Math Is Fun Forum

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

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.

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

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igloo myrtilles fourmis

irspow

Member

Registered: 2005-11-24

Posts: 1,055

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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I am at an age where I have forgotten more than I remember, but I still pretend to know it all.

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?

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

irspow

Member

Registered: 2005-11-24

Posts: 1,055

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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I am at an age where I have forgotten more than I remember, but I still pretend to know it all.