Question #150750
The surface area of a sphere inscribed in a regular tetrahedron is 144 cm2

22. What is the altitude of the tetrahedron?.
a) 20 cm
b) 24 cm
c) 28 cm
d) 48 cm
1
Expert's answer
2020-12-15T18:19:52-0500

Solution. The radius of a sphere inscribed in a regular tetrahedron


r=a612r=\frac{a\sqrt{6}}{12}

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=4π6a2144=πa26S=4\pi r^2=4\pi \frac{6a^2}{144}=\frac{\pi a^2}{6}

Therefore


a=6Sπa=\sqrt{\frac{6S}{\pi}}

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=a63=636Sπ=2Sπh=\frac{a\sqrt{6}}{3}=\frac{\sqrt{6}}{3}\sqrt{\frac{6S}{\pi}}=2\sqrt{\frac{S}{\pi}}

h=2144π=24π13.54cmh=2\sqrt{\frac{144}{\pi}}=\frac{24}{\sqrt{\pi}}\approx 13.54cm

Answer. 13.54


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