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To calculate the volume of a pyramid (not tetrahedron!) you've to use the formula $\frac{1}{3}B\cdot H,$ where $B$ is the area of the base and $H$ is the height.

My question is: why 1/3? Is a pyramid one-third of a cuboid?

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volume of pyramid –  pedja Jan 20 '12 at 9:51
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See also my answer and Kaestur Hakarl's answer on a question about the 1/3 in the volume formula for cones, particularly the animation in Kaestur's answer. –  Isaac Jan 20 '12 at 13:50
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4 Answers 4

While using calculus to derive this is a bit heavy-handed, consider this: $V= \int A(h) \,\mathrm{d}h$, where $A$ is the area of a cross-section parallel to the base, but at distance $h$ from the apex. Since the length of the sides of the cross-section grows $\propto h$, $A(h)\propto h^2$. But $B=A(H)$, so $A(h)=\frac{h^2}{H^2}B$.

Evaluate the integral: $$ V= \int_0^H A(h)\,\mathrm{d}h = \int_0^H B \frac{h^2}{H^2}\,\mathrm{d}h = \frac{1}{3}BH$$

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Dear Ansgar: +1! Nice answer. It shows that the factor $1/3$ is due to the fact that we have a cone, and has nothing to do with the shape of the base. This is also explained in this Wikipedia entry. –  Pierre-Yves Gaillard Jan 20 '12 at 12:33
    
Although the OP may not know calculus, a simpler and rougher version of this argument can be given without calculus. Suppose we build a pyramid by putting 1 cubical block at the top, a 2x2 square of such blocks underneath that, then 3x3, and so on, down to the base which is $n\times n$. Then the volume is $V_n=1^2+2^2+3^2+\ldots+n^2$. We make a guess that $V_n$ may be a polynomial. In that case, we must have $V_n-V_{n-1}=n^2$. The only way this can happen is if $V_n$ is a third-order polynomial $V_n=kn^3+\ldots$, wher $k$ is unknown. Requiring $kn^3-k(n-1)^3=n^2+\ldots$ gives $k=1/3$. –  Ben Crowell Jan 20 '12 at 17:18
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The difficult part is to show that the volume is given by such a simple formula involving only the area of the base and the height. All proofs of this formula have to use a limiting argument of some sort.

As for the factor ${1\over3}$, this is easy: A cube is the union of 6 congruent pyramids with a face as base and the midpoint of the cube as peak.

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Sweet and simple! I think the generalization to other values of $H^2/B$ follows straightforwardly from simple geometrical scaling, and the generalization to skewed pyramids is pretty obvious if you imagine slicing the pyramid into many thin slices parallel to the base. The fact that the base doesn't have to be a square follows because you can subdivide any base into many small squares, each one forming the base of a pyramid with the same apex. –  Ben Crowell Jan 20 '12 at 17:05
    
@Ben Crowell: That's what I said, you need a limiting argument of some sort. This has to do with Hilbert's third problem; see here: en.wikipedia.org/wiki/Hilbert's_third_problem –  Christian Blatter Jan 20 '12 at 19:04
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Consider a triangular prism. It can be divided into three equivalent pyramids, and so the volume of a triangular pyramid is $1/3$ of the volume of a triangular prism with the same base and height (see figure on the left). A pyramid whose base has $n$ sides may be divided into $n-2$ tetrahedrons. It follows that its volume is $1/3$ of the volume of a prism with the same base and height (see figure on the right).

enter image description here

Explanation concerning the sketch on the left in response to Byron Schmuland's comment. Removing the front pyramid we are left with two equivalent pyramids (one on the left and the other on the right) whose height is the distance from the upper front corner to the prism back face and whose bases are the two halves of the back face. On the other hand the front pyramid is equivalent to the right pyramid (equal bases and equal heights).

Reference: F. G. - M., §496 and §497 of Cours de Géométrie Élémentaire, 1917.

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This is Euclid XII-6 and XII-7: aleph0.clarku.edu/~djoyce/java/elements/bookXII/propXII6.html –  Ben Crowell Jan 20 '12 at 17:09
    
@BenCrowell: This method is described in the reference. –  Américo Tavares Jan 20 '12 at 17:22
    
@AméricoTavares I'm having some difficulty in seeing that the middle pyramid on the left has the same volume as the other two. Is there a geometrical explanation that I'm missing? –  Byron Schmuland Jan 22 '12 at 0:21
    
@Byron Schmuland:If you remove the front pyramid you are left with two equivalent pyramids (one on the left and the other on the right) whose height is the distance from the upper front corner to the prism back face and whose bases are the two halves of the back face. On the other hand the front pyramid is equivalent to the right pyramid (equal bases and equal heights). –  Américo Tavares Jan 22 '12 at 2:03
    
@AméricoTavares Ah! I see it now. Thanks. –  Byron Schmuland Jan 22 '12 at 2:09
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The volume of a cuboid is $B\cdot H$. So, yes, a pyramid is a third of a cuboid, since $3(\frac{1}{3}B\cdot H)=B\cdot H$.

A way to see this is: http://www.korthalsaltes.com/model.php?name_en=three%20pyramids%20that%20form%20a%20cube

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