11.5. All edges of the tetrahedron are equal to $12 \mathrm{~cm}$. Can it be placed in a box that has the shape of a rectangular parallelepiped with sides $9 \mathrm{~cm}, 13$ cm and 15 cm $?$
Answer: Yes.

Consider a cube $A B C D A_{1} B_{1} C_{1} D_{1}$ with side length $a$. The tetrahedron $A C B_{1} D_{1}$ has all edges equal to $\sqrt{2} a$. Therefore, a tetrahedron with an edge length of 12 can fit into a cube with a side length of $6 \sqrt{2}$. Since $6 \sqrt{2}<9$, the cube, along with the tetrahedron, fits into the box.

![](https://cdn.mathpix.com/cropped/2024_05_06_389d7ed6a0aa309fac57g-2.jpg?height=357&width=394&top_left_y=758&top_left_x=1436)