Tetrahedron

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
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Let the side of the cube be 1 unit long.
Surface area of a face of the cube =1^2
Surface are of the cube =6\times 1^2
Length of a diagonal of the cube =\sqrt 2
Each face of the regular tetrahedron is an equilateral triangle of side length \sqrt 2.
Altitude of triangle
\dfrac{\sqrt 2}{2}\sqrt 3=\dfrac{\sqrt 6}{2}
Surface area of a face of the tetrahedron
\dfrac{1}{2}\sqrt 2\dfrac{\sqrt 6}{2}=\dfrac{\sqrt 3}{2}
Surface are of tetrahedron
4\dfrac{\sqrt 3}{2}=2\sqrt 3
Ratio of surface area of cube to surface area of tetrahedron
\dfrac{6}{2\sqrt 3}=\dfrac{6\sqrt 3}{6}=\dfrac{\sqrt 3}{1}

Answer: \sqrt 3 to 1

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About mvtrinh

Retired high school math teacher.
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