Question

一个圆内切于一个正方形，接着一个正方形内接于这个圆，最后一个圆内切于这个正方形。较小圆的面积与较大正方形的面积之比是多少？

A circle is inscribed in a square, then a square is inscribed in this circle, and finally, a circle is inscribed in this square. What is the ratio of the area of the smaller circle to the area of the larger square?

$\dfrac{\pi}{16}$

$\dfrac{\pi}{8}$

$\dfrac{3\pi}{16}$

$\dfrac{\pi}{4}$

$\dfrac{\pi}{2}$

答案：B

知识点：圆面积正方形（几何）面积比

难度评级：1240

小提示：

设较小圆半径为 $r$ ，然后向外推。

Let the smaller circle have radius $r$ and work outward

大提示：

较大圆的半径是较小正方形对角线的一半，即 $\sqrt2\,r$ 。

The larger circle’s radius is half the diagonal of the smaller square, $\sqrt2\,r$

解答：

设较小圆半径为 $r$ ，则它的面积为 $\pi r^2$ 。

外切于这个小圆的较小正方形边长为 $2r$ ，其对角线 $2\sqrt2\,r$ 是较大圆的直径。所以较大圆半径为 $\sqrt2\,r$ 。

外切于较大圆的较大正方形边长为 $2\sqrt2\,r$ ，面积为 $8r^2$ 。

所求比为 $\dfrac{\pi r^2}{8r^2} = \dfrac{\pi}{8}\text{。}$

所以正确答案是 B。

Let the smaller circle have radius $r,$ so its area is $\pi r^2.$

The smaller square, which circumscribes this circle, has side $2r,$ and its diagonal $2\sqrt2\,r$ is the diameter of the larger circle. So the larger circle has radius $\sqrt2\,r.$

The larger square circumscribes the larger circle, so it has side $2\sqrt2\,r$ and area $8r^2.$

The desired ratio is $\dfrac{\pi r^2}{8r^2} = \dfrac{\pi}{8}.$

Thus, B is the correct answer.

2005 AMC 10B 第 7 题

答案：B

难度评级：1240

小提示：

大提示：

解答：

其他年份的第 7 题