How did $1/3$ comes from in the volume of tetrahedron?

Question

The volume of tetrahedron is given by $$\frac{1}{3}(\text{Area of base})(\text{vertical height})$$

Similar formula is applicable for the volume of a cone.

I know that a right circular cone can be enveloped in a right circular cylinder with common base and height but can't still corelated their volumes.

Can someone explain or prove where did this $\frac13$ come in the formula of volume of tetrahedron. Thanks

Do you want a formal proof or an intuitive explanation/geometric argument? — Saha, Feb 26 '22 at 22:12
A similar question: How to calculate the volume of an arbitrary pyramid without calculus? — David K, Feb 26 '22 at 22:36
Does this answer your question? Volume of Pyramid among many others. — Nij, Feb 26 '22 at 23:52

Saha · Answer 1 · 2022-02-26T23:38:27.100

Here is a geometric intuition where a cube is divided into three pyramids with the same base:

And because three of them fit into a cube with the same "height" and "base" it is a third of that Volume ($V=b*h/3$).

This can also be proven using calculus but I think this is rather what you asked for.

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