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Difference between revisions of "1981 AHSME Problems/Problem 9"

(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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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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<asy>
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import three;
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unitsize(1cm);
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size(200);
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currentprojection=orthographic(1/3,-1,1/2);
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draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--cycle,black);
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draw((0,0,0)--(0,0,1),black);
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draw((0,1,0)--(0,1,1),black);
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draw((1,1,0)--(1,1,1),black);
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draw((1,0,0)--(1,0,1),black);
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draw((0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle,black);
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draw((0,0,0)--(1,1,1),black);
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label("$P$",(0, 0, 0),NW);
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label("$Q$",(1, 1, 1),NE);
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</asy>
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<math> \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 </math>
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== 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>
 
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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==See also==
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{{AHSME box|year=1981|num-b=7|num-a=9}}
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{{MAA Notice}}

Revision as of 12:40, 28 June 2025

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

See also

1981 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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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