| 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 . [3]
The surface area of an elongated triangular cupola 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 , its surface and volume can be formulated as: [2]
It has the three-dimensional same symmetry as the triangular cupola, the cyclic group of order 6. Its dihedral angle can be calculated by adding the angle of a triangular cupola and a hexagonal prism: [4]