### Exercise:

*Can you find a zonohedron with 20 faces ? *

### Answer:

*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. *

### Further:

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.