document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=262981>Question 262981</A>: <I><SPAN STYLE=\"word-break: break-all;\"> There are two circles, one circle is inscribed and another circle is circumscribed over a square. What is the ratio of area of inner to outer circle?  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #193730 by </B><A HREF=/tutors/aboutme.mpl?userid=ajmisra><B>ajmisra(4)</B></A><img src=\"/images/stars/iconNewId_16x16.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ajmisra><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=193730\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> It quite easy to solve this one..<BR>\n" );
document.write( "Let each of the sides of the square be 'x'<BR>\n" );
document.write( "Then, by pythagoras theorem we know the diagonal of the spuare would be <BR>\n" );
document.write( "square root of (x^2 + x^2) i.e. (sq root of 2)times 'x'\r<BR>\n" );
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document.write( "Now the inscribed circle is within the bounds of the sqaure, hence has a radius of half the sqaure's sides; i.e radius of inscribed circle is 'x/2'<BR>\n" );
document.write( "Thus area of inscribed circle is 'pie' times (x/2)^2\r<BR>\n" );
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document.write( "For the circumscribed circle the diameter of the circle would be the diagonal of the square which we know to be (sq root of 2)times 'x'<BR>\n" );
document.write( "Hence radius of this circumscribed circle would be 1/2 the diameter\r<BR>\n" );
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document.write( "Thus area of this circumscribed circle is 'pie' times {1/2 [(sq root of 2)times 'x']}^2\r<BR>\n" );
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document.write( "Taking the required ratio we have,\r<BR>\n" );
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document.write( "Inscribed Circle Area/Circumscribed Circle Area =\r<BR>\n" );
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document.write( "{pie' times (x/2)^2} /{'pie' times {1/2 [(sq root of 2)times 'x']}^2}\r<BR>\n" );
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document.write( "i.e. cancelling 'pie' in numerator and denominator we have<BR>\n" );
document.write( " {(x^2)/4} / [(sq root of 2)^2 times x^2]/4<BR>\n" );
document.write( "further removing the common '4' denominators we have<BR>\n" );
document.write( "x^2 / (sq root of 2)^2 times x^2\r<BR>\n" );
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document.write( "Further reducing would give us<BR>\n" );
document.write( "x^2 / 2 times x^2 (since (sq root of 2)^2 is simply 2)\r<BR>\n" );
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document.write( "Again reducing getting rid of x^2 from numerator and denominator gives us:<BR>\n" );
document.write( "1/2\r<BR>\n" );
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document.write( "Thus, <BR>\n" );
document.write( "Inscribed Circle Area/Circumscribed Circle Area = 1/2 as required! </SPAN>\n" );
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