## Task B-4.7.

Given is a parallelogram $A B C D$. Points $E$ and $F$ are the feet of the perpendiculars drawn from vertex $D$ to the sides $\overline{A B}$ and $\overline{B C}$, respectively. If $\cos \varangle E D F=\frac{1}{3},|D E|=20,|D F|=32$, what is the area of the quadrilateral $D E B F$?
## First Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_5ec97620b8cc23e4b297g-34.jpg?height=417&width=936&top_left_y=2010&top_left_x=477)

Let $\varangle E D F = \varphi$. Since $D E$ is perpendicular to $A B$ and $C D$, $\varangle F D C = 90^{\circ} - \varphi$. 1 point

Then $\varangle D C B = \varangle B A D = \varphi$. 1 point

From triangle $DFC$ we have

$$
\cos \varphi = \frac{|C F|}{|C D|} = \frac{1}{3}
$$

which implies

$$
|C F| = \frac{1}{3} |C D|
$$

and by applying the Pythagorean theorem,

$$
\begin{gathered}
32^{2} + \left(\frac{1}{3} |C D|\right)^{2} = |C D|^{2} \\
\frac{8}{9} |C D|^{2} = 32^{2}
\end{gathered}
$$

which gives

$$
|C D| = 24 \sqrt{2}
$$

Similarly, from triangle $DEA$ we get

$$
\cos \varphi = \frac{|A E|}{|A D|} = \frac{1}{3}
$$

which implies

$$
\begin{gathered}
|A E| = \frac{1}{3} |A D| \\
20^{2} + \left(\frac{1}{3} |A D|\right)^{2} = |A D|^{2} \\
\frac{8}{9} |A D|^{2} = 20^{2}
\end{gathered}
$$

which gives

$$
|A D| = 15 \sqrt{2}
$$

It follows that:

$$
\begin{gathered}
|C F| = \frac{1}{3} |C D| = 8 \sqrt{2} \\
|B F| = 15 \sqrt{2} - 8 \sqrt{2} = 7 \sqrt{2} \\
|A E| = \frac{1}{3} |A D| = 5 \sqrt{2} \\
|B E| = 24 \sqrt{2} - 5 \sqrt{2} = 19 \sqrt{2}
\end{gathered}
$$

The area of quadrilateral $DEBF$ is calculated as the sum of the areas of triangles $DEB$ and $BFD$:

$$
P = \frac{|D E| \cdot |B E|}{2} + \frac{|D F| \cdot |B F|}{2} = \frac{20 \cdot 19 \sqrt{2}}{2} + \frac{32 \cdot 7 \sqrt{2}}{2} = 302 \sqrt{2}
$$

Note: The lengths of the sides of the parallelogram can also be calculated using the similarity of triangles $AED$ and $CFD$. Regardless of the method of calculation, the correct length of one side of the parallelogram is worth 2 points, and 4 points for both.

Furthermore, in quadrilateral $DEBF$, $\varangle FBE = 180^{\circ} - \varphi$, so the area of the quadrilateral can be calculated as:

$$
\begin{gathered}
P_{DEBF} = P_{EFD} + P_{EBF} = \frac{|D E| \cdot |D F|}{2} \sin \varphi + \frac{|E B| \cdot |B F|}{2} \sin (180^{\circ} - \varphi) = \\
\frac{20 \cdot 32}{2} \sin \varphi + \frac{19 \sqrt{2} \cdot 7 \sqrt{2}}{2} \sin \varphi = 302 \sqrt{2}
\end{gathered}
$$

Calculating the lengths $|BE|$ and $|BF|$ is worth 2 points, and the final area is worth 2 points.