document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=415244>Question 415244</A>: <I><SPAN STYLE=\"word-break: break-all;\"> what is the perimeter of a rectangular field if it is 528 and the length is 5 times longer than it is wide. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #291162 by </B><A HREF=/tutors/aboutme.mpl?userid=htmentor><B>htmentor(1343)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=htmentor><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=htmentor><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=291162\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Perimeter = the sum of all 4 sides<BR>\n" );
document.write( "Since it is a rectangle sum = 2l + 2w, where l = length and w = width<BR>\n" );
document.write( "Thus 2l + 2w = 528 (1)<BR>\n" );
document.write( "l = 5w (2)<BR>\n" );
document.write( "Substituting for l in (1) gives 2*5w + 2w = 528<BR>\n" );
document.write( "Or 12w = 528 <BR>\n" );
document.write( "Solving for w gives w = 528/12 = 44<BR>\n" );
document.write( "And therefore l = 5*44 = 220<BR>\n" );
document.write( "Check: 2*220 + 2*44 = 440 + 88 = 528 </SPAN>\n" );
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