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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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    See also [my answer](http://math.stackexchange.com/a/633/72) and [Kaestur Hakarl's answer](http://math.stackexchange.com/a/644/72) on a question about the 1/3 in the volume formula for cones, particularly the animation in Kaestur's answer.2012-01-20

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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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    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 $k=1/3$.2012-01-20
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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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    @AméricoTavares Ah! I see it now. Thanks.2012-01-22
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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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    @Ben Crowell: That's what I said, you need a limiting argument of some sort. This has to do with Hilbert's third $p$roblem;$ $see$ $here: htt$p$://en.wiki$p$edia.org/wiki/Hilbert's_third_problem2012-01-20
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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