Example 65. One edge of a triangular pyramid is equal to $a$, and each of the others is equal to $b$. Let's find the volume of the pyramid.
Solution. Without dwelling on the construction of the image (Fig. 74) and other stages of the solution, let us note that if we solve this example using the formula \( V = \frac{1}{3} S_{\text{base}} \cdot H \), then finding \( H \) would require quite complex calculations. We will take a different path. We will construct a section of the pyramid by a plane passing through the edge \( AB \) perpendicular to the edge \( SC \). Taking \( |AB| = a \), we get:

\[
\begin{gathered}
V = \frac{1}{3} S_{\triangle ABM} \cdot |SM| + \\
+ \frac{1}{3} S_{\triangle ABM} \cdot |CM| = \\
= \frac{1}{3} S_{\triangle ABM} \cdot |SC| = \frac{b}{3} \cdot S_{\triangle ABM}
\end{gathered}
\]

108

(Note that constructing the section in this case is a positional problem, since \(\triangle SAC\) and \(\triangle SBC\) are equilateral, i.e., medians \(AM\) and \(BM\) are images of perpendiculars to the edge \(SC\), and therefore \(\triangle ABM\) is an image of the section perpendicular to the edge \(SC\).)

Since \( S_{\triangle ABM} = \frac{1}{2} |AB| \cdot |MD| \), where \([MD]\) is the median of the isosceles \(\triangle ABM\), in which \( |AM| = |BM| = \frac{b \sqrt{3}}{2} \), then

\[
S_{\triangle ABM} = \frac{1}{2} a \sqrt{\frac{3 b^2}{4} - \frac{a^2}{4}} = \frac{a}{4} \sqrt{3 b^2 - a^2}
\]

Therefore, \( V = \frac{a b}{12} \sqrt{3 b^2 - a^2} \).

We omit the investigation in this example (do it yourself).

## Problems for Independent Solution

177. In the cube \(ABCD A_1 B_1 C_1 D_1\), with edge length \(a\), point \(K\) is the midpoint of edge \(D_1 C_1\). Find the distances from points \(A_1\), \(A\), and \(C_1\) to the plane passing through points \(B\), \(K\), and \(D\).
178. Through the diagonal of the lower base