document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=971545>Question 971545</A>: <I><SPAN STYLE=\"word-break: break-all;\"> the diagonal of a square is 6 meters more than its sides. What is the length of a side in meters </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #593978 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=593978\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> S=side<BR>\n" );
document.write( "S sqrt2=diagonal 45-45-90 triangle \r<BR>\n" );
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document.write( "S +6=s sqrt 2\r<BR>\n" );
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document.write( "6 = s sqrt 2-s\r<BR>\n" );
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document.write( "S (sqrt 2. - 1)=6\r<BR>\n" );
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document.write( "S=6/(sqrt2 - 1)\r<BR>\n" );
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document.write( "S=14.49 m approximately \r<BR>\n" );
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document.write( "S sqrt2 = 20.49 m approximately \r<BR>\n" );
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