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.
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