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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2015 AMC 8 (Problems • Answer Key • Resources)
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

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

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