Difference between revisions of "2015 AMC 12A Problems/Problem 2"
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==Solution== | ==Solution== | ||
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| + | Letting <math>x</math> be the third side, then by the triangle inequality, <math>20-15 < x < 20+15</math>, or <math>5 < x < 35</math>. Therefore the perimeter must be greater than 40 but less than 70. 72 is not in this range, so <math>\boxed{\textbf{(E)} \, 72}</math> is our answer. | ||
== See Also == | == See Also == | ||
{{AMC12 box|year=2015|ab=A|num-b=1|num-a=3}} | {{AMC12 box|year=2015|ab=A|num-b=1|num-a=3}} | ||
Revision as of 11:02, 5 February 2015
Problem
Two of the three sides of a triangle are 20 and 15. Which of the following numbers is not a possible perimeter of the triangle?
$\textbf{(A)}\ 52\qquad\textbf{(B)}\ 57\qquad\textbf{(C)}\ 62\qquad\textbf{(D)}}\ 67\qquad\textbf{(E)}\ 72$ (Error compiling LaTeX. Unknown error_msg)
Solution
Letting 
be the third side, then by the triangle inequality, 
, or 
. Therefore the perimeter must be greater than 40 but less than 70. 72 is not in this range, so 
is our answer.
See Also
| 2015 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 1 |
Followed by Problem 3 |
| 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 AMC 12 Problems and Solutions | |
