Points $A$ and $B$ are on side $\overline{YZ}$ of rectangle $WXYZ$ such that $\overline{WA}$ and $\overline{WB}$ trisect $\angle ZWX$. If $WX = 2$ and $XY = 3$, then what is the area of rectangle $WXYZ$?
WX * XY = 2 * 3 = 6 = [ WXYZ]
+48 
