Four of the 8 vertices of a cube are vertices of a regular tetrahedron. Find the ratio of the surface area of the cube to the surface area of the tetrahedron.

- 2
- 3
- √2
- √3
- 2

First, take a minute to enjoy the fact that a tetrahedron can be inscribed so neatly in a cube.

OK. Let the edge of the cube =

The tetrahedron’s four faces are equilateral triangles, e.g. △BCD, whose sides, e.g. CD, are

SA(cube) = 6

SA(tetrahedron) = 4 x area of face BCD.

The area of an equilateral triangle of side

So area of △BCD = (

So SA (tetrahedron) = 4 x (

*s*.The tetrahedron’s four faces are equilateral triangles, e.g. △BCD, whose sides, e.g. CD, are

*s*√2 (△CDE is a 45-right triangle).SA(cube) = 6

*s*^{2}SA(tetrahedron) = 4 x area of face BCD.

The area of an equilateral triangle of side

*a*is (*a*^{2}√3)/4. Here,*a*=*s*√2.So area of △BCD = (

*s*√2)^{2}√3)/4 = (2*s*^{2}√3)/4 = (*s*^{2}√3)/2.So SA (tetrahedron) = 4 x (

*s*^{2}√3)/2 = 2*s*^{2}√3.So the ratio of surface areas is:

6s^{2} |
= | 3 | = | √3 | . |

2s^{2}√3 |
√3 | 1 |

The answer is (d).