## Task B-1.3.

The lengths of the sides of the parallelogram are $10 \mathrm{~cm}$ and $8 \mathrm{~cm}$. Calculate the area of the shaded part in the figure if the area of the entire parallelogram is $40 \mathrm{~cm}^{2}$ and $|E B|=|B F|=|G D|=|D H|=1 \mathrm{~cm}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_6dc7744662d888032875g-02.jpg?height=257&width=694&top_left_y=1907&top_left_x=1138)
## First Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_6dc7744662d888032875g-03.jpg?height=417&width=963&top_left_y=288&top_left_x=478)

Triangles $G D H$ and $E B F$ are congruent by the S-S-S theorem, and then triangles $G H S$ and $E F S$ are congruent by the S-A-S theorem. Therefore, the shaded quadrilaterals $D H S G$ and $B F S E$ are congruent and have the same area, and point $S$ is the midpoint of diagonal $B D$.

Let $x$ be half the height of the parallelogram $A B C D$ to side $\overline{C D}$, and $y$ be half the height to side $\overline{A D}$.

The area of parallelogram $P D Q S$ is one-fourth of the area of the entire parallelogram, which is $10 \mathrm{~cm}^{2}$. Since its heights are $x$ and $y$, we have:

$10=5 \cdot x \Longrightarrow x=2 \mathrm{~cm}$.

$10=4 \cdot y \Longrightarrow y=2.5 \mathrm{~cm}$.

The areas of triangles $G D S$ and $D H S$ are:

$$
\begin{gathered}
P(G D S)=\frac{|G D| \cdot x}{2}=\frac{1 \cdot 2}{2}=1 \mathrm{~cm}^{2} \\
P(D H S)=\frac{|D H| \cdot y}{2}=\frac{1 \cdot 2.5}{2}=1.25 \mathrm{~cm}^{2}
\end{gathered}
$$

Thus, the desired area is

$$
2(P(G D S)+P(D H S))=2(1+1.25)=4.5 \mathrm{~cm}^{2}
$$