Well, the lateral area of a cone is [pi]rl, where r is the radius of the base of the cone and l is the slant heighth of the cone. But I did a simple proof, where, if you take each infinitely thin line on the lateral edge, and add them together, you'll get the lateral area. Well, since they go around perimeter times, why isn't the formula 2[pi]rl? Thanks for any help.

Why work in 3 dimensions? The same paradox exists in 2 dimensions as well.

Everyone knows the area of a circle with radius R is Pi R^2.
But how about this?
Take each infinitely thin radius R, and add them all together (permieter times - 2 Pi R times), you'll get 2 Pi R^2.

Oops?

The answer:
The addition of "infinitely thin lines" is simply not defined by multiplication like that.
It works only for "straight" shapes such as rectangles, but not "round" shapes like circles or cones.
If you want to calculate an area in that manner, you'll have to use calculus (integration).
When you do that, you'll always find the correct result.

Example: Finding the area inside a circle.

Draw a circle around the point (0, 0) with radius R (so it goes through (-R, 0) and (R, 0)).
The formula for the top part of the circle is:

y = sqrt(R^2 - x^2)

Therefore, the area of the top part of the circle (i.e. the area of half a circle) is:

S/2 = § (-R,R) sqrt(R^2 - x^2) dx

Where § (-R, R) means integral from -R to R.

That integral can be worked out by using substitution. I will leave the gory details to you if you wish to battle it later. I will leave a hint at the bottom of the post.*
The integral evaluates as:

S/2 = Pi R^2 / 2
Therefore:
S = Pi R^2

Now, take a look at your cone.
You are trying to find the surface area of a 3 dimensional object.
Instead of finding it directly, try to "unroll" the cone as if it's made of paper.
What shape do you get?

Yes, that's right. You got a partial circle!
The radius is L, since the center (the vertex of the ex-cone...) is L units away from any point on the circle (points on the base of the ex-cone...).
The perimeter is 2Pi R, since it's the entire perimeter of the previous cone's base.

If it were a full circle, the permieter would be 2Pi L and the area Pi L^2.
But the permieter is only 2Pi R, in other words, R/L of a full circle.
So the area is R/L of a full circle as well, or R/L * Pi L^2 = Pi L R.

Satisfied yet? Any comments/questions/mistakes I made? Please reply :)
[hr]
* The promised integral hint.
Try the following substitution:
x = R sin(t)
The following should work as well:
x = R cos(t)
This is a common trick in integral evaluation.

[quote]
<Insert everything Yoni said>
[/quote]
(not really quoting to save space)

I'm just waiting for the guy to post a reply in the sense of "dude, I was only joking. jeez" or "LOL, it was only hoax!".
::)

I understand, thanks for the help. Oh, and to avoid starting another thread, is anyone able to prove Herons formula algebraically or geometrically (not triginometrically)? I started a proof, and got fairly far (I managed to derive a formula with S), but I was not able to finish.

Sure.

Assume the triangle has sides a, b, c, opposite the vertices A, B, C respectively.

Drop a height from C upon c.
Call that height "h" and the point where it intersects with AB "D".

Let x = BD, then c-x = AD.

From the Pythagorean theorem:

x^2 + h^2 = a^2
(c-x)^2 + h^2 = b^2

Where we have to solve for x and h.
The solution for h (I'll let you work this out yourself):

h = sqrt(a^2 - (a^2 - b^2 + c^2)^2 / 4c^2)

The area of the entire triangle is:

S = hc / 2

Or:

S = sqrt(a^2 c^2 / 4 - (a^2 - b^2 + c^2)^2 / 16)

All that's left to do is prove:

a^2 c^2 / 4 - (a^2 - b^2 + c^2)^2 / 16 = s(s-a)(s-b)(s-c)

And that's not difficult, only long and annoying. Simply open all parentheses on both sides and verify that you got the same expression.

I don't know of an elegant non-trigonometric way to prove this offhand.

Here is a simple, elegant way using non-complicated trigonometry:

http://planetmath.org/encyclopedia/ProofOfHeronsFormula.html

Edit: Btw, best math forum post in a long time :)

Wow..... Thanks. And I thought I was good at math. The Pythagorean Theorem isn't much to prove, I suppose.

I'm still waiting for ITAKal89's punchline ::)

I'm a bit confused about the proof, probably because I'm not in the mood to solve a formula with 5 variables. Anyway, my punch line? O_oo_O

Okay, I kind of tried the Herons formula proof on my own, and I got x^2 = h^2. Is that right? =/

I am having serious trouble solving for x, lol. I now, officially, hate Algebra.

[quote author=ITAKal89 link=board=36;threadid=6916;start=0#msg61292 date=1085274204]
I now, officially, hate Algebra.
[/quote]

That'll do fine as a punchline, thank you =)

Now, could you show me all of the Algebra please?

It's not in 5 variables, it's in 2 variables (h, x) and has some parameters (a, b, c).

Here's a little algebra.

x^2 + h^2 = a^2
(c-x)^2 + h^2 = b^2

Decrease the second equation from the first:

x^2 - (c-x)^2 = a^2 - b^2

x^2 - c^2 + 2cx - x^2 = a^2 - b^2

-c^2 + 2cx = a^2 - b^2

2cx = a^2 - b^2 + c^2

x = (a^2 - b^2 + c^2)/2c

Solving from h should be immediate (plug above x into first equation...) Hint: h = sqrt(a^2 - x^2)... Check against the formula for h in my previous post.

And pay no attention to Netcooler, he forgot to take his medication :(

Well, that seemed a bit blatantly obvious. Either exhaustion or lack of will power prevented me from seeing that. I tried substitution, for the entire thing, which is probably why it took me so long to try and solve for x. >_> ¬_¬' <_<That's my punchline.

S = hc / 2

Or:

S = sqrt(a^2 c^2 / 4 - (a^2 - b^2 + c^2)^2 / 16)


I do not understand how you went from hc/2 to sqrt(a^2 c^2 / 4 - (a^2 - b^2 + c^2)^2 / 16) using substitution.

By plugging in the value I found for h and entering the "c/2" into the sqrt as "c^2/4"?

A quick question: If I draw a scalene triangle, and draw angle bisectors from each of the vertex angles to the sides of the triangle, will the point where they all intersect each other be the center point for the circle inscribed in the circle, and will the 3 sides of the triangle be tangent lines?

That's right, except "circle inscribed in the triangle*" (incircle).

http://mathworld.wolfram.com/Incircle.html