In the figure, there is a square $ABCD$ with a side length of $10 \, \text{cm}$ and a smaller square $EFGH$, whose vertices $E, F, G, H$ lie on the diagonals $AC, BD$ of square $ABCD$ (see figure). The area that lies inside square $ABCD$ and outside square $EFGH$ is shaded gray. The line $p$, which is parallel to $AB$ and is at a distance of $6.5 \, \text{cm}$, passes through square $EFGH$ and divides the gray area into two parts. The area of one of these parts is $13.8 \, \text{cm}^2$ larger than the area of the other. Calculate the length of side $EF$.

(L. Šimünek)

![](https://cdn.mathpix.com/cropped/2024_04_17_de96de15b4a98a6160b2g-3.jpg?height=580&width=591&top_left_y=635&top_left_x=767)
Parallel lines $p$ and $D C$ are $3.5 \mathrm{~cm}$ apart. At a distance of $3.5 \mathrm{~cm}$ from line $A B$, draw line $q$, which passes through the square $E F G H$.

![](https://cdn.mathpix.com/cropped/2024_04_17_de96de15b4a98a6160b2g-3.jpg?height=574&width=591&top_left_y=1409&top_left_x=767)

Lines $p$ and $q$ define two identical rectangles from the gray area, which together represent the given difference in areas. Let the area of one of them be $S$, its sides $a$ and $b$ (see figure). We calculate $S, b$, and then $a$:

$$
\begin{aligned}
S & =13.8: 2=6.9\left(\mathrm{~cm}^{2}\right), \\
b & =10-2 \cdot 3.5=3(\mathrm{~cm}), \\
a & =S: b=6.9: 3=2.3(\mathrm{~cm}) .
\end{aligned}
$$

The length of side $E F$ is $10-2 a$, i.e., $10-2 \cdot 2.3=5.4(\mathrm{~cm})$.

Evaluation. 2 points for the auxiliary line; 3 points for the length of side $a$; 1 point for the length of side $E F$.