document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=636224>Question 636224</A>: <I><SPAN STYLE=\"word-break: break-all;\"> One side of the rectangle is 1 dm less than it's diagonal. If the perimeter of the rectangle is 62 dm, find the area of the rectangle. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #400832 by </B><A HREF=/tutors/aboutme.mpl?userid=solver91311><B>solver91311(24713)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=solver91311><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=solver91311><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=400832\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font face=\"Times New Roman\" size=\"+2\">\r<BR>\n" );
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document.write( "Let <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE x\"> be the \"one side\" of the rectangle.  Then <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE x\ +\ 1\"> is the diagonal.\r<BR>\n" );
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document.write( "With a tip 'o the hat to Mr. Pythagoras, we can say that the other side of the rectangle is:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \sqrt{(x\ +\ 1)^2\ -\ x^2}\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \sqrt{x^2\ +\ 2x\ +\ 1\ -\ x^2}\">\r<BR>\n" );
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document.write( "So the missing side of the rectangle is <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \sqrt{2x\ +\ 1}\">\r<BR>\n" );
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document.write( "The perimeter of a rectangle is 2 times the length plus 2 times the width so:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ 2x\ +\ 2\sqrt{2x\ +\ 1}\ =\ 62\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ \sqrt{2x\ +\ 1}\ =\ 31\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ \sqrt{2x\ +\ 1}\ =\ 31\ -\ x\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ 2x\ +\ 1\ =\ 961\ -\ 62x\ +\ x^2\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 64x\ +\ 960\ =\ 0\">\r<BR>\n" );
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document.write( "Solve the quadratic (it factors).  One of the roots will be extraneous and you can tell by checking them in the original perimeter equation.  Discard the extraneous root.  Then calculate <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \sqrt{2x\ +\ 1}\">.  Multiply the length times the width to get the area.\r<BR>\n" );
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document.write( "John<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE e^{i\pi}\ +\ 1\ =\ 0\"><BR>\n" );
document.write( "My calculator said it, I believe it, that settles it<BR>\n" );
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