document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=154402>Question 154402</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I need to find the surface area of a christmas tree I made for my summer project. The cone shape is 22 inches tall and has a diameter of 7.5 inches.  There is also a base for the cone shape.  I also have a cylinder shape for the tree trunk wich is 3 inches in diameter and 5 inches in height.  Please show me the calculations.  I am using 2TTR(R+H) for both shapes but I do not think it is right and then I am not sure how to do the base of the cone. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #113709 by </B><A HREF=/tutors/aboutme.mpl?userid=ankor@dixie-net.com><B>ankor@dixie-net.com(22740)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ankor@dixie-net.com><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=ankor@dixie-net.com><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=113709\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> I need to find the surface area of a Christmas tree I made for my summer project. The cone shape is 22 inches tall and has a diameter of 7.5 inches. There is also a base for the cone shape. I also have a cylinder shape for the tree trunk which is 3 inches in diameter and 5 inches in height. Please show me the calculations. I am using 2TTR(R+H) for both shapes but I do not think it is right and then I am not sure how to do the base of the cone.<BR>\n" );
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document.write( "First just find the surface area of the cone<BR>\n" );
document.write( "The formula for that: SA = <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;\" > + <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=pi%2Ar%2Al ALIGN=MIDDLE  ALT=\"pi%2Ar%2Al\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "In your problem r = 3.75 (half the diameter),<BR>\n" );
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document.write( "l is the slanted distance of the cone, <BR>\n" );
document.write( "that will be the hypotenuse of a triangle formed by r and h<BR>\n" );
document.write( "l = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sqrt%283.75%5E2+%2B+22%5E2%29 ALIGN=MIDDLE  ALT=\"sqrt%283.75%5E2+%2B+22%5E2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "l = 22.3173 inches<BR>\n" );
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document.write( "SA = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=pi%2A3.75%5E2 ALIGN=MIDDLE  ALT=\"pi%2A3.75%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" > + <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=pi%2A3.75%2A22.3173 ALIGN=MIDDLE  ALT=\"pi%2A3.75%2A22.3173\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "SA = 44.1786 + 262.9195<BR>\n" );
document.write( "SA = 307.0981 sq inches<BR>\n" );
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document.write( "Add the trunk: <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=pi%2Ad+%2A+h ALIGN=MIDDLE  ALT=\"pi%2Ad+%2A+h\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Trunk = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=pi%2A3%2A5 ALIGN=MIDDLE  ALT=\"pi%2A3%2A5\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Trunk = 47.1239<BR>\n" );
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document.write( "Total; 307.0981 + 47.1239 = 354.222 sq/in (this includes the circular surface area of the bottom of the trunk)\r<BR>\n" );
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