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 $WXYZ$?
This problem actually has a very simple answer.
Let's note the two sidelengths given, We have \(WX = 2\) and \(XY=3\)
In order to find the area of WXYZ, we must find the height and base.
The base of the rectange is \(WX\) and the height of rectangle is \(XY\)
The problem gave all the dimensions we need to already solve this problem. We have
\(WX \cdot XY = 2 \cdot 3 = 6\)
So our final answer is 6.
Thanks! :)