11.5. A rectangular parallelepiped with edge lengths $\sqrt{70}, \sqrt{99}, \sqrt{126}$ is orthogonally projected onto all possible planes. Find the maximum value of the projection area. Justify your answer.
Solution: The projection represents a hexagon with pairwise parallel sides (some angles may be degenerate). It is divided into three parallelograms, so its area is twice the area of the triangle $M N P-$ see the figure.

This area - the area of the projection of the triangle - is maximal when the projection is made perpendicular to the projected triangle. In this case, the triangle is equal to its projection, and the area of the projection, consequently, is equal to the area of the triangle itself. The lengths of the sides of the triangle are easily found using the Pythagorean theorem: $\sqrt{70+99}=13, \sqrt{70+126}=14$ and $\sqrt{99+126}=15$. It remains to find the area of the triangle with sides $13,14,15$. By Heron's formula, it is $\sqrt{21 \cdot(21-13) \cdot(21-14) \cdot(21-15)}=84$. Then the maximum area

![](https://cdn.mathpix.com/cropped/2024_05_06_c5135f02f0293a986456g-8.jpg?height=611&width=916&top_left_y=126&top_left_x=570)

of the projection is 168.

Answer: 168.

Recommendations for checking:

| is in the work | points |
| :--- | :--- |
| Correct and fully justified answer | 7 points |
| Arithmetic errors in the solution (possibly leading to an incorrect answer) | deduct 1 point |
| There is an unproven statement about the relationship between the area of the triangle $M N P$ and the maximum projection area; with this, the problem is solved. | 5 points |
| The lengths of segments or areas of figures are incorrectly (using incorrect formulas) found; all steps of the solution are performed and a numerical value of the projection area (possibly incorrect) is obtained | 3 points |
| The diagonals of all faces of the parallelepiped (or other parameters of the triangle $M N P$ that allow finding its area) are correctly found | 2 points |
| Answer without justification and/or incorrect answer | 0 points |