document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=879092>Question 879092</A>: <I><SPAN STYLE=\"word-break: break-all;\"> a circle is inscribed in a square which is circumscribed by another circle. If the diagonal of the square is 2x, find the ratio of the area of the large circle to the area of the small circle? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #530678 by </B><A HREF=/tutors/aboutme.mpl?userid=ankor@dixie-net.com><B>ankor@dixie-net.com(22740)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ankor@dixie-net.com><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=ankor@dixie-net.com><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=530678\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> a circle is inscribed in a square which is circumscribed by another circle.<BR>\n" );
document.write( " If the diagonal of the square is 2x, find the ratio of the area of the large circle to the area of the small circle?<BR>\n" );
document.write( ":<BR>\n" );
document.write( "Draw this out, observe that the diameter of the large circle is equal to the<BR>\n" );
document.write( "diagonal of the square, 2x, therefore the x = the radius of the large circle<BR>\n" );
document.write( ":<BR>\n" );
document.write( "The side of the square is equal to the diameter of the small circle. Find the<BR>\n" );
document.write( "side (s) of the square using the diagonal of the square; pythag.<BR>\n" );
document.write( "s^2 + s^2 = (2x)^2<BR>\n" );
document.write( "2s^2 = 4x^2<BR>\n" );
document.write( "divide both sides by 2<BR>\n" );
document.write( "s^2 = 2x^2<BR>\n" );
document.write( "s = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sqrt%282x%5E2%29 ALIGN=MIDDLE  ALT=\"sqrt%282x%5E2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "s = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%2Asqrt%282%29 ALIGN=MIDDLE  ALT=\"x%2Asqrt%282%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >, the diameter of the small circle<BR>\n" );
document.write( "therefore<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28x%2Asqrt%282%29%29%2F2 ALIGN=MIDDLE  ALT=\"%28x%2Asqrt%282%29%29%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" > = the radius of the small circle<BR>\n" );
document.write( ":<BR>\n" );
document.write( "\" find the ratio of the area of the large circle to the area of the small circle?\"<BR>\n" );
document.write( "Cancel pi, <BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28pi%2Ax%5E2%29%2F%28pi%28%28x%2Asqrt%282%29%29%2F2%29%5E2%29 ALIGN=MIDDLE  ALT=\"%28pi%2Ax%5E2%29%2F%28pi%28%28x%2Asqrt%282%29%29%2F2%29%5E2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5E2%2F%28%282x%5E2%29%2F4%29%29 ALIGN=MIDDLE  ALT=\"x%5E2%2F%28%282x%5E2%29%2F4%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5E2%2F%28x%5E2%2F2%29 ALIGN=MIDDLE  ALT=\"x%5E2%2F%28x%5E2%2F2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5E2+%2A+%282%2Fx%5E2%29 ALIGN=MIDDLE  ALT=\"x%5E2+%2A+%282%2Fx%5E2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > = 2<BR>\n" );
document.write( "Area of large circle to small is 2:1<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );