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Difference between revisions of "2015 AMC 10B Problems/Problem 13"
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==Problem 13==
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==Problem==
  
 
The line <math>12x+5y=60</math> forms a triangle with the coordinate axes. What is the sum of the lengths of the altitudes of this triangle?
 
The line <math>12x+5y=60</math> forms a triangle with the coordinate axes. What is the sum of the lengths of the altitudes of this triangle?

Revision as of 23:27, 3 March 2015

Problem

The line $12x+5y=60$ forms a triangle with the coordinate axes. What is the sum of the lengths of the altitudes of this triangle?

$\textbf{(A) } 20 \qquad\textbf{(B) } \dfrac{360}{17} \qquad\textbf{(C) } \dfrac{107}{5} \qquad\textbf{(D) } \dfrac{43}{2} \qquad\textbf{(E) } \dfrac{281}{13}$

Solution

We find the x-intercepts and the y-intercepts to find the intersections of the axes and the line. If $x=0$, then $y=12$. If $y$ is $0$, then $x=5$. Our three vertices are $(0,0)$, $(5,0)$, and $(0,12)$. Two of our altitudes are $5$ and $12$. Since the area of the triangle is $30$, our final altitude has to be $30$ divided by the hypotenuse. By the Pythagorean Theorem, our hypotenuse is $13$, so the sum of our altitudes is $\boxed{\textbf{(E)} 281/13}$.

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