Given the edge length a . The surface area of a truncated tetrahedron A is the sum of 4 regular hexagons and 4 equilateral triangles' area, and its
volume V is: \begin{align} A &= 7\sqrt{3}a^2 &&\approx 12.124a^2, \\ V &= \tfrac{23}{12}\sqrt{2}a^3 &&\approx 2.711a^3. \end{align} The dihedral angle of a truncated tetrahedron between triangle-to-hexagon is approximately 109.47°, and that between adjacent hexagonal faces is approximately 70.53°. The densest packing of the truncated tetrahedron is believed to be \Phi = \frac{207}{208} , as reported by two independent groups using
Monte Carlo methods by and . Although no mathematical proof exists that this is the best possible packing for the truncated tetrahedron, the high proximity to the unity and independence of the findings make it unlikely that an even denser packing is to be found. If the truncation of the corners is slightly smaller than that of a truncated tetrahedron, this new shape can be used to fill space completely. The truncated tetrahedron is an
Archimedean solid, meaning it is a highly symmetric and semi-regular polyhedron, and two or more different
regular polygonal faces meet in a vertex. The truncated tetrahedron has the same
three-dimensional group symmetry as the regular tetrahedron, the
tetrahedral symmetry \mathrm{T}_\mathrm{h} . The polygonal faces that meet for every vertex are one equilateral triangle and two regular hexagons, and the
vertex figure is denoted as 3 \cdot 6^2 . Its
dual polyhedron is
triakis tetrahedron, a
Catalan solid, shares the same symmetry as the truncated tetrahedron. == Related polyhedra ==