Difference between revisions of "2015 AMC 8 Problems/Problem 8"
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<math>\textbf{(A) }24\qquad\textbf{(B) }29\qquad\textbf{(C) }43\qquad\textbf{(D) }48\qquad \textbf{(E) }57</math> | <math>\textbf{(A) }24\qquad\textbf{(B) }29\qquad\textbf{(C) }43\qquad\textbf{(D) }48\qquad \textbf{(E) }57</math> | ||
| − | + | ===Solution=== | |
| + | We know from the triangle inequality that the last side, <math>c</math>, fulfills <math>c<5+19=24</math>. Adding <math>5+19</math> to both sides of the inequality, we get <math>c+5+19<48</math>. However, we know that <math>c+5+19</math> is the perimeter of our triangle, so we get <math>\boxed{\textbf{(D) }~48}</math> as our answer. | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2015|num-b=7|num-a=9}} | {{AMC8 box|year=2015|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 17:33, 25 November 2015
What is the smallest whole number larger than the perimeter of any triangle with a side of length 
and a side of length 
?
Solution
We know from the triangle inequality that the last side, 
, fulfills 
. Adding 
to both sides of the inequality, we get 
. However, we know that 
is the perimeter of our triangle, so we get 
as 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
