SOLUTION: A sphere is inscribed in a regular tetrahedron. If the length of an altitude of the tetrahedron is 36, what is length of a radius of the sphere

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Question 548597: A sphere is inscribed in a regular tetrahedron. If the length of an altitude of the tetrahedron is 36, what is length of a radius of the sphere
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
The radius (r) of an inscribed sphere (aka insphere) of a regular tetrahedron with edge length = a, is given by:
r+=+%28sqrt%286%29%28a%2F12%29%29
The edge length is not given but we do know the altitude or height (h) of the tetrahedron so we can compute the edge length as follows:
h+=+a%2Asqrt%286%29%2F3 and h is given as h+=+36 so...
a%2Asqrt%286%29%2F3+=+36
a+=+108%2Fsqrt%286%29 Now we can find r:
r+=+%28sqrt%286%29%28a%2F3%29%29 Substitute a+=+108%2Fsqrt%286%29
r+=+%28sqrt%286%29%28108%2F%2812%2Asqrt%286%29%29%29%29
r+=+9