# Simplex volume in terms of the radius of the inscribed sphere?

It is well known that for the area of a triangle $A$ we have $$A=r\cdot s,$$ where $s$ is the semiperimeter, and $r$ is the radius of the inscribed circle.

Is there an analogue for the higher-dimensional case. In other words, can I express the volume of a $d$-simplex in terms of the radius of its inscribed sphere and the volume of its boundary? If such a formula exists, what are the references?

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## 1 Answer

The same holds, that in $\mathbb{R}^n$, $A = r \frac {V_b}{n}$, where $A$ is the volume of your simplex, and $V_b$ is the volume of your boundary.

To see why this works, simply show that a simplex with base of volume $B$ and height $r$ has volume $r \frac {B}{n}$, and then add up over all faces.

Hint: The constant $\frac {1}{n}$ comes from $\int x^{n-1}\, dx = \frac {x^n}{n}$.

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I see, one just decomposes the simplex into smaller simplices, spanned by the insphere center and the faces. However, the $d$-simplex volume should be $r\frac{B}{d!}$. – A.Schulz Jan 16 '13 at 16:47
@A.Schulz I disagree with the $d!$. As a reminder, the tetrahedron has volume $\frac {1}{3} A \cdot h$. You might be confusing this with the volume of the standard $n-$simplex, which is$\frac {1}{n!}$. This comes about because it has a base in $n-1$ dimensions with volume $\frac {1}{(n-1)!}$, and a height of 1, so according to the formula it has volume $\frac {1}{n!}$. – Calvin Lin Jan 16 '13 at 16:59
You are right. I was confused by the standard formula for the simplex volume. – A.Schulz Jan 16 '13 at 17:38