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

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 $30\%$ 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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 1: / Line 1: @@
 ==Problem==
 The lengths of the sides of a triangle in inches are three consecutive integers. The length of the shortest side is <math>30\%</math> of the perimeter. What is the length of the longest side?
 <math> \textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11 </math>
-==Solution 1==
+==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 longest 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:
 <math>n = 0.3(3n+3) \Rightarrow n = 0.9n+0.9 \Rightarrow 0.1n = 0.9 \Rightarrow n=9</math>. The longest side is then <math>n+2 = 11</math>. Thus, answer choice <math>\boxed{\textbf{(E)}\ 11}</math> is correct.
 ==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=20</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.
+==Video by MathTalks==
+https://www.youtube.com/watch?v=6hRHZxSieKc
 ==See Also==
-{{AMC8 box|year=2010|before=num-b=12|num-a=14}}
+{{AMC8 box|year=2010|num-b=12|num-a=14}}
+{{MAA Notice}}

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