document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=309153>Question 309153</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Please show me how to find the surface of a cone that has a diameter of 6 cm and a slant height of 8 cm. Thanks </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #221058 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=221058\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font face=\"Garamond\" size=\"+2\">\r<BR>\n" );
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document.write( "Depends.  Do you need the Lateral Surface Area or the Total Surface Area?  The difference is that one includes the area of the base and one does not.\r<BR>\n" );
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document.write( "The base is nothing more than a circle of radius that is one-half of the given diameter.\r<BR>\n" );
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document.write( "The Lateral Surface area is found by multiplying <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large \pi\"> times the radius times the slant height.  The Total Surface Area is the Lateral Surface Area plus the area of the base, <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large \pi r^2\">\r<BR>\n" );
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document.write( "So:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ SA_L\ =\ \pi rh\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ SA_T\ =\ SA_L\ +\ \pi r^2\ =\ \pi rh\ +\ \pi r^2\">\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" );
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