Rhombus $ABCD$ has side length $4$ and $\angle B = 150^\circ$. Region $R$ consists of all points inside the rhombus that are closer to vertex $B$ than any of the other three vertices. What is the area of $R$?

1. Let $A=(-4,0), B=(0,0), C=(2\sqrt3,2), D=(2\sqrt3-4,2)$. The perpendicular bisectors defining the region meet at $H=(-2,0)$, $E=(-2,4-2\sqrt3)$, $F=(2\sqrt3-2,2\sqrt3-2)$, and $G=(\sqrt3,1)$, so $R=EHBGF$.
2. Split the pentagon along $BE$ and $BF$. The triangle areas are $[BHE]=4-2\sqrt3$, $[BEF]=8\sqrt3-12$, and $[BFG]=4-2\sqrt3$.
3. Thus $[R]=(4-2\sqrt3)+(8\sqrt3-12)+(4-2\sqrt3)=4(\sqrt3-1)$.