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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.

Martin Sleziak
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Sol Lederman
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4 Answers4

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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.

Gabe Cunningham
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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.

Ross Millikan
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    There is a visual demonstration of this in this SE answer: http://math.stackexchange.com/questions/623/why-is-the-volume-of-a-cone-one-third-of-the-volume-of-a-cylinder – Derek Jennings Nov 04 '10 at 14:35
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    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 – Sol Lederman Nov 04 '10 at 15:08
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    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. – Joaquin Brandan Jul 11 '20 at 22:58
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    @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. – Ross Millikan Jul 11 '20 at 23:54
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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.

dfeuer
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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.

Ayman Omar
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