Elongated triangular cupola
Type	Johnson J17 – J18 – J19
Faces	4 triangles 9 squares 1 hexagon
Edges	27
Vertices	15
Vertex configuration	6(42.6) 3(3.4.3.4) 6(3.43)
Symmetry group	C3v
Dual polyhedron	-
Properties	convex
Net

In geometry, the elongated triangular cupola is a polyhedron constructed from a hexagonal prism by attaching a triangular cupola. It is an example of a Johnson solid.

The elongated triangular cupola is constructed from a hexagonal prism by attaching a triangular cupola onto one of its bases, a process known as the elongation.^[1] This cupola covers the hexagonal face so that the resulting polyhedron has four equilateral triangles, nine squares, and one regular hexagon.^[2] A convex polyhedron in which all of the faces are regular polygons is the Johnson solid. The elongated triangular cupola is one of them, enumerated as the eighteenth Johnson solid ${\displaystyle J_{18}}$.^[3]

The surface area of an elongated triangular cupola ${\displaystyle A}$ is the sum of all polygonal face's area. The volume of an elongated triangular cupola can be ascertained by dissecting it into a cupola and a hexagonal prism, after which summing their volume. Given the edge length ${\displaystyle a}$, its surface and volume can be formulated as:^[2] $${\displaystyle {\begin{aligned}A&={\frac {18+5{\sqrt {3}}}{2}}a^{2}&\approx 13.330a^{2},\\V&={\frac {5{\sqrt {2}}+9{\sqrt {3}}}{6}}a^{3}&\approx 3.777a^{3}.\end{aligned}}}$$

It has the three-dimensional same symmetry as the triangular cupola, the cyclic group ${\displaystyle C_{3\mathrm {v} }}$ of order 6. Its dihedral angle can be calculated by adding the angle of a triangular cupola and a hexagonal prism:^[4]

• the dihedral angle of an elongated triangular cupola between square-to-triangle is that of a triangular cupola between those: 125.3°;
• the dihedral angle of an elongated triangular cupola between two adjacent squares is that of a hexagonal prism, the internal angle of its base 120°;
• the dihedral angle of a hexagonal prism between square-to-hexagon is 90°, that of a triangular cupola between square-to-hexagon is 54.7°, and that of a triangular cupola between triangle-to-hexagonal is an 70.5°. Therefore, the elongated triangular cupola between square-to-square and triangle-to-square, on the edge where a triangular cupola is attached to a hexagonal prism, is 90° + 54.7° = 144.7° and 90° + 70.5° = 166.5° respectively.

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