document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1012985>Question 1012985</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The perimeter of a rectangle is 32cm. What is the shortest possible diagonal 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 #629139 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=629139\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> sides are x and 16-x<BR>\n" );
document.write( "They add to 16 which is half the perimeter.<BR>\n" );
document.write( "square them to x^2 and (16-x)^2, and that equals D^2, where D is the diagonal.<BR>\n" );
document.write( "x^2+256-32x+x^2=D^2<BR>\n" );
document.write( "D^2=2x^2-32x+256<BR>\n" );
document.write( "The minimum is the vertex and x=-b/2a at the vertex.  That is 32/4=8<BR>\n" );
document.write( "The diagonal is shortest when the rectangle is a square, with sides 8.  The diagonal length squared is 128, so the diagonal is 8 sqrt(2), from a 45-45-90 right triangle.<BR>\n" );
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