11.010. In a cube with an edge length of $a$, the center of the upper face is connected to the vertices of the base. Find the total surface area of the resulting pyramid.
## Solution.

From the condition, it is clear that the height of the pyramid is equal to the height of the cube, i.e., $a$ (Fig. 11.9). $K$ is the midpoint of $AD$, $HH_{1}$ is the height. Then $HK=\frac{a}{2}$; $H_{1}K=\sqrt{HH_{1}^{2}+HK^{2}}=\sqrt{a^{2}+\left(\frac{a}{2}\right)^{2}}=\frac{a \sqrt{5}}{2}$. The semi-perimeter of the base $p=\frac{1}{2} \cdot 4a=2a$. The lateral surface area $S_{\text {lat }}=H_{1}K \cdot p=\frac{a \sqrt{5}}{2} \cdot 2a=a^{2} \sqrt{5}$. The base area $S_{\text {base }}=a^{2}$. The total surface area

$$
S=S_{\text {lat }}+S_{\text {base }}=a^{2}+a^{2} \sqrt{5}=a^{2}(1+\sqrt{5})
$$

Answer: $\quad a^{2}(1+\sqrt{5})$.