1970 AMC 12 第 2 题

Question

一个正方形和一个圆的周长相等。圆的面积与正方形面积之比是：

A square and a circle have equal perimeters. The ratio of the area of the circle to the area of the square is:

\frac{4}{\pi}

\frac{\pi}{\sqrt2}

\frac{4}{1}

\frac{\sqrt2}{\pi}

\frac{\pi}{4}

设它们共同的周长为

p

Let the common perimeter be

p

用

p

分别表示圆的半径和正方形的边长，再将两个面积相除

Express the circle radius and square side in terms of

p

, then divide their areas

若共同的周长为

p

，则圆的半径为

\frac{p}{2\pi}

，正方形的边长为

\frac{p}{4}

。因此

\frac{\pi\left(\frac{p}{2\pi}\right)^2}{\left(\frac p4\right)^2} =\frac{\frac{p^2}{4\pi}}{\frac{p^2}{16}}=\frac4\pi\text{。}

因此，正确答案是 A。

If the common perimeter is

p,

then the circle has radius

\frac{p}{2\pi}

and the square has side

\frac{p}{4}.

Therefore

\frac{\pi\left(\frac{p}{2\pi}\right)^2}{\left(\frac p4\right)^2} =\frac{\frac{p^2}{4\pi}}{\frac{p^2}{16}}=\frac4\pi.

Therefore, the correct answer is A.

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