document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=259915>Question 259915</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A square is inscribed in a circle. If the side of the square is 4 what is the area and circumference of the circle. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #191377 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><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=191377\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A square is inscribed in a circle. If the side of the square is 4 what is the area and circumference of the circle.<BR>\n" );
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document.write( "Draw the picture and you will see that the <BR>\n" );
document.write( "diagonal of the square is the diameter <BR>\n" );
document.write( "of the circle.<BR>\n" );
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document.write( "Using Pythagoras you get<BR>\n" );
document.write( "diagonal = 4*sqrt(2)<BR>\n" );
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document.write( "Circumference of the circle = (pi)(diameter)<BR>\n" );
document.write( "= 4sqrt(2)(pi)<BR>\n" );
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document.write( "This is approximately 17.77 units<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "stan H. </SPAN>\n" );
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