2 items are filed under this topic.
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The sides of a circle |
2018-04-10 |
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From Reid: Hello,
I recently was wondering about whether or not a circle has an infinite number of sides, and I ended up searching it on your website. I saw that you guys found the question to be too ambiguous or something of that nature, and I thought about your process of reasoning involving vertices and such. I soon realized that I may have come up with the solution to the question, but I want to confirm it with you guys. Allow me to explain:
A circle, unlike any other typical 2 dimensional polygon, can sustain an infinite number of straight lines coming in contact with only one point on it. A square, for example, cannot. A square only has 4 locations that can sustain a such a line, each of those being its corners. The flat edge of a square cannot support a tangential line, because the line would either be crossing the edge of the square or coming into contact with multiple points along its edge.
This concept is consistent in every 2 dimensional polygon: pentagons sustain 5 locations for tangential lines, hexagons 6, nonagons 9, etc.
The reason a circle has an infinite number of sides is simply the fact that it must have an infinite number of "corners", assuming it can be defined as a polygon like any of the shapes I described above.
Corners can only exist assuming there is two sides coming into contact with one another.
If there is an infinite number of corners in the circle, which is apparent due to the above reasoning, there must be an infinite number of sides coming into contact with each other.
Thus, a circle is a polygon that consists of an infinite number of sides coming into contact with each other.
Please review this reasoning and let me know if it is solid.
Thank you! Answered by Penny Nom. |
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The sides of a circle |
2004-01-07 |
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From Helena: My name is Helena and I am 10 years old. On a resent math exam I was asked
the question" How many sides does a circle have?" and I wrote down none. The
teacher said the answer was one side. Answered by Chris Fisher. |
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