document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=280316>Question 280316</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A square has twice the perimeter of a circle . the ratio of the area of the square to the area of the circle is\r<BR>\n" );
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document.write( "A 4  b pi  c 1/pi  d  16/pi   e none of these \r<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 #203773 by </B><A HREF=/tutors/aboutme.mpl?userid=dabanfield><B>dabanfield(803)</B></A><img src=\"/images/stars/stars-11.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=dabanfield><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=203773\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A square has twice the perimeter of a circle . the ratio of the area of the square to the area of the circle is <BR>\n" );
document.write( "A 4 b pi c 1/pi d 16/pi e none of these \r<BR>\n" );
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document.write( "Let x be a side of the square.  The perimeter of the square is then 4*x. The perimeter (i.e. circumference) of the circle then is (1/2)*4x = 2x. The circumference is also given by pi*diameter = pi*2*r where r is the radius. \r<BR>\n" );
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document.write( "So we have pi*2*r = 2*x and<BR>\n" );
document.write( "r = 2x/2pi = x/pi\r<BR>\n" );
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document.write( "The area of the square is x*x = x^2<BR>\n" );
document.write( "The area of the circle with radius r is pi*r^2\r<BR>\n" );
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document.write( "The ratios of the two areas is area of square/area of circle.\r<BR>\n" );
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document.write( "Using the fact that r = x/pi we the ratios become:\r<BR>\n" );
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document.write( "x^2/(pi*(x/pi)^2) =\r<BR>\n" );
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document.write( "x^2/(pi*(x^2/pi^2))=\r<BR>\n" );
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document.write( "(x^2/x^2)/(pi/pi^2)=<BR>\n" );
document.write( "1/pi </SPAN>\n" );
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