document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=173197>Question 173197</A>: <I><SPAN STYLE=\"word-break: break-all;\"> fIND THE SURFACE AREA OF A RIGHT CONE WITH A A BASE RADIUS OF 6 CM AND HEIGHT OF 6 TIMES THE SQUARE ROOT OF 3 CM? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #128133 by </B><A HREF=/tutors/aboutme.mpl?userid=colliefan><B>colliefan(242)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=colliefan><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=128133\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Area of the base is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=pi+%2A+r%5E2 ALIGN=MIDDLE  ALT=\"pi+%2A+r%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Area of the cone is pi * r * s, where r is the radius and s the distance from the apex of the cone to the base. \r<BR>\n" );
document.write( "\n" );
document.write( "The area of the whole figure then is A = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=pi+%2A+r%5E2++%2B+pi+%2A+r+%2A+s ALIGN=MIDDLE  ALT=\"pi+%2A+r%5E2++%2B+pi+%2A+r+%2A+s\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "You are given the value of r, but not of s. You can calculate though because it is the hypotenuse of a right triangle with s and h as its legs. From the Pythagorean theorem, <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=a%5E2+%2B+b%5E2+=+c%5E2 ALIGN=MIDDLE  ALT=\"a%5E2+%2B+b%5E2+=+c%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=6%5E2+%2B+%286sqrt%283%29%29%5E2+=+s%5E2 ALIGN=MIDDLE  ALT=\"6%5E2+%2B+%286sqrt%283%29%29%5E2+=+s%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=36+%2B+36%2A3+=+s%5E2 ALIGN=MIDDLE  ALT=\"36+%2B+36%2A3+=+s%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=144+=+s%5E2 ALIGN=MIDDLE  ALT=\"144+=+s%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "s = 12\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Substitute the values of s and r in the above equation to calculate the area\r<BR>\n" );
document.write( "\n" );
document.write( "A = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=pi+%2A+r%5E2++%2B+pi+%2A+r+%2A+s ALIGN=MIDDLE  ALT=\"pi+%2A+r%5E2++%2B+pi+%2A+r+%2A+s\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "\n" );
document.write( "If you factor the Pi from both terms:<BR>\n" );
document.write( "A = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=Pi+%2A+%28r%5E2+%2B+rs%29 ALIGN=MIDDLE  ALT=\"Pi+%2A+%28r%5E2+%2B+rs%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "A = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=Pi+%286%5E2+%2B+6%2A12%29 ALIGN=MIDDLE  ALT=\"Pi+%286%5E2+%2B+6%2A12%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "A = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=Pi+%28108%29 ALIGN=MIDDLE  ALT=\"Pi+%28108%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "A = 339.28 sq cm\r<BR>\n" );
document.write( "\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );