1952 AMC 12 第 39 题

Question

若一个长方形的周长为

p

，对角线长为

d

，则其长与宽之差为：

If the perimeter of a rectangle is

p

and its diagonal is

d,

the difference between the length and width of the rectangle is:

\dfrac{\sqrt{8d^2-p^2}}2

\dfrac{\sqrt{8d^2+p^2}}2

\dfrac{\sqrt{6d^2-p^2}}2

\dfrac{\sqrt{6d^2+p^2}}2

\dfrac{\sqrt{8d^2-p^2}}4

设两条边长为

L

和

W

，并分别写出关于

L+W

与

L^2+W^2

的方程

Let the side lengths be

L

and

W

, and write equations for

L+W

and

L^2+W^2

写出

2(L^2+W^2)

，再减去

(L+W)^2

，得到

(L-W)^2

Write

2(L^2+W^2)

, then subtract

(L+W)^2

to obtain

(L-W)^2

由周长和对角线可得

\begin{aligned} L+W&=\frac p2,\\ L^2+W^2&=d^2 \end{aligned}\text{。}

因此

\begin{aligned} (L-W)^2 &=2(L^2+W^2)\\ &\quad{}-(L+W)^2\\ &=2d^2-\frac{p^2}{4}\\ &=\frac{8d^2-p^2}{4} \end{aligned}\text{。}

取非负平方根，得到

L-W=\frac{\sqrt{8d^2-p^2}}2\text{。}

因此，正确答案是 A。

The perimeter and diagonal give

\begin{aligned} L+W&=\frac p2,\\ L^2+W^2&=d^2. \end{aligned}

Hence

\begin{aligned} (L-W)^2 &=2(L^2+W^2)\\ &\quad{}-(L+W)^2\\ &=2d^2-\frac{p^2}{4}\\ &=\frac{8d^2-p^2}{4}. \end{aligned}

Taking the nonnegative square root gives

L-W=\frac{\sqrt{8d^2-p^2}}2.

Thus, the correct answer is A.

其他年份的第 39 题