Difference between revisions of "2011 AMC 10B Problems/Problem 14"

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== Problem 14 ==
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== Problem==
  
 
A rectangular parking lot has a diagonal of <math>25</math> meters and an area of <math>168</math> square meters. In meters, what is the perimeter of the parking lot?
 
A rectangular parking lot has a diagonal of <math>25</math> meters and an area of <math>168</math> square meters. In meters, what is the perimeter of the parking lot?
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\end{align*}</cmath>
 
\end{align*}</cmath>
 
The perimeter of a rectangle is <math>2 (a + b) = 2 (31) = \boxed{\textbf{(C)} 62}</math>
 
The perimeter of a rectangle is <math>2 (a + b) = 2 (31) = \boxed{\textbf{(C)} 62}</math>
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== See Also==
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{{AMC10 box|year=2011|ab=B|num-b=13|num-a=15}}

Revision as of 16:14, 4 June 2011

Problem

A rectangular parking lot has a diagonal of $25$ meters and an area of $168$ square meters. In meters, what is the perimeter of the parking lot?

$\textbf{(A)}\ 52 \qquad\textbf{(B)}\ 58 \qquad\textbf{(C)}\ 62 \qquad\textbf{(D)}\ 68 \qquad\textbf{(E)}\ 70$

Solution

Let the sides of the rectangular parking lot be $a$ and $b$. Then $a^2 + b^2 = 625$ and $ab = 168$. Add the two equations together, then factor. \begin{align*} a^2 + 2ab + b^2 &= 625 + 168 \times 2\\ (a + b)^2 &= 961\\ a + b &= 31 \end{align*} The perimeter of a rectangle is $2 (a + b) = 2 (31) = \boxed{\textbf{(C)} 62}$

See Also

2011 AMC 10B (ProblemsAnswer KeyResources)
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 10 Problems and Solutions
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