Answer to Question #150750 in Geometry for solid mensuration

Question

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

Answer 1

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

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

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

$$S=4\pi r^2=4\pi \frac{6a^2}{144}=\frac{\pi a^2}{6}$$

Therefore

$$a=\sqrt{\frac{6S}{\pi}}$$

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:

$$h=\frac{a\sqrt{6}}{3}=\frac{\sqrt{6}}{3}\sqrt{\frac{6S}{\pi}}=2\sqrt{\frac{S}{\pi}}$$

$$h=2\sqrt{\frac{144}{\pi}}=\frac{24}{\sqrt{\pi}}\approx 13.54cm$$

Answer. 13.54