document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=121968>Question 121968</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The length of a rectangle is 1cm longer than its width.  If the diagonal of the rectangle is 4cm, what are the dimensions(the length and width) of the rectangle?\r<BR>\n" );
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document.write( "x + (x+1)= 4\r<BR>\n" );
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document.write( "Am I on the right track in solving this problem? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #89542 by </B><A HREF=/tutors/aboutme.mpl?userid=oscargut><B>oscargut(2103)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=oscargut><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=oscargut><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=89542\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let L=length and W=width\r<BR>\n" );
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document.write( "then L=W+1 \r<BR>\n" );
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document.write( "and using pitagoras theorem L^2+W^2=4^2\r<BR>\n" );
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document.write( "then  (W+1)^2+W^2=16<BR>\n" );
document.write( "then   2W^2+2W-15=0\r<BR>\n" );
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document.write( "so W= (-2+sqrt(124))/4 or W=(-2-sqrt(124))/4 the second solution is negative so<BR>\n" );
document.write( "W= (-2+sqrt(124))/4 using L=W+1 you can calculate L </SPAN>\n" );
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