document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=311312>Question 311312</A>: <I><SPAN STYLE=\"word-break: break-all;\"> the areas of two regular hexagons are 54x square root of 3 and 150x the square root of 3. Waht is the ratio of their perimeters?<BR>\n" );
document.write( "thank you </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #222644 by </B><A HREF=/tutors/aboutme.mpl?userid=fractalier><B>fractalier(6550)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=fractalier><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=fractalier><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=222644\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Since the ratio of the areas of two similar polygons (regular in this case) is equal to the square of the ratio of their sides (and thus their perimeters), we can take the square root of these ratios to find the answer...so...<BR>\n" );
document.write( "54x*sqrt(3) / 150x*sqrt(3) reduced becomes<BR>\n" );
document.write( "9 / 25<BR>\n" );
document.write( "If we take the square root of this we get,<BR>\n" );
document.write( "3 to 5, the answer.<BR>\n" );
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