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

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By MathTalks
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Revision as of 11:55, 14 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.

See Also

Video:

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

By MathTalks


2010 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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