Triangular cupola
Type	Johnson J2 – J3 – J4
Faces	4 triangles 3 squares 1 hexagon
Edges	15
Vertices	9
Vertex configuration	${\displaystyle {\begin{aligned}&6\times (3\times 4\times 6)\,+\\&3\times (3\times 4\times 3\times 4)\end{aligned}}}$
Symmetry group	${\displaystyle C_{3v}}$
Properties	convex
Net

In geometry, the triangular cupola is the cupola with hexagon as its base and triangle as its top. If the edges are equal in length, the triangular cupola is the Johnson solid. It can be seen as half a cuboctahedron. The triangular cupola can be applied to construct many polyhedrons.

The triangular cupola has 4 triangles, 3 squares, and 1 hexagon as their faces; the hexagon is the base and one of the four triangles is the top. If all of the edges are equal in length, the triangles and the hexagon becomes regular.^[1][2] The dihedral angle between each triangle and the hexagon is approximately 70.5°, that between each square and the hexagon is 54.7°, and that between square and triangle is 125.3°.^[3] A convex polyhedron in which all of the faces are regular is a Johnson solid, and the triangular cupola is among them, enumerated as the third Johnson solid ${\displaystyle J_{3}}$.^[2]

Given that ${\displaystyle a}$ is the edge length of a triangular cupola. Its surface area ${\displaystyle A}$ can be calculated by adding the area of four equilateral triangles, three squares, and one hexagon:^[1] $${\displaystyle A=\left(3+{\frac {5{\sqrt {3}}}{2}}\right)a^{2}\approx 7.33a^{2}.}$$ Its height ${\displaystyle h}$ and volume ${\displaystyle V}$ is:^[4][1] $${\displaystyle {\begin{aligned}h&={\frac {\sqrt {6}}{3}}a\approx 0.82a,\\V&=\left({\frac {5}{3{\sqrt {2}}}}\right)a^{3}\approx 1.18a^{3}.\end{aligned}}}$$

It has an axis of symmetry passing through the center of its both top and base, which is symmetrical by rotating around it at one- and two-thirds of a full-turn angle. It is also mirror-symmetric relative to any perpendicular plane passing through a bisector of the hexagonal base. Therefore, it has pyramidal symmetry, the cyclic group ${\displaystyle C_{3\mathrm {v} }}$ of order 6.^[3]

The triangular cupola can be found in the construction of many polyhedrons. An example is the cuboctahedron in which the triangular cupola may be considered as its hemisphere.^[5] A construction that involves the attachment of its base to another polyhedron is known as augmentation; attaching it to prisms or antiprisms is known as elongation or gyroelongation.^[6][7] Some of the other Johnson solids constructed in such a way are elongated triangular cupola ${\displaystyle J_{18}}$, gyroelongated triangular cupola ${\displaystyle J_{22}}$, triangular orthobicupola ${\displaystyle J_{27}}$, elongated triangular orthobicupola ${\displaystyle J_{35}}$, elongated triangular gyrobicupola ${\displaystyle J_{36}}$, gyroelongated triangular bicupola ${\displaystyle J_{44}}$, augmented truncated tetrahedron ${\displaystyle J_{65}}$.^[8]

The triangular cupola may also be applied in constructing truncated tetrahedron, although it leaves some hollows and a regular tetrahedron as its interior. Cundy (1956) constructed such polyhedron in a similar way as the rhombic dodecahedron constructed by attaching six square pyramids outwards, each of which apices are in the cube's center. That being said, such truncated tetrahedron is constructed by attaching four triangular cupolas rectangle-by-rectangle; those cupolas in which the alternating sides of both right isosceles triangle and rectangle have the edges in terms of ratio ${\textstyle 1:{\frac {1}{2}}{\sqrt {2}}}$. The truncated octahedron can be constructed by attaching eight of those same triangular cupolas triangle-by-triangle.^[9]

• Weisstein, Eric W., "Triangular cupola" ("Johnson solid") at MathWorld.