AMC8 2004

AMC8 2004 · Q25

AMC8 2004 · Q25. It mainly tests Pythagorean theorem, Area & perimeter.

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?

两个 $4 \times 4$ 正方形垂直相交，平分它们的相交边，如图所示。圆的直径是两个交点之间的线段。从正方形中移除圆后形成的阴影区域的面积是多少？

(A) $16 - 4\pi$ $16 - 4\pi$

(B) $16 - 2\pi$ $16 - 2\pi$

(D) $28 - 2\pi$ $28 - 2\pi$

(E) $32 - 2\pi$ $32 - 2\pi$

Answer

Correct choice: (D)

正确答案：（D）

Solution

(D) The overlap of the two squares is a smaller square with side length 2, so the area of the region covered by the squares is $2(4\times 4)-(2\times 2)=32-4=28.$ The diameter of the circle has length $\sqrt{2^2+2^2}=\sqrt{8}$, the length of the diagonal of the smaller square. The shaded area created by removing the circle from the squares is $28-\pi\left(\frac{\sqrt{8}}{2}\right)^2=28-2\pi.$

（D）两个正方形的重叠部分是一个边长为 2 的小正方形，因此正方形覆盖的区域面积为 $2(4\times 4)-(2\times 2)=32-4=28.$ 圆的直径长度为 $\sqrt{2^2+2^2}=\sqrt{8}$，即小正方形对角线的长度。将圆从正方形覆盖区域中去除后得到的阴影面积为 $28-\pi\left(\frac{\sqrt{8}}{2}\right)^2=28-2\pi.$

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