Difference between revisions of "2015 AMC 8 Problems/Problem 8"

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<math>\textbf{(A) }24\qquad\textbf{(B) }29\qquad\textbf{(C) }43\qquad\textbf{(D) }48\qquad \textbf{(E) }57</math>
 
<math>\textbf{(A) }24\qquad\textbf{(B) }29\qquad\textbf{(C) }43\qquad\textbf{(D) }48\qquad \textbf{(E) }57</math>
  
 
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===Solution===
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We know from the triangle inequality that the last side, <math>c</math>, fulfills <math>c<5+19=24</math>. Adding <math>5+19</math> to both sides of the inequality, we get <math>c+5+19<48</math>. However, we know that <math>c+5+19</math> is the perimeter of our triangle, so we get <math>\boxed{\textbf{(D) }~48}</math> as our answer.
 
==See Also==
 
==See Also==
  
 
{{AMC8 box|year=2015|num-b=7|num-a=9}}
 
{{AMC8 box|year=2015|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:33, 25 November 2015

What is the smallest whole number larger than the perimeter of any triangle with a side of length $5$ and a side of length $19$?

$\textbf{(A) }24\qquad\textbf{(B) }29\qquad\textbf{(C) }43\qquad\textbf{(D) }48\qquad \textbf{(E) }57$

Solution

We know from the triangle inequality that the last side, $c$, fulfills $c<5+19=24$. Adding $5+19$ to both sides of the inequality, we get $c+5+19<48$. However, we know that $c+5+19$ is the perimeter of our triangle, so we get $\boxed{\textbf{(D) }~48}$ as our answer.

See Also

2015 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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 AJHSME/AMC 8 Problems and Solutions

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