+295 Points $A$ and $B$ are on side $\overline{YZ}$ of rectangle $WXYZ$ such that $\overline{WA}$ and $\overline{WB}$ trisect $\angle ZWX$. If $BY = 2$ and $AZ = 4$, then what is the area of rectangle $WXYZ$?
+15140 What is the area of rectangle ZWXY?
Let ZW be the baseline and WX be the height of the rectangle ZWXY, and ZW be x.
Then:
\(\overline{WX}=\dfrac{x-2}{tan\ 60°}\\ \overline{AX}=\dfrac{tan\ 30°(x-2)}{tan\ 60°}=\dfrac{x-2}{3}\\ \)
\(\triangle ZAY\\ (x-\overline{AX})^2+\overline{WX}^2=4^2 \\ (x-\dfrac{x-2}{3})^2+(\dfrac{x-2}{tan\ 60°})^2=16\\ (\dfrac{2x}{3}+\dfrac{2}{3})^2+(\dfrac{x}{tan\ 60°}-\dfrac{2}{\tan\ 60°})^2=16\\ x\in \{-4,\dfrac{32}{7}\}\ \color{blue}WolframAlpha\)
\(A_{ZWXY}=x\cdot \dfrac{x-2}{tan\ 60°}=\dfrac{32}{7}\cdot \dfrac{\frac{32}{7}-2}{tan\ 60°}\\ \color{blue}A_{ZWXY}=6.7868\)
The area of rectangle ZWXY is 6.7868.
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