Difference between revisions of "2014 AMC 12B Problems/Problem 14"
(Created page with "==Solution== Let the side lengths of the rectangular box be <math> x, y</math> and <math>z</math>. From the information we get <cmath> 4(x+y+z) = 48 \Rightarrow x+y+z = 12 <...") |
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+ | ==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? | ||
+ | |||
+ | <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]] | ||
+ | |||
==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: | ||
− | + | \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*} | ||
− | <cmath> | + | Hence <cmath>x^2+y^2+z^2 = 50</cmath> |
− | + | <cmath>4 \sqrt{x^2+y^2+z^2} = \boxed{\textbf{(D)}\ 20\sqrt 2} </cmath> | |
− | + | == See also == | |
− | + | {{AMC12 box|year=2014|ab=B|num-b=13|num-a=15}} | |
+ | {{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?
Solution
Let the side lengths of the rectangular box be and . From the information we get
The sum of all the lengths of the box's interior diagonals is
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
See also
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 |
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