2014 AMC 12B Problems/Problem 14

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:

$144 =(x+y+z)^2 = x^2+y^2+x^2 + 2(xy+yz+xz)$

$144 = x^2+y^2+x^2 + 94$

Hence $x^2+y^2+x^2 = 50$

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