11.013. The center of the upper base of a cube with an edge equal to $a$ is connected to the midpoints of the sides of the lower base, which are also connected in sequential order. Calculate the total surface area of the resulting pyramid.
## Solution.

Since the edge of the cube is $a$, the side of the base of the pyramid $S A B C D$ is $\frac{a \sqrt{2}}{2}$ (Fig. 11.11). Considering that $O K=\frac{1}{2} A D=\frac{a \sqrt{2}}{4}$, we find the slant height of the pyramid: $S K=\sqrt{S O^{2}+O K^{2}}=\sqrt{a^{2}+\frac{a^{2}}{8}}=\frac{3 a \sqrt{2}}{4}$. Therefore, $S_{\text {lateral }}=4 \cdot \frac{1}{2} \cdot \frac{a \sqrt{2}}{2} \cdot \frac{3 a \sqrt{2}}{4}=\frac{3 a^{2}}{2}, S_{\text {total }}=\left(\frac{a \sqrt{2}}{2}\right)^{2}+S_{\text {lateral }}=2 a^{2}$. Answer: $2 a^{2}$.