[Example 4.4.3] Three $12 \mathrm{~cm} \times 12 \mathrm{~cm}$ squares are each divided into two pieces, $A$ and $B$, by a line connecting the midpoints of two adjacent sides, as shown in Figure 4-7. The six pieces are then glued to the outside of a regular hexagon, as shown in Figure 4-8, and folded into a polyhedron. Find the volume of this polyhedron.
The polyhedron folded as shown in Figure 4-9. Observing this figure, we can see that the three dihedral angles at the vertices marked $P, Q, R, S$ are all $90^{\circ}$. This indicates that the polyhedron is part of a cube, as shown in Figure 4-10, i.e., by completing the polyhedron in Figure 4-9 to form a cube with an edge length of 12 cm. A cross-section through the midpoints of the 6 relevant edges of the cube in Figure 4-10 yields the polyhedron in Figure 4-9. Therefore, the volume of the polyhedron is half the volume of the cube in Figure 4-10.
$$
V=\frac{1}{2} \cdot 12^{3}=864\left(\mathrm{~cm}^{3}\right)
$$

A key step in solving this problem is to complete the irregular polyhedron into a cube.