11.033. In a triangular pyramid, the lateral edges are mutually perpendicular and have lengths $\sqrt{70}, \sqrt{99}$, and $\sqrt{126}$ cm. Find the volume and the area of the base of the pyramid.
Solution.

Given $\angle A P C=\angle A P B=\angle B P C=90^{\circ}, A P=\sqrt{70}, P B=\sqrt{99}$, $P C=\sqrt{126}$ (Fig. $11.28, a$ ). By the Pythagorean theorem, $A B=\sqrt{P A^{2}+P B^{2}}=\sqrt{70+99}=13, B C=\sqrt{P B^{2}+P C^{2}}=\sqrt{99+126}=15$,

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-671.jpg?height=504&width=422&top_left_y=92&top_left_x=173)

Fig. 11.29

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-671.jpg?height=420&width=406&top_left_y=174&top_left_x=749)

Fig. 11.30

$A C=\sqrt{P A^{2}+P C^{2}}=\sqrt{70+126}=14$. Then the area of the base $S_{A B C}$ can be found using Heron's formula

$S_{A B C}=\sqrt{21(21-13)(21-14)(21-15)}=84\left(\mathrm{~cm}^{2}\right)$. To find the volume of the pyramid, we will flip it so that the base is the face $P A B$ (Fig. 11.28, b). This will not change its volume. Then $C P=\sqrt{126}$ is the height of the resulting pyramid, since $\triangle P A B$ is a right triangle, its area $S_{P A B}=\frac{1}{2} P A \cdot P B=\frac{1}{2} \sqrt{70} \cdot \sqrt{99}$. The volume

$$
V=\frac{1}{3} C P \cdot S_{P A B}=\frac{1}{3} \sqrt{126} \cdot \frac{1}{2} \cdot \sqrt{70} \cdot \sqrt{99}=21 \sqrt{55}\left(\mathrm{~cm}^{3}\right)
$$

Answer: $21 \sqrt{55} \mathrm{~cm}^{3}, 84 \mathrm{~cm}^{2}$.