document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1047288>Question 1047288</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The width of a rectangle is 5 less than twice its length. If the area of the rectangle is 70 cm2, what is the length of the diagonal? <BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #662820 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=662820\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The diagonal will be the hypotenuse of a right triangle where the legs are one length and one width.<BR>\n" );
document.write( "length=L<BR>\n" );
document.write( "width=(2L-5)<BR>\n" );
document.write( "Area is their product or 2L^2-5L=70<BR>\n" );
document.write( "2L^2-5L-70<BR>\n" );
document.write( "L=(1/4)*5+/-sqrt (25+560); sqrt 585=24.19<BR>\n" );
document.write( "L=(1/4)*(29.19), use only positive root<BR>\n" );
document.write( "L=7.30 cm<BR>\n" );
document.write( "Width=14.60-5=9.60<BR>\n" );
document.write( "Their product is 70.08, within rounding. If I use 7.2967 and 9.5934, I get 70.0002<BR>\n" );
document.write( "The sqrt of the hypotenuse = sqrt of sum of 7.3^2 and 9.6^2<BR>\n" );
document.write( "That equals sqrt (145.45)=12.06 cm </SPAN>\n" );
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