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Difference between revisions of "1981 AHSME Problems/Problem 9"
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== Problem 9 ==
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In the adjoining figure, <math>PQ</math> is a diagonal of the cube. If <math>PQ</math> has length <math>a</math>, then the surface area of the cube is
 
In the adjoining figure, <math>PQ</math> is a diagonal of the cube. If <math>PQ</math> has length <math>a</math>, then the surface area of the cube is
  
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== Solution ==
 
== Solution ==
  
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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> The surface area of the cube is therefore <math>6(\frac{a}{\sqrt{3}})^2=6 \cdot \frac{a^2}{3}=\boxed{2a^2},</math> which is answer choice <math>\boxed{\text{A}}.</math>
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Let <math>s</math> be the side length of the cube.  By the Pythagorean theorem, any diagonal of any face of the cube has length <math>s\sqrt{2}</math>, and thus any diagonal of the cube has length <math>a = s \sqrt{3}</math>. So <math>s = \frac{a}{\sqrt{3}}</math>, and therefore the surface area of the cube is <math>6\left(\frac{a}{\sqrt{3}}\right)^2=6 \cdot \frac{a^2}{3}=\boxed{(\textbf{A})\ 2a^2}</math>.
  
 
==See also==
 
==See also==
  
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{{AHSME box|year=1981|num-b=7|num-a=9}}
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{{AHSME box|year=1981|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 16:17, 18 August 2025

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

Let $s$ be the side length of the cube. By the Pythagorean theorem, any diagonal of any face of the cube has length $s\sqrt{2}$, and thus any diagonal of the cube has length $a = s \sqrt{3}$. So $s = \frac{a}{\sqrt{3}}$, and therefore the surface area of the cube is $6\left(\frac{a}{\sqrt{3}}\right)^2=6 \cdot \frac{a^2}{3}=\boxed{(\textbf{A})\ 2a^2}$.

See also

1981 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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