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 $WXYZ$?
You are given the dimensions of the rectangle directly, the area is of course WX⋅XY=2⋅3=6.
+1266 
