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I'm interested to know if anyone can point me to a non-calculus way of seeing that the $\text{volume of a pyramid} = \frac{1}{3}\times(\text{area of base})\times(\text{height})$. Yes, I've googled.

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  • $\begingroup$ There are many kinds of pyramids. Do you have any particular type in mind? For example, one might take a convex n-gon and take the convex hull with a point that does not lie in the plane of the n-gon. $\endgroup$ Commented Nov 4, 2010 at 21:51
  • $\begingroup$ @Joseph: He meant the one with a square base I suppose. $\endgroup$ Commented Nov 4, 2010 at 22:58

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I actually find it easiest to take a cube and divide it into 6 pyramids: just connect all of the corners of a cube to the midpoint, and for each of the 6 faces of the cube, you get a pyramid having that face as its base. If the cube has a side length s, then each pyramid has an s by s base and height s/2, and its volume is 1/6 the volume of the cube, which is 1/3 base times height.

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The easiest intuitive one I know is to take a cube and divide it into three pyramids. Take one corner and form the pyramid with each of the three faces that don't touch that corner. They are congruent, so have volume 1/3*base*height as the base is a square the side of the cube and the height is the side of the cube as well. Then appeal to affine transformations to say this applies to all pyramids.

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    $\begingroup$ There is a visual demonstration of this in this SE answer: math.stackexchange.com/questions/623/… $\endgroup$ Commented Nov 4, 2010 at 14:35
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    $\begingroup$ I needed to visualize this so I found a template for a paper model cut it out and now I see it! Thanks. www.ms.uky.edu/algebracubed/lessons/Dissecting_the_Cube_lesson.pdf $\endgroup$ Commented Nov 4, 2010 at 15:08
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    $\begingroup$ Could you expand on "appeal to affine transformations to say this applies to all pyramids"? I did did some reaserch but I could not find a good fit. $\endgroup$ Commented Jul 11, 2020 at 22:58
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    $\begingroup$ @JoaquinBrandan: Affine transformations include translation, rotation, and scaling. You can take any quadrilateral pyramid to any other with one. If you put the base in the $xy$ plane the area of the base will scale with the product of the $x$ and $y$ scales. The height will scale with the $z$ scaling. The volume will scale with the product of the three. This way if we have it for one pyramid we have it for them all. $\endgroup$ Commented Jul 11, 2020 at 23:54
  • $\begingroup$ Hi, can you explain this statement "Take one corner and form the pyramid with each of the three faces that don't touch that corner. "? What do you mean by "with each of the three faces that don't touch that corner. "? Sorry, I cannot picture this in my head. Thanks $\endgroup$ Commented May 8 at 22:46
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Proof:

Put together 6 pyramids to make a cube. You will see that volume of whole cube can be considered as sum of 2 equal smaller volumes (area by half-height of cube). Its also the height of pyramid multiplieds by area. The small volume is also the sum of 3 pyramids, because 3 is half of 6. So single pyramid is 1/3rd of this multiplication.

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    $\begingroup$ How is this different from Gabe's answer? $\endgroup$ Commented Nov 5, 2010 at 0:23
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    $\begingroup$ oh, sorry, did not mean to duplicate it. I did not see someone beat me on it $\endgroup$
    – user3140
    Commented Nov 5, 2010 at 1:18
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Carrying off on Gabe's answer I would like to add a little bit of detail:

Quick summary:

You have 6 pyramids in a cube.

Cube details:

The base of the cube = $\textrm{s} \cdot \textrm{s} = \textrm{s}^2$

The height of the cube = s

The volume of the cube = $\textrm{s} \cdot \textrm{s} \cdot \textrm{s} = \textrm{s}^3$

Square-based pyramid details:

The base of each square pyramid is the same as the base of the cube = $\textrm{s} \cdot \textrm{s} = \textrm{s}^2$

The height of each pyramid is half the height of the cube = $\frac{\textrm{s}}{2}$

The volume of the square-based pyramid is as mentioned 1/6 the cube = $\frac{\textrm{s}^{3}}{6}$

Solving:

Now if we cut the cube in half at the midpoint from the top of the cube we can isolate one of the pyramids inside the half cube.

Volume of half cube = length $\cdot$ width $\cdot$ height = $\textrm{s} \cdot \textrm{s} \cdot \frac{\textrm{s}}{2} = \frac{\textrm{s}^{3}}{2}$

Now to find the volume of the pyramid:

$$\frac{\textrm{s}^{3}}{2} - \frac{\textrm{s}^{3}}{6} = \frac{\textrm{s}^{3}}{3}$$

Hope this helped.

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Here is a visual explanation::

enter image description here

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