My podcast co-host Dave Gale asked an interesting question: why is the volume of a pyramid $V=\frac{1}{3}x^2h$? Dave being Dave, he doesn’t want to mess around with 3D geometry, so I figured I should find an explanation a smart GCSE student would be able to grasp.

I’m going to start with a cube - the explanation extends neatly to a cuboid, but a cube is the easiest place to start. I’m going to draw a line from each corner of the cube to the corner directly opposite, making four lines in all; they all cross at the centre of the cube, so I’m going to say they’re actually eight edges, splitting each long diagonal in half.

You can see, if you squint a bit, that if you take any edge of the cube, there’s a triangle made up of it and the two half-diagonals attached to it - connecting it to the centre. Looking instead at any face of the cube, its four edges are each attached to a triangle, and all of the triangles meet at the centre of the cube, making a pyramid - a square with four triangles attached to it that meet at a point. There are six of these pyramids altogether and - here’s the kicker - they’re all the same volume (because they’re congruent).

So, if the sides of the cube are $x$ units long, the volume of the pyramid is $\frac{x^3}{6}$. But wait - the height of each pyramid is $\frac{x}{2}$, so the volume of the pyramid is indeed $\frac{1}{3}{x^2h}$. So… we’re done, right?

Not quite - I’ve just shown it works for a particular pyramid. Luckily, it’s very easy to extend to any right pyramid (meaning that the centre is above the middle of the square): you just stretch it out. If you enlarge any shape in one dimension by a given scale factor, the volume increases by the same scale factor - so no matter what the height of your right pyramid, the same formula holds.

### But wait… there’s more!

It’s not just cuboids this argument works for. You can also shear the cuboid - meaning you move the top horizontally but leave the base alone. (The shape you end up with is a parallelapiped - a solid with faces made of parallelograms, possibly the best name of all the shapes). And, it turns out, shearing a cuboid doesn’t change the height of its centre, or the volume of the pyramids inside it ((For the pedants: at least two of the ‘pyramids’ have parallelograms for bases now, but the ones attached to the base and the top are still pyramids.)) That means, the top of the pyramid doesn’t have to be above the centre for the formula to work; it can be above a corner or even not above the base. Doesn’t matter: still works.

### Are you done yet?

Nope. Let’s cut our wonky pyramid in half, through the top and two opposite points on the base. That gives us a tetrahedron - and its area is $\frac{1}{3}{Ah}$, where $A$ is the area of its base. That’s a very interesting fact, because you can make any pointy polyhedron - a base with triangles meeting at the top - out of tetrahedra by cutting the base up into triangles and joining the triangles to the top. That means that the volume of any pointy polyhedron is the sum of the volumes of the tetrahedra, which works out to be $\frac{1}{3}{Ah}$, where $A$ is the total area of the base.

Now for the coup de grâce: if the base is a regular polygon, and you keep adding sides, you can get as close as you like to a circle ((I’ll not prove that here)). That means, the polyhedron you get is as close as you like to a cone, which must have the volume of $\frac{1}{3}{Ah}$ because it’s made up of tetrahedra. But the area of the base is $\pi r^2$, so the volume of a cone is $\frac{1}{3}\pi r^2 h$. Ka-pow!