Difference between revisions of "2015 AMC 8 Problems/Problem 8"
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===Solution=== | ===Solution=== | ||
| − | By the triangle inequality rule, the last side has to be 23. 5 + 19 + 23 = 47. Therefore D | + | By the triangle inequality rule, the last side has to be 23. 5 + 19 + 23 = 47. Therefore <math>\qquad\textbf{(D) }48</math> |
| + | is correct. | ||
==See Also== | ==See Also== | ||
Revision as of 23:28, 8 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
By the triangle inequality rule, the last side has to be 23. 5 + 19 + 23 = 47. Therefore 
is correct.
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. 
