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

Revision as of 10:14, 3 March 2015

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

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

2014 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

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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?
-<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>
+<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>
 [[Category: Introductory Geometry Problems]]

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

Revision as of 10:14, 3 March 2015

Problem 14

Solution

See also