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

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===Solution===
 
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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 <math>\box{43}</math>.
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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.

Revision as of 20:52, 4 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

As per the tringle inequality, the sum of the length of any 2 sides is greater than the largest side. So, let $x$ be the third side and $19$ be the largest side. $5+x > 19$. This gives the smallest value of $x$ as $15$. The perimeter would be $5 + 15 + 19 = 39$ The smallest side which would be greater than the perimeter would be 43.

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