Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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...")
 
Line 1: Line 1:
 +
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
 +
 +
<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>
 +
 +
<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>
 +
 +
== Solution ==
 +
 
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>
 +
 +
==See also==
 +
 +
{{AHSME box|year=1981|num-b=7|num-a=9}}
 +
{{MAA Notice}}

Revision as of 13:40, 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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1981_AHSME_Problems/Problem_9&oldid=252650"