document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=173453>Question 173453</A>: <I><SPAN STYLE=\"word-break: break-all;\"> if the perimeter of a rectangle is 20 feet anf the diagonal is 2 square root 13 feet, then what are the length and width? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #128311 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=128311\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The perimeter of a rectangle is given by <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=P=2l%2B2w ALIGN=MIDDLE  ALT=\"P=2l%2B2w\"   DISABLED_style= \"max-width:90%; height:auto;\" >, so we know that <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2l%2B2w=20 ALIGN=MIDDLE  ALT=\"2l%2B2w=20\"   DISABLED_style= \"max-width:90%; height:auto;\" >.\r<BR>\n" );
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document.write( "Solving for <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=l ALIGN=MIDDLE  ALT=\"l\"   DISABLED_style= \"max-width:90%; height:auto;\" >:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2l%2B2w=20 ALIGN=MIDDLE  ALT=\"2l%2B2w=20\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2l=20-2w ALIGN=MIDDLE  ALT=\"2l=20-2w\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=l=10-w ALIGN=MIDDLE  ALT=\"l=10-w\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "The measure of a diagonal of a rectangle in terms of the length and width is given by the Pythagorean Theorem:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=d=sqrt%28l%5E2%2Bw%5E2%29 ALIGN=MIDDLE  ALT=\"d=sqrt%28l%5E2%2Bw%5E2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > where <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=d ALIGN=MIDDLE  ALT=\"d\"   DISABLED_style= \"max-width:90%; height:auto;\" > is the measure of the diagonal.\r<BR>\n" );
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document.write( "But we know that:  <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sqrt%28l%5E2%2Bw%5E2%29=2sqrt%2813%29 ALIGN=MIDDLE  ALT=\"sqrt%28l%5E2%2Bw%5E2%29=2sqrt%2813%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Square both sides:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=l%5E2%2Bw%5E2=4%2A13=52 ALIGN=MIDDLE  ALT=\"l%5E2%2Bw%5E2=4%2A13=52\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Now substitute the expression for <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=l ALIGN=MIDDLE  ALT=\"l\"   DISABLED_style= \"max-width:90%; height:auto;\" > in terms of <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=w ALIGN=MIDDLE  ALT=\"w\"   DISABLED_style= \"max-width:90%; height:auto;\" > developed earlier:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%2810-w%29%5E2%2Bw%5E2=4%2A13=52 ALIGN=MIDDLE  ALT=\"%2810-w%29%5E2%2Bw%5E2=4%2A13=52\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Expand, collect terms, and put into standard form for a quadratic:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=100-20w%2Bw%5E2%2Bw%5E2=52 ALIGN=MIDDLE  ALT=\"100-20w%2Bw%5E2%2Bw%5E2=52\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2w%5E2-20w%2B48=0 ALIGN=MIDDLE  ALT=\"2w%5E2-20w%2B48=0\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Divide through by <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2 ALIGN=MIDDLE  ALT=\"2\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=w%5E2-10w%2B24=0 ALIGN=MIDDLE  ALT=\"w%5E2-10w%2B24=0\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Leaving you with a factorable quadratic to solve.  Hint:  <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-4%2A-6=24 ALIGN=MIDDLE  ALT=\"-4%2A-6=24\"   DISABLED_style= \"max-width:90%; height:auto;\" > and <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-4-6=-10 ALIGN=MIDDLE  ALT=\"-4-6=-10\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Since you are solving for the width, pick the smaller of the two roots as your width and calculate the length from <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=l=10-w ALIGN=MIDDLE  ALT=\"l=10-w\"   DISABLED_style= \"max-width:90%; height:auto;\" > (which you will find is the other root of your quadratic) </SPAN>\n" );
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