Let's unfold the net of a tetrahedron from a $30 \mathrm{~cm}$ side square $A B C D$. This square needs to be divided into 4 triangles, which form the tetrahedron (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_9429d59cc9c9383fdb65g-1.jpg?height=383&width=396&top_left_y=253&top_left_x=859)

Figure 1

One vertex of the tetrahedron must coincide with one vertex of the square (otherwise we couldn't divide the square into 4 triangles); let this vertex be $A$. The other two vertices of the triangle fall on the $B C$ and $C D$ sides of the square. When we fold up the triangles, for a tetrahedron to form, the edges that are folded together must be of equal length, i.e., $B E=E C$, $C F=F D$, so $E$ is 15 cm from vertex $B$ (and of course from $C$ as well). Similarly, $C F=F D=15 \mathrm{~cm}$. The tetrahedron is thus bounded by an isosceles right triangle with a leg of $15 \mathrm{~cm}$, two right triangles with legs of $15 \mathrm{~cm}$ and $30 \mathrm{~cm}$, and the isosceles triangle $A E F$.

The volume of the tetrahedron is given by one-third of the product of the base area and the height corresponding to the base.

Now let's consider the base to be the isosceles right triangle $E C F$.

![](https://cdn.mathpix.com/cropped/2024_05_02_9429d59cc9c9383fdb65g-1.jpg?height=458&width=352&top_left_y=1083&top_left_x=881)

Figure 2

Since in the triangle $A B E$, $\angle A B E=90^{\circ}$ and in the triangle $A F D$, $\angle A D F=90^{\circ}$, the line $A B$ is perpendicular to the base plane $E B F$, so $A B=30 \mathrm{~cm}$ is the height of the tetrahedron.

$$
V_{\text {tetrahedron }}=\frac{\frac{15 \cdot 15}{2} \cdot 30}{3}=1125 \mathrm{~cm}^{3}
$$

Orsolya Rábai (Eger, Gárdonyi G. Gymnasium, 9th grade)
We cut a tetrahedron-shaped cardboard box along the three edges starting from one of its vertices, then we lay out the "separated" faces onto the plane of the opposite face. This way, we get a square with a side length of $30 \mathrm{~cm}$. What is the volume of the tetrahedron?