11.215 Find the volume of a triangular pyramid, the sides of the base of which are equal to $a, b$ and $c$, if each of these sides is equal to the lateral edge that does not intersect with it.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-268.jpg?height=321&width=453&top_left_y=1356&top_left_x=116)

Through each edge of the given pyramid $A B C D$, we draw a plane parallel to the opposite edge. The resulting body $T=A N D K L B M C$ is a parallelepiped, and $V_{A B C D}=\frac{1}{3} V_{T}$ (this was proven in the solution to problem 11.202).

Moreover, by the condition, $A B=C D \Rightarrow A B=L N$, i.e., the faces of $T$ are rectangles $\Rightarrow V_{T}=K A \cdot K C \cdot K D$. Let $A C=a, C D=b, A D=c$. Then, by the Pythagorean theorem, $a^{2}=K A^{2}+K C^{2}, b^{2}=K C^{2}+K D^{2}, c^{2}=K A^{2}+K D^{2} \Rightarrow$

$\Rightarrow K A^{2}=\frac{a^{2}+c^{2}-b^{2}}{2}, K C^{2}=\frac{a^{2}+b^{2}-c^{2}}{2}, K D^{2}=\frac{b^{2}+c^{2}-a^{2}}{2}$.

From this, we have

Answer: $\frac{1}{3} \sqrt{\frac{a^{2}+b^{2}-c^{2}}{2}} \sqrt{\frac{a^{2}+c^{2}-b^{2}}{2}} \sqrt{\frac{b^{2}+c^{2}-a^{2}}{2}}$.