Prove slant surface of a cone is always a circular sector

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In the elementary proof of the slant surface area of a cone ##A=\pi r s##, where ##s## is the slant height, it is assumed that the net of a cone is a circular sector. In other words, if we cut the slant surface of a cone from its apex to its base along a straight line, the resulting surface can always be flatten out (onto a 2D plane without crumbling).

How do we prove that the resulting surface can always be flatten out?

This is no true for a hemisphere, for example.
 

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Svein
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  1. You can draw a straight line from the apex to every point in the base. Thus the slanting side of the cone is a collection of straight lines, and can therefore be placed in a plane.
  2. On a cone, the distance from the apex to the base is the same all around the cone. That means if you slit the cone from the apex to the base, the distance from the cone to the base is still the same at all points in the base - even if the base is cut somewhere. Therefore, the flattened image must be a part of a circle.
 
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2. On a cone, the distance from the apex to the base is the same all around the cone. That means if you slit the cone from the apex to the base, the distance from the cone to the base is still the same at all points in the base - even if the base is cut somewhere. Therefore, the flattened image must be a part of a circle.
What is "the distance from the cone to the base"? And "the base is cut somewhere"?
 
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Svein
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What is "the distance for the cone to the base"? And "the base is cut somewhere"?
  1. Sorry, sloppy checking. It should be: "the distance from the apex of the cone to the base"
  2. You specified "if we cut the slant surface of a cone from its apex to its base along a straight line". Therefore you must necessarily cut the base somewhere along that line.
 

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