A question says:
A sphere is inscribed in a regular tetrahedron. If the length of an altitude of the tetrahedron is 36, what is the length of a radius of the sphere?
I'm not sure where to start.
This is what I think so far:
- I think that the sphere touches the "center" of each of the tetrahedron's sides.
- Halfway down one of the tetrahedron's sides, where it meets the "altitude line" perpendicularly, is the radius of the sphere.
Apparently, the answer's 9.
Best Answer
The center of the tetrahedron divides each of the four heights (or medians) in the ratio $1:3$ (in an equilateral triangle the corresponding ratio is $1:2$). The smaller part is also the radius of the inscribed sphere. Therefore the radius of this sphere is one quarter of the height or $9$ in your case.