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 $\frac{60+156+65}{13}=\boxed{\textbf{(E)} \dfrac{281}{13}}$ .

2015 AMC 10B (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 AMC 10 Problems and Solutions

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

Page

Toolbox

Search

2015 AMC 10B Problems/Problem 13

Problem

Solution

See Also