AMC12 2015 B

AMC12 2015 B · Q11

AMC12 2015 B · Q11. It mainly tests Pythagorean theorem, Area & perimeter.

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 $AD$ 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 $5 + 12 + \frac{60}{13} = \frac{281}{13}$.

将三角形的顶点标记为 $A = (0, 0)$，$B = (5, 0)$，$C = (0, 12)$。由勾股定理 $BC = 13$。两条高是 $5$ 和 $12$。设 $AD$ 为第三条高。这个三角形的面积是 $30$，所以 $\frac{1}{2} \cdot AD \cdot BC = 30$。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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