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Elongated square pyramid
Elongated square pyramid
Type	Johnson; J7 – J8 – J9
Faces	4 triangles; 1+4 squares
Edges	16
Vertices	9
Vertex configuration	; ;
Symmetry group
Dihedral angle (degrees)	triangle-to-triangle: 109.47°; square-to-square: 90°; triangle-to-square: 144.74°;
Properties	convex, composite
Net

In geometry, the elongated square pyramid is a convex polyhedron constructed from a cube by attaching an equilateral square pyramid onto one of its faces. It is an example of Johnson solid.

Construction

The elongated square bipyramid is a composite, since it can constructed by attaching two equilateral square pyramids onto the faces of a cube that are opposite each other, a process known as elongation.^[1]^[2] This construction involves the removal of those two squares and replacing them with those pyramids, resulting in eight equilateral triangles and four squares as their faces.^[3] A convex polyhedron in which all of its faces are regular is a Johnson solid, and the elongated square bipyramid is one of them, denoted as $J_{15}$ , the fifteenth Johnson solid.^[4]

Properties

Given that $a$ is the edge length of an elongated square pyramid. The height of an elongated square pyramid can be calculated by adding the height of an equilateral square pyramid and a cube. The height of a cube is the same as the edge length of a cube's side, and the height of an equilateral square pyramid is $(1/{\sqrt {2}})a$ . Therefore, the height of an elongated square bipyramid is:^[5] $a+{\frac {1}{\sqrt {2}}}a=\left(1+{\frac {\sqrt {2}}{2}}\right)a\approx 1.707a.$ Its surface area can be calculated by adding all the area of four equilateral triangles and four squares:^[3] $\left(5+{\sqrt {3}}\right)a^{2}\approx 6.732a^{2}.$ Its volume is obtained by slicing it into an equilateral square pyramid and a cube, and then adding them:^[3] $\left(1+{\frac {\sqrt {2}}{6}}\right)a^{3}\approx 1.236a^{3}.$

The elongated square pyramid has the same three-dimensional symmetry group as the equilateral square pyramid, the cyclic group $C_{4v}$ of order eight. Its dihedral angle can be obtained by adding the angle of an equilateral square pyramid and a cube:^[6]

The dihedral angle of an elongated square bipyramid between two adjacent triangles is the dihedral angle of an equilateral triangle between its lateral faces, $\arccos(-1/3)\approx 109.47^{\circ }$ ,
The dihedral angle of an elongated square bipyramid between two adjacent squares is the dihedral angle of a cube between those, $\pi /2=90^{\circ }$ ,
The dihedral angle of an equilateral square pyramid between square and triangle is $\arctan \left({\sqrt {2}}\right)\approx 54.74^{\circ }$ . Therefore, the dihedral angle of an elongated square bipyramid between triangle-to-square, on the edge where the equilateral square pyramids attach the cube, is $\arctan \left({\sqrt {2}}\right)+{\frac {\pi }{2}}\approx 144.74^{\circ }.$