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

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== Solution 2 (Quick) ==
 
== Solution 2 (Quick) ==
  
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We see the answer choices or the perimeter are integers. Therefore, the sides of the rectangle are most likely integers that satisfy <math>a^2+b^2=25^2</math>. In other words, <math>(a,b,25)</math> is a set of Pythagorean triples. Guess and checking, we have <math>(7,24,25)</math> as the triplet, as the area is <math>7 \cdot 24 = 168</math> as requested. Therefore, the perimeter is <math>2(7+24)=\boxed{\textbf{(C}} 62}</math>
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We see the answer choices or the perimeter are integers. Therefore, the sides of the rectangle are most likely integers that satisfy <math>a^2+b^2=25^2</math>. In other words, <math>(a,b,25)</math> is a set of Pythagorean triples. Guess and checking, we have <math>(7,24,25)</math> as the triplet, as the area is <math>7 \cdot 24 = 168</math> as requested. Therefore, the perimeter is <math>2(7+24)=\boxed{\textbf{(C)} 62}</math>.
  
 
== See Also==
 
== See Also==

Revision as of 19:13, 29 December 2019

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 $2 (a + b) = 2 (31) = \boxed{\textbf{(C)} 62}$

Solution 2 (Quick)

We see the answer choices or the perimeter are integers. Therefore, the sides of the rectangle are most likely integers that satisfy $a^2+b^2=25^2$. In other words, $(a,b,25)$ is a set of Pythagorean triples. Guess and checking, we have $(7,24,25)$ as the triplet, as the area is $7 \cdot 24 = 168$ as requested. Therefore, the perimeter is $2(7+24)=\boxed{\textbf{(C)} 62}$.

See Also

2011 AMC 10B (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 10 Problems and Solutions

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