Solution. The radius of a sphere inscribed in a regular tetrahedron
r=12a6 where a is edge length.
According to the condition of the problem the surface area of a sphere inscribed in a regular tetrahedron is 144 cm2
S=4πr2=4π1446a2=6πa2Therefore
a=π6S 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=3a6=36π6S=2πS
h=2π144=π24≈13.54cm Answer. 13.54
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