document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1149201>Question 1149201</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A regular hexagon is inscribed in a circle. Find the ratio of the area of the hexagon to the area of the circle. \r<BR>\n" );
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document.write( "Example of diagram: https://imgur.com/a/z81W87V </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #770548 by </B><A HREF=/tutors/aboutme.mpl?userid=greenestamps><B>greenestamps(13200)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=greenestamps><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=greenestamps><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=770548\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "Draw the three diagonals of the hexagon that connect opposite vertices, dividing the hexagon into 6 equilateral triangles.<br><BR>\n" );
document.write( "The side length of the hexagon, and therefore the side length of each equilateral triangle, is the same as the radius of the circle.<br><BR>\n" );
document.write( "Area of hexagon = area of 6 equilateral triangles with radius r = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=6%2A%28r%5E2%2Asqrt%283%29%2F4%29 ALIGN=MIDDLE  ALT=\"6%2A%28r%5E2%2Asqrt%283%29%2F4%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "Area of circle = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=pi%28r%5E2%29 ALIGN=MIDDLE  ALT=\"pi%28r%5E2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "Ratio of areas = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%283r%5E2%2Asqrt%283%29%2F2%29%2F%28pi%2Ar%5E2%29+=+%283%2Asqrt%283%29%29%2F%282pi%29 ALIGN=MIDDLE  ALT=\"%283r%5E2%2Asqrt%283%29%2F2%29%2F%28pi%2Ar%5E2%29+=+%283%2Asqrt%283%29%29%2F%282pi%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
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