Editing Elongated triangular bipyramid (section)

== Properties ==
[[File:3D Johnson J14.stl|thumb|3D model of an elongated triangular bipyramid]]
The surface area of an elongated triangular bipyramid <math> A </math> is the sum of all polygonal face's area: six equilateral triangles and three squares. The volume of an elongated triangular bipyramid <math> V </math> can be ascertained by slicing it off into two tetrahedrons and a regular triangular prism and then adding their volume. The height of an elongated triangular bipyramid <math> h </math> is the sum of two tetrahedrons and a regular triangular prism' height. Therefore, given the edge length <math> a </math>, its surface area and volume is formulated as:{{r|berman|pye}}
<math display="block"> \begin{align}
 A &= \left(\frac{3\sqrt{3}}{2} + 3\right)a^2 \approx 5.598a^2, \\
 V &= \left(\frac{\sqrt{2}}{6} + \frac{\sqrt{3}}{2} \right) a^3 \approx 0.669a^3, \\
 h &= \left(\frac{2\sqrt{6}}{3} + 1 \right)\cdot a \approx \cdot 2.633a.
\end{align}
</math>

It has the same [[Point groups in three dimensions|three-dimensional symmetry group]] as the triangular prism, the [[dihedral group]] <math> D_{3 \mathrm{h}} </math> of order twelve. The [[dihedral angle]] of an elongated triangular bipyramid can be calculated by adding the angle of the tetrahedron and the triangular prism:{{r|johnson}}
* the dihedral angle of a tetrahedron between two adjacent triangular faces is <math display="inline"> \arccos \left(\frac{1}{3}\right) \approx 70.5^\circ </math>;
* the dihedral angle of the triangular prism between the square to its bases is <math display="inline"> \frac{\pi}{2} = 90^\circ </math>, and the dihedral angle between square-to-triangle, on the edge where tetrahedron and triangular prism are attached, is <math display="inline"> \arccos \left(\frac{1}{3}\right) + \frac{\pi}{2} \approx 160.5^\circ </math>;
* the dihedral angle of the triangular prism between two adjacent square faces is the internal angle of an equilateral triangle <math display="inline"> \frac{\pi}{3} = 60^\circ </math>.