1951 AMC 12 第 48 题

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所有题目均经美国数学协会(MAA)官方合法授权使用。

48.

内接于半圆的正方形面积与内接于整圆的正方形面积之比为:

The area of a square inscribed in a semicircle is to the area of the square inscribed in the entire circle as:

1:21:2

2:32:3

2:52:5

3:43:4

3:53:5

答案:C
知识点:正方形(几何)面积比
难度评级:1470
小提示:

设圆的半径为 RR,内接于半圆的正方形边长为 ss

Let the circle have radius RR and the semicircle-square have side ss

大提示:

对于内接于半圆的正方形,由一个上顶点可得 (s2)2+s2=R2(\frac{s}{2})^2+s^2=R^2

For the semicircle-square, a top vertex gives (s2)2+s2=R2(\frac{s}{2})^2+s^2=R^2

解答:

对于内接于半圆的正方形,将其底边放在直径上。一个上顶点到圆心的水平距离为 s2\frac{s}{2},竖直距离为 ss,所以 (s2)2+s2=R2 \left(\frac s2\right)^2+s^2=R^2\text{,}得到 s2=4R25s^2=\frac{4R^2}{5}。内接于整圆的正方形对角线长为 2R2R,故面积为 2R22R^2。所求比为 4R252R2=25 \frac{\frac{4R^2}{5}}{2R^2}=\frac25\text{。}

因此,正确答案是 C

For the square in the semicircle, put its base on the diameter. A top vertex has horizontal distance s2\frac{s}{2} from the center and vertical distance s,s, so (s2)2+s2=R2, \left(\frac s2\right)^2+s^2=R^2, giving s2=4R25.s^2=\frac{4R^2}{5}. A square inscribed in the full circle has diagonal 2R,2R, hence area 2R2.2R^2. The ratio is 4R252R2=25. \frac{\frac{4R^2}{5}}{2R^2}=\frac25.

Thus, the correct answer is C.

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