2015 AMC 8 Problems/Problem 8

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 $c+5+19<48$ , and because $c+5+19$ is the perimeter of our triangle, $\boxed{\textbf{(D)}~48}$ is our answer.

2015 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2015 AMC 8 Problems/Problem 8

Solution

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