11.213 Calculate the volume of a triangular pyramid, in which two opposite edges are 4 and 12 cm, and each of the other edges is 7 cm.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-267.jpg?height=310&width=304&top_left_y=1063&top_left_x=161)

Let in the given pyramid $F A B C$, $F B=12, A C=4$, and the other edges are equal to 7. If $K$ is the midpoint of $A C$, then $F K$ and $B K$ are the heights of the isosceles triangles $\triangle A F C$ and $\triangle A B C$. Since $\triangle A F C = \triangle A B C$, we have $F K = B K = \sqrt{F C^2 - K C^2} = \sqrt{45}$. If $K M$ is the height of $\triangle F B K$, then $K M = \sqrt{F K^2 - F M^2} = 3$, and $S_{F B K} = 18$. By Theorem 11,

$$
A C \perp F B K \Rightarrow V_{F A B C} = \frac{1}{3} A C \cdot S_{F B K} = 24
$$

Answer: $24 \text{ cm}^3$.