2.36. The base of the pyramid is a parallelogram $ABCD$ with an area of $m^2$ and such that $BD \perp AD$; the dihedral angles at the edges $AD$ and $BC$ are $45^{\circ}$, and at the edges $AB$ and $CD$ are $60^{\circ}$. Find the lateral surface area and volume of the pyramid.
2.36. Let $A D$ be denoted as $x$, and $B D$ as $y$ (Fig. 2.31). Then $S_{\text {base }}=x y=m^{2}$. Since $B D \perp A D$, it follows that $S D \perp A D$, i.e., $\angle S D B$ is the dihedral angle and, by the condition, $\angle S D B=45^{\circ}$. Through point $O$, draw $M K \perp A B$; we have $\angle S M K=\angle S K M=60^{\circ}$. From $\triangle S O D$ it follows that

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-475.jpg?height=561&width=390&top_left_y=112&top_left_x=74)

Fig. 2.31 that $S O=O D=\frac{y}{2}$, and from $\triangle S O M$ that $O M=\frac{S O}{\sqrt{3}}=\frac{y}{2 \sqrt{3}}$; therefore, $S_{\text {base }}=M K \cdot A B=\frac{y \sqrt{x^{2}+y^{2}}}{\sqrt{3}}=m^{2}$. But $y=\frac{m^{2}}{x}$ and, thus, $\sqrt{x^{2}+\frac{m^{4}}{x^{2}}} \cdot \frac{m^{2}}{x \sqrt{3}}=m^{2}$, or $x^{4}+m^{4}=3 x^{4}$, from which $x=\frac{m}{\sqrt[4]{2}}, y=m \sqrt[4]{2}$. Now we find

$$
V=\frac{1}{3} S_{\text {base }} \cdot S O=\frac{1}{3} m^{2} \cdot \frac{1}{2} m \sqrt[4]{2}=\frac{1}{6} m^{3} \sqrt[4]{2}
$$

It remains to find $S_{\text {lateral }}=A D \cdot S D+A B \cdot S M$, where $A D=x, S D=$

$$
\begin{aligned}
& =O D \sqrt{2}=\frac{y \sqrt{2}}{2}, A B=\sqrt{x^{2}+y^{2}}, S M=2 O M=\frac{y}{\sqrt{3}} . \text { Therefore, } \\
& S_{60 \mathrm{x}}=\frac{x y \sqrt{2}}{2}+\sqrt{\frac{x^{2} y^{2}+y^{4}}{3}}=\frac{m^{2} \sqrt{2}}{2}+\sqrt{\frac{m^{4}}{3}+\frac{2 m^{4}}{3}}=\frac{1}{2} m^{2}(2+\sqrt{2})
\end{aligned}
$$

Answer: $\frac{m^{2}(2+\sqrt{2)}}{2} ; \frac{m^{3} \sqrt[4]{2}}{6}$.