The triangular cupola has four
triangles, three
squares, and one
hexagon as its 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 = \left(3+\frac{5\sqrt{3}}{2} \right) a^2 \approx 7.33a^2. Its height h and volume V is: \begin{align} h &= \frac{\sqrt{6}}{3} a\approx 0.82a, \\ V &= \left(\frac{5}{3\sqrt{2}}\right)a^3 \approx 1.18a^3. \end{align} It has an
axis of symmetry passing through the center of its both top and base, which is symmetrical by rotating around it at one- and two-thirds of a full-turn angle. It is also mirror-symmetric relative to any perpendicular plane passing through a bisector of the hexagonal base. Therefore, it has
pyramidal symmetry, the
cyclic group C_{3\mathrm{v}} of order 6. == Related polyhedra ==