Four identical pieces, in the shape of right triangles, were arranged in two different ways, as shown in the given figures. The squares $A B C D$ and $E F G H$ have sides respectively equal to $3 \mathrm{~cm}$ and $9 \mathrm{~cm}$. Determine the measure of the side of the square $I J K L$.
![](https://cdn.mathpix.com/cropped/2024_05_01_280b19a87fa67143351eg-011.jpg?height=432&width=1036&top_left_y=1326&top_left_x=592)
Solution 1: Let $x$ and $y$ be the larger and smaller legs, respectively, of the right triangle. Since the side of square $A B C D$ measures $3 \mathrm{~cm}$, we have $x-y=3$. On the other hand, since the side of $E F G H$ measures $9 \mathrm{~cm}$, we have $x+y=9$. Solving the system, we find $x=6$ and $y=3$. Therefore, the side of square $I J K L$, which is the hypotenuse of the right triangle, measures $\sqrt{6^{2}+3^{2}}=\sqrt{45}=3 \sqrt{5} \mathrm{~cm}$, by the Pythagorean Theorem.
![](https://cdn.mathpix.com/cropped/2024_05_01_280b19a87fa67143351eg-050.jpg?height=400&width=1030&top_left_y=1744&top_left_x=590)

Solution 2: The squares $I J K L$ and $M N O P$ have sides that are the hypotenuses of the given right triangles, so they have the same area.

By superimposing the two figures and making these two squares coincide, we find eight triangles and conclude that $8 \times$ the area of the triangle is equal to the area of $E F G H$ minus the area of $A B C D$, that is, it is equal to $9^{2}-3^{2}=72$. Therefore, the area of each triangle is $9 \mathrm{~cm}^{2}$. From the figure, we see that the area of $I J K L$ is equal to $4 \times$ the area of the triangle plus the area of $A B C D$, that is, it is equal to $4 \times 9+9=45$.

![](https://cdn.mathpix.com/cropped/2024_05_01_280b19a87fa67143351eg-050.jpg?height=414&width=419&top_left_y=2306&top_left_x=1458)

Therefore, the side of square $I J K L$ measures $\sqrt{45}=3 \sqrt{5} \mathrm{~cm}$.