document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=612509>Question 612509</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A right circular cone has a radius 12 inches and a height of 25 inches. What is the lateral area of the cone? It says the answer is 12 pi to the square root of 769. How is that?  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #385540 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=385540\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font face=\"Times New Roman\" size=\"+2\">\r<BR>\n" );
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document.write( "The Lateral Surface Area of a right circular cone is given by:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ L.A.\ =\ \pi{rh_s}\">\r<BR>\n" );
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document.write( "Where <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE r\"> is the radius of the base and <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE h_s\"> is the slant height.\r<BR>\n" );
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document.write( "But you are given the height which is the measure of the distance from the center of the base to the apex of the cone.  You need the slant height which is the measure from a point on the circle that describes the base and the apex.  The radius and height form the legs of a right triangle and the slant height is the hypotenuse.  Calculate the slant height using Pythagoras.  Then plug in to the Lateral Area formula above.\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" );
document.write( "My calculator said it, I believe it, that settles it<BR>\n" );
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