Difference between revisions of "2015 AMC 12A Problems/Problem 2"

 
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Two of the three sides of a triangle are 20 and 15. Which of the following numbers is not a possible perimeter of the triangle?
 
Two of the three sides of a triangle are 20 and 15. Which of the following numbers is not a possible perimeter of the triangle?
  
<math> \textbf{(A)}\ 52\qquad\textbf{(B)}\ 57\qquad\textbf{(C)}\ 62\qquad\textbf{(D)}}\ 67\qquad\textbf{(E)}\ 72</math>
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<math> \textbf{(A)}\ 52\qquad\textbf{(B)}\ 57\qquad\textbf{(C)}\ 62\qquad\textbf{(D)}\ 67\qquad\textbf{(E)}\ 72 </math>
  
 
==Solution==
 
==Solution==
The third side must be less than 20 + 15 = 35 by the Triangle Inequality, and so the perimeter must be less than 20 + 15 + 35 = 70. Clearly, <math>\boxed{E}</math> must be our answer.
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Letting <math>x</math> be the third side, then by the triangle inequality, <math>20-15 < x < 20+15</math>, or <math>5 < x < 35</math>. Therefore the perimeter must be greater than 40 but less than 70. 72 is not in this range, so <math>\boxed{\textbf{(E)} \, 72}</math> is our answer.
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== See Also ==
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{{AMC12 box|year=2015|ab=A|num-b=1|num-a=3}}

Latest revision as of 12:13, 29 March 2015

Problem

Two of the three sides of a triangle are 20 and 15. Which of the following numbers is not a possible perimeter of the triangle?

$\textbf{(A)}\ 52\qquad\textbf{(B)}\ 57\qquad\textbf{(C)}\ 62\qquad\textbf{(D)}\ 67\qquad\textbf{(E)}\ 72$

Solution

Letting $x$ be the third side, then by the triangle inequality, $20-15 < x < 20+15$, or $5 < x < 35$. Therefore the perimeter must be greater than 40 but less than 70. 72 is not in this range, so $\boxed{\textbf{(E)} \, 72}$ is our answer.

See Also

2015 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All AMC 12 Problems and Solutions