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

Perhaps this result is true.

I just worked it out with my

silly new calculus of a hexagon.(this thread)

This result is for any regular

polygon (all sides equal and all angles equal).

It works for a square of length 5.

And for a pentagon of side length 5, it says 43.01, but I don't know if it is right.

*Last edited by John E. Franklin (2005-12-21 10:40:35)*

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

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

Yes, that's right. I recognise that as a known formula. Well done for working it out!

Why did the vector cross the road?

It wanted to be normal.

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

I would be interested in finding info on that formula.

Do you know of any links.

I'll check around first...

...I found it at one web page, but no explanation...

This looks even more general!!

*Last edited by John E. Franklin (2005-12-21 11:30:43)*

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

I learnt the formula from 'Murderous Maths - The Fiendish Angletron', but it didn't give any explanation as to where it came from. It preferred to tell an odd story about supersin, cosgirl and tandog.

*Last edited by mathsyperson (2005-12-21 12:09:59)*

Why did the vector cross the road?

It wanted to be normal.

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

I went there and I found this:

Even Tandog has a trick formula for polygon areas. There's also an extract from Shakespeare's long forgotten play Henry X11 part 5 and finally... the secret identites of the superheros are revealed.

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

The Shakespeare thing was basically the trigonometric way of working put the area of a triangle, put into the form of an old-fashioned poem. I would put it here, but as my room is currently a complete mess I can't seem to find the book at the moment.

Edit: I found it!

*If a triangle has no right angle,And the area needs to be seen,Multiply half by two of the sides,And the sin of the angle between.*

There's a little treat for anyone who likes browsing through old topics.

*Last edited by mathsyperson (2006-01-07 05:15:13)*

Why did the vector cross the road?

It wanted to be normal.

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

John, if you take any polygon and draw lines from the vertices to the center you will get n equal triangles with bases the length s and a vertice angle of 360/n°.

Since the height of any given triangle is s/2tan(180/n) the area of any triangle there will be;

A = (s/2) [s/2tan(180/n)] = s²/(4tan[180/n])

So the total area is number of sides times the area of any of the triangles which gives the formula you have above. But there was no calculus needed.

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

Don't you mean (s^2/4)*****tan(360/2n), irs?

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

Woops, I was looking at the reciprocal tangent, ignore that.

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