document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=74597>Question 74597</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The base of a regular square pyramid is inscribed in the base of a cylinder.  The height of the cylinder is triple the height of the pyramid.  Find the ratio of the volume of the pyramid to the volume of the cylinder.<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #53747 by </B><A HREF=/tutors/aboutme.mpl?userid=scott8148><B>scott8148(6628)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=scott8148><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=scott8148><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=53747\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> For a square inscribed in a circle, the diagonal of the square is twice the radius of the circle.\r<BR>\n" );
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document.write( "area of circle is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=pi%2Ar%5E2 ALIGN=MIDDLE  ALT=\"pi%2Ar%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "area of square is  <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2r%5E2 ALIGN=MIDDLE  ALT=\"2r%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "the volume of a pyramid is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28%28area-of-base%29%2A%28height%29%29%2F3 ALIGN=MIDDLE  ALT=\"%28%28area-of-base%29%2A%28height%29%29%2F3\"   DISABLED_style= \"max-width:90%; height:auto;\" > in this case<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%282r%5E2h%29%2F3 ALIGN=MIDDLE  ALT=\"%282r%5E2h%29%2F3\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "the volume of a cylinder is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28area-of-base%29%2A%28height%29 ALIGN=MIDDLE  ALT=\"%28area-of-base%29%2A%28height%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > in this case <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=pi%2Ar%5E2%2A3h ALIGN=MIDDLE  ALT=\"pi%2Ar%5E2%2A3h\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "the ratio of the volumes is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28%282r%5E2h%29%2F3%29%2F%28pi%2Ar%5E2%2A3h%29 ALIGN=MIDDLE  ALT=\"%28%282r%5E2h%29%2F3%29%2F%28pi%2Ar%5E2%2A3h%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "multiplying the top and bottom of the fraction by 3 and cancelling the h and the <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=r%5E2 ALIGN=MIDDLE  ALT=\"r%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" > leaves <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2%2F%289pi%29 ALIGN=MIDDLE  ALT=\"2%2F%289pi%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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