Elongated triangular cupola

{{Short description|Polyhedron with triangular cupola and hexagonal prism}}

{{Infobox polyhedron

|image=elongated_triangular_cupola.png

|type=Johnson
{{math|gyroelongated square bipyramidJ{{sub|18}}elongated square cupola}}

|faces=4 triangles
9 squares
1 hexagon

|edges=27

|vertices=15

|symmetry={{math|C{{sub|3v}}}}

|vertex_config={{math|6(4{{sup|2}}.6)
3(3.4.3.4)
6(3.4{{sup|3}})}}

|dual=-

|properties=convex

|net=Johnson solid 18 net.png

}}

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.{{r|rajwade}} This cupola covers the hexagonal face so that the resulting polyhedron has four equilateral triangles, nine squares, and one regular hexagon.{{r|berman}} 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} .{{r|francis}}

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:{{r|berman}}

\begin{align}

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{align}

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:{{r|johnson}}

  • 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.

References

{{Reflist|refs=

{{citation

| last = Berman | first = Martin

| year = 1971

| title = Regular-faced convex polyhedra

| journal = Journal of the Franklin Institute

| volume = 291

| issue = 5

| pages = 329–352

| doi = 10.1016/0016-0032(71)90071-8

| mr = 290245

}}.

{{citation|first=Darryl|last=Francis|title=Johnson solids & their acronyms|journal=Word Ways|date=August 2013|volume=46|issue=3|page=177|url=https://go.gale.com/ps/i.do?id=GALE%7CA340298118}}.

{{citation

| last = Johnson | first = Norman W. | authorlink = Norman W. Johnson

| year = 1966

| title = Convex polyhedra with regular faces

| journal = Canadian Journal of Mathematics

| volume = 18

| pages = 169–200

| doi = 10.4153/cjm-1966-021-8

| mr = 0185507

| s2cid = 122006114

| zbl = 0132.14603| doi-access = free

}}.

{{citation

| last = Rajwade | first = A. R.

| title = Convex Polyhedra with Regularity Conditions and Hilbert's Third Problem

| series = Texts and Readings in Mathematics

| year = 2001

| url = https://books.google.com/books?id=afJdDwAAQBAJ&pg=PA84

| publisher = Hindustan Book Agency

| page = 84–89

| isbn = 978-93-86279-06-4

| doi = 10.1007/978-93-86279-06-4

| url-access = subscription

}}.

}}