In rectangle $ABCD$, angle $C$ is trisected by $\overline{CF}$ and $\overline{CE}$, where $E$ is on $\overline{AB}$, $F$ is on $\overline{AD}$, $BE=6$ and $AF=2$. Which of the following is closest to the area of the rectangle $ABCD$?

$\text{(A)}\ 110\qquad\text{(B)}\ 120\qquad\text{(C)}\ 130\qquad\text{(D)}\ 140\qquad\text{(E)}\ 150$
Since $\angle C = 90^\circ$, each of the three smaller angles is $30^\circ$, and $\triangle BEC$ and $\triangle CDF$ are both $30-60-90$ triangles.

Defining the variables as illustrated above, we have $x = 6\sqrt{3}$ from $\triangle BEC$
Then $x-2 = 6\sqrt{3} - 2$, and $y = \sqrt{3} (6 \sqrt{3} - 2) = 18 - 2\sqrt{3}$.
The area of the rectangle is thus $xy = 6\sqrt{3}(18 - 2\sqrt{3}) = 108\sqrt{3} - 36$.
Using the approximation $\sqrt{3} \approx 1.7$, we get an area of just under $147.6$, which is closest to answer $\boxed{\text{E}}$.  (The actual area is actually greater, since $\sqrt{3} > 1.7$).