1958 AMC 12 第 36 题

Question

一个三角形的三边长分别为

30

、

70

和

80

个单位。向长度为

80

的边作高，则该边被分成的两段中较长的一段为：

The sides of a triangle are

30,

70,

and

80

units. If an altitude is dropped upon the side of length

80,

the larger segment cut off on this side is:

62

63

64

65

66

设高把长度为

80

的边分成

x

和

80-x

Let the altitude divide the side of length

80

into

x

and

80-x

令高的平方的两个表达式相等

Equate the two expressions for the square of the altitude

设与长度为

30

的边相邻的线段为

x

。若高为

h

，则

\begin{aligned} 30^2-x^2&=h^2,\\ h^2&=70^2-(80-x)^2 \end{aligned}\text{。}

消去

x^2

并求解得

x=15

。另一段为

80-15=65

，它是较长的一段。

所以正确答案为 D。

Let

x

be the segment adjacent to the side of length

30.

If the altitude has length

h,

then

\begin{aligned} 30^2-x^2&=h^2,\\ h^2&=70^2-(80-x)^2. \end{aligned}

Cancelling

x^2

and solving gives

x=15.

The other segment is

80-15=65,

which is the larger one.

Thus, the correct answer is D.

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