$18 \cdot 63$ rectangle $A B C D$ has $\angle C$ trisected by $C E, C F$, where $E$ is on $A B$ and $F$ is on $A D$, and $A F=2, B E$ $=6$. Then $S_{A B C D}$ is closest to
(A) 110 .
(B) 120
(C) 130 .
(D) 140 .
(E) 150 .
[Solution] From the given, $\angle E C B=\frac{1}{3} \angle D C B=30^{\circ}$,
then
$$
\begin{array}{l}
C B=\sqrt{3} B E=6 \sqrt{3}, \therefore D F=6 \sqrt{3}-2 . \\
\text { Similarly, we can find } \quad C D=\sqrt{3} D F=18-2 \sqrt{3} . \\
\therefore \quad S_{A B C D}=C B \cdot C D=108 \sqrt{3}-36 \approx 150 .
\end{array}
$$

Therefore, the answer is $(E)$.