2.9. In a regular tetrahedron $S A B C$, a section is constructed by a plane passing through the edge $A C$ and a point $K$ on the edge $S B$, such that $B K: K S=2: 1$. Find the volume of the truncated pyramid $K A B C$, if the edge of the tetrahedron is $a$.
2.9. The desired volume is found using the formula $V_{K A B C}=\frac{1}{3} S_{\triangle A B C} \cdot K N$ (Fig. 2.9), where $S_{\triangle A B C}=\frac{a^{2} \sqrt{3}}{4}$. Since $S O \| K N$, then $B N: N O=B K: K S=2: 1$, from which $B N=\frac{2}{3} B O$. But $B O=\frac{a \sqrt{3}}{3}$, i.e.

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-460.jpg?height=402&width=389&top_left_y=674&top_left_x=765)

Fig. 2.9 $B N=\frac{2 a \sqrt{3}}{9}$ and, therefore,

$$
K N=\sqrt{B K^{2}-B N^{2}}=\sqrt{\frac{4 a^{2}}{9}-\frac{4 a^{2}}{27}}=\frac{2 a \sqrt{6}}{9}
$$

Finally, we get

$$
V=\frac{1}{3} \cdot \frac{a^{2} \sqrt{3}}{4} \cdot \frac{2 a \sqrt{6}}{9}=\frac{a^{3} \sqrt{2}}{18}
$$

Answer: $\frac{a^{3} \sqrt{2}}{18}$.