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

Revision as of 23:17, 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

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.

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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 ===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.
+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.
+==See Also==
+{{AMC8 box|year=2015|num-b=7|num-a=9}}
+{{MAA Notice}}

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

Revision as of 23:17, 4 November 2020

Solution

See Also