2007 iTest Problems/Problem 28: Difference between revisions
Created page with "== Problem == The space diagonal (interior diagonal) of a cube has length 6. Find the <math>\textit{surface area}</math> of the cube. == Solution ==" |
Rockmanex3 (talk | contribs) Solution to Problem 28 |
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== Solution == | == Solution == | ||
<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); | |||
draw((0,0,0)--(0,0,1)); | |||
draw((0,1,0)--(0,1,1)); | |||
draw((1,1,0)--(1,1,1)); | |||
draw((1,0,0)--(1,0,1)); | |||
draw((0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle); | |||
draw((0,0,0)--(1,0,0)--(1,1,0)--cycle); | |||
draw((0,0,0)--(1,1,0)--(1,1,1)--cycle,blue); | |||
label("$s\sqrt{2}$",(0.5,0.5,0),SE); | |||
label("$s$",(1,1,0.5),E); | |||
label("$6$",(0.5,0.5,0.5),SE); | |||
</asy> | |||
Finding the space diagonal of a cube requires a side length and a face diagonal. Using the [[Pythagorean Theorem]], | |||
<cmath>s^2 + 2s^2 = 36</cmath> | |||
<cmath>3s^2 = 36</cmath> | |||
<cmath>s^2 = 12</cmath> | |||
Since the area of one face is <math>12</math>, the surface area of the cube is <math>\boxed{72}</math>. | |||
==See Also== | |||
{{iTest box|year=2007|num-b=27|num-a=29}} | |||
[[Category:Intermediate Geometry Problems]] | |||
Latest revision as of 23:08, 15 June 2018
Problem
The space diagonal (interior diagonal) of a cube has length 6. Find the
of the cube.
Solution
Finding the space diagonal of a cube requires a side length and a face diagonal. Using the Pythagorean Theorem,
Since the area of one face is
, the surface area of the cube is
.
See Also
| 2007 iTest (Problems, Answer Key) | ||
| Preceded by: Problem 27 |
Followed by: Problem 29 | |
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