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

Revision as of 12:20, 21 February 2014

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?

$\textbf{(A)}\ 8\sqrt{3}\qquad\textbf{(B)}\ 10\sqrt{2}\qquad\textbf{(C)}\ 16\sqrt{3}\qquad\textbf{(D)}}\ 20\sqrt{2}\qquad\textbf{(E)}\ 40\sqrt{2}$ (Error compiling LaTeX. Unknown error_msg)

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

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 26: / Line 26: @@
 Hence <cmath>x^2+y^2+x^2 = 50</cmath>
+<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>
+{{AMC12 box|year=2014|ab=B|num-b=13|num-a=15}}
+{{MAA Notice}}

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

Revision as of 12:20, 21 February 2014

Problem 14

Solution