2004 AMC 8 第 25 题

先试着解答 2004 AMC 8 第 25 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2004 AMC 8 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

25.

两个 4×44 \times 4 正方形成直角相交,并且每个正方形都平分另一个正方形被其穿过的两条边,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 4×44 \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?
16−4π16-4\pi
16−2π16-2\pi
28−4π28-4\pi
28−2π28-2\pi
32−2π32-2\pi
答案:D
知识点:圆面积容斥原理
小提示:
求两个正方形并集覆盖的面积
Find the area covered by the union of the two squares.
大提示:
圆的直径是 2×22\times2 重叠正方形的对角线
The circle diameter is the diagonal of the 2×22\times2 overlap square.
解答:
两个 4×44\times4 正方形总面积为 3232,但它们的重叠部分是一个 2×22\times2、面积为 44 的正方形。因此两个正方形并集的面积为 32−4=2832-4=28。
圆的直径是这个 2×22\times2 重叠正方形的对角线,所以直径为 222\sqrt2,半径为 2\sqrt2。
圆面积为 π(2)2=2π\pi(\sqrt2)^2=2\pi,所以阴影面积为 28−2π28-2\pi。
所以正确答案是 D。
The two 4×44\times4 squares have total area 32,32, but their overlap is a 2×22\times2 square with area 4.4. Thus the area covered by the union of the two squares is 32−4=28.32-4=28.
The circle’s diameter is the diagonal of that 2×22\times2 overlap square, so the diameter is 222\sqrt2 and the radius is 2.\sqrt2.
The circle area is π(2)2=2π,\pi(\sqrt2)^2=2\pi, so the shaded area is 28−2π.28-2\pi.
Thus, D is the correct answer.
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