smc

Drop perpendiculars from $A$ and $B$ to $CD$, thus forming a 30-60-90 right triangle, a 45-45-90 right triangle, and a 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, $CD=CE+EF+FD=3+5+\sqrt{3}=8+\sqrt{3},\boxed{8+\sqrt{3}}$.