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"Inradius" means radius of largest sphere that is tangent to all faces.

For example:

Cube - Surface area $= 6a^2$, Inradius $= a/2$, Volume $= a^3$.

Sphere - Surface area $= 4\pi r^2$, Inradius $= r$, Volume $= 4/3 \pi r^3$.

I know this won't work for all solids, but for most (including regular pyramids) it works.

Is there a reason why this formula works? In particular where did the 1/3 come from?

Note: I am familiar with the 2-D analogue $A=rs$.

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The 2-D analogue works because you can draw edges from the incenter to the sides of a polygon, cutting it into triangles. These triangles have a height of $r$ and their bases sum to the perimeter, so their total area is half the inradius times the perimeter.

In 3-D, you can do the same thing. By connecting the incenter to the edges of a polyhedron, you cut the polyhedron into pyramids which all have a height of $r$ and whose bases make up the surface area of the polyhedron. The volume of a pyramid is the height times one-third the area of the base, so the total volume is the inradius times one-third of the surface area. This works for any polyhedron that has an inscribed sphere touching all of its faces.

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  • $\begingroup$ Thank you. Is there an intuitive reason why the volume of a pyramid has a factor of 1/3? $\endgroup$
    – John Smith
    Commented Jul 23, 2016 at 12:25
  • $\begingroup$ @JohnSmith You could look at the answers to this question or this question. $\endgroup$
    – f''
    Commented Jul 23, 2016 at 15:48

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