document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1013636>Question 1013636</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The ratio for the diagonal and length of a rectangle is 8:5. Find the area of that rectangle it it is inscribed in a circle whose area is 120 sq. cm. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #629974 by </B><A HREF=/tutors/aboutme.mpl?userid=fractalier><B>fractalier(6550)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=fractalier><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=fractalier><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=629974\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> If a rectangle is inscribed in a circle, then the diagonal of the rectangle is the diameter of the circle.<BR>\n" );
document.write( "We can find the diameter from the area...<BR>\n" );
document.write( "A = (pi)r^2 = (pi)(d^2/4) so that<BR>\n" );
document.write( "120 = (3.14/4)(d^2)  which leads to<BR>\n" );
document.write( "d^2 = 152.87<BR>\n" );
document.write( "d = 12.36 cm<BR>\n" );
document.write( "Now the length is 5/8 of that or<BR>\n" );
document.write( "(5/8)(12.36) = 7.73 cm<BR>\n" );
document.write( "All we need is the width...we use Pythagoras for that<BR>\n" );
document.write( "w^2 + 7.73^2 = 152.87 (which was d^2) so that<BR>\n" );
document.write( "w = 9.65 cm  (longer than the length?)<BR>\n" );
document.write( "The area of the rectangle A = LW =<BR>\n" );
document.write( "(7.73)(9.65) = 74.6 sq. cm. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );