Triangular cupola explained

Type:	Johnson
Edges:	15
Vertices:	9
Symmetry:	C _{3v}
Vertex Config:	\begin{align} & 6 x ( 3 x 4 x 6 ) + \\ & 3 x ( 3 x 4 x 3 x 4 ) \end{align}
Net:	Triangular cupola (symmetric net).svg

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.

Properties

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. 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°. 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

J _{3} .

Given that

a is the edge length of a triangular cupola. Its surface area A can be calculated by adding the area of four equilateral triangles, three squares, and one hexagon:$$A\; =\; \backslash left(3+\backslash frac\; \backslash right)\; a^2\; \backslash approx\; 7.33a^2.$$Its height h and volume V is:C _{3v} of order 6.

Related polyhedra

J _{18} , gyroelongated triangular cupola J _{22} , triangular orthobicupola J _{27} , elongated triangular orthobicupola J _{35} , elongated triangular gyrobicupola J _{36} , gyroelongated triangular bicupola J _{44} , augmented truncated tetrahedron J _{65} .

The triangular cupola may also be applied in constructing truncated tetrahedron, although it leaves some hollows and a regular tetrahedron as its interior. 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