# Radius of insphere in a Tetrahedron

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1. Mar 7, 2015

### Matejxx1

1. The problem statement, all variables and given/known data

What is the largest possible radius of a sphere which is inscribed in a regular tetrahedron
a=10 ( this is the side of the tetrahedron)
r=?
r=5*√6/6
2. Relevant equations

3. The attempt at a solution
So first I calculated the Height of pyramid
a2=(2/3*va)2+h2
h=√(a2-(2/3*a*√3/2)2)
h=√(a2-(4*a2*3)/4)
h=√(a2-a2/3)
h=√(a2*2/3)
h=a*√(2/3)
and then I though since we have 2 similar triangles we could write the relation as
r/(h-r)=(2*va)/3a
r/(h-r)=2*a*√3/(3*a*2)
r/(h-r)=√3/3
r=(√3h-√3r)/3
3r=√3h-√3r
3r+√3r=√3*h
r(3+√3)=√3h
r=√3*h/(3+√3)
r=√3*a*√(2/3)/(3+√3)
r=a*√2/(3+√3)
r=a*√2+(3-√3)/(9-3)
r=(3a*√2-a√6)/6
r=2,99 which is wrong the right answer is
r=5*√6/6 but I don't know how to get it my only guess is that the relation is not correct but I don't know what else to try
any help would be really appreciated

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2. Mar 7, 2015

### HallsofIvy

Staff Emeritus
I don't get either of those answers. I get that the radius of the largest sphere, that can be inscribed in a tetrahedron of edge length a, is a/4.

3. Mar 7, 2015

### Matejxx1

I figured out how to do it with trial and error :P
By the way another question if I may.
I see most people posting equations like this

$$\frac{4s^3+4s^2+72}{s + 3}\;$$

Which is easier to read then (4s3+4s2+72)/(s+3)
mind telling me if you know how to do it ?

4. Mar 7, 2015