Editing Truncated octahedron (section)

=== As an Archimedean solid ===
A truncated octahedron is constructed from a [[regular octahedron]] by cutting off all vertices. This resulting polyhedron has six squares and eight hexagons, leaving out six [[square pyramid]]s. Setting the edge length of the regular octahedron equal to <math> 3a </math>, it follows that the length of each edge of a square pyramid (to be removed) is <math> a </math> (the square pyramid has four [[Equilateral square pyramid|equilateral]] triangles as faces, the first [[Johnson solid]]). From the equilateral square pyramid's property, its volume is <math display="inline"> \tfrac{\sqrt{2}}{6}a^3 </math>. Because six equilateral square pyramids are removed by truncation, the volume of a truncated octahedron <math> V </math> is obtained by subtracting the volume of those six from that of a regular octahedron:{{r|berman}}
<math display="block"> V = \frac{\sqrt{2}}{3} (3a)^3 - 6 \cdot \frac{\sqrt{2}}{6} a^3 = 8\sqrt{2}a^3 \approx 11.3137 a^3. </math>
The surface area of a truncated octahedron <math> A </math> can be obtained by summing all polygonals' area, six squares and eight hexagons. Considering the edge length <math> a </math>, this is:{{r|berman}}
<math display="block"> A = (6 + 12\sqrt{3})a^2 \approx 26.7846a^2. </math>

[[File:Truncated octahedron.stl|thumb|left|3D model of a truncated octahedron]]
The truncated octahedron is one of the thirteen [[Archimedean solid]]s. In other words, it has a highly symmetric and semi-regular polyhedron with two or more different regular polygonal faces that meet in a vertex.{{r|diudea}} The [[dual polyhedron]] of a truncated octahedron is the [[tetrakis hexahedron]]. They both have the same three-dimensional symmetry group as the regular octahedron does, the [[octahedral symmetry]] <math> \mathrm{O}_\mathrm{h} </math>.{{r|kocakoca}} A square and two hexagons surround each of its vertex, denoting its [[vertex figure]] as <math> 4 \cdot 6^2 </math>.{{r|williams}}

The dihedral angle of a truncated octahedron between square-to-hexagon is <math display="inline"> \arccos(-1/\sqrt{3}) \approx 125.26^\circ </math>, and that between adjacent hexagonal faces is <math display="inline"> \arccos (-1/3) \approx 109.47^\circ </math>.{{r|johnson}}

The [[Cartesian coordinates]] of the vertices of a truncated octahedron with edge length 1 are all permutations of<ref>{{cite journal
 | last = Chieh | first = C.
 | bibcode = 1979AcCrA..35..946C
 | date = November 1979
 | doi = 10.1107/s0567739479002114
 | issue = 6
 | journal = Acta Crystallographica Section A
 | pages = 946–952
 | publisher = International Union of Crystallography (IUCr)
 | title = The Archimedean truncated octahedron, and packing of geometric units in cubic crystal structures
 | url = https://scholar.archive.org/work/lfzy6gofjvc55nygiq36od3cwi
 | volume = 35}}</ref>
<math display="block">
\bigl(\pm\sqrt{2}, \pm\tfrac{\sqrt{2}}{2}, 0\bigr).
</math>