Finding the circumradius of a regular tetrahedron

Question

Given a regular tetrahedron of edge length $a$, how do I prove that the circumradius of the tetrahedron is equal to $\frac{\sqrt 6}{4}a$?

Answer 1

In Elements XIII, 13 Euclid proves that the square of the diameter of the circumsphere is one and a half times the square of the side of the tetrahedron.

So$$d^2=\frac{3}{2}a^2$$Then if $r$ is the radius$$4r^2=\frac{3}{2}a^2$$making$$r^2=\frac{3}{8}a^2=\frac{6}{16}a^2$$and$$r=\frac{\sqrt 6}{4}a$$