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

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===Solution===
 
===Solution===
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By the triangle inequality rule, the last side has to be 23. 5 + 19 + 23 = 47. Therefore D, 48
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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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is correct.
 
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===Solution 2===
 
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By the triangle inequality rule, the last side has to be less than the sum of the other 2 sides. If we assign the value p for the final side length, then p < 5 + 19. This simplifies to p < 24. If we add 5 + 19 to both sides of the inequality, then we get p + 5 + 19 < 24 + 5 + 19. If you simplify the right side, you will get p + 5 + 19 < 48. Notice that p + 5 + 19 is the perimeter of the triangle, and the inequality shows that the perimeter is less than 48, so our answer is <math>{\textbf{(D) }48}</math>. Re-posted by Sagadu. Original by someone else. Whoever took down the original, please stop.
 
  
 
==See Also==
 
==See Also==

Revision as of 01:29, 11 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

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 (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

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