jmc

We can use analytic geometry for this problem. Let us start by giving $D$ the coordinate $(0,0)$, $A$ the coordinate $(0,1)$, and so forth. $\overline{AC}$ and $\overline{EB}$ can be represented by the equations $y=-x+1$ and $y=2x-1$, respectively. Solving for their intersection gives point $F$ coordinates $\left(\frac{2}{3},\frac{1}{3}\right)$. Now, $△$$EFC$'s area is simply $\frac{\frac{1}{2}\cdot \frac{1}{3}}{2}$ or $\frac{1}{12}$. This means that pentagon $ABCEF$'s area is $\frac{1}{2}+\frac{1}{12}=\frac{7}{12}$ of the entire square, and it follows that quadrilateral $AFED$'s area is $\frac{5}{12}$ of the square. The area of the square is then $\frac{45}{\frac{5}{12}}=9\cdot 12=\boxed{108}$.