Answer to Question #150156 in Geometry for solid

Question #150156
The surface area of a sphere inscribed in a regular tetrahedron is 144pi centimeter squared.
1. What is the radius of the sphere?

2. What is the altitude of the tetrahedron?

3. If a sphere is inscribed in a cube of side 12cm, what is the volume of the sphere?
1
Expert's answer
2020-12-16T18:53:44-0500

1 . The surface area of a sphere of radius rr is

S=4S=4πr2\pi r^2

then,

r=S4πr=\sqrt{\frac{S}{4\pi}}

Given S=144πS=144\pi cm2cm^2

r=144πcm24π=122=6r=\sqrt{\frac{144\pi cm^2}{4\pi}} = \frac{12}{2} = 6 cmcm

2 . For a regular tetrahedron of edge length aa :

Altitude h=23ah=\sqrt{\frac{2}{3}}a

a=2r6a=2r\sqrt6 =126=12\sqrt6

h=23×126=24h=\sqrt\frac{2}{3}\times12\sqrt6 = 24

3 . The diameter of the sphere would be equal to the side of the square .

d=12cmd=12cm

then,

r=6cmr=6cm

So,

V=43×πr3V=\frac{4}{3} \times \pi r^3

V=43×π×216V=\frac{4}{3} \times \pi \times216

V=904.32V=904.32


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