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Difference between revisions of "1952 AHSME Problems/Problem 39"

(Created page with "== Problem == If the perimeter of a rectangle is <math>p</math> and its diagonal is <math>d</math>, the difference between the length and width of the rectangle is: <math>\tex...")
 
(→‎Solution)
 
Line 10: Line 10:
  
 
== Solution ==
 
== Solution ==
−
<math>\fbox{}</math>
+
<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);
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label("$x$",E); label("$x$",F); label("$y$",G); label("$y$",H);
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</asy>
 +
Let the sides of the rectangle be x and y. WLOG, assume x>y. Then, <math>2x+2y=p \Rightarrow x+y=\frac{p}{2}</math>.
 +
 
 +
By pythagorean theorem, <math>x^2 + y^2 =d^2</math>.
 +
 
 +
Since <math>x+y=\frac{p}{2}</math>, <math>(x+y)^2=\frac{p^2}{4} \Rightarrow x^2+2xy+y^2=\frac{p^2}{4}</math>.
 +
 
 +
Rearranging to solve for <math>2xy</math> gives <math>2xy = \frac{p^2}{4}-d^2</math>.
 +
 
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Rearranging <math>(x-y)^2</math> in terms of the defined variables becomes <math>(x-y)^2 = d^2 - (\frac{p^2}{4}-d^2) </math>.
 +
 
 +
In order to get (x-y), we have to take the square root of the expression and simplify.
 +
 
 +
<math>(x-y)=\sqrt{2d^2-\frac{p^2}{4}} \Rightarrow (x-y)=\sqrt{\frac{8d^2-p^2}{4}}</math> <math>\Rightarrow</math> <math>(x-y)=\frac{\sqrt{8d^2+p^2}}{2}</math> <math>\Rightarrow</math>
 +
<math>\fbox{A}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 21:36, 15 April 2020

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.

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

See also

1952 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 38 		Followed by
Problem 40
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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