2014 AMC 12B Problems/Problem 14

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:

\[144 =(x+y+z)^2 =  x^2+y^2+z^2 + 2(xy+yz+xz)\]

\[144 =  x^2+y^2+z^2 + 94\]

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

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

Solution 2

Let $a, b, c$ be the side lengths of the rectangular prism. So we have $2(ab + ac + bc) = 94$ and $a + b + c= 12$. Note that the interior diagonal is just $\sqrt{a^2 + b^2 + c^2}$. Since $a^2 + b^2 + c^2 = (a+b+c)^2 - 2(ab + ac + bc) = 144 - 94 = 50$. So the length of the interior diagonal is just $5\sqrt{2}$. Since we have 4 of the same interior diagonals that answer is $20\sqrt{2}$ which is D. ~coolmath_2018

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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