1952 AMC 12 第 37 题

Question

在一个半径为

8

英寸的圆中，画两条相距

8

英寸且长度相等的平行弦。圆内位于两弦之间的部分面积为：

Two equal parallel chords are drawn

8

inches apart in a circle of radius

8

inches. The area of that part of the circle that lies between the chords is:

21\dfrac13\pi-32\sqrt3

32\sqrt3+21\dfrac13\pi

32\sqrt3+42\dfrac23\pi

16\sqrt3+42\dfrac23\pi

42\dfrac23\pi

知识点：圆弦扇形面积

长度相等的平行弦到圆心距离相同，所以每条弦到圆心的距离都是

4

英寸

Equal parallel chords lie the same distance from the center, so each is

4

inches from it

从整圆面积中减去两个全等的外侧弓形面积

Subtract the two congruent outer circular segments from the whole circle

两条等弦关于圆心对称，各距圆心

4

英寸。对任一弦，圆心处的半角满足

\cos\theta=\frac{4}{8}=\frac{1}{2}

，所以

\theta=60^\circ

。每个外侧弓形的面积等于一个

120^\circ

扇形面积减去等腰三角形面积：

\begin{aligned} &\frac{120}{360}\pi(8)^2 -\frac12(8)^2\sin120^\circ\\ &\qquad=\frac{64\pi}{3}-16\sqrt3 \end{aligned}\text{。}

因此，两弦之间的面积为

\begin{aligned} &64\pi-2\left(\frac{64\pi}{3}-16\sqrt3\right)\\ &\qquad=\frac{64\pi}{3}+32\sqrt3\\ &\qquad=21\frac13\pi+32\sqrt3 \end{aligned}\text{。}

因此，正确答案是 B。

The equal chords are symmetrically

4

inches from the center. For either chord, the half-angle at the center satisfies

\cos\theta=\frac{4}{8}=\frac{1}{2},

so

\theta=60^\circ.

Each outer segment is a

120^\circ

sector minus the isosceles triangle:

\begin{aligned} &\frac{120}{360}\pi(8)^2 -\frac12(8)^2\sin120^\circ\\ &\qquad=\frac{64\pi}{3}-16\sqrt3. \end{aligned}

The area between the chords is therefore

\begin{aligned} &64\pi-2\left(\frac{64\pi}{3}-16\sqrt3\right)\\ &\qquad=\frac{64\pi}{3}+32\sqrt3\\ &\qquad=21\frac13\pi+32\sqrt3. \end{aligned}

Thus, the correct answer is B.

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