geometrypolyhedraterminology
What is the name of this convex polyhedron?
$(V,E,F)=(14,36,24)$.
The top and bottom vertices are degree-$6$, spanning
hexagons, which are zigzag connected in the band between the
two hexagons. The faces are approximate isosceles triangles
in this physical model.
I believe this is what's referred to as a Cuboctahedron-Rhombic Dodecahedron Compound.
According to Mathworld on compound polyhedron:
A polyhedron compound is an arrangement of a number of interpenetrating polyhedra, either all the same or of several distinct types, usually having visually attractive symmetric properties. Compounds of multiple Platonic and Archimedean solids can be especially attractive, as can compounds of these solids and their duals. [Note, the two intersecting figures are in fact duals.] [Italics, boldface, and brackets mine.]
See Mathworld on Cuboctahedron-Rhombic Dodecahedron Compound, and in particular, I believe the figure at very bottom, left: the Cuboctahedron-Rhombic Dodecahedron Compound- depicts the figure you've provided above. It's a compound of two (dual) Archimedean polyhedra: the cuboctahedron (pictured immediately below, to left) and the rhombic dodecahedron (image immediately below, to right).
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EDIT: and actually, I think the figure below (right) is what you're after: it is described as the "Cuboctahedron-Rhombic Dodecahedron solid which is common to both polyhedra"; i.e. their intersection. Each figure (above) is the dual of the other. No specific name is given to the convex figure of intersection (below, to right), save for referring to it as a "Cuboctahedron-Rhombic Dodecahedron convex solid."
The page linked is worth a visit, since you can see the figures below, rotate them, and the page provides additional information on the polyhedra, the surface area, volume, etc. (edges, vertices,...). There are also links to other polyhedral compounds.
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Let $k$ be the number of edges per face; then the total number of edges is $e=\frac{kr}{2}$ because every edge is adjacent to two faces. We know $r=12$ and $e=30$, so $k=...$
Best Answer
I believe it is a gyroelongated hexagonal bipyramid.
https://en.wikipedia.org/wiki/Gyroelongated_bipyramid
The example shown in the Wikipedia article uses equilateral triangles, which results in coplanar faces around the degree-6 vertices, but as you mention, the model uses isosceles triangles.