18.102 As shown in the figure, quadrilateral $A B C D$ is a trapezoid, $A B / / D C, A B=5$, $B C=3 \sqrt{2}, \angle B C D=45^{\circ}$ and $\angle C D A=60^{\circ}$, then the length of $D C$ is
(A) $7+\frac{2}{3} \sqrt{3}$.
(B) 8 .
(C) $9 \frac{1}{2}$.
(D) $8+\sqrt{3}$.
(E) $8+3 \sqrt{3}$.
[Solution] As shown in the figure, perpendicular lines are drawn from $A, B$ to $DC$, intersecting $DC$ at $F, E$ respectively. $\triangle BEC$ is an isosceles right triangle, so
$$
B E=E C=3 \text{. }
$$

Since $ABEF$ is a rectangle, we have $FE=5$,
$$
A F=3 \text{. }
$$
$\triangle AFD$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ special right triangle, so
$$
D F=\frac{1}{\sqrt{3}} A F=\sqrt{3}, \quad D C=8+\sqrt{3} .
$$

Therefore, the answer is (D).