[ Linear dependence of vectors ] [ Angles between lines and planes $]$

The side of the base of a regular triangular pyramid is equal to $a$. The lateral edge forms an angle of $60^{\circ}$ with the base plane. Find the area of the lateral surface of the pyramid.
## First solution method.

Let $A B C P$ be the given regular triangular pyramid with vertex $P, A B=B C=A C=a, M$ - the center

of the equilateral triangle $A B C, K$ - the midpoint of $A B, \angle P A M=\angle P B M=\angle P C M=60^{\circ}$. Since the pyramid is regular, $P M$ is its height, and $P K$ is its apothem. Let $S$ be the lateral surface area of the pyramid $A B C P$. From the right triangles $Р A M$ and $P K M$, we find that

$$
\begin{gathered}
P M=A M \operatorname{tg} \angle P A M=\frac{2}{3} \cdot \frac{a \sqrt{3}}{2} \cdot \operatorname{tg} 60^{\circ}=\frac{a \sqrt{3}}{3} \cdot \sqrt{3}=a \\
P K=\sqrt{P M^{2}+K M^{2}}=\sqrt{a^{2}+\left(\frac{a \sqrt{3}}{6}\right)^{2}}=\frac{a \sqrt{13}}{2 \sqrt{3}}
\end{gathered}
$$

Since the lateral faces of the regular pyramid are equal isosceles triangles and $P K$ is the height of triangle $A B P$, we have

$$
\begin{aligned}
S=S_{\triangle A B P}+ & S_{\triangle B C P}+S_{\triangle A C P}=3 S_{\triangle A B P}=3 \cdot \frac{1}{2} A B \cdot P K= \\
& =3 \cdot \frac{1}{2} a \cdot \frac{\frac{a \sqrt{13}}{2 \sqrt{3}}}{}=\frac{3 a^{2} \sqrt{13}}{4 \sqrt{3}}=\frac{a^{2} \sqrt{39}}{4} .
\end{aligned}
$$

## Second solution method.

Let $A B C P$ be the given regular triangular pyramid with vertex $P, A B=B C=A C=a, M$ - the center

of the equilateral triangle $A B C, K$ - the midpoint of $A B, \angle P A M=\angle P B M=\angle P C M=60^{\circ}$. Since the pyramid is regular, $P M$ is its height. Since $P K \perp A B$ and $M K \perp A B$, the angle $P K M$ is the linear angle of the dihedral angle between the plane of the lateral face $A B P$ and the plane of the base $A B C$. From the right triangle $P K M$, we find that

$$
\begin{gathered}
\operatorname{tg} \beta=\operatorname{tg} \angle P K M=\frac{P^{P} M}{M K}=\frac{\frac{a}{b}}{2 \sqrt{3}}=2 \sqrt{3}, \\
\cos \beta=\frac{1}{\sqrt{1+g^{2} \beta}}=\frac{1}{\sqrt{13},}
\end{gathered}
$$

Let $S$ be the lateral surface area of the given pyramid. The orthogonal projection of the lateral surface of the pyramid onto the plane of the base is the triangle $A B C$, therefore

![](https://cdn.mathpix.com/cropped/2024_05_06_808033c11c6fe49d1738g-23.jpg?height=151&width=572&top_left_y=1711&top_left_x=743)

Answer

$\frac{\mathbf{a}^{2} \sqrt{39}}{4}$