The figure is composed of eight congruent squares and is divided by segments into five differently colored parts. Point $X$ is the midpoint of segment $K J$, point $Y$ is the midpoint of segment $E X$, and segment $B Z$ is congruent with $B C$. The area of the black part of the figure is $7.5 \mathrm{~cm}^{2}$.

Determine the areas of the remaining four parts.

(E. Semerádová and M. Dillingerová)

![](https://cdn.mathpix.com/cropped/2024_04_17_a6c4eb9df3478319268bg-3.jpg?height=457&width=548&top_left_y=1368&top_left_x=754)
In the figure, we highlight eight identical squares and use them to express the areas of individual parts:

![](https://cdn.mathpix.com/cropped/2024_04_17_a6c4eb9df3478319268bg-3.jpg?height=456&width=552&top_left_y=2005&top_left_x=752)

The white part forms half of a square.

The dark gray part can be divided into a square and a right triangle $Z U L$, whose legs are equal to one and two sides of the square. This triangle has the same area as the square $\left(\frac{1}{2} \cdot 1 \cdot 2=1\right)$, so the area of the dark gray part is the same as the area of 2 squares.

The black part can be divided into three right triangles: triangles $Z E L$ and $E K X$ are congruent with the previously mentioned triangle $Z U L$, and triangle $Z D C$ forms half of a square. The area of the black part is thus the same as the area of $\frac{5}{2}$ squares $\left(1+1+\frac{1}{2}=\frac{5}{2}\right)$.

The light gray part can be divided into a square and triangle $G X Y$. Points $X$ and $E$ are vertices of the squares, and point $Y$ is the midpoint of segment $E X$, which we interpret as the diagonal of rectangle $K E V X$. Segment $H Y$ is thus congruent to half of a side of the square. This segment is the height of triangle $G X Y$ on side $G X$, which is equal to three sides of the square. Triangle $G X Y$ has the same area as $\frac{3}{4}$ of a square $\left(\frac{1}{2} \cdot \frac{1}{2} \cdot 3=\frac{3}{4}\right)$, so the area of the light gray part is the same as the area of $\frac{7}{4}$ squares $\left(\frac{3}{4}+1=\frac{7}{4}\right)$.

The remaining part can be expressed as the difference of the above parts,

$$
8-2-\frac{1}{2}-\frac{5}{2}-\frac{7}{4}=3-\frac{7}{4}=\frac{5}{4}
$$

so the area of the hatched part is the same as the area of $\frac{5}{4}$ squares.

Now we use the fact that the area of the black part is $7.5 \mathrm{~cm}^{2}$. This corresponds to $\frac{5}{2}$ squares, so the area of one square is $3 \mathrm{~cm}^{2}\left(\frac{5}{2} \cdot 3=7.5\right)$. From this, we conclude that the area of the white part is $1.5 \mathrm{~cm}^{2}\left(\frac{1}{2} \cdot 3=1.5\right)$, the area of the dark gray part is $6 \mathrm{~cm}^{2}(2 \cdot 3=6)$, the area of the light gray part is $5.25 \mathrm{~cm}^{2}\left(\frac{7}{4} \cdot 3=5.25\right)$, and the area of the hatched part is $3.75 \mathrm{~cm}^{2}\left(\frac{5}{4} \cdot 3=3.75\right)$.

Evaluation. 4 points for expressing the areas of all parts in terms of squares; 2 points for calculating the areas in $\mathrm{cm}^{2}$.