## Radius of Sphere at the Centre of a Tetrahedron

The distance AM in the below is
$x sin 60 = \frac{x \sqrt{3}}{2}$
.

The centre of the triangle ABC is
$\frac{2}{3}$
of the distance from A nto M ie
$\frac{2}{3} \times \frac{x \sqrt{2}}{3}= \frac{x}{\sqrt{3}}$
.
The centre of the sphere at the centre of the tetrahedron is vertically above the centre O of the triangle ABC and
$\frac{1}{3}$
of the height of the tetrahedron above the base, since a cross section of the tetrahedron seen from the side is a triangle, and the centre of a triangle is
$\frac{1}{3}$
the distance from the centre of an edge to a vertex.
The height OT of the triangle is
$\sqrt{x^2 - (\frac{x}{\sqrt{3}})^2} = x \sqrt{\frac{2}{3}}$

The radius of the tetrahedron is
$\frac{1}{3} \times x \sqrt{\frac{2}{3}} = \frac{x}{3} \sqrt{\frac{2}{3}}$