Difference between revisions of "2008 AMC 12A Problems/Problem 8"
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What is the [[volume]] of a [[cube]] whose [[surface area]] is twice that of a cube with volume 1? | What is the [[volume]] of a [[cube]] whose [[surface area]] is twice that of a cube with volume 1? | ||
| − | <math>\ | + | <math>\mathrm{(A)}\ \sqrt{2}\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ 2\sqrt{2}\qquad\mathrm{(D)}\ 4\qquad\mathrm{(E)}\ 8</math> |
==Solution== | ==Solution== | ||
A cube with volume <math>1</math> has a side of length <math>\sqrt[3]{1}=1</math> and thus a surface area of <math>6 \cdot 1^2=6</math>. | A cube with volume <math>1</math> has a side of length <math>\sqrt[3]{1}=1</math> and thus a surface area of <math>6 \cdot 1^2=6</math>. | ||
| − | A cube whose surface area is <math>6\cdot2=12</math> has a side of length <math>\sqrt{\frac{12}{6}}=\sqrt{2}</math> and a volume of <math>(\sqrt{2})^3 = 2\sqrt{2} \Rightarrow C</math>. | + | A cube whose surface area is <math>6\cdot2=12</math> has a side of length <math>\sqrt{\frac{12}{6}}=\sqrt{2}</math> and a volume of <math>(\sqrt{2})^3=2\sqrt{2}\Rightarrow\mathrm{(C)}</math>. |
==See Also== | ==See Also== | ||
{{AMC12 box|year=2008|ab=A|num-b=7|num-a=9}} | {{AMC12 box|year=2008|ab=A|num-b=7|num-a=9}} | ||
Revision as of 23:34, 25 April 2008
Problem
What is the volume of a cube whose surface area is twice that of a cube with volume 1?
Solution
A cube with volume 
has a side of length ![$\sqrt[3]{1}=1$](http://latex.artofproblemsolving.com/3/c/6/3c65bfc5db8bea84fcd769f69191c1c492e6718f.png)
and thus a surface area of 
.
A cube whose surface area is 
has a side of length 
and a volume of 
.
See Also
| 2008 AMC 12A (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 | |
| All AMC 12 Problems and Solutions | |
