Elongated square pyramid
Type	Johnson J7 – J8 – J9
Faces	4 triangles 1+4 squares
Edges	16
Vertices	9
Vertex configuration	${\displaystyle 4\times (4^{3})}$ ${\displaystyle 1\times (3^{4})}$ ${\displaystyle 4\times (3^{2}\times 4^{2})}$
Symmetry group	${\displaystyle C_{4v}}$
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.

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 ${\displaystyle J_{15}}$, the fifteenth Johnson solid.^[4]

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

The elongated square pyramid has the same three-dimensional symmetry group as the equilateral square pyramid, the cyclic group ${\displaystyle C_{4v}}$ 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, ${\displaystyle \arccos(-1/3)\approx 109.47^{\circ }}$,
• The dihedral angle of an elongated square bipyramid between two adjacent squares is the dihedral angle of a cube between those, ${\displaystyle \pi /2=90^{\circ }}$,
• The dihedral angle of an equilateral square pyramid between square and triangle is ${\displaystyle \arctan \left({\sqrt {2}}\right)\approx 54.74^{\circ }}$. 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 $${\displaystyle \arctan \left({\sqrt {2}}\right)+{\frac {\pi }{2}}\approx 144.74^{\circ }.}$$

• Elongated square bipyramid

• Weisstein, Eric W., "Johnson solid" ("Elongated square pyramid") at MathWorld.