Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,585

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)*

**igloo** **myrtilles** **fourmis**

Offline

**irspow****Member**- Registered: 2005-11-24
- Posts: 456

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.

Offline

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

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

Why did the vector cross the road?

It wanted to be normal.

Offline

**irspow****Member**- Registered: 2005-11-24
- Posts: 456

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.

Offline

Pages: **1**