11.039. In a cube, the centers of the bases are connected to the centers of the side faces. Calculate the surface area of the resulting octahedron if the edge of the cube is $a$.
Solution.

$$
S=8 \cdot S_{\triangle B E C}, \text { where } C E=\sqrt{\frac{a^{2}}{4}+\frac{a^{2}}{4}}=\frac{a \sqrt{2}}{2} ; C E=B E=B C \text { (Fig. 11.34), }
$$

therefore $\angle B E C=\angle E C B=\angle C B E=60^{\circ}$ and

$$
S_{\triangle B E C}=\frac{1}{2} \frac{a \sqrt{2}}{2} \cdot \frac{a \sqrt{2}}{2} \sin 60^{\circ}=\frac{a^{2} \sqrt{3}}{8}, S=8 \cdot \frac{a^{2} \sqrt{3}}{8}=a^{2} \sqrt{3}
$$

Answer: $a^{2} \sqrt{3}$.