In the figure below, the sides of rectangle $A B C D$ have been extended so that $E B=2 A B$, $A H=3 A D$, $D G=2 D C$, and $F C=3 B C$. Find the ratio between the areas of quadrilateral $E H G F$ and rectangle $A B C D$.

![](https://cdn.mathpix.com/cropped/2024_05_01_80f7f1185898b39e7876g-06.jpg?height=400&width=1170&top_left_y=2073&top_left_x=343)
Solution

The area of quadrilateral $E F G H$ can be decomposed into four triangles and one rectangle. If $S=A_{A B C D}$, we have

$$
\begin{aligned}
A_{E A H} & =(E A \cdot A H) / 2 \\
& =(A B \cdot 3 A D) / 2 \\
& =3 S / 2 \\
A_{D H G} & =(D H \cdot D G) / 2 \\
& =(2 A D \cdot 2 D C) / 2 \\
& =2 S \\
A_{F C G} & =(F C \cdot C G) / 2 \\
& =(3 B C \cdot D C) / 2 \\
& =3 S / 2 \\
A_{F B E} & =(F B \cdot B E) / 2 \\
& =(2 B C \cdot 2 A B) / 2 \\
& =2 S
\end{aligned}
$$

Therefore,

$$
\begin{aligned}
\frac{A_{E F G H}}{A_{A B C D}} & =\frac{3 S / 2+2 S+3 S / 2+2 S+S}{S} \\
& =8
\end{aligned}
$$