jmc

Obviously, there are no formulas to find the area of such a messed up shape, but we do recognize some shapes we do know how to find the area of. If we continue segment $\overline{FE}$ until it reaches the right side at $G$, we create two rectangles - one on the top and one on the bottom. We know how to find the area of a rectangle, and we're given the sides! We can easily find that the area of $ABGF$ is $6\times 5=30$. For the rectangle on the bottom, we do know the length of one of its sides, but we don't know the other. Note that $GC+GB=9$, and $GB=AF=5$, so we must have $$GC+5=9\Rightarrow GC=4$$ The area of the bottom rectangle is then $$(DC)(GC)=4\times 4=16$$ Finally, we just add the areas of the rectangles together to get $16+30=46$. $\boxed{46}$