document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=321119>Question 321119</A>: <I><SPAN STYLE=\"word-break: break-all;\"> If one side of a square is increased by 8 cm and an adjacent side decreased by 2 cm a rectangle is formed whose permeter is 40 cm.  Find the length of a side of the square. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #229964 by </B><A HREF=/tutors/aboutme.mpl?userid=mananth><B>mananth(16946)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=mananth><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=mananth><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=229964\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let the length of the side of square be x<BR>\n" );
document.write( "x+8, x-2 are the sides of the rectangle.<BR>\n" );
document.write( "Perimeter = 2L+2W<BR>\n" );
document.write( "2(x-2)+2(x+8)=40<BR>\n" );
document.write( "2x-4+2x+16=40<BR>\n" );
document.write( "4x+12=404x=28<BR>\n" );
document.write( "x=7 cm the length of side of square </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );