Difference between revisions of "2015 AMC 8 Problems/Problem 8"
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===Solution=== | ===Solution=== | ||
| − | 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)}~ | + | 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. | }</math> is our answer. | ||
Revision as of 07:25, 1 November 2019
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 
, and because 
is the perimeter of our triangle, $\boxed{\textbf{(D)}~48
}$ (Error compiling LaTeX. Unknown error_msg) 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
