jmc

Three right triangles lie outside $△AMN$. Their areas are $\frac{1}{4}$, $\frac{1}{4}$, and $\frac{1}{8}$ for a total of $\frac{5}{8}$ of the rectangle. The area of $△AMN$ is $\frac{3}{8}(72)=\boxed{27}$.

OR

Let the rectangle have sides of $2a$ and $2b$ so that $4ab=72$ and $ab=18$.Three right triangles lie outside triangle $AMN$, and their areas are $\frac{1}{2}(2a)(b)$, $\frac{1}{2}(2b)(a)$, $\frac{1}{2}(a)(b)$, for a total of $\frac{5}{2}(ab)=\frac{5}{2}(18)=45$. The area of triangle $AMN$ is $72-45=\boxed{27}$.