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 = 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) = 6s2
SA(tetrahedron) = 4 x area of face BCD.
The area of an equilateral triangle of side a is (a2√3)/4. Here, a = s√2.
So area of △BCD = (s√2)2 √3)/4 = (2s2 √3)/4 = (s2√3)/2.
So SA (tetrahedron) = 4 x (s2√3)/2 = 2s2√3.
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) = 6s2
SA(tetrahedron) = 4 x area of face BCD.
The area of an equilateral triangle of side a is (a2√3)/4. Here, a = s√2.
So area of △BCD = (s√2)2 √3)/4 = (2s2 √3)/4 = (s2√3)/2.
So SA (tetrahedron) = 4 x (s2√3)/2 = 2s2√3.
So the ratio of surface areas is:
| 6s2 | = | 3 | = | √3 | . |
| 2s2√3 | √3 | 1 |
The answer is (d).