Difference between revisions of "2007 iTest Problems/Problem 27"
(Created page with "== Problem == The face diagonal of a cube has length <math>4</math>. Find the value of n given that <math>n\sqrt2</math> is the <math>\textit{volume}</math> of the cube. == Sol...") |
Rockmanex3 (talk | contribs) (Solution to Problem 27) |
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The face diagonal of a cube has length <math>4</math>. Find the value of n given that <math>n\sqrt2</math> is the <math>\textit{volume}</math> of the cube. | The face diagonal of a cube has length <math>4</math>. Find the value of n given that <math>n\sqrt2</math> is the <math>\textit{volume}</math> of the cube. | ||
− | == Solution == | + | ==Solution== |
+ | <asy> | ||
+ | |||
+ | draw((0,0)--(10,0)--(10,10)--(0,10)--(0,0)); | ||
+ | draw((10,0)--(0,10)); | ||
+ | label("$4$",(5,5),NE); | ||
+ | label("$s$",(5,0),S); | ||
+ | label("$s$",(0,5),W); | ||
+ | |||
+ | </asy> | ||
+ | |||
+ | If the length of the face diagonal of the cube is <math>4</math>, then by using 45-45-90 triangles, the side length of the cube is <math>2 \sqrt{2}</math>. Thus, the volume of the cube is <math>(2 \sqrt{2})^3 = 16 \sqrt{2}</math>, so <math>n = \boxed{16}</math>. | ||
+ | |||
+ | ==See Also== | ||
+ | {{iTest box|year=2007|num-b=26|num-a=28}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] |
Latest revision as of 18:15, 10 June 2018
Problem
The face diagonal of a cube has length . Find the value of n given that is the of the cube.
Solution
If the length of the face diagonal of the cube is , then by using 45-45-90 triangles, the side length of the cube is . Thus, the volume of the cube is , so .
See Also
2007 iTest (Problems, Answer Key) | ||
Preceded by: Problem 26 |
Followed by: Problem 28 | |
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