2011 AMC 8 第 25 题

Question

一个半径为

1

的圆内切于一个正方形，并外接于另一个正方形，如图所示。圆内阴影面积与两个正方形之间阴影面积的比，最接近下列哪个分数？

A circle with radius

1

is inscribed in a square and circumscribed about another square as shown. Which fraction is closest to the ratio of the circle’s shaded area to the shaded area between the two squares?

\dfrac{1}{2}

1

\dfrac{3}{2}

2

\dfrac{5}{2}

答案：A

知识点：圆面积正方形（几何）勾股定理

难度评级：1660

小提示：

所求比值是

\frac{\pi-2}{2}

，再与选项比较

The desired ratio is

\frac{\pi-2}{2}

, then compare to the choices.

大提示：

求圆面积和两个正方形的面积

Find the circle area and both square areas.

解答：

圆内阴影面积等于圆面积减去小正方形面积。内正方形的边长可由勾股定理求得：

\sqrt{1^2 + 1^2} = \sqrt{2}\text{。}

因此内正方形面积为

\sqrt{2}^2 = 2

。圆内阴影面积为

1^2\pi - 2 = \pi - 2

。

外正方形面积为

2^2 = 4

，所以两个正方形之间的阴影面积为

4 - 2 = 2

。

所求比值为

\dfrac{\pi - 2}{2} \approx \dfrac{3.14 - 2}{2} \approx \dfrac{1}{2}\text{。}

所以正确答案是 A。

The circle’s shaded area is equal to the area of the circle minus the area of the smaller square. The side length of the inner square can be calculated using the Pythagorean Theorem to get

\sqrt{1^2 + 1^2} = \sqrt{2}.

Therefore, the area of the inner square is

\sqrt{2}^2 = 2.

The circle’s shaded area is then

1^2\pi - 2 = \pi - 2.

The area of the outside square is

2^2 = 4,

so the shaded area between the two squares is

4 - 2 = 2.

The desired fraction is

\dfrac{\pi - 2}{2} \approx \dfrac{3.14 - 2}{2} \approx \dfrac{1}{2}.

Thus, A is the correct answer.

2011 AMC 8 第 25 题

答案：A

难度评级：1660

小提示：

大提示：

解答：

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