Solution. The radius of a sphere inscribed in a regular tetrahedron
r = a 6 12 r=\frac{a\sqrt{6}}{12} r = 12 a 6 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 = 4 π r 2 = 4 π 6 a 2 144 = π a 2 6 S=4\pi r^2=4\pi \frac{6a^2}{144}=\frac{\pi a^2}{6} S = 4 π r 2 = 4 π 144 6 a 2 = 6 π a 2 Therefore
a = 6 S π a=\sqrt{\frac{6S}{\pi}} a = π 6 S 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 6 3 = 6 3 6 S π = 2 S π h=\frac{a\sqrt{6}}{3}=\frac{\sqrt{6}}{3}\sqrt{\frac{6S}{\pi}}=2\sqrt{\frac{S}{\pi}} h = 3 a 6 = 3 6 π 6 S = 2 π S
h = 2 144 π = 24 π ≈ 13.54 c m h=2\sqrt{\frac{144}{\pi}}=\frac{24}{\sqrt{\pi}}\approx 13.54cm h = 2 π 144 = π 24 ≈ 13.54 c m Answer. 13.54
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