1984 AMC 12 Problem 8

Drop perpendiculars from $A$ and $B$ to $DC$

The right-hand triangle is $45^\circ$-$45^\circ$-$90^\circ,$ and the left-hand triangle is $30^\circ$-$60^\circ$-$90^\circ$

Drop perpendiculars from $A$ and $B$ to $DC.$ Since $BC=3\sqrt2$ and the angle at $C$ is $45^\circ,$ both the height and the right horizontal offset are $3.$ The left $30^\circ$-$60^\circ$-$90^\circ$ triangle has height $3,$ so its horizontal offset is $\sqrt3.$ Therefore $$DC=\sqrt3+AB+3=8+\sqrt3.$$ The correct answer is D.