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#1 2006-02-19 05:26:53
- stormswimmer
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- Registered: 2006-01-27
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Irregular Polygons
How do you calculate the the area and perimeter of an irregular polygon knowing the length of sides
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#2 2006-02-19 05:54:58
- irspow
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- Registered: 2005-11-24
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Re: Irregular Polygons
The perimeter would simply be the sum of the lengths of all sides. To get the area you either need to break down the irregular polygon into regular polygons and sum their areas or integrate the equations that describe the upper and lower bounderies of the shape in question.
Regular polygons are quite easy as there exist formulas to calculate the area or perimeters of any regular polygon. Irregular polygons need to be examined in each particular case.
I am at an age where I have forgotten more than I remember, but I still pretend to know it all.
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#3 2006-02-19 08:31:13
- MathsIsFun
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Re: Irregular Polygons
If you only need to solve one (or a few), then break into triangles as irspow suggested. But there are methods to do it automatically using a conmputer algorithm.
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
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#4 2006-02-19 08:51:23
- irspow
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Re: Irregular Polygons
For any regular polygon;
A = s²/(4tan[180/n]); where s is the length of a side and n is the number of sides. (This is using degrees)
P = ns;
I didn't know that there were algorithms for irregular polygons. Do these use the techniques similar to integration or are they different?
I am at an age where I have forgotten more than I remember, but I still pretend to know it all.
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#5 2006-02-19 10:15:03
- stormswimmer
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Re: Irregular Polygons
I was given this formula
1 / 2 (x0 * y1 + x1 * y2 + ... + xn-1 * y0 - y0 * x1 - y1 * x2 - ... - yn-1 * x0)
but it wasn't complete can anyone explain this to me
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#6 2006-02-19 11:11:00
- MathsIsFun
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Re: Irregular Polygons
Have a look at my reply to this topic. I think your formula is based on that idea, just simplified a lot.
In my illustration there, every shape's area would be the x-width times the average height, or: (x1-x0) × (y0+y1)/2
Just add up all the areas (taking care to make them negative areas when heading in the negative x direction) and you will end up with the "net" area. If you go the wrong direction around you will just get the negative net area.
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
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