In quadrilateral $ABCD$, $m\angle B=m\angle C=120^\circ$, $AB=4$, $BC=6$, $CD=7$. Find the area of $ABCD$.

1. Extend $AB$ beyond $B$ and $DC$ beyond $C$ until they meet at $P$. Since $\angle PBC=\angle PCB=60^\circ$, triangle $PBC$ is equilateral with side $BC$, and $\angle P=60^\circ$. Then $[ABCD]=[PAD]-[PBC]=\frac{\sqrt3}4\big((AB+BC)(CD+BC)-BC^2\big)=\frac{\sqrt3}4\big(AB\cdot BC+BC\cdot CD+CD\cdot AB\big)$.
2. Here $AB\cdot BC+BC\cdot CD+CD\cdot AB=24+42+28=94$.
3. The area is $\boxed{\frac{47\sqrt3}2}$.