1952 AMC 12 第 15 题

Question

一个三角形的三边之比为

6:8:9

。则：

The sides of a triangle are in the ratio

6:8:9.

Then:

该三角形是钝角三角形

The triangle is obtuse

三个角之比为

6:8:9

The angles are in the ratio

6:8:9

该三角形是锐角三角形

The triangle is acute

最大边所对的角是最小边所对角的两倍

The angle opposite the largest side is double the angle opposite the smallest side

以上答案均不正确

None of these

比较最大边的平方与另外两边平方之和

Compare the square of the largest side with the sum of the squares of the other two

此处

9^2

小于

6^2+8^2

Here

9^2

is less than

6^2+8^2

该三角形的最大边为

9

，且

9^2=81\lt 6^2+8^2=100\text{。}

由勾股不等式的逆定理，边

9

所对的角为锐角。由于它是最大角，所以所有角都是锐角。

因此，正确答案是 C。

For a triangle with largest side

9,

9^2=81\lt 6^2+8^2=100.

By the converse of the Pythagorean inequality, the angle opposite side

9

is acute. Since it is the largest angle, every angle is acute.

Thus, the correct answer is C.

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