1955 AMC 12 第 14 题

Question

矩形

R

的长比正方形

S

的边长多

10

%，矩形的宽比正方形的边长少

10

%。两者面积之比

R:S

为：

The length of rectangle

R

is

10

percent more than the side of square

S.

The width of the rectangle is

10

percent less than the side of the square. The ratio of the areas,

R:S,

is:

99:100

101:100

1:1

199:200

201:200

设正方形的边长为

s

Let the square’s side length be

s

矩形的面积为

(1.1s)(0.9s)

The rectangle’s area is

(1.1s)(0.9s)

若正方形的边长为

s

，则矩形的长和宽分别为

1.1s

和

0.9s

。因此

\begin{aligned} \frac{[R]}{[S]} &=\frac{(1.1s)(0.9s)}{s^2}\\ &=0.99=\frac{99}{100} \end{aligned}\text{。}

因此，正确答案是 A。

If the square side is

s,

then the rectangle has dimensions

1.1s

and

0.9s.

Therefore

\begin{aligned} \frac{[R]}{[S]} &=\frac{(1.1s)(0.9s)}{s^2}\\ &=0.99=\frac{99}{100}. \end{aligned}

Thus, the correct answer is A.

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