document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1005728>Question 1005728</A>: <I><SPAN STYLE=\"word-break: break-all;\"> the length of a rectangle is twice the width. If the perimeter is 20cm, 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 #621876 by </B><A HREF=/tutors/aboutme.mpl?userid=fractalier><B>fractalier(6550)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=fractalier><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=fractalier><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=621876\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Call length and width, L and W.  We have<BR>\n" );
document.write( "L = 2W and<BR>\n" );
document.write( "perimeter P = 2L + 2W = 20<BR>\n" );
document.write( "Substitute the first into the second equation and get<BR>\n" );
document.write( "2(2W) + 2W = 20<BR>\n" );
document.write( "4W + 2W = 20<BR>\n" );
document.write( "6W = 20<BR>\n" );
document.write( "W = 20/6 = 10/3 = 3 1/3 cm<BR>\n" );
document.write( "L = 6 2/3 cm </SPAN>\n" );
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