Regular tetrahedron

Statement

Given a regular tetrahedron with edge $\ell$, calculate the height, the radius of the inscribed sphere and the volume.

Solution

The base is an equilateral triangle: its height is $h_b = \tfrac{\sqrt{3}}{2}\ell$ and the apothem (distance from the centre to a side) is $m = \tfrac{h_b}{3} = \tfrac{\sqrt{3}}{6}\ell$. The foot $H$ of the height falls at the centroid, at a distance $2m = \tfrac{\sqrt{3}}{3}\ell$ from each base vertex.

Height of the tetrahedron (Pythagoras on edge and projection): $h = \sqrt{\ell^2 - (2m)^2} = \sqrt{\ell^2 - \tfrac{\ell^2}{3}} = \ell\sqrt{\tfrac{2}{3}} = \boxed{\dfrac{\sqrt{6}}{3}\,\ell}.$ Volume: $V = \tfrac{1}{3}S_b\, h = \tfrac{1}{3}\cdot\tfrac{\sqrt{3}}{4}\ell^2\cdot \tfrac{\sqrt{6}}{3}\ell = \boxed{\dfrac{\sqrt{2}}{12}\,\ell^3}.$ Radius of the inscribed sphere: for a regular tetrahedron $r = \tfrac{1}{4}h$ holds, hence $r = \tfrac{1}{4}\cdot\tfrac{\sqrt{6}}{3}\ell = \boxed{\dfrac{\sqrt{6}}{12}\,\ell}.$

Links

Topics: Synthetic geometry in space
Concepts: Pyramid · Regular polyhedra · Sphere · Volume
Skills: Calculating · Using formulae
Exercise types: Geometric problem