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You are not logged in. #1 20051222 09:33:36
polygon areaPerhaps this result is true. 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 (20051222 09:40:35) igloo myrtilles fourmis #2 20051222 10:08:39
Re: polygon areaYes, 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. #3 20051222 10:18:50
Re: polygon areaI would be interested in finding info on that formula. Last edited by John E. Franklin (20051222 10:30:43) igloo myrtilles fourmis #4 20051222 11:09:32
Re: polygon areaI 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 (20051222 11:09:59) Why did the vector cross the road? It wanted to be normal. #5 20051223 07:24:50
Re: polygon areaI went there and I found this: igloo myrtilles fourmis #6 20051223 07:35:20
Re: polygon areaThe Shakespeare thing was basically the trigonometric way of working put the area of a triangle, put into the form of an oldfashioned 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. Last edited by mathsyperson (20060108 04:15:13) Why did the vector cross the road? It wanted to be normal. #7 20060109 09:57:56
Re: polygon areaJohn, 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°. #8 20060529 12:12:57
Re: polygon areaDon't you mean (s^2/4)*tan(360/2n), irs? igloo myrtilles fourmis #9 20060529 12:15:19
Re: polygon areaWoops, I was looking at the reciprocal tangent, ignore that. igloo myrtilles fourmis 