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Difference between revisions of "2008 iTest Problems/Problem 38"

(Solution to Problem 38)
 
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==Problem==
 
==Problem==
  
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The volume of a certain rectangular solidis <math>216\text{ cm}^3</math>, its total surface area is <math>288\text{ cm}^2</math>, and its three dimensions are in  
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The volume of a certain rectangular solid is <math>216\text{ cm}^3</math>, its total surface area is <math>288\text{ cm}^2</math>, and its three dimensions are in  
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geometric progression. Find the sum of the lengths in cm of all the edges of this solid.  
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geometric progression. Find the sum of the lengths in cm of all the edges of this solid.
  
 
==Solution==
 
==Solution==

Latest revision as of 16:08, 15 March 2021

Problem

The volume of a certain rectangular solid is $216\text{ cm}^3$, its total surface area is $288\text{ cm}^2$, and its three dimensions are in geometric progression. Find the sum of the lengths in cm of all the edges of this solid.

Solution

Let the three side lengths be $\tfrac{a}{r}$, $a$, and $ar$. Because the volume of the solid is $216\text{ cm}^3$, \[\frac{a}{r} \cdot a \cdot ar = 216\] \[a = 6\] The surface area of the solid is $288\text{ cm}^2$, so \[2(\frac{a^2}{r} + a^2r + a^2) = 288\] Note that the sum of the side lengths of the cube is $4(\tfrac{6}{r} + 6 + 6r)$ and that the equation above has a similar form. \[2(\frac{36}{r} + 36r + 36) = 288\] \[2(\frac{6}{r} + 6r + 6) = 48\] \[4(\frac{6}{r} + 6r + 6) = 96\] The sum of all the edges of the cube is $\boxed{96}$ centimeters.

See Also

2008 iTest (Problems)
Preceded by:
Problem 37 		Followed by:
Problem 39
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