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Radius of insphere in a Tetrahedron

  1. Mar 7, 2015 #1
    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
     

    Attached Files:

  2. jcsd
  3. Mar 7, 2015 #2

    HallsofIvy

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    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.
     
  4. Mar 7, 2015 #3
    Hey, Thanks for the answer!
    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

    [tex] \frac{4s^3+4s^2+72}{s + 3}\; [/tex]

    Which is easier to read then (4s3+4s2+72)/(s+3)
    mind telling me if you know how to do it ?
     
  5. Mar 7, 2015 #4

    LCKurtz

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