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Area of a Cone


Date: Sat, 5 Nov 1994 02:44:19 -0500 (EST)
From: Gary Bornstein
Subject: Cone Area

Hi doc,

Can you help a desperate college student?
What is the area of a cone when given the height and the angle
at the convergence point.

Appreciate your help.
Gary


Date: Sat, 5 Nov 1994 09:53:20 -0500 (EST)
From: Dr. Sydney
Subject: Re: Cone Area

Dear Gary,

        Thanks for writing to us with your question.  I assume you are
talking about surface area when you ask about the area of a cone.  Well, we
know that one formula for finding the surface area of a cone is A=(Pi)rs,
where r=radius of the cone at the base and s=the length of the side of the
cone.  We can then derive a formula for finding surface area given only the
height of the cone and the angle.  I think it is helpful to look at a picture
here, and since my drawing capabilities aren't so hot over email (really, 
they aren't so hot anyway...) maybe it would be good for you to draw a 
picture of the cone and label the height, h; the length of the side of the 
cone, s; the radius of the cone, r; and the angle, theta.  Let's examine the 
triangle  formed at the cross-section if we cut the cone in half right 
through the tip so we are left with two symmetrical pieces.  We see that 
we have a triangle with height h, base 2r, where the other sides have 
length s.  If we cut this triangle in half, we get a right triangle with 
height h, base r, hypotenuse s, with angle theta between sides of length h 
and s. It looks something like this:
        
        |\
        |theta  
       h|  \s
        |   \
        |    \
        |_____\ 
           r

Now we need to put r and s in terms of h and theta.  Here it is helpful to
remember your trig formulas like cos (theta) = adjacent/hypotenouse, tan
(theta)=opposite/hypotenuse.

Using these formulas we can say cos (theta)=h/s and tan (theta)= r/h.  If
we solve each of these equations for s and for r, respectively, all we have
to do then is plug them into the original equation, A=(Pi)rs, and we will
have a formula for surface area in terms of h and theta.  

I hope this is helpful.  If you are at all confused or don't know where to
go from here, please feel free to write back.  

--Sydney


Date: Sat, 5 Nov 1994 13:22:42 -0500 (EST)
From: Gary Bornstein
Subject: Re: Cone Area

Thanks for your response, but I'm still a little confused.

It's all starting to come back to me, but doesn't tan(theta)=opp/adj, as in
SOH CAH TOA?  This is throwing me off for the part below.

Please clarify,
GARY


From: Dr. Sydney
Subject: Re: Cone Area
Date: Sat, 5 Nov 1994 13:53:15 -0500 (EST)

Dear Gary:

        Sorry! I accidentally typed tan (theta) = opposite/hypotenuse when I
really did mean tan (theta) = opposite/adjacent.  Thanks for the
correction.  What I typed in the bottom paragraphs still holds, though, I
think.  r=opposite and h=adjacent so tan (theta) = r/h.  Do you see where to
go from here?  If not, feel free to write back.  Thanks for correcting the
mistake!

--Sydney


Date: Sat, 5 Nov 1994 14:28:48 -0500 (EST)
From: Gary Bornstein
Subject: Re: Cone Area

No problem. It got me thinking. I haven't done this stuff in years. I think
I got it now, but could you verify.

        AREA CONE= (PI*H^2)/COS(THETA)

based on:
                   r=htan(theta), s=h/cos(theta)

gary 


From: Dr. Sydney
Subject: Re: Cone Area
Date: Sat, 5 Nov 1994 14:54:37 -0500 (EST)

Dear Gary:

        You have the right relations between s,r and h, theta:

                r=htan(theta) 

                 s=h/cos(theta) 

If you plug this into the original equation for surface area:

                 A=(Pi)rs

you would get:

A=(Pi)(h tan(theta))(h/cos(theta))

Simplify this to get:

A=(Pi)(h^2)tan(theta)
  -------------------  You can stop here and use this formula.
       cos(theta)

Does that make sense to you?  Please feel free to write back if you have 
any questions.  Thanks for writing us!

--Sydney


Date: Sun, 6 Nov 94 16:24:25 EST
From: "John Conway"
Subject: Re:  Cone Area

The curved surface area of a cone unrolls to form a sector of a circle
of radius  S (the slant height of the cone).  The area of this whole
circle is therefore pi.S^2.

The perimeter of the circle is 2.pi.S, but the perimeter of the base-
circle of the original cone is only 2.pi.R,  where  R is its radius.

So we only have a proportion

         (2.pi.R)/(2.pi.S)   of  pi.S^2,    


   that is,     pi.R.S.

   for the curved part of the cone surface.  

Since the base circle has area pi.R^2, the formula for the
whole area is

            pi.R(R+S).

I don't remember either of these formulae, and I don't think
anyone else should!  I DO remember the pi.R^2 and  2.pi.R  formulae,
and I do think everyone else should!   (I also remember quite a lot of
formulae that I DON'T think anyone else should, but that's beside the
point.)

          The method of finding out such things is what we should
be teaching (and remembering).

            John Conway.

   PS - perhaps people involved in the manufacture of funnels and
ice-cream cones should be allowed to remember the formulae.
    
Associated Topics:
High School Geometry
High School Higher-Dimensional Geometry
High School Trigonometry

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