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

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<math> \textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11 </math>
 
<math> \textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11 </math>
  
==Solution 1==
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==Solution 1(algebra solution)==
 
Let <math>n</math>, <math>n+1</math>, and <math>n+2</math> be the lengths of the sides of the triangle. Then the perimeter of the triangle is <math>n + (n+1) + (n+2) = 3n+3</math>. Using the fact that the length of the smallest side is <math>30\%</math> of the perimeter, it follows that:
 
Let <math>n</math>, <math>n+1</math>, and <math>n+2</math> be the lengths of the sides of the triangle. Then the perimeter of the triangle is <math>n + (n+1) + (n+2) = 3n+3</math>. Using the fact that the length of the smallest side is <math>30\%</math> of the perimeter, it follows that:
  
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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 19:54, 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 $30\%$ 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.

Video by MathTalks

https://www.youtube.com/watch?v=6hRHZxSieKc




See Also

2010 AMC 8 (ProblemsAnswer KeyResources)
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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