2015 AMC 8 Problems/Problem 8

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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
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All AJHSME/AMC 8 Problems and Solutions

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