## Tuesday, September 29, 2020

### Proof of the volume of a pyramid and cone formulae

Let's take a square based pyramid, one which is arranged such that it fits in the corner of a cube like this:
The volume of this kind of pyramid is a third of that of the cube. This is because 3 such pyramids will perfectly fit in the cube, as shown below:
In fact you don't even need the pyramid to fit in a cube, it can fit in a cuboid instead. Three such rectangular based pyramids will always exactly fit the cuboid. Now the next natural question is about pyramids that are arranged differently, such as when the top point is in the center of the base, as shown below:
You will notice that no matter how the pyramid is arranged, it always consists of an infinite number of stacked squares, starting from the square at the base of the pyramid and getting progressively smaller until the top point is reached. Since the base and top point of both pyramids are exactly the same, the only thing that is changing is where these square slices are horizontally. So the same set of squares in the orange pyramid are used in the green pyramid with the only difference being where they are placed. This means that the same volume is being used by both pyramids so the same formula is used for both. Therefore, the formula for the volume of a rectangular based pyramid is: $$V = \frac{1}{3}hA$$ where $h$ is the height of the pyramid and $A$ is the area of the base. Now what about when the base is not a square? Will a circle work? A circular based pyramid is a cone:
The cone is exactly the same as the square based pyramid except that instead of consisting of a stack of squares consists of a stack of circles (discs, to be exact). Now if the circles fit exactly inside those of a square based pyramid, then the area of the circle is going to be a fraction of the area of the square:
In terms of the radius of the circle ($r$), that fraction is: $$\frac{\pi r^2}{(2r)^2} = \frac{\pi r^2}{4r^2} = \frac{\pi}{4}$$ Now keeping in mind that this fraction will always be the same for any square in the stack of squares forming the pyramid (from the base to the top point), this means that the volume of the cone must be the same fraction of the volume of the pyramid. Therefore: $$V = \frac{\pi}{4}\frac{1}{3}h(2r)^2 = \frac{1}{3}h\pi r^2$$ In fact, the equation did not change from the previous one because it's still using the base area except that the base area is now of a circle instead of a rectangle. What about a trangular based pyramid (tetrahedron)? Same concept. The stack is made of triangles instead of rectangles or circles:
The fraction of the area occupied by the triangle over the area occupied by the square is half: $$\frac{l^2/2}{l^2} = \frac{1}{2}$$ where $l$ is the side length of the square. So now the volume will be half that of the square based pyramid: $$V = \frac{1}{2}\frac{1}{3}h(l)^2 = \frac{1}{3}h\frac{l^2}{2} = \frac{1}{6}hl^2$$ Again we see that the base area equation still holds and must hold for any shaped base pyramid (where the base shape is linearly reduced to a point along the height of the pyramid).