As one knows (for example by reading Euclid's Elements) proving the formula for the volume of a (generic) pyramid is tricky. Basically, the proof involves a decomposition into infinitely many pieces or to know the effect of linear transformations on the volume (which again relies on decomposing in infinitely many pieces) or the principle of Cavallieri (which again...).
However there are a few pyramids for which the volume is computable with finitely many cuts from another known body (polygonal prisma ).
On the other hand it is known (see the wiki artile on the Dehn invariant) that some pyramids cannot be dissected back to a prisma.
Question: apart from the examples below, which volumes of pyramids can be computed starting only with the volume of (polygonal) prisma and using finitely many dissections and gluing?
Here are the examples:
a- start with a cube. Form six pyramids, each of which has a face of the cube as a basis and the top at the center of the cube. Since they have equal volume, you can deduce the volume of those pyramids.
b- start with a cube. slice it into 3 pyramids
c- start with the pyramid obtained in b. You can put two, three or four of them together to make a new pyramid.
d- start again with the pyramid obtained in b. You can slice it again in two isometric pyramids (whose basis are right triangles)
e- start with the pyramid obtained in d and put two or three of them together to make a new pyramid. If you put four together, you get back the pyramid from a.
f- glue together some of the pyramids obtained in d and b
g- you can start with a trigonal trapezohedron whose faces are all rhombic. Now this is not a prism, but the shear (w.r.t. the basis) of a prism (with basis a rhombus). Shear of prisms can be cut into pieces to get back the original prism. So we know the volume of that guy. Because of his symmetry group you can cut it into three pieces (like b).
images taken from http://images.math.cnrs.fr/Aires-et-volumes-decoupage-et,848.html?lang=fr