Difference between revisions of "2014 AMC 12B Problems/Problem 14"

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A rectangular box has a total surface area of 94 square inches. The sum of the lengths of all its edges is 48 inches. What is the sum of the lengths in inches of all of its interior diagonals?
 
A rectangular box has a total surface area of 94 square inches. The sum of the lengths of all its edges is 48 inches. What is the sum of the lengths in inches of all of its interior diagonals?
  
<math> \textbf{(A)}\ 8\sqrt{3}\qquad\textbf{(B)}\ 10\sqrt{2}\qquad\textbf{(C)}\ 16\sqrt{3}\qquad\textbf{(D)}}\ 20\sqrt{2}\qquad\textbf{(E)}\ 40\sqrt{2} </math>
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<math> \textbf{(A)}\ 8\sqrt{3}\qquad\textbf{(B)}\ 10\sqrt{2}\qquad\textbf{(C)}\ 16\sqrt{3}\qquad\textbf{(D)}\ 20\sqrt{2}\qquad\textbf{(E)}\ 40\sqrt{2} </math>
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[[Category: Introductory Geometry Problems]]
  
 
==Solution==
 
==Solution==
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<cmath> 4(x+y+z) = 48 \Rightarrow x+y+z = 12 </cmath>
 
<cmath> 4(x+y+z) = 48 \Rightarrow x+y+z = 12 </cmath>
 
  
 
<cmath>2(xy+yz+xz) = 94 </cmath>
 
<cmath>2(xy+yz+xz) = 94 </cmath>
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Squaring the first expression, we get:  
 
Squaring the first expression, we get:  
  
<cmath>144 =(x+y+z)^2 =  x^2+y^2+x^2 + 2(xy+yz+xz)</cmath>
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\begin{align*}
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144 =(x+y+z)^2 &=  x^2+y^2+z^2 + 2(xy+yz+xz) \\
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&=  x^2+y^2+z^2 + 94.
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\end{align*}
  
<cmath>144 =  x^2+y^2+x^2 + 94</cmath>
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Hence <cmath>x^2+y^2+z^2 = 50</cmath>
 
 
Hence <cmath>x^2+y^2+x^2 = 50</cmath>
 
  
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<cmath>4 \sqrt{x^2+y^2+z^2} =  \boxed{\textbf{(D)}\ 20\sqrt 2} </cmath>
  
<cmath>4 \sqrt{x^2+y^2+z^2} =  \boxed{\textbf{(D)}\ 20\sqrt 2} </cmath>
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== See also ==
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{{AMC12 box|year=2014|ab=B|num-b=13|num-a=15}}
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{{MAA Notice}}

Latest revision as of 12:27, 18 June 2024

Problem 14

A rectangular box has a total surface area of 94 square inches. The sum of the lengths of all its edges is 48 inches. What is the sum of the lengths in inches of all of its interior diagonals?

$\textbf{(A)}\ 8\sqrt{3}\qquad\textbf{(B)}\ 10\sqrt{2}\qquad\textbf{(C)}\ 16\sqrt{3}\qquad\textbf{(D)}\ 20\sqrt{2}\qquad\textbf{(E)}\ 40\sqrt{2}$

Solution

Let the side lengths of the rectangular box be $x, y$ and $z$. From the information we get

\[4(x+y+z) = 48 \Rightarrow x+y+z = 12\]

\[2(xy+yz+xz) = 94\]

The sum of all the lengths of the box's interior diagonals is

\[4 \sqrt{x^2+y^2+z^2}\]

Squaring the first expression, we get:

\begin{align*} 144 =(x+y+z)^2 &= x^2+y^2+z^2 + 2(xy+yz+xz) \\ &= x^2+y^2+z^2 + 94. \end{align*}

Hence \[x^2+y^2+z^2 = 50\]

\[4 \sqrt{x^2+y^2+z^2} =  \boxed{\textbf{(D)}\ 20\sqrt 2}\]

See also

2014 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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All AMC 12 Problems and Solutions

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