document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=463746>Question 463746</A>: <I><SPAN STYLE=\"word-break: break-all;\"> a circle is inscribed in a square and there's another small circle at the down part of the circle in the right side, how to get the area of the small square  and circumference? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #317881 by </B><A HREF=/tutors/aboutme.mpl?userid=robertb><B>robertb(5830)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=robertb><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=robertb><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=317881\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let s = the length of the edge of the square, and let r = radius of the smaller circle.  ==> the radius of the circle inscribed in the square is s/2.\r<BR>\n" );
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document.write( "Then <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28s%2F2%29%5E2+%2B+%28s%2F2%29%5E2+=+%28s%2F2+%2B+r+%2B+sqrt%282%29%2Ar%29%5E2+=+%28s%2F2+%2B+%281%2Bsqrt%282%29%29r%29%5E2 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "by the Pythagorean theorem.\r<BR>\n" );
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document.write( "<==> <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%281%2Bsqrt%282%29%29%5E2r%5E2++%2B+s%281%2Bsqrt%282%29%29r+-+s%5E2%2F4+=+0 ALIGN=MIDDLE  ALT=\"%281%2Bsqrt%282%29%29%5E2r%5E2++%2B+s%281%2Bsqrt%282%29%29r+-+s%5E2%2F4+=+0\"   DISABLED_style= \"max-width:90%; height:auto;\" >, after simplification.\r<BR>\n" );
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document.write( "Applying the quadratic formula, we get\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=r+=+%28-s%281%2Bsqrt%282%29%29+%2B-+sqrt%28+2s%5E2%281%2Bsqrt%282%29%29%5E2+%29%29%2F%282%2A%281%2Bsqrt%282%29+%29%5E2%29+ ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" >.\r<BR>\n" );
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document.write( "r = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28s%2F2%29%28%28sqrt%282%29+-+1%29%2F%28sqrt%282%29+%2B+1%29%29 ALIGN=MIDDLE  ALT=\"%28s%2F2%29%28%28sqrt%282%29+-+1%29%2F%28sqrt%282%29+%2B+1%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >, or <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-s%2F2 ALIGN=MIDDLE  ALT=\"-s%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" >.\r<BR>\n" );
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document.write( "Eliminate the second value for r.  (Why?)\r<BR>\n" );
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document.write( "From here just apply the formulas <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A+=+pi%2Ar%5E2 ALIGN=MIDDLE  ALT=\"A+=+pi%2Ar%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" > and <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=C+=+2%2Api%2Ar ALIGN=MIDDLE  ALT=\"C+=+2%2Api%2Ar\"   DISABLED_style= \"max-width:90%; height:auto;\" > to get the area and circumference of the small circle, respectively. </SPAN>\n" );
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