document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1005117>Question 1005117</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The length of a rectangle is 3cm more than the width. if the width is increased by 2cm. and the length decreased by1cm., the figure becomes a square with an area of 81 sq. cms. what is the original perimeter of the rectangle </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #621422 by </B><A HREF=/tutors/aboutme.mpl?userid=addingup><B>addingup(3677)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=addingup><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=addingup><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=621422\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> (w+3-1)(w+2)= 81<BR>\n" );
document.write( "(w+2)(w+2)= 81 now FOIL<BR>\n" );
document.write( "w^2 First<BR>\n" );
document.write( "2w Outer<BR>\n" );
document.write( "2w Inner<BR>\n" );
document.write( "4 Last<BR>\n" );
document.write( "Now we have:<BR>\n" );
document.write( "w^2+2w+2w+4<BR>\n" );
document.write( "= w^2+4w+4= 81 write the left side as a square<BR>\n" );
document.write( "(w+2)^2= 81 take the square root of both sides:<BR>\n" );
document.write( "w+2= 9 or w+2= -9<BR>\n" );
document.write( "We can't use the negative, toss it out. Our answer is:<BR>\n" );
document.write( "w= 9 This is the width of the rectangle, and the length is:<BR>\n" );
document.write( "l= 9+3= 12<BR>\n" );
document.write( "Perimeter: 2l+2w<BR>\n" );
document.write( "2(12)+2(9)= 24+18= 42 is the original perimeter of the rectangle \r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( " <BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );