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1981 AHSME Problems/Problem 9

Revision as of 21:14, 12 March 2018 by Coolmath34 (talk | contribs) (Created page with "Because the space diagonal of a cube with side length <math>s</math> is <math>s \sqrt{3},</math> the side length of the cube in this problem is <math>\frac{a}{\sqrt{3}}.</math...")
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Because the space diagonal of a cube with side length $s$ is $s \sqrt{3},$ the side length of the cube in this problem is $\frac{a}{\sqrt{3}}.$ The surface area of the cube is therefore $6(\frac{a}{\sqrt{3}})^2=6 \cdot \frac{a^2}{3}=\boxed{2a^2},$ which is answer choice $\boxed{\text{A}}.$

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