2010 AMC 8 Problems/Problem 13

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

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.

2010 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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All AJHSME/AMC 8 Problems and Solutions

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2010 AMC 8 Problems/Problem 13

Contents

Problem

Solution 1

Solution 2

See Also