Difference between revisions of "2010 AMC 8 Problems/Problem 13"
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==Solution 2== | ==Solution 2== | ||
Since the length of the shortest side is a whole number and is equal to <math>\frac{3}{10}</math> of the perimeter, it follows that the perimeter must be a multiple of <math>10</math>. Adding the two previous integers to each answer choice, we see that <math>11+10+9=30</math>. Thus, answer choice <math>\boxed{\textbf{(E)}\ 11}</math> is correct. | Since the length of the shortest side is a whole number and is equal to <math>\frac{3}{10}</math> of the perimeter, it follows that the perimeter must be a multiple of <math>10</math>. Adding the two previous integers to each answer choice, we see that <math>11+10+9=30</math>. Thus, answer choice <math>\boxed{\textbf{(E)}\ 11}</math> is correct. | ||
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+ | ==Video by MathTalks== | ||
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+ | https://www.youtube.com/watch?v=6hRHZxSieKc | ||
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==See Also== | ==See Also== | ||
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{{AMC8 box|year=2010|num-b=12|num-a=14}} | {{AMC8 box|year=2010|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 20:54, 22 January 2023
Problem
The lengths of the sides of a triangle in inches are three consecutive integers. The length of the shortest side is of the perimeter. What is the length of the longest side?
Solution 1(algebra solution)
Let ,
, and
be the lengths of the sides of the triangle. Then the perimeter of the triangle is
. Using the fact that the length of the smallest side is
of the perimeter, it follows that:
. The longest side is then
. Thus, answer choice
is correct.
Solution 2
Since the length of the shortest side is a whole number and is equal to of the perimeter, it follows that the perimeter must be a multiple of
. Adding the two previous integers to each answer choice, we see that
. Thus, answer choice
is correct.
Video by MathTalks
https://www.youtube.com/watch?v=6hRHZxSieKc
See Also
2010 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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 |
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