Tetrahedron in Cube

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.

  1. 2
  2. 3
  3. 2
  4. 3
  5. 2

Show/Hide Solution

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 = s.
The tetrahedron’s four faces are equilateral triangles, e.g. △BCD, whose sides, e.g. CD, are s2 (△CDE is a 45-right triangle).
SA(cube) = 6s2
SA(tetrahedron) = 4 x area of face BCD.
The area of an equilateral triangle of side a is (a23)/4. Here, a = s2.
So area of △BCD = (s2)23)/4 = (2s23)/4 = (s23)/2.
So SA (tetrahedron) = 4 x (s23)/2 = 2s23.

So the ratio of surface areas is:

6s2  =  3  =  3 .
2s23 3 1

The answer is (d).

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