The area of polygon $ABCDEF$, in square units, is
$\text{(A)}\ 24 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 46 \qquad \text{(D)}\ 66 \qquad \text{(E)}\ 74$
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 $6\times5 = 30$. 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 $16 + 30 = 46$.
$\boxed{\text{C}}$