26. The center of the cube is reflected off all its faces, and the resulting 6 points together with the vertices of the cube determine 14 vertices of a polyhedron, all 12 faces of which are congruent rhombuses. (This polyhedron is called a rhombic dodecahedron—natural monocrystals of garnet have this shape.) Calculate the surface area and volume of the constructed polyhedron if the edge of the original cube has length $a$.
26. The twelve faces of the rhombic dodecahedron (Fig. 100) are rhombuses, one diagonal of which is equal to $a$, and the other is $-a \sqrt{2}$ (why?). The area of each rhombus is $a^{2} \frac{\sqrt{2}}{2}$,

![](https://cdn.mathpix.com/cropped/2024_05_21_5cdadc99c18e78f1502ag-167.jpg?height=503&width=494&top_left_y=1873&top_left_x=358)

Fig. 100 and the surface area of the entire dodecahedron is $6 a^{2} \sqrt{2}$. Furthermore, since each of the 6 pyramids constructed on the faces of the cube is symmetric to the pyramid with the same base and vertex at the center of the cube, the sum of the volumes of these six pyramids is equal to the volume of the cube, and the volume of the dodecahedron is twice the volume of the cube, i.e., equal to $2 a^{3}$.