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

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===Solution===
 
===Solution===
We kk;k
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By the triangle inequality rule, the last side has to be 23. 5 + 19 + 23 = 47. Therefore D, 48 is correct.
  
 
==See Also==
 
==See Also==

Revision as of 20:36, 5 November 2019

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

By the triangle inequality rule, the last side has to be 23. 5 + 19 + 23 = 47. Therefore D, 48 is correct.

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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