From a square with a side of $6 \mathrm{~cm}$, we cut off congruent isosceles right triangles from each vertex so that the area of the square is reduced by $32 \%$. What is the length of the legs of these triangles?

(M. Krejčová)
The area of a square with a side of $6 \mathrm{~cm}$ is $36 \mathrm{~cm}^{2}$. The cut-off parts together have an area of $0.32 \cdot 36 = 11.52 \left(\mathrm{~cm}^{2}\right)$. If we denote the leg of the cut-off triangle by $x$, then the area of each such triangle is $\frac{1}{2} x^{2}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_c7e802ccb04b770c89dfg-4.jpg?height=363&width=386&top_left_y=1315&top_left_x=869)

Together, we get the equation, which we can easily solve:

$$
\begin{aligned}
4 \cdot \frac{x^{2}}{2} & =11.52 \\
x^{2} & =5.76 \\
x & =2.4
\end{aligned}
$$

The legs of the cut-off right triangles have a length of $2.4 \mathrm{~cm}$.