1981 AHSME Problems/Problem 9

Problem 9

In the adjoining figure, $PQ$ is a diagonal of the cube. If $PQ$ has length $a$ , then the surface area of the cube is

$[asy] import three; unitsize(1cm); size(200); currentprojection=orthographic(1/3,-1,1/2); draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--cycle,black); draw((0,0,0)--(0,0,1),black); draw((0,1,0)--(0,1,1),black); draw((1,1,0)--(1,1,1),black); draw((1,0,0)--(1,0,1),black); draw((0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle,black); draw((0,0,0)--(1,1,1),black); label("$P$",(0, 0, 0),NW); label("$Q$",(1, 1, 1),NE); [/asy]$

$\textbf{(A)}\ 2a^2\qquad\textbf{(B)}\ 2\sqrt{2}a^2\qquad\textbf{(C)}\ 2\sqrt{3}a^2\qquad\textbf{(D)}\ 3\sqrt{3}a^2\qquad\textbf{(E)}\ 6a^2$

Solution

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 $\boxed{\text{A}}.$

1981 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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All AHSME Problems and Solutions

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1981 AHSME Problems/Problem 9

Problem 9

Solution

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