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

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===Solution===
 
===Solution===
As per the tringle inequality, the sum of the length of any 2 sides is greater than the largest side. So, let <math>x</math> be the third side and <math>19</math> be the largest side. <math>5+x > 19</math>. This gives the smallest value of <math>x</math> as <math>15</math>. The perimeter would be <math>5 + 15 + 19 = 39</math> The smallest side which would be greater than the perimeter would be 43.
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We know from the triangle inequality that the last side, <math>s</math>, fulfills <math>s<5+19=24</math>. Adding <math>5+19</math> to both sides of the inequality, we get <math>s+5+19<48</math>, and because <math>s+5+19</math> is the perimeter of our triangle, <math>\boxed{\textbf{(D)}\ 48}</math> is our answer.
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==See Also==
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{{AMC8 box|year=2015|num-b=7|num-a=9}}
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{{MAA Notice}}

Revision as of 00:17, 5 November 2020

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.

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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