1989 AMC 12 第 13 题

Question

两条宽度为

1

的带状区域以角

\alpha

交叠，如图所示。交叠部分（阴影）的面积为

Two strips of width

1

overlap at an angle of

\alpha

as shown. The area of the overlap (shown shaded) is

\sin\alpha

\frac1{\sin\alpha}

\frac1{1-\cos\alpha}

\frac1{\sin^2\alpha}

\frac1{(1-\cos\alpha)^2}

交叠部分是高为

1

的平行四边形

The overlap is a parallelogram with altitude

1

若其倾斜边长为

L

，则

L\sin\alpha=1

If its slanted side has length

L,

then

L\sin\alpha=1

交叠部分是平行四边形。取倾斜的一边为底，其垂直高为斜带的宽

1

。若底长为

L

，其竖直分量等于水平带的宽，所以

L\sin\alpha=1

。因此

L=\frac{1}{\sin\alpha}

，面积为

L\cdot1=\frac{1}{\sin\alpha}

。

所以正确答案是 B。

The overlap is a parallelogram. Taking a slanted side as its base, the perpendicular height is the width

1

of the slanted strip. If that base has length

L,

its vertical component is the width of the horizontal strip, so

L\sin\alpha=1.

Hence

L=\frac{1}{\sin\alpha},

and the area is

L\cdot1=\frac{1}{\sin\alpha}.

Thus the correct answer is B.

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