document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=246702>Question 246702</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The area of a rectangle is 54 square feet. If the width is 2/3 of the length. What is the perimeter of the rectangle ?<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #180094 by </B><A HREF=/tutors/aboutme.mpl?userid=richwmiller><B>richwmiller(17219)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=richwmiller><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=richwmiller><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=180094\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> area is found by multiplying length by width<BR>\n" );
document.write( "LW=A<BR>\n" );
document.write( "W=2/3L<BR>\n" );
document.write( "54 ft^2=LW=L*(2/3*L)=(2/3)*(L^2)<BR>\n" );
document.write( "54*(3/2)=L^2<BR>\n" );
document.write( "divide 54 by 2<BR>\n" );
document.write( "27*3=L^2<BR>\n" );
document.write( "81=L^2<BR>\n" );
document.write( "9=L<BR>\n" );
document.write( "W=(2/3)*9=6\r<BR>\n" );
document.write( "\n" );
document.write( "Perimeter=2L+2W<BR>\n" );
document.write( "substitute 9 for L and 6 for W\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );