# 2015 AMC 10B Problems/Problem 13

## 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 $12$, and since it is a 5-12-13 right triangle, the hypotenuse is $13$. Since the area of the triangle is $30$, so our final altitude is $\frac{30(2)}{13}=\frac{60}{13}$. The sum of our altitudes is $\frac{60+156+65}{13}=\boxed{\textbf{(E)} \dfrac{281}{13}}$. Note that there is no need to calculate the final answer after we know that the third altitude has length $\frac{60}{13}$ since $E$ is the only choice with a denominator of $13$.

## See Also

 2015 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 12 Followed byProblem 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 AMC 10 Problems and Solutions

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

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