Question

直线 $12x+5y=60$ 与坐标轴围成一个三角形。这个三角形三条高的长度之和是多少？

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?

$20$

$\dfrac{360}{17}$

$\dfrac{107}{5}$

$\dfrac{43}{2}$

$\dfrac{281}{13}$

直线与坐标轴的截距为 $5$ 和 $12$ ，所以三角形是直角三角形，两条直角边为五和十二，斜边为 $13$ 。

两条直角边对应的高分别为 $5$ 和 $12$ 。用面积公式 $A = \frac{bh}2$ ，其中 $b$ 是底、 $h$ 是高。

三角形面积为 $\frac {12\cdot 5}2 = 30$ 设斜边上的高为 $h$ ，则 $30 = \frac{13h}2$ 所以这条高为 $\frac{60}{13}$ 。

三条高之和为 $12+5+\dfrac{60}{13} = \dfrac{281}{13} .$

所以正确答案是 E。

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 .