document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1201954>Question 1201954</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A rectangle whose length is twice as long as its width is inscribed in a circle of area π. What is the area of the rectangle?\r<BR>\n" );
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document.write( "(A) 2/5<BR>\n" );
document.write( "(B) 4/5<BR>\n" );
document.write( "(C) 8/5<BR>\n" );
document.write( "(D) 5/4<BR>\n" );
document.write( "(E) 5/8 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #836542 by </B><A HREF=/tutors/aboutme.mpl?userid=htmentor><B>htmentor(1343)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=htmentor><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=htmentor><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=836542\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A circle with an area of pi has a radius of 1. Now consider the inscribed rectangle. The diagonal of this rectangle is the hypotenuse of a right triangle with the length and width as the other two sides. The hypotenuse of the rectangle is equal to 2, the diameter of the circle. Since the length is twice the width, by the Pythagorean theorem, we have 2^2 = W^2 + (2W)^2 = 5W^2.<BR>\n" );
document.write( "Thus W = sqrt(4/5) = 2/sqrt(5)<BR>\n" );
document.write( "Therefore, the area is W*L = 4/sqrt(5)*2/sqrt(5) = 8/5 </SPAN>\n" );
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