The figure is composed of five identical squares and is divided by segments into three differently colored parts. The area of the gray part is $0.6 \mathrm{~cm}^{2}$ larger than the area of the white part.

Determine the area of the entire figure.

(E. Semerádová)

![](https://cdn.mathpix.com/cropped/2024_04_17_1b1f73b26d19cdf7d5a3g-3.jpg?height=308&width=305&top_left_y=580&top_left_x=881)
We can imagine the gray part as a right-angled triangle with legs equal to two and three sides of the square, from which one square is removed. The area of this part is therefore the same as the area of 2 squares $\left(\frac{1}{2} \cdot 2 \cdot 3-1=2\right)$.

The white part forms a triangle, one side of which is equal to the side of the square and the height to this side is equal to three sides of the square. The area of this part is therefore the same as the area of $\frac{3}{2}$ of a square $\left(\frac{1}{2} \cdot 1 \cdot 3=\frac{3}{2}\right)$.

The difference in the areas of the gray and white parts corresponds to half of a square, which is given as $0.6 \mathrm{~cm}^{2}$ in the problem. Thus, one square has an area of $1.2 \mathrm{~cm}^{2}$ and the area of the entire figure is $6 \mathrm{~cm}^{2}(5 \cdot 1.2=6)$.

Evaluation. 2 points for expressing the areas of the gray and white parts; 2 points for calculating the area in $\mathrm{cm}^{2}$.