11.128. A cube is inscribed in a regular quadrilateral pyramid such that four of its vertices are on the apothems of the pyramid and four are in the plane of the base. All edges of the pyramid are equal, each having a length of $a$. Calculate the total surface area and volume of the cube.
Solution.

Let the edge of the cube $E F=x$ (Fig. 11.24). Consider $\triangle B S C$ (equilateral $): S E=\frac{a \sqrt{3}}{2} \cdot$ From $\triangle S O E\left(\angle S O E=90^{\circ}\right)$ :

$$
S O=\sqrt{S E^{2}-O E^{2}}=\sqrt{\frac{3 a^{2}}{4}-\frac{a^{2}}{4}}=\frac{a \sqrt{2}}{2} ;
$$

$F E=O E-O F=\frac{a}{2}-\frac{x \sqrt{2}}{2} \cdot \Delta E_{1} F E \sim \Delta S O E$.

Then $\frac{E_{1} F}{S O}=\frac{F E}{O E} ; E_{1} F \cdot O E=S O \cdot F E, x \cdot \frac{a}{2}=\frac{a \sqrt{2}}{2}\left(\frac{a}{2}-\frac{x \sqrt{2}}{2}\right) x=\frac{a \sqrt{2}}{4}$. The total surface area of the cube $S=6 x^{2}=\frac{3}{4} a^{2}$, its volume $\quad V=x^{3}=\frac{a^{3} \sqrt{2}}{32}$.

Answer: $\frac{3}{4} a^{2} ; \frac{a^{3} \sqrt{2}}{32}$.