Elongated triangular cupola

Short description: 18th Johnson solid (14 faces)

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.

Construction

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. This cupola covers the hexagonal face so that the resulting polyhedron has four equilateral triangles, nine squares, and one regular hexagon. 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 $J_{18}$.

Properties

The surface area of an elongated triangular cupola $A$ 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 $a$, its surface and volume can be formulated as: $$\begin{array}{rlr}A& =\frac{18+5\sqrt{3}}{2}{a}^2& \approx 13.330a^2,\\ V& =\frac{5\sqrt{2}+9\sqrt{3}}{6}{a}^3& \approx 3.777a^3.\end{array}$$

File:J18 elongated triangular cupola.stl It has the three-dimensional same symmetry as the triangular cupola, the cyclic group $C_{3\mathrm{v}}$ of order 6. Its dihedral angle can be calculated by adding the angle of a triangular cupola and a hexagonal prism:

• 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.

External links

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