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#1 2005-12-21 05:16:48

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,552

hexagon area with calculus

Make sure you integrate perpendicular to the  side of the polygon.
I mean the x-axis cuts a side of the polygon in half.
The 12 in the following equation is the perimeter.

The reason for the :sqrt 3 in the denominator is because all of the
thousands of hexagons inside one another are similar or proportional.
So you can look at the biggest one, and when x is :sqrt 3, then
you want 12dx for that skinny donut area.

The thing that interests me is what happens at all the vertices?
It is surprising it works.

View Image: mathisfun14.jpg

Last edited by John E. Franklin (2005-12-21 08:35:42)


igloo myrtilles fourmis

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#2 2006-01-08 10:03:11

irspow
Member
Registered: 2005-11-24
Posts: 455

Re: hexagon area with calculus

John, you do not need calculus for finding the area of polygons.

   Area = ns² / (4tan[180/n]) for anypolygon with equal sides numbering greater than 3

  n = number of sides

  s = length of side

  This is used with an assumption of degrees being used and not radians for the trigonomic function.  I am sure that you can figure out how to convert this to use for radians.

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#3 2006-01-08 10:32:43

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: hexagon area with calculus

John already worked that out all by himself, here. Isn't he clever? big_smile


Why did the vector cross the road?
It wanted to be normal.

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#4 2006-01-08 10:49:20

irspow
Member
Registered: 2005-11-24
Posts: 455

Re: hexagon area with calculus

I know that John is clever.  He is always trying to work things out on his own.  A great way of really understanding the logic behind many concepts.  I will provide the proof over on the other thread because he seemed to want that there.

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