💎 Polyhedra 3D Lab — Platonic, Duals & Truncations

Explore the five Platonic solids and their Archimedean cousins in shaded 3D. Morph any solid into its truncation, overlay its dual, and get exact volume, surface area, in/mid/circumradius and dihedral angle for any edge length, plus a live check of Euler's formula V − E + F = 2.

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Shown
V − E + F
Faces
Volume
Surface area
Circumradius R
Midradius ρ
Inradius r
Dihedral angle(s)
Dual
Angular defect (Descartes)
Fills its circumsphere
Sphericity Ψ
Mass
Plato’s element · found in
How this works — why there are only five

A Platonic solid has identical regular faces with the same number meeting at every corner. The angles around a vertex must add to less than 360°, which allows only five: tetrahedron {3,3}, cube {4,3}, octahedron {3,4}, dodecahedron {5,3} and icosahedron {3,5} (Schläfli symbol {face sides, faces per vertex}). Swapping the two numbers gives the dual — put a vertex at each face centre.

Truncation slices every corner off. At the “uniform” point (the slider snaps a gentle marker there) every edge is equal and you get the Archimedean solid — truncated tetrahedron, truncated cube, truncated icosahedron (the football / C₆₀) and so on. At 100 % the cut reaches the edge midpoints and you get the rectified solid: the cuboctahedron or icosidodecahedron.

Every measurement is computed from the actual 3D mesh on screen, not looked up: volume by the divergence theorem (sum of signed tetrahedra from the centre), area from the polygons, radii from the vertex, edge-midpoint and face-plane distances, dihedral angles from neighbouring face normals — then scaled to your edge length. So the numbers stay correct at every truncation level, where edges come in two lengths and “a” refers to the original solid's edge.

Worked example: a regular tetrahedron with a = 1 cm has V = a³/(6√2) = 0.11785 cm³, A = √3 a² = 1.7321 cm², dihedral 70.529°.