document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1166618>Question 1166618</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The perimeter of a rectangle is 48 cm and its area is 135 cm^2. Find the length and the breadth/width of the rectangle.\r<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #791132 by </B><A HREF=/tutors/aboutme.mpl?userid=greenestamps><B>greenestamps(13200)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=greenestamps><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=greenestamps><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=791132\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "This one is not quite as obvious as the other similar problem (rectangle area 130 with length 3 more than the width; answer length 13, width 10).<br><BR>\n" );
document.write( "However, note in the solution from the other tutor that in solving the problem using formal algebra you have to factor a quadratic equation by finding two numbers whose sum is 24 and whose product is 135.<br><BR>\n" );
document.write( "But that is exactly what the original problem asks you to do -- so the formal algebra doesn't actually get you anywhere towards solving the problem.<br><BR>\n" );
document.write( "You know length plus width is 24; find two numbers with a sum of 24 that have a product of 135.<br><BR>\n" );
document.write( "Obviously one of the numbers has to end in \"5\"; a very little bit of trial and error finds the dimensions to be 15 and 9.<br><BR>\n" );
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