1952 AMC 12 Problem 39

Question

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

Let the side lengths be

L

and

W

, and write equations for

L+W

and

L^2+W^2

Write

2(L^2+W^2)

, then subtract

(L+W)^2

to obtain

(L-W)^2

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.

Problem 39 in other years