Example 76. The side of the base of a regular triangular pyramid is equal to $a$; the height dropped from the vertex of the base to the opposite lateral face is equal to $b$. Let's find the volume of the pyramid.

Construction of the image. Let the quadrilateral $S A B C$ with its diagonals be the image of the given pyramid (Fig. 85). This image is complete. Considering $\triangle A B C$ as the image of a regular triangle, we use two parameters, considering $[S O]$ as the image of the height of the pyramid, we also use two parameters. Construct $[S D]$ - the median of $\triangle S B C$ - the image of the apothem of the lateral face of the pyramid. Let $[A K]$ be the image of the perpendicular dropped from the vertex $A$ of the given pyramid to the apothem of the opposite lateral face. As it is not difficult to show, the segment $A K$ will then also be the image of the height dropped from the vertex of the base to the lateral face. This metric condition, imposed on the image, entails the use of another parameter. Thus, all five parameters have been used on the image. However, the condition that in the original $\left|A_{0} B_{0}\right|=a$ and $\left|A_{0} K_{0}\right|=b$ has remained unaccounted for. The fact that these data are literal gives some freedom; we will consider that on the constructed image $|A B|:|A K|=a: b$. Thus, we impose another metric condition on the image and, consequently, use another parameter. But then the image becomes metrically overdefined. Nevertheless, with the fulfillment of a certain dependence between $a$ and $b$ (we will obtain this dependence in the study) the constructed image is correct.

Since by the meaning of the problem the height $A_{0} K_{0}$ exists, then from $\Delta S_{0} A_{0} D_{0}$, where $\left|A_{0} K_{0}\right|=b$ and $\left|A_{0} D_{0}\right|=\frac{a \sqrt{3}}{2}$, it can be calculated that $\left|D_{0} K_{0}\right|:\left|S_{0} K_{0}\right|=\frac{3 a^{2}-4 b^{2}}{4 b^{2}-2 a^{2}}$. 130

Choosing different (admissible) values of $a$ and $b$, we will obtain different values of the ratio $\left|D_{0} K_{0}\right|:\left|S_{0} K_{0}\right|$, i.e., and different positions of the point $K_{0}$. For example, setting $\left\{\begin{array}{l}a=10, \\ b=8,\end{array}\right.$ we would get that $\frac{3 a^{2}-4 b^{2}}{4 b^{2}-2 a^{2}}=\frac{11}{14}$. With such a specific assignment of values of $a$ and $b$, as we see, the position of the point $K_{0}$ will be quite determined, since $\left|D_{0} K_{0}\right|:\left|S_{0} K_{0}\right|=11: 14$. This means that the arbitrariness in the image of the point $K_{0}$ is no longer permissible.

Given: $\quad S A B C-$ regular pyramid, $S$-vertex, $[A K]-$ height dropped to the face $S B C,|A B|=a,|A K|=b$.

To find: $V-$ volume of the pyramid.
Solution. Since $V=\frac{1}{3} S_{\triangle A B C} \cdot|S O|$, where $S_{\triangle A B C}=\frac{a^{2} \sqrt{3}}{4}$, it is sufficient to calculate the length of the height $S O$.

From the similarity of right triangles $S O D$ and $A D K$, we get:

$$
\frac{|S O|}{|A K|}=\frac{|O D|}{|D K|}
$$

from which $|S O|=\frac{|A K| \cdot|O D|}{|D K|}$, where $|A K|=b,|O D|=\frac{1}{3}|A D|=\frac{a \sqrt{3}}{6}$, $|D K|=\sqrt{|A D|^{2}-|A K|^{2}}=\frac{\sqrt{3 a^{2}-4 b^{2}}}{2}$.

Thus,

$$
|S O|=\frac{a b \sqrt{3}}{3 \sqrt{3 a^{2}-4 b^{2}}} \text {. }
$$

Then

$$
V=\frac{a^{3} b}{12 \sqrt{3 a^{2}-4 b^{2}}}
$$

Investigation. By the meaning of the problem, $[A K]$ is the height dropped onto the face $S B C$ (onto the face, not the plane of the face!). It is clear that if the given pyramid is so "low" that $\widehat{A S D}=$ $=90^{\circ}$, then $K=S$, and, consequently, the segments $A K$ and $S A$ coincide. If the pyramid is even "lower," then the height dropped onto the face $S B C$ does not exist at all.

Assuming $\widehat{A S D}=90^{\circ}$, from the right $\triangle A S D$, where $|D O|=$ $=\frac{a \sqrt{3}}{6},|A O|=\frac{a \sqrt{3}}{3}$, we find:

$$
|S O|=\sqrt{\frac{a \sqrt{3}}{6} \cdot \frac{a \sqrt{3}}{3}}=\frac{a \sqrt{6}}{6}
$$

Thus, for the height $A K$ to exist, it is necessary that the values of $a$ and $b$ satisfy the inequality

$$
\frac{a b \sqrt{3}}{3 \sqrt{3 a^{2}-4 b^{2}}} \geqslant \frac{a \sqrt{6}}{6}
$$

5*

![](https://cdn.mathpix.com/cropped/2024_05_21_318d1c34bf6979fc9790g-133.jpg?height=780&width=462&top_left_y=244&top_left_x=360)

Fig. 86 or the equivalent system of inequalities:

$$
\frac{a \sqrt{2}}{2} \leqslant b \leqslant \frac{a \sqrt{3}}{2}
$$

At the same time, it is clear that by the meaning of the problem $|A K|>0$.

Thus,

$$
V=\frac{a^{3} b}{12 \sqrt{3 a^{2}-4 b^{2}}}, \text { where } \frac{a \sqrt{2}}{2} \leqslant b<\frac{a \sqrt{3}}{2}
$$