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Difference between revisions of "2008 AMC 12A Problems/Problem 8"

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Alternatively, we can use the fact that the surface area of a cube is directly proportional to the square of its side length. Therefore, if the surface area of a cube increases by a factor of <math>2</math>, its side length must increase by a factor of <math>\sqrt{2}</math>, meaning the new side length of the cube is <math>1 * \sqrt{2} = \sqrt{2}</math>. So, its volume is <math>({\sqrt{2}})^3 = 2\sqrt{2} \Rightarrow\mathrm{(C)}</math>.
 
Alternatively, we can use the fact that the surface area of a cube is directly proportional to the square of its side length. Therefore, if the surface area of a cube increases by a factor of <math>2</math>, its side length must increase by a factor of <math>\sqrt{2}</math>, meaning the new side length of the cube is <math>1 * \sqrt{2} = \sqrt{2}</math>. So, its volume is <math>({\sqrt{2}})^3 = 2\sqrt{2} \Rightarrow\mathrm{(C)}</math>.
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== Video Solution by OmegaLearn ==
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https://youtu.be/MOcX5BFbcwU?t=54
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~ pi_is_3.14
  
 
==See Also==
 
==See Also==

Latest revision as of 04:49, 16 January 2023

Problem

What is the volume of a cube whose surface area is twice that of a cube with volume 1?

$\mathrm{(A)}\ \sqrt{2}\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ 2\sqrt{2}\qquad\mathrm{(D)}\ 4\qquad\mathrm{(E)}\ 8$

Solution

A cube with volume $1$ has a side of length $\sqrt[3]{1}=1$ and thus a surface area of $6 \cdot 1^2=6$.

A cube whose surface area is $6\cdot2=12$ has a side of length $\sqrt{\frac{12}{6}}=\sqrt{2}$ and a volume of $(\sqrt{2})^3=2\sqrt{2}\Rightarrow\mathrm{(C)}$.


Alternatively, we can use the fact that the surface area of a cube is directly proportional to the square of its side length. Therefore, if the surface area of a cube increases by a factor of $2$, its side length must increase by a factor of $\sqrt{2}$, meaning the new side length of the cube is $1 * \sqrt{2} = \sqrt{2}$. So, its volume is $({\sqrt{2}})^3 = 2\sqrt{2} \Rightarrow\mathrm{(C)}$.

Video Solution by OmegaLearn

https://youtu.be/MOcX5BFbcwU?t=54

~ pi_is_3.14

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

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