1980 AHSME Problems/Problem 16
Problem
Four of the eight vertices of a cube are the vertices of a regular tetrahedron. Find the ratio of the surface area of the cube to the surface area of the tetrahedron.
Solution
We assume the side length of the cube is . The side length of the tetrahedron is
, so the surface area is
. The surface area of the cube is
, so the ratio of the surface area of the cube to the surface area of the tetrahedron is
See also
1980 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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