Can you find a zonohedron with 20 faces ?
Look at the rhombic triacontahedron and choose any one of its six equatorial zones of ten parallelegrams. Remove the zone of faces, and bring the remaining the top and bottom portions together. (They will join neatly; the edges are parallel because the removed faces are parallelegrams.) Removing ten from thirty leaves a zonohedron with twenty faces called the rhombic icosahedron.
You can keep going and remove a zone of eight parallelegrams from the result to get a 12-sided zonohedron. This is sometimes called the rhombic dodecahedron of the second kind. It's faces are golden rhombi (in which the ratio of the lengths of the diagnals is the golden ratio), so it has different face angles than the standard rhombic dodecahedron (in which the ratio of the lengths of the diagonals is the square root of 2.)
Removing a zone of 6 parallelegrams from either rhombic dodecahedron leaves a 6-sided parallelepiped. From the rhombic dodecahedon of the second kind, depending on which zone is removed, the result might be either pointy or flat. (From the ordinary rhombic dodecahedron, the result is always flat.)
Generally, one can reduce any zonohedron by such steps, or build it up by the reverse steps. As a consequence, an n-zone zonohedron can be dissected into n(n-1)(n-2)/6 parallelepipeds, i.e., one for each way to pick three from the n directions. For example, with n=6, the rhombic triacontahedron can be dissected into 20 parallelepipeds.