document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=513206>Question 513206</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I have a circle with a square in the middle, the outside of the square is shaded around the circle. How do I find the area? x2+y2=36 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #342910 by </B><A HREF=/tutors/aboutme.mpl?userid=solver91311><B>solver91311(24713)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=solver91311><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=solver91311><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=342910\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font face=\"Times New Roman\" size=\"+2\">\r<BR>\n" );
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document.write( "I'm presuming that <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large x^2\ +\ y^2\ =\ 36\"> is the equation of your circle centered at the origin.  The equation of any circle centered at the origin is <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large x^2\ +\ y^2\ =\ r^2\"> where <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large r\"> is the radius.\r<BR>\n" );
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document.write( "Hence, the radius of your circle must be 6.  If you construct a line segment from the center of the circle to one of the vertices of the square, you will note that you have also constructed a radius and that the measure of this segment must also be 6.  Next, construct another segment from the center to an adjacent vertex.  Now you should have an isosceles right triangle where the legs of the triangle each measure 6.  That means the hypotenuse of the isosceles right triangle, which is also one side of the square, must measure <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large 6\sqrt{2}\">.\r<BR>\n" );
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document.write( "The area of the square is then <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large (6\sqrt{2})^2\ =\ 36\,\cdot\,2\ =\ 72\">\r<BR>\n" );
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document.write( "The area of the circle is <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large \pi{r^2}\ =\ 36\pi\">.\r<BR>\n" );
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document.write( "Then the shaded portion which I take to be the part of the diagram that is inside the circle but outside of the square is the difference between the area of the circle and the area of the square:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ 36\pi\ -\ 72\ =\ 36(\pi\ -\ 2)\">\r<BR>\n" );
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document.write( "Is the exact answer.  You can use your calculator for a numeric approximation if you need it.\r<BR>\n" );
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document.write( "John<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE e^{i\pi} + 1 = 0\"><BR>\n" );
document.write( "My calculator said it, I believe it, that settles it<BR>\n" );
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