1952 AHSME Problems/Problem 39

Problem

If the perimeter of a rectangle is $p$ and its diagonal is $d$ , the difference between the length and width of the rectangle is:

$\textbf{(A)}\ \frac {\sqrt {8d^2 - p^2}}{2} \qquad \textbf{(B)}\ \frac {\sqrt {8d^2 + p^2}}{2} \qquad \textbf{(C)}\ \frac{\sqrt{6d^2-p^2}}{2}\qquad\\ \textbf{(D)}\ \frac {\sqrt {6d^2 + p^2}}{2} \qquad \textbf{(E)}\ \frac {8d^2 - p^2}{4}$

Solution

$[asy] pair A,B,C,D,E,F,G,H; A=(0,0); B=(10,0); C=(10,5); D=(0,5); E=(5,5.5); F=(5,-0.5); G=(-0.5,2.75); H=(10.5,2.75); draw(A--B--C--D--cycle); draw (B--D); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",(-0.5,5),N); label("$x$",E); label("$x$",F); label("$y$",G); label("$y$",H); [/asy]$ Let the sides of the rectangle be x and y. WLOG, assume x>y. Then, $2x+2y=p \Rightarrow x+y=\frac{p}{2}$ .

By pythagorean theorem, $x^2 + y^2 =d^2$ .

Since $x+y=\frac{p}{2}$ , $(x+y)^2=\frac{p^2}{4} \Rightarrow x^2+2xy+y^2=\frac{p^2}{4}$ .

Rearranging to solve for $2xy$ gives $2xy = \frac{p^2}{4}-d^2$ .

Rearranging $(x-y)^2$ in terms of the defined variables becomes $(x-y)^2 = d^2 - (\frac{p^2}{4}-d^2)$ .

In order to get (x-y), we have to take the square root of the expression and simplify.

$(x-y)=\sqrt{2d^2-\frac{p^2}{4}} \Rightarrow (x-y)=\sqrt{\frac{8d^2-p^2}{4}}$ $\Rightarrow$ $(x-y)=\frac{\sqrt{8d^2+p^2}}{2}$ $\Rightarrow$ $\fbox{A}$ .

1952 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 38	Followed by Problem 40
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1952 AHSME Problems/Problem 39

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