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Difference between revisions of "2010 AMC 8 Problems/Problem 13"

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==Solution 2==
 
==Solution 2==
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Since the length of the shortest side is a whole number and is equal to 3/10 of the perimeter, it follows that the perimeter must be a multiple of 10 Adding the two previous integers to each answer choice, we see that 11+10+9=30 Thus, answer choice E:11 is correct
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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.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2010|num-b=12|num-a=14}}
 
{{AMC8 box|year=2010|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:24, 15 August 2021

Problem

The lengths of the sides of a triangle in inches are three consecutive integers. The length of the shortest side is $30\%$ of the perimeter. What is the length of the longest side?

$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$

Solution 1(algebra solution)

Let $n$, $n+1$, and $n+2$ be the lengths of the sides of the triangle. Then the perimeter of the triangle is $n + (n+1) + (n+2) = 3n+3$. Using the fact that the length of the smallest side is $30\%$ of the perimeter, it follows that:

$n = 0.3(3n+3) \Rightarrow n = 0.9n+0.9 \Rightarrow 0.1n = 0.9 \Rightarrow n=9$. The longest side is then $n+2 = 11$. Thus, answer choice $\boxed{\textbf{(E)}\ 11}$ is correct.

Solution 2

Since the length of the shortest side is a whole number and is equal to $\frac{3}{10}$ of the perimeter, it follows that the perimeter must be a multiple of $10$. Adding the two previous integers to each answer choice, we see that $11+10+9=30$. Thus, answer choice $\boxed{\textbf{(E)}\ 11}$ is correct.

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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