2.1. The sine of the dihedral angle at the lateral edge of a regular quadrilateral pyramid is $\frac{15}{17}$. Find the area of the lateral surface of the pyramid if the area of its diagonal section is $3 \sqrt{34}$.
Answer: 68.

Solution. Let the linear angle of the dihedral angle given in the problem be denoted as $\alpha$. This angle is always obtuse, so $\cos \alpha=-\frac{8}{17}$.

The projection of the lateral face onto the diagonal section is a triangle, the area of which is half the area of this section. Since the dihedral angle between the lateral face and the diagonal section is $\frac{\alpha}{2}$, we have $S_{\text {face }} \cdot \cos \frac{\alpha}{2}=\frac{1}{2} S_{\text {section }}$. Therefore, $S_{\text {lateral }}=4 \cdot \frac{S_{\text {section }}}{2 \cos \frac{\alpha}{2}}=\frac{2 \sqrt{2} S_{\text {section }}}{\sqrt{1+\cos \alpha}}=68$.