Question

两个 $4 \times 4$ 正方形成直角相交，并平分相交的边，如图所示。圆的直径是两个交点之间的线段。从两个正方形中去掉圆后形成的阴影区域面积是多少？

Two $4 \times 4$ squares intersect at right angles, bisecting their intersecting sides, as shown. The circle's diameter is the segment between the two points of intersection. What is the area of the shaded region created by removing the circle from the squares?

$16-4\pi$

$16-2\pi$

$28-4\pi$

$28-2\pi$

$32-2\pi$

答案：D

知识点：圆面积容斥原理

难度评级：1580

解答：

两个 $4\times4$ 正方形总面积为 $32$ ，但它们的重叠部分是一个 $2\times2$ 、面积为 $4$ 的正方形。因此两个正方形并集的面积为 $32-4=28$ 。

圆的直径是这个 $2\times2$ 重叠正方形的对角线，所以直径为 $2\sqrt2$ ，半径为 $\sqrt2$ 。

圆面积为 $\pi(\sqrt2)^2=2\pi$ ，所以阴影面积为 $28-2\pi$ 。

所以正确答案是 D。

The two $4\times4$ squares have total area $32$ , but their overlap is a $2\times2$ square with area $4$ . Thus the area covered by the union of the two squares is $32-4=28$ .

The circle’s diameter is the diagonal of that $2\times2$ overlap square, so the diameter is $2\sqrt2$ and the radius is $\sqrt2$ .

The circle area is $\pi(\sqrt2)^2=2\pi$ , so the shaded area is $28-2\pi$ .

Thus, D is the correct answer.

2004 AMC 8 第 25 题

答案：D

难度评级：1580

解答：

其他年份的第 25 题