document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=101992>Question 101992</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The lenght of a rectangle is one inch less than twice its width. The diagonal of the rectangle is two inches more than its length. Find the area 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 #74159 by </B><A HREF=/tutors/aboutme.mpl?userid=edjones><B>edjones(8007)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=edjones><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=edjones><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=74159\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> L=2W-1<BR>\n" );
document.write( "c=diagonal and is the hypotenuse of a triangle with sides L and W<BR>\n" );
document.write( "c=2W+1<BR>\n" );
document.write( "a^2+b^2=c^2<BR>\n" );
document.write( "(2W-1)^2+W^2=(2W+1)^2<BR>\n" );
document.write( "4W^2-4W+1+W^2=4W^2+4W+1<BR>\n" );
document.write( "W^2-8W=0<BR>\n" );
document.write( "W(W-8)=0<BR>\n" );
document.write( "W=0, W=8 Since the rectangle can't have a width of 0 W=0 is not an answer.<BR>\n" );
document.write( "W=8 is the only answer.<BR>\n" );
document.write( "L=15<BR>\n" );
document.write( "c=17<BR>\n" );
document.write( "A=15*8=120sq inches (Answer)<BR>\n" );
document.write( "check:<BR>\n" );
document.write( "8^2+15^2=64+225=289<BR>\n" );
document.write( "17^2=289<BR>\n" );
document.write( "Ed<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );