Difference between revisions of "2015 AMC 8 Problems/Problem 8"
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===Solution 2=== | ===Solution 2=== | ||
| − | 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>D 48</math>. Re-posted by Sagadu. Original by someone else. | + | 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 01:21, 11 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 D, 48 is correct.
Solution 2
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 
. Re-posted by Sagadu. Original by someone else. Whoever took down the original, please stop.
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. 
