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

## Problem

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, $s$, fulfills $s<5+19=24$. Adding $5+19$ to both sides of the inequality, we get $s+5+19<48$, and because $s+5+19$ is the perimeter of our triangle, $\boxed{\textbf{(D)}\ 48}$ is our answer.