11.112. 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 the volume of the pyramid.
Solution.

Let $MO$ be the height of the pyramid $MABCD$ (Fig. 11.10). Through point $O$ in the plane of the base, draw a line $EF$ ($E$ belongs to $AB$, $F$ belongs to $CD$), perpendicular to the parallel lines $AB$ and $CD$. $OF$ and $OE$ are the projections of $MF$ and $ME$ on the plane of the base, respectively. Then $MF \perp CD$, $ME \perp AB$, $\angle MFO$ and $\angle MEO$ are the linear angles of the dihedral angles at the edges $CD$ and $AB$, respectively, $\angle MFO = \angle MEO = 60^\circ$. Therefore, $\triangle MOF = \triangle MOE$ by the leg and acute angle, and thus $OF = OE$, i.e., point $O$ is equidistant from the sides $AB$ and $CD$ of the parallelogram.

Therefore, $O$ lies on the line parallel to $AB$ and $CD$ and passing through the midpoints of the other two sides of the parallelogram. Similarly, since the dihedral angles at the edges $AD$ and $BC$ are also equal, point $O$ is equidistant from the sides $AD$ and $BC$ of the parallelogram and lies on the line parallel to $AD$ and $BC$ and passing through the midpoints of the other two sides of the parallelogram. Thus, $O$ is the point of intersection of the diagonals of the parallelogram $ABCD$.

Since $OD \perp AD$ and $OB \perp BC$, then $MD \perp AD$, $MB \perp BC$, and $\angle MDO$ and $\angle MBO$ are the linear angles of the dihedral angles at the edges $AD$ and $BC$, respectively, $\angle MDO = \angle MBO = 45^\circ \cdot S_{\triangle AOB} = S_{\triangle DOC} = S_{\triangle BOC} = \frac{m^2}{4}$. Since $\triangle AOB$ and $\triangle DOC$ are the projections of $\triangle AMB$ and $\triangle DMC$ on the plane of the base, $\triangle AOD$ and $\triangle BOC$ are the projections of $\triangle AMD$ and $\triangle BMC$ on the plane of the base, then $S_{\triangle AMB} = S_{\triangle DMC} = \frac{m^2}{2}$, $S_{\triangle AMD} = S_{\triangle BMC} = \frac{m^2}{2\sqrt{2}}$. The lateral surface area of the pyramid:

$$
S = 2\left(S_{\triangle AMB} + S_{\triangle AMD}\right) = 2\left(\frac{m^2}{2} + \frac{m^2}{2\sqrt{2}}\right) = \frac{m^2(\sqrt{2} + 1)}{\sqrt{2}} = \frac{m^2(\sqrt{2} + 2)}{2}
$$

Let $MO = H$. From $\triangle MOE (\angle MOE = 90^\circ): OE = H \operatorname{ctg} 60^\circ = \frac{H}{\sqrt{3}}$.

From $\triangle MOB (\angle MOB = 90^\circ): OB = H \operatorname{ctg} 45^\circ = H$.

From $\triangle OEB (\angle OEB = 90^\circ): EB^2 = OB^2 - OE^2 = H^2 - \frac{H^2}{3} = \frac{2H^2}{3}$.

$BD = 2 \cdot OB = 2H \cdot \triangle BEO \sim \triangle BDA$ - by two angles.

Then $\frac{S_{\triangle BEO}}{S_{\triangle BDA}} = \frac{EB^2}{BD^2} = \frac{\frac{2H^2}{3}}{4H^2} = \frac{1}{6} ; S_{\triangle BEO} = \frac{1}{6} S_{\triangle BDA} = \frac{1}{6} \cdot \frac{1}{2} m^2 = \frac{m^2}{12}$.

But $S_{\triangle BEO} = \frac{1}{2} OE \cdot EB = \frac{1}{2} \cdot \frac{H}{\sqrt{3}} \cdot H \sqrt{\frac{2}{3}} = \frac{H^2 \sqrt{2}}{6}$. Then $\frac{H^2 \sqrt{2}}{6} = \frac{m^2}{12}$; $H^2 = \frac{m^2 \sqrt{2}}{4} ; H^2 = \frac{m_4^2}{2}$. The volume of the pyramid $V = \frac{1}{3} m^2 \cdot H = \frac{m^3 \sqrt[4]{2}}{6}$.

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