2014 AMC 10A 第 9 题

Question

一个直角三角形的两条直角边也是高，长度分别为

2\sqrt3

和

6

。这个三角形的第三条高有多长？

The two legs of a right triangle, which are altitudes, have lengths

2\sqrt3

and

6.

How long is the third altitude of the triangle?

1

2

3

4

5

用两条直角边求三角形面积。

Use the two given legs to find the area

把斜边作为底边来表示第三条高。

Use the hypotenuse as the base for the third altitude

三角形面积为

\dfrac{1}{2} \cdot 2\sqrt{3} \cdot 6 = 6\sqrt{3}\text{。}

斜边长为

\sqrt{(2\sqrt{3})^2 + 6^2} = \sqrt{48} = 4\sqrt{3}\text{。}

设斜边上的高为

h

，

\dfrac{1}{2} \cdot h \cdot 4\sqrt{3} = 6\sqrt{3}

h = 3\text{。}

所以正确答案是 C。

We get that the area of the triangle is

\dfrac{1}{2} \cdot 2\sqrt{3} \cdot 6 = 6\sqrt{3}.

The length of the hypotenuse is

\sqrt{(2\sqrt{3})^2 + 6^2} = \sqrt{48} = 4\sqrt{3}.

Dropping the altitude,

h,

from the vertex to the hypotenuse, we get that

\dfrac{1}{2} \cdot h \cdot 4\sqrt{3} = 6\sqrt{3}

h = 3.

Thus, C is the correct answer.

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