Example 67. The base of the pyramid is an equilateral triangle with a side length of $a$. One of the lateral faces, which is perpendicular to the plane of the base, is also an equilateral triangle. Find the area of the lateral surface of the pyramid.

Construction of the image. Let the quadrilateral $S A B C$ with its diagonals be the image of the given pyramid

![](https://cdn.mathpix.com/cropped/2024_05_21_318d1c34bf6979fc9790g-115.jpg?height=620&width=674&top_left_y=247&top_left_x=357)

Fig. 76

pyramid (Fig. 76). This image is complete, $p=5$.

Given: $S A B C$ - pyramid, $S$ - vertex, $|A B|=|B C|=$ $=|A C|=|S B|=|S C|=a$, $(S B C) \perp(A B C)$.

Find: $\quad S_{\text {lateral }}$ - area of the lateral surface.
Solution. $S_{\text {side }}=S_{\triangle S A B}+$ $+S_{\triangle S A C}+S_{\triangle S B C}$. But $\triangle S A B \cong$ $\cong \triangle S A C$ (by three sides), i.e.

$$
S_{\text {side }}=2 S_{\triangle S A B}+S_{\triangle S B C}
$$

Since $S_{\triangle S A B}=\frac{1}{2}|A B| \times$ $\times|S K|$, we will first find $|S K|$ - the length of the height of $\triangle S A B$. Constructing a perpendicular to a given line is, as is known, a metric construction in general. In the considered example, this construction needs to be performed on a metrically defined image.

We will construct the height $S K$ as follows:

1) $[S D]$ - median of $\triangle S B C$
2) $[C M]$ - median of $\triangle A B C$,
3) $[D K] \|[C M]$
4) $[S K]$.

Since $[S D]$ is the median of an equilateral triangle, $[S D] \perp[B C]$. But $(S B C) \perp(A B C)$. Then $[S D] \perp(A B C)$, and therefore, $[D K]$ is the projection of segment $S K$ onto the plane $A B C$. But $[D K] \|[C M]$, and $[C M]$ is the median of the equilateral $\triangle A B C$, i.e., $[C M] \perp[A B]$.

Therefore, $[D K] \perp[A B]$, and thus, $[S K] \perp[A B]$. From the right $\triangle S D K$ we have:

$$
\begin{gathered}
|S K|=\sqrt{\left.S D\right|^{2}+|D K|^{2}}, \text { where }|S D|=\frac{a \sqrt{3}}{2}, \\
|D K|=\frac{1}{2}|C M|=\frac{a \sqrt{3}}{4}
\end{gathered}
$$

Thus, $|S K|=\frac{a \sqrt{15}}{4}$. We obtain:

$$
S_{\triangle S A B}=\frac{a^{2} \sqrt{15}}{8} \text { and } S_{\triangle S B C}=\frac{a^{2} \sqrt{3}}{4}
$$

Therefore, $S_{\text {side }}=\frac{a^{2}}{4}(\sqrt{\overline{15}}+\sqrt{ } \sqrt{3})$.

Investigation. Clearly, by the meaning of the problem, $|S K|>$ $|S D|$ and therefore $S_{\text {side }}>3 S_{\triangle S B C}$, i.e., $\frac{a^{2}}{4}(\sqrt{15}+\sqrt{3})>\frac{3 a^{2} \sqrt{3}}{4}$. As it is not difficult to verify, this inequality is true.

Thus,

$$
S_{6 \circ \kappa}=\frac{a^{2}}{4}(\sqrt{15}+\sqrt{3})
$$