AMC10 2015 B

AMC10 2015 B · Q13

AMC10 2015 B · Q13. It mainly tests Area & perimeter, Coordinate geometry.

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?

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

(A) 20 20

(B) $\frac{360}{17}$ $\frac{360}{17}$

(D) $\frac{43}{2}$ $\frac{43}{2}$

(E) $\frac{281}{13}$ $\frac{281}{13}$

Answer

Correct choice: (E)

正确答案：（E）

Solution

Label the vertices of the triangle $A = (0,0)$, $B = (5,0)$, and $C = (0,12)$. By the Pythagorean Theorem $BC = 13$. Two altitudes are 5 and 12. Let $\overline{AD}$ be the third altitude. The area of this triangle is 30, so $\frac{1}{2} \cdot AD \cdot BC = 30$. Therefore $AD = \frac{60}{13}$. The sum of the lengths of the altitudes is $5 + 12 + \frac{60}{13} = \frac{281}{13}$.

将三角形的顶点标为$A = (0,0)$、$B = (5,0)$和$C = (0,12)$。由勾股定理$BC = 13$。两个高是5和12。设$\overline{AD}$是第三条高。这个三角形的面积是30，因此$\frac{1}{2} \cdot AD \cdot BC = 30$。因此$AD = \frac{60}{13}$。高的长度之和是$5 + 12 + \frac{60}{13} = \frac{281}{13}$。

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