Question

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?

Answer 1

1 . The surface area of a sphere of radius $r$ is

$S=4$$\pi r^2$

then,

$r=\sqrt{\frac{S}{4\pi}}$

Given $S=144\pi$ $cm^2$

$r=\sqrt{\frac{144\pi cm^2}{4\pi}} = \frac{12}{2} = 6$ $cm$

2 . For a regular tetrahedron of edge length $a$ :

Altitude $h=\sqrt{\frac{2}{3}}a$

$a=2r\sqrt6$ $=12\sqrt6$

$h=\sqrt\frac{2}{3}\times12\sqrt6 = 24$

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

$d=12cm$

then,

$r=6cm$

So,

$V=\frac{4}{3} \times \pi r^3$

$V=\frac{4}{3} \times \pi \times216$

$V=904.32$