I've read previous answers that state that the volume of a pyramid is $\frac{1}{3}$ (base $\times$ height). One way to visualize the volume of a square-based pyramid is to envision a cube where every surface is the base of a pyramid. There would be 6 congruent pyramids in the cube and therefore, the volume of any one pyramid would be $\frac{1}{6}$ the volume of the cube. Therefore shouldn't the volume of the pyramid be $\frac{1}{6}$ (base $\times$ height), not $\frac{1}{3}$ (base $\times$ height)? Just a little confused and would appreciate any insight.
[Math] Volume of a Square-based Pyramid
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Best Answer
If we call the length of each side of your cube x, then the height of each of the six enclosed pyramids would be x/2. So the volume of each pyramid would be given by:
$$V=\frac{1}{3}\times basearea\times height=\frac{1}{3}\times x^2\times \frac{x}{2}=\frac{x^3}{6}$$
So everything is consistent. :)