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 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

**Answer**: to

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