document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=41008>Question 41008</A>: <I><SPAN STYLE=\"word-break: break-all;\"> width is 10 feet less than length.  the perimeter is 100 feet.  how do i find the area? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #26428 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=26428\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> width is 10 feet less than length.  the perimeter is 100 feet.  how do i find the area?\r<BR>\n" );
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document.write( "Let the length be \"x\".<BR>\n" );
document.write( "Then the width is \"x-10\"\r<BR>\n" );
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document.write( "Perimeter=2(l+w)=100 ft.<BR>\n" );
document.write( "l+w=50<BR>\n" );
document.write( "x+x-10=50<BR>\n" );
document.write( "2x=60<BR>\n" );
document.write( "x=30 (length of the rectangle)<BR>\n" );
document.write( "x-10=20 (width of the rectangle)<BR>\n" );
document.write( "Area = 20*30=600 sq ft.<BR>\n" );
document.write( "Cheers,<BR>\n" );
document.write( "Stan H. </SPAN>\n" );
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