## Task B-3.4.

A square with side $a=6 \sqrt{2}$ is the common base of two regular pyramids. A sphere can be described around this resulting body, passing through all its vertices. If the height of one pyramid is 9, determine the total volume of this body.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_7713364b1a15726862ddg-13.jpg?height=445&width=457&top_left_y=2122&top_left_x=728)

Consider the intersection of the sphere and the plane $A V_{2} V_{1}$. We will get a circle passing through the vertices $A, V_{2}, C, V_{1}$ with diameter $\overline{V_{2} V_{1}}$. Therefore, the triangle $A V_{2} V_{1}$ is a right triangle.

The diagonal of the given square is $d=a \sqrt{2}=12$.

We will apply the Pythagorean theorem to the right triangle $A V_{2} V_{1}$ and calculate the height of the second pyramid:

$$
\frac{d}{2}=\sqrt{v_{1} \cdot v_{2}} \Rightarrow 6=\sqrt{9 v_{2}} \Rightarrow v_{2}=4
$$

The volume of the resulting body is equal to

$$
\begin{aligned}
& V=\frac{1}{3} a^{2} v_{1}+\frac{1}{3} a^{2} v_{2}=\frac{1}{3} a^{2}\left(v_{1}+v_{2}\right) \\
& \Rightarrow V=\frac{1}{3} \cdot 72 \cdot 13=312
\end{aligned}
$$