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Volume & Surface Area of a Regular Pentagonal Pyramid

Updated on January 22, 2014
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TR Smith is a product designer and former teacher who uses math in her work every day.

Regular Pentagonal Pyramid (source: author)
Regular Pentagonal Pyramid (source: author)

A regular pentagonal pyramid is a solid geometric shape whose base is a regular pentagon. It has six faces: five equal sized isosceles triangles and one regular pentagon. If you know the side length of the pentagon and the length of the edge running from the base to the top vertex of the pyramid, then you can calculate the pyramid's surface area and volume by plugging these two values into two formulas. The equations for volume and surface area are given below, along with some examples.

Volume Formula for a Regular Pentagonal Pyramid

The volume of any pyramid is the area of the base times one third of the height. To compute the volume of a pyramid with a regular pentagon base, we must find the area of the base and the height.

If the side length of the pentagonal base is P and the length of the edge from the base to the apex is L, then the exact height of the pyramid is

H = sqrt[ L^2 - (P^2)/(2cos(54))^2 ]
= sqrt[ L^2 - (P^2)(5 + sqrt(5))/10 ]

The exact area of the base is

A = (P^2)*sqrt(25 +10*sqrt(5))/4

Using these two ingredients, the exact volume of a regular pentagonal pyramid is

V = A*H/3

If you don't want to use the exact values for the height and base area, you can use the simplified equations

H ≈ sqrt(L^2 - 0.7236*P^2)
A ≈ 1.7205*P^2

Surface Area Formula for a Regular Pentagonal Pyramid

The total surface area of a pentagonal pyramid is the sum of the areas of the five triangular faces plus the area of the base. In the case of a pyramid with a regular pentagon base, the five triangles are identical with side lengths L, L, and P. The area of each of these isosceles triangles is

T = P*sqrt(4L^2 - P^2)/4

If you multiply this by 5 and add the area of the pentagonal base, you get the total exact surface area

S = 5P*sqrt(4L^2 - P^2)/4 + (P^2)*sqrt(25 +10*sqrt(5))/4

Example 1

A regular pentagonal pyramid has a base whose sides measure 13.5 cm. The length of the edge from the base to the apex is 21.7 cm. In other words, P = 13.5 and L = 21.7. What is its volume and surface area?

Using the simplified formulas for height and base area we get

H ≈ sqrt(21.7^2 - 0.7236*13.5^2)
≈ sqrt(339.0139)
≈ 18.4123301 cm

A ≈ 1.7205*13.5^2
≈ 313.561125 cm^2

V ≈ (313.561125*18.4123301)/3
≈ 1924.4636 cm^3

Using the formula for the area of the triangular faces, we get

T = 13.5*sqrt(4*21.7^2 - 13.5^2)/4
≈ 139.208420071 cm^2

S ≈ 5*139.208420071 + 313.561125
≈ 1009.6032 cm^2

Example 2

A regular pentagonal pyramid has a volume of 237. If the side length of the base is half the length of the the triangular faces, what are P and L?

This problem tells us that L = 2P. If we plug P and L = 2P into the volume formula, we get

237 = sqrt[4P^2 - (P^2)(5 + sqrt(5))/10]*(P^2)*sqrt(25 +10*sqrt(5))/12
= (P^3)sqrt[ (35 - sqrt(5))(25 + 10*sqrt(5))/10 ] / 12
≈ 1.03806781406 * P^3

Now we can solve for P:

237/1.03806781406 ≈ P^3
228.308783675 ≈ P^3
6.11187138749 ≈ P

This means P is approximately 6.112 and L is approximately 12.224.

Dodecahedron crystal
Dodecahedron crystal

Applications to Dodecahedra

A dodecahedron is a 3-d figure with 12 equal regular pentagonal faces. If you constuct a line segment from each vertex of the dodecahedron to the geometric center point, you can partition the dodecahedron into 12 equal pentagonal pyramids.

The volume of a dodecahedron with an edge length of 1 is exactly (15 + 7*sqrt(5))/4. Therefore, the volume of each of these pentagonal pyramids is (15 + 7*sqrt(5))/48.

The area of the base of one of these pyramids is exactly sqrt(25 + 10*sqrt(5))/4 using the formula for pentagonal area. Using the the fact that the volume of a pyramid equals one third of the base area times the height, we can deduce that the height of one of these pyramids is exactly

(15 + 7*sqrt(5))/4*sqrt(25 + 10*sqrt(5)).


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      WR 10 months ago

      If the triangles are equilateral so that every edge of the pentagonal pyramid has a length of 1, what is the volume?

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