ on the bottom side, with neighbors $B=(b,12)$ on the top side and $D=(18,d)$ on the right side.

1. $AB=13$ gives $(b-a)^2+144=169$, so $b-a=-5$ ($B$ is to the left of $A$, as in the figure).
2. $AB\perp AD$ gives $(b-a)(18-a)+12d=0$, so $12d=5(18-a)$; and $AD=13$ gives $(18-a)^2+d^2=169$. Hence $(18-a)^2\left(1+\frac{25}{144}\right)=169$, $(18-a)^2=144$, $a=6$, $d=5$, $b=1$.
3. So $A=(6,0)$, $B=(1,12)$, $D=(18,5)$, and the fourth vertex is $C=B+D-A=(13,17)$, outside the rectangle.

The edge $DC$ crosses the rectangle's top side at $X=(181/12,12)$. The overlap is quadrilateral $A DXB$. By the shoelace formula, its area is
\[
\frac12\left|30+\frac{1687}{12}+169-72\right|=\boxed{\frac{3211}{24}}.
\]