document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=79074>Question 79074</A>: <I><SPAN STYLE=\"word-break: break-all;\"> If the sides of a square are decreased by 3 cm, the area is decreased by 81 cm2.  What were the dimensions of the original square? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #56754 by </B><A HREF=/tutors/aboutme.mpl?userid=checkley75><B>checkley75(3666)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=checkley75><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=56754\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> x^2=(x-3)^2+81<BR>\n" );
document.write( "x^2=x^2-6x+9+81<BR>\n" );
document.write( "6x=90<BR>\n" );
document.write( "x=90/6<BR>\n" );
document.write( "x=15 cm. for the original length of the sides of the square.<BR>\n" );
document.write( "proof<BR>\n" );
document.write( "15^2=(15-3)^2+81<BR>\n" );
document.write( "225=12^2+81<BR>\n" );
document.write( "225=144+81<BR>\n" );
document.write( "225=225<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );