Four of the nonadjacent vertices of a cube form a regular tetrahedron. What is the ratio of the surface area of the cube to the surface area of the tetrahedron?
Source: NCTM Mathematics Teacher, December 2005
Let the side of the cube be unit long.
Surface area of a face of the cube
Surface are of the cube
Length of a diagonal of the cube
Each face of the regular tetrahedron is an equilateral triangle of side length .
Altitude of triangle
Surface area of a face of the tetrahedron
Surface are of tetrahedron
Ratio of surface area of cube to surface area of tetrahedron