Editing Rhombicuboctahedron (section)

=== Measurement and metric properties ===
The surface area of a rhombicuboctahedron <math> A </math> can be determined by adding the area of all faces: 8 equilateral triangles and 18 squares. The volume of a rhombicuboctahedron <math> V </math> can be determined by slicing it into two square cupolas and one octagonal prism. Given that the edge length <math> a </math>, its surface area and volume is:{{sfnp|Berman|1971|p=336|loc=See table IV, the Properties of regular-faced convex polyhedra, line 13.}}
<math display="block"> \begin{align}
 A &= \left(18+2\sqrt{3}\right)a^2 &\approx 21.464a^2,\\
 V &= \frac{12+10\sqrt{2}}{3}a^3 &\approx 8.714a^3.
\end{align} </math>

The optimal [[Packing density|packing fraction]] of rhombicuboctahedra is given by
<math display="block"> \eta = \frac{4}{3} \left( 4\sqrt{2} - 5 \right). </math>
It was noticed that this optimal value is obtained in a [[Bravais lattice]] by {{harvtxt|de Graaf|van Roij|Dijkstra|2011}}.{{sfnp|de Graaf|van Roij|Dijkstra|2011}} Since the rhombicuboctahedron is contained in a [[rhombic dodecahedron]] whose [[inscribed sphere]] is identical to its inscribed sphere, the value of the optimal packing fraction is a corollary of the [[Kepler conjecture]]: it can be achieved by putting a rhombicuboctahedron in each cell of the [[rhombic dodecahedral honeycomb]], and it cannot be surpassed, since otherwise the optimal packing density of spheres could be surpassed by putting a sphere in each rhombicuboctahedron of the hypothetical packing which surpasses it.{{cn|date=May 2024}}

The [[dihedral angle]] of a rhombicuboctahedron can be determined by adding the dihedral angle of a square cupola and an octagonal prism:{{sfnp|Johnson|1966}}
* the dihedral angle of a rhombicuboctahedron between two adjacent squares on both the top and bottom is that of a square cupola 135°. The dihedral angle of an octagonal prism between two adjacent squares is the internal angle of a [[regular octagon]] 135°. The dihedral angle between two adjacent squares on the edge where a square cupola is attached to an octagonal prism is the sum of the dihedral angle of a square cupola square-to-octagon and the dihedral angle of an octagonal prism square-to-octagon 45° + 90° = 135°. Therefore, the dihedral angle of a rhombicuboctahedron for every two adjacent squares is 135°.
* the dihedral angle of a rhombicuboctahedron square-to-triangle is that of a square cupola between those 144.7°. The dihedral angle between square-to-triangle, on the edge where a square cupola is attached to an octagonal prism is the sum of the dihedral angle of a square cupola triangle-to-octagon and the dihedral angle of an octagonal prism square-to-octagon 54.7° + 90° = 144.7°. Therefore, the dihedral angle of a rhombicuboctahedron for every square-to-triangle is 144.7°.

A rhombicuboctahedron has the [[Rupert property]], meaning there is a polyhedron with the same or larger size that can pass through its hole.<ref>{{multiref
|{{harvp|Hoffmann|2019}}
|{{harvp|Chai|Yuan|Zamfirescu|2018}}
}}</ref>