jmc

Each of the sides of the square is divided into two segments by a vertex of the rectangle. Call the lengths of these two segments $r$ and $s$. Also, let $C$ be the foot of the perpendicular dropped from $A$ to the side containing the point $B$. Since $AC=r+s$ and $BC=|r-s|$, $$(r+s{)}^2+(r-s{)}^2=12^2,$$ from the Pythagorean theorem. This simplifies to $2r^2+2s^2=144$, since the terms $2rs$ and $-2rs$ sum to 0. The combined area of the four removed triangles is $\frac{1}{2}{r}^2+\frac{1}{2}{s}^2+\frac{1}{2}{r}^2+\frac{1}{2}{s}^2=r^2+s^2$. From the equation $2r^2+2s^2=144$, this area is $144/2=\boxed{72}$ square units.