AMC 8 2004 Problem 25

AMC8 2004 年第 25 题

圆★★★★☆

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?

如图所示，两个 4ｘ4 正方形以直角相交，将其相交的边平分。圆的直径是两个交点之间的线段。从正方形中去除圆之后所得到的阴影部分的面积是多少？

A. $16-4\pi$
B. $16-2\pi$
C. $28-4\pi$
D. $28-2\pi$
E. $32-2\pi$

正确答案 · Correct Answer

解析 · Solution

解法1

常规解法 · 并集减圆两个 4×4 方相交，重叠是边长 2 的小正方形（边被互相平分），面积 4。并集 =16+16-4=28。圆直径 = 小方对角线

=2\sqrt2

，半径

\sqrt2

，圆面积

\pi(\sqrt2)^2=2\pi

。阴影

=28-2\pi

。

Solution 1

Elementary · union minus circle The two squares overlap in a square of side 2 (each bisects the other's sides), area 4, so their union is 16+16-4=28. The circle's diameter is the diagonal of the small square,

2\sqrt2

, so its radius is

\sqrt2

and area

2\pi

. The shaded region is

28-2\pi

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