A right-angled triangle has an area of $36 \mathrm{~m}^{2}$. Inside it, a square is placed such that two sides of the square are parts of two sides of the triangle, and one vertex of the square is at one-third of the longest side.

Determine the area of this square.

(E. Novotná)
The only possible placement of the square within the triangle is as shown in the following image. Additional division of the sides into thirds and connecting the corresponding points with dashed lines divides the given triangle into 9 congruent smaller triangles:

![](https://cdn.mathpix.com/cropped/2024_04_17_768244f2338d28a7fb01g-4.jpg?height=311&width=600&top_left_y=1329&top_left_x=728)

The square consists of four smaller triangles, so the ratio of the area of the square to the area of the given triangle is 4:9. The triangle has an area of $36 \mathrm{~cm}^{2}$, so the area of the square is

$$
\frac{4}{9} \cdot 36=4 \cdot 4=16\left(\mathrm{~cm}^{2}\right)
$$

Notes. The area of a right triangle is half the area of the completed rectangle. The previous idea of dividing into thirds can also be developed according to the following image:

![](https://cdn.mathpix.com/cropped/2024_04_17_768244f2338d28a7fb01g-4.jpg?height=315&width=600&top_left_y=2124&top_left_x=728)

From the above, it follows that the legs of the given triangle are in the ratio $1: 2$.

Within the given triangle, several similar triangles can be discovered. This observation can also be used in solving the problem.