AMC10 2014 A

AMC10 2014 A · Q9

AMC10 2014 A · Q9. It mainly tests Triangles (properties), Pythagorean theorem.

The two legs of a right triangle, which are altitudes, have lengths $2\sqrt{3}$ and 6. How long is the third altitude of the triangle?

一个直角三角形的两条直角边（同时也是高）长度分别为 $2\sqrt{3}$ 和 6。这个三角形的第三条高有多长？

(A) 1 1

(B) 2 2

(D) 4 4

(E) 5 5

Answer

Correct choice: (C)

正确答案：（C）

Solution

The area of the triangle is $\frac{1}{2} \cdot 2\sqrt{3} \cdot 6 = 6\sqrt{3}$. By the Pythagorean Theorem, the hypotenuse has length $4\sqrt{3}$. The desired altitude has length $\frac{6\sqrt{3}}{\frac{1}{2} \cdot 4\sqrt{3}} = 3$.

三角形的面积为 $\frac{1}{2} \cdot 2\sqrt{3} \cdot 6 = 6\sqrt{3}$。由勾股定理，斜边长为 $4\sqrt{3}$。所求的高为 $\frac{6\sqrt{3}}{\frac{1}{2} \cdot 4\sqrt{3}} = 3$。

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