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.


Solution 2

$s$, fulfills $s<5+19=24$, meaning that s<24. The greatest value of s can be 23. 23+5+19 is 47. 48 is our nearest answer choice so its $\boxed{\textbf{(D)}\ 48}$.-TheNerdWhoIsNerdy.

Video Solution (HOW TO THINK CRITICALLY!!!)

https://youtu.be/rxXHhn5PB9w

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Video Solution

https://youtu.be/zUiKAoX2D_Q

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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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