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While learning how to calculate the surface area of a cylinder, I saw that one of the methods relies on knowing that if we cut a cone, it is the sector of a circle. As shown in the image below:

$\hskip2in$ enter image description here enter image description here

I wasn't able to intuitively see this without getting scissors and paper and doing it myself. I'm wondering if there's any trick or mathematical method that can easily prove this fact without resorting to a "crafting" approach.

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    $\begingroup$ Just think of a regular pyramid, and let the number of vertices in the base to increase indefinitely. $\endgroup$ Apr 22, 2021 at 21:17
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    $\begingroup$ You can maybe be interested by this question and the answer I gave showing that certain aspects of the cone's geometry (geodesic triangles) can be studied on the planar version $\endgroup$
    – Jean Marie
    Apr 22, 2021 at 21:22

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Fix an angle $0 < \theta < 2\pi$, and suppose a sector of incident angle $\theta$ can be rolled up yielding a surface of rotation. Let $C_{S}$ denote the resulting surface of rotation for a sector in a disk of radius $S$. We want to see that $C_{S}$ is part of the lateral surface of a cone.

Because this picture has no intrinsic unit of length, scaling the sector about its center by a positive factor $t$ and then rolling up gives $C_{S}$ scaled (in space) by a factor of $t$ about the vertex. Consequently, $C_{S}$ must be a cone, the only scale-invariant non-planar surface of rotation.

To go a bit higher-tech, "rolling up" is an example of what differential geometers call a "local isometry". Since a plane sector is Euclidean, the question amounts—by Gauss's Theorema Egregium—to finding all flat surfaces of rotation. It's easy to show (portions of) planes, cones, and cylinders are the only examples. Intuitively, the generating curve has to be a line because one "principal curvature" has to be $0$, and the only possibilities are therefore that the line either meets the axis of rotation perpendicularly (plane) or non-perpendicularly (cone), or else is parallel to the axis (cylinder). Conversely, each of these surfaces is intrinsically flat.

The expository article Paper Surface Geometry might be of interest (though regrettably it's behind a paywall unless you have institutional access).

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    $\begingroup$ Very good answer. Besides, I just found an interesting A. M. Monthly article explaining the property of conics in this "flattened perspective". $\endgroup$
    – Jean Marie
    Apr 23, 2021 at 14:49
  • $\begingroup$ Thanks for the Apostol-Mnatsakanian article link (and +1 to your comment about unwrapping a triangle on a cone) $\endgroup$ Apr 23, 2021 at 22:59
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Think of a pyramid with a regular polygon as the base (hexagon in the figure).

enter image description here

The net of the lateral surface is made of an assembly of identical triangles, which are inscribed in a circular sector.

If you increase the number of sides, the surface tends to a cone and the net to that sector. The aperture arc of the net has a length equal to the perimeter of the base.

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A comment NOT an answer: First of all, I like your question. If you consider the tip of the cone as a center of a circle with a radius $l$ (the slant height of the cone), then the sector is not hard to visualize. In fact, this kind of "crafting" approach is very common in some branches of mathematics, for example, geometric topology. The so-called "Cut and Paste" technique is very helpful to identify many complicated surfaces. What I want to tell you this technique is not unnatural at all. However, if you do not like this method, you can surely use Calculus to find the surface area of a cone. I personally don't like to use Calculus to compute some area or volume if I can compute them from the scratch.

Finally, like you, I am also waiting for a cool answer from some smart people here.

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