jmc

To quickly solve this multiple choice problem, make the (not necessarily valid, but very convenient) assumption that $ABCD$ can have any dimension. Give the rectangle dimensions of $AB=CD=12$ and $BC=AD=6$, which is the easiest way to avoid fractions. Labelling the right midpoint as $M$, and the bottom midpoint as $N$, we know that $DN=NC=6$, and $BM=MC=3$. $[△ADN]=\frac{1}{2}\cdot 6\cdot 6=18$ $[△MNC]=\frac{1}{2}\cdot 3\cdot 6=9$ $[△ABM]=\frac{1}{2}\cdot 12\cdot 3=18$ $[△AMN]=[\square ABCD]-[△ADN]-[△MNC]-[△ABM]$ $[△AMN]=72-18-9-18$ $[△AMN]=27$, and the answer is $\boxed{27}$