2. The length of a rectangular parallelepiped is 16, and the width is 12. If the centers of three of its faces, which share a common vertex, are connected, a triangle with an area of 30 is formed. Find the height of the parallelepiped.
2. Answer: 7.2. Solution. Let $A B C D A_{1} B_{1} C_{1} D_{1}$ be a parallelepiped, $A B=12, A D=16, A A_{1}=x, X$ be the center of the face $A A_{1} D_{1} D, Y$ be the center of the face

| 0 | 1 | -1 | 0 |
| :---: | :---: | :---: | :---: |
| -1 | 0 | 0 | 1 |
| 0 | -1 | 1 | 0 |
| 1 | 0 | 0 | -1 |

$C C_{1} D_{1} D, Z$ be the center of the face $A_{1} B_{1} C_{1} D_{1}$. Then, by the Pythagorean theorem, we get $X Y=10, X Z=\sqrt{\frac{x^{2}}{4}+36}, Y Z=\sqrt{\frac{x^{2}}{4}+64}$. From the condition, it follows that the height $Z H$ of the acute-angled triangle (as can be seen from the cosine theorem) $X Y Z$ is 6. Using the Pythagorean theorem again, we get $Y H=\frac{x}{2}, X H=\sqrt{\frac{x^{2}}{4}+28} \Rightarrow 10=\frac{x}{2}+\sqrt{\frac{x^{2}}{4}+28} \Rightarrow x=7.2$.

Grading criteria. Correct solution - 7 points. No justification for the acute-angledness of the triangle with vertices at the centers of the faces (and thus, the position of the height of the triangle is not justified) - 5 points. The irrational equation is correctly formulated but not solved - 1 point. In all other cases - 0 points.