2015 AMC 10B Problem 13

Question

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?

20

\dfrac{360}{17}

\dfrac{107}{5}

\dfrac{43}{2}

\dfrac{281}{13}

The intercepts are

5

and

12

Use area to find the altitude to the hypotenuse

The triangle is a right triangle with legs of

12

and

5.

This makes the hypotenuse

13.

Two of the altitudes are then

12

and

5.

Also, for any side,

A = \frac{bh}2

where

b

is the base and

h

is the altitude.

The area is

\frac {12\cdot 5}2 = 30,

so the other altitude

h

can be found with

30 = \frac{13h}2.

Thus, this altitude is

\frac{60}{13}.

Therefore, the sum is

12+5+\dfrac{60}{13} = \dfrac{281}{13} .

Thus, the correct answer is E .

Problem 13 in other years