In rectangle $ADEH$, points $B$ and $C$ trisect $\overline{AD}$, and points $G$ and $F$ trisect $\overline{HE}$. In addition, $AH=AC=2$, and $AD=3$. What is the area of quadrilateral $WXYZ$ shown in the figure? $\textbf{(A) } \frac{1}{2}\qquad$ $\textbf{(B) } \frac{\sqrt{2}}{2}\qquad$ $\textbf{(C) } \frac{\sqrt{3}}{2}\qquad$ $\textbf{(D) } \sqrt{2} \qquad$ $\textbf{(E) } \frac{2\sqrt{3}}{3}\qquad$
Solution
Use coordinates $H=(0,0)$, $E=(3,0)$, $A=(0,2)$, and $D=(3,2)$. Since $B,C$ and $G,F$ trisect the corresponding sides, $B=(1,2)$, $C=(2,2)$, $G=(1,0)$, and $F=(2,0)$.
The four drawn lines have equations $HC:y=x$, $EB:y=3-x$, $AF:y=2-x$, and $DG:y=x-1$. Their intersections are $$W=(HC\cap EB)=(3/2,3/2),\quad X=(DG\cap EB)=(2,1),$$ $$Y=(AF\cap DG)=(3/2,1/2),\quad Z=(AF\cap HC)=(1,1).$$
The diagonals $WY$ and $XZ$ are perpendicular, and each has length $1$. Hence $$[WXYZ]=\frac12(WY)(XZ)=\frac12.$$ The answer is $\textbf{(A)}$.
In rectangle $ADEH$, $AB=CD=HG=FE=2$, $BC=GF=1$, and $DE=AH=3$. Segments $AF$, $DG$, $HC$, and $EB$ intersect to form quadrilateral $WXYZ$, as shown. What is the area of $WXYZ$?
$\textbf{(A) } \frac{1}{2}\qquad$ $\textbf{(B) } \frac{\sqrt{2}}{2}\qquad$ $\textbf{(C) } \frac{\sqrt{3}}{2}\qquad$ $\textbf{(D) } \sqrt{2} \qquad$ $\textbf{(E) } \frac{2\sqrt{3}}{3}\qquad$ 
