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Square Pyramid


J01SquarePyramidJ01Net

A square pyramid is a pyramid with a square base. It is a pentahedron.

The lateral edge length e and slant height s of a right square pyramid of side length a and height h are

e=sqrt(h^2+1/2a^2)
(1)
s=sqrt(h^2+1/4a^2).
(2)

The corresponding surface area and volume are

S=a(a+sqrt(a^2+4h^2))
(3)
V=1/3a^2h.
(4)
SquarePyramidCube

The volume of a square pyramid in the special case h=a/2 can be found immediately from the cube dissection illustrated above, giving

 V=1/6a^3.
(5)

If the four triangles of the square pyramid are equilateral, so that all edges of the square pyramid have the same lengths, then the right square pyramid is the polyhedron known as Johnson solid J_1.

The square pyramid J_1 of edge length a has height

 h=1/2sqrt(2)a,
(6)

and so the lateral edge length and slant height are

e=a
(7)
s=1/2sqrt(3)a.
(8)

The surface area and volume are therefore

S=(1+sqrt(3))a^2
(9)
V=1/6sqrt(2)a^3.
(10)
PyramidSphere

Consider a hemisphere placed on the base of a square pyramid (having side lengths a and height h). Further, let the hemisphere be tangent to the four apex edges. Then what is the volume of the hemisphere that is interior the pyramid (Cipra 1993)?

From Fig. (a), the circumradius of the base is a/sqrt(2). Now find h in terms of r and a. Fig. (b) shows a cross section cut by the plane through the pyramid's apex, one of the base's vertices, and the base center. This figure gives

b=sqrt(1/2a^2-r^2)
(11)
c=sqrt(h^2-r^2),
(12)

so the slant height is

 s=sqrt(h^2+1/2a^2)=b+c=sqrt(1/2a^2-r^2)+sqrt(h^2-r^2).
(13)

Solving for h gives

 h=(ra)/(sqrt(a^2-2r^2)).
(14)

We know, however, that the hemisphere must be tangent to the sides, so r=a/2, and

 h=(1/2a)/(sqrt(a^2-1/2a^2))a=(1/2)/(sqrt(1/2))a=1/2sqrt(2)a.
(15)

Fig. (c) shows a cross section through the center, apex, and midpoints of opposite sides. The Pythagorean theorem once again gives

 l=sqrt(1/4a^2+h^2)=sqrt(1/4a^2+1/2a^2)=1/2sqrt(3)a.
(16)

We now need to find x and y.

 sqrt(1/4a^2-x^2)+d=l.
(17)

But we know l and h, and d is given by

 d=sqrt(h^2-x^2),
(18)

so

 sqrt(1/4a^2-x^2)+sqrt(1/2a^2-x^2)=1/2sqrt(3)a.
(19)

Solving gives

 x=1/6sqrt(6)a,
(20)

so

 y=sqrt(r^2-x^2)=a/(2sqrt(3)).
(21)

We can now find the volume of the spherical cap as

 V_(cap)=1/6piH(3A^2+H^2),
(22)

where

A=y=a/(2sqrt(3))
(23)
H=r-x=a(1/2-1/(sqrt(6))),
(24)

so

 V_(cap)=1/6pia^3(1/2-7/(6sqrt(6))).
(25)

Therefore, the volume within the pyramid is

V_(inside)=2/3pir^3-4V_(cap)
(26)
=pia^3(7/(9sqrt(6))-1/4).
(27)

This problem appeared in the Japanese scholastic aptitude test (Cipra 1993).


See also

Johnson Solid, Pentahedron, Pentagonal Pyramid, Pyramid, Spherical Cap, Square Pyramidal Number, Triangular Pyramid

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References

Cipra, B. "An Awesome Look at Japan Math SAT." Science 259, 22, 1993.

Cite this as:

Weisstein, Eric W. "Square Pyramid." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/SquarePyramid.html

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