11.062. The base of the pyramid is a square. Two lateral faces are perpendicular to the base plane, while the other two are inclined to it at an angle of $45^{\circ}$. The middle-sized lateral edge is equal to $l$. Find the volume and the total surface area of the pyramid.
Solution.

Given $S C=l, \angle S B C=90^{\circ}, \angle S C B=45^{\circ}$ (Fig. 11.57), hence $S B=B C=\frac{l}{\sqrt{2}}$. We have $V=\frac{1}{3} B C^{2} \cdot S B=\frac{l^{3} \sqrt{2}}{12}$. The total surface area $-S_{\text {full }}=S_{\text {base }}+2 S_{\triangle S A B}+2 S_{\triangle S A D}$, since $S_{\triangle S A B}=S_{\triangle S B C}$, $S_{\triangle S A D}=S_{\triangle S C D}$. But $S_{\triangle S A B}=\frac{1}{2} A B \cdot S B=\frac{1}{2}\left(\frac{l}{\sqrt{2}}\right)^{2}=\frac{l^{2}}{4}$,

$$
S_{\triangle S A D}=\frac{1}{2} A D \cdot S A=\frac{1}{2} \cdot \frac{l}{\sqrt{2}} \cdot l=\frac{l^{2} \sqrt{2}}{4}
$$

Thus,

$$
S_{\text {full }}=\frac{l^{2}}{2}+\frac{l^{2}}{2}+\frac{l^{2} \sqrt{2}}{2}=\frac{l^{2}(2+\sqrt{2})}{2}
$$

Answer: $\frac{l^{3} \sqrt{2}}{12}, \frac{l^{2}(2+\sqrt{2})}{2}$.