11.136. The side of the base of a regular triangular pyramid is equal to $a$. A plane is drawn through one of the base edges, perpendicular to the opposite lateral edge and dividing this edge in the ratio $m: n$, counting from the vertex of the base. Determine the total surface area of the pyramid.
Solution.

The full surface area of the pyramid can be found using the formula

$$
S_{\text {full }}=\frac{a^{2} \sqrt{3}}{4}+\frac{1}{2} \cdot 3 a \cdot S D \text { (Fig. 11.33). Since } \triangle B O S \sim \triangle B K D \text { (right-angled }
$$

triangles with a common angle), we have $\frac{B D}{B S}=\frac{B K}{B O}$,

$$
\frac{a \sqrt{3}}{2}: B S=B K: \frac{a \sqrt{3}}{3}, a^{2}=2 B K \cdot B S
$$

By the condition $\frac{B K}{K S}=\frac{m}{n}, K S=\frac{n}{m} B K, B S=B K+K S=\frac{n+m}{m} \cdot B K$, $B K=\frac{m}{m+n} B S, B S^{2}=\frac{(m+n) a^{2}}{2 m}$. In $\triangle S O D$ we have $S D^{2}=S O^{2}+O D^{2} ;$
from this, considering that $S O^{2}=B S^{2}-B O^{2}$, we find $S D^{2}=\frac{(m+n) a^{2}}{2 m}-$ $-\frac{a^{2}}{3}+\frac{a^{2}}{12}=\frac{3 m a^{2}+6 n a^{2}}{12 m}=\frac{(m+2 n) a^{2}}{4 m}$.

We obtained $S_{\text {full }}=\frac{a^{2} \sqrt{3}}{4}+\frac{1}{2} \cdot 3 a \cdot \frac{a}{2} \sqrt{\frac{m+2 n}{m}}=\frac{a^{2} \sqrt{3}}{4}\left(1+\sqrt{\frac{3(m+2 n)}{m}}\right)$.

Answer: $\frac{a^{2} \sqrt{3}}{4}\left(1+\sqrt{\frac{3(m+2 n)}{m}}\right)$.