Elongated triangular bipyramid

Elongated triangular bipyramid
Type	Johnson J13 – J14 – J15
Faces	6 triangles 3 squares
Edges	15
Vertices	8
Vertex configuration	2(33) 6(32.42)
Symmetry group	D3h, [3,2], (*322)
Rotation group	D3, [3,2]+, (322)
Dual polyhedron	Triangular bifrustum
Properties	convex
Net

In geometry, the elongated triangular bipyramid (or dipyramid) or triakis triangular prism a polyhedron constructed from a triangular prism by attaching two tetrahedrons to its bases. It is an example of Johnson solid.

The elongated triangular bipyramid is constructed from a triangular prism by attaching two tetrahedrons onto its bases, a process known as the elongation.^[1] These tetrahedrons cover the triangular faces so that the resulting polyhedron has nine faces (six of them are equilateral triangles and three of them are squares), fifteen edges, and eight vertices.^[2] A convex polyhedron in which all of the faces are regular polygons is the Johnson solid. The elongated bipyramid is one of them, enumerated as the fourteenth Johnson solid ${\displaystyle J_{14}}$.^[3]

The surface area of an elongated triangular bipyramid ${\displaystyle A}$ is the sum of all polygonal face's area: six equilateral triangles and three squares. The volume of an elongated triangular bipyramid ${\displaystyle V}$ 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 ${\displaystyle h}$ is the sum of two tetrahedrons and a regular triangular prism' height. Therefore, given the edge length ${\displaystyle a}$, its surface area and volume is formulated as:^[2][4] $${\displaystyle {\begin{aligned}A&=\left({\frac {3{\sqrt {3}}}{2}}+3\right)a^{2}\approx 5.598a^{2},\\V&=\left({\frac {\sqrt {2}}{6}}+{\frac {\sqrt {3}}{2}}\right)a^{3}\approx 0.669a^{3},\\h&=\left({\frac {2{\sqrt {6}}}{3}}+1\right)\cdot a\approx \cdot 2.633a.\end{aligned}}}$$

It has the same three-dimensional symmetry group as the triangular prism, the dihedral group ${\displaystyle D_{3\mathrm {h} }}$ 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:^[5]

• the dihedral angle of a tetrahedron between two adjacent triangular faces is ${\textstyle \arccos \left({\frac {1}{3}}\right)\approx 70.5^{\circ }}$;
• the dihedral angle of the triangular prism between the square to its bases is ${\textstyle {\frac {\pi }{2}}=90^{\circ }}$, and the dihedral angle between square-to-triangle, on the edge where tetrahedron and triangular prism are attached, is ${\textstyle \arccos \left({\frac {1}{3}}\right)+{\frac {\pi }{2}}\approx 160.5^{\circ }}$;
• the dihedral angle of the triangular prism between two adjacent square faces is the internal angle of an equilateral triangle ${\textstyle {\frac {\pi }{3}}=60^{\circ }}$.

The nirrosula, an African musical instrument woven out of strips of plant leaves, is made in the form of a series of elongated bipyramids with non-equilateral triangles as the faces of their end caps.^[6]

• Weisstein, Eric W., "Elongated triangular bipyramid" ("Johnson solid") at MathWorld.