## Tetrahedron

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

SOLUTION

Let the side of the cube be $1$ unit long.
Surface area of a face of the cube $=1^2$
Surface are of the cube $=6\times 1^2$
Length of a diagonal of the cube $=\sqrt 2$
Each face of the regular tetrahedron is an equilateral triangle of side length $\sqrt 2$.
Altitude of triangle
$\dfrac{\sqrt 2}{2}\sqrt 3=\dfrac{\sqrt 6}{2}$
Surface area of a face of the tetrahedron
$\dfrac{1}{2}\sqrt 2\dfrac{\sqrt 6}{2}=\dfrac{\sqrt 3}{2}$
Surface are of tetrahedron
$4\dfrac{\sqrt 3}{2}=2\sqrt 3$
Ratio of surface area of cube to surface area of tetrahedron
$\dfrac{6}{2\sqrt 3}=\dfrac{6\sqrt 3}{6}=\dfrac{\sqrt 3}{1}$

Answer: $\sqrt 3$ to $1$