Figure $ABCD$ is a trapezoid with $AB||DC$, $AB=5$, $BC=3\sqrt{2}$, $\angle BCD=45^\circ$, and $\angle CDA=60^\circ$. The length of $DC$ is
$\mathrm{(A) \ }7+\frac{2}{3}\sqrt{3} \qquad \mathrm{(B) \ }8 \qquad \mathrm{(C) \ } 9 \frac{1}{2} \qquad \mathrm{(D) \ }8+\sqrt{3} \qquad \mathrm{(E) \ } 8+3\sqrt{3}$
Drop [perpendiculars](https://artofproblemsolving.com/wiki/index.php/Perpendicular) from $A$ and $B$ to $CD$, thus forming a [30-60-90 right triangle](https://artofproblemsolving.com/wiki/index.php/Right_triangle#Special_right_triangles), a [45-45-90  right triangle](https://artofproblemsolving.com/wiki/index.php/Right_triangle#Special_right_triangles), and a [rectangle](https://artofproblemsolving.com/wiki/index.php/Rectangle), as shown in the diagram. Since $ABEF$ is a rectangle, we have $EF=AB=5$. Also, from the properties of 45-45-90 right triangles, we have $EB=EC=\frac{BC}{\sqrt{2}}=\frac{3\sqrt{2}}{\sqrt{2}}=3$. Again since $ABEF$ is a rectangle, $AF=3$, too. From the properties of 30-60-90 right triangles, we have $DF=\frac{AF}{\sqrt{3}}=\frac{3}{\sqrt{3}}=\sqrt{3}$. Therefore, $CD=CE+EF+FD=3+5+\sqrt{3}=8+\sqrt{3}, \boxed{\text{D}}$.