The surface area of a sphere inscribed in a regular tetrahedron is 144π cm2144\pi\space cm^2144π cm2
1. What is the radius of the sphere?
Surface area of a sphere S=4πr2S=4\pi r^2S=4πr2
r−r-r− radius of sphere
r=S4π=144π4π=36=6.r=\sqrt{\frac{S}{4\pi}}=\sqrt{\frac{144\pi}{4\pi}}=\sqrt{36}=6.r=4πS=4π144π=36=6.
Answer: radius of the sphere is 6 cm.6\space cm.6 cm.
2. What is the altitude of the tetrahedron?
For a regular tetrahedron of edge length a:
Altitude h=23ah=\sqrt{\frac{2}{3}}ah=32a
a=2r6=126a=2r\sqrt{6}=12\sqrt{6}a=2r6=126
h=23126=24.h=\sqrt{\frac{2}{3}}12\sqrt{6}=24.h=32126=24.
Answer:altitude of the tetrahedron is 24 cm.24\space cm.24 cm.
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