If we take a Dodecahedron and remove the constraint that the pentagonal faces have to be regular pentagons, we get a solid called a Tetartoid. If we take the dual of that, we will end up with the corresponding irregular Icosahedron.
Similarly, there seems to be an irregular analogue for the tetrahedron. However, I've never seen one for the cube and octahedron (which happen to be dual of each other). Is it even possible to have irregular analogues of those two? The criterion is that the solid should be convex with all faces identical and the same number of faces meeting at each vertex. The cube analogue would have six faces which were all quadrilaterals while the octahedron analogue would have eight faces which were all scalene triangles.
EDIT: Actually the Icosahedral object might not necessarily be composed of identical triangles.