Area of an Irregular PolygonDate: 03/29/2001 at 11:37:19 From: Erin Cooper Subject: Irregular polygons Can you tell me the formula for finding the area of irregular polygons? Date: 03/29/2001 at 12:23:44 From: Doctor Peterson Subject: Re: Irregular polygons Hi, Erin. It depends very much on what you know about the polygon, and what kinds of math you know. There is a nice formula if you know the coordinates of the vertices, which you can find here by clicking on Two Dimensions: Polygons: http://mathforum.org/dr.math/faq/formulas/faq.analygeom_2.html Another nice formula works if all the vertices are at "lattice points" (integer coordinates) and you can count the number of lattice points in the polygon; search our archives at for "lattice points" to find Pick's rule. http://mathforum.org/mathgrepform.html Yet another set of formulas would be used by a surveyor, someone who measures the lengths of the sides and the angles between them and uses trigonometry; these can be found here by selecting Relations in Oblique Triangles: http://mathforum.org/dr.math/faq/formulas/faq.trig.html At your level, I would expect that things are not quite so irregular that you need these sorts of formulas. Perhaps your polygons have all right angles, and you can break them down into right triangles and rectangles. If you need help with some specific problems, write back and show me a sample problem and how far you can get in solving it. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 03/29/2001 at 20:05:37 From: David Cooper Subject: Re: Irregular polygons Thank you for helping with my math problem. Here is a problem that we were given at school. 28.5 13.5 +----------------+ +-------+ | | |13.8 | | +---------+ | |12.0 13.0 |26.5 | | | | +----------------------------------+ 55.0 Thank you, Erin Date: 03/29/2001 at 22:29:15 From: Doctor Peterson Subject: Re: Irregular polygons Hi, Erin. I'm not sure how to interpret the 12.0, since it looks as if the two top edges are level. I'll assume the picture should be more like this: 13.5 +-------+ | | 28.5 | | +----------------+ |13.8 | | | | |26.5 | +---------+ | |12.0 13.0 | | | | | +----------------------------------+ 55.0 You can generally approach this sort of problem either by cutting the shape into smaller pieces and adding their areas, or by starting with a larger shape and subtracting pieces from it. Here's one way to do it by addition: 13.5 +-------+ | | 28.5 | | +----------------+ |13.8 | | | | |26.5 | +---------+ | |12.0 | 13.0 | | | | | | | | | | +----------------+---------+-------+ 55.0 You just have to figure out the height of that middle section, which you can do by subtracting 13.8 from 26.5. (Look at the right-hand rectangle to see why.) Here's another way, using subtraction: 13.5 +--------------------------+-------+ | | | | 28.5 | | +----------------+---------+13.8 | | | | |26.5 | +---------+ | |12.0 13.0 | | | | | +----------------------------------+ 55.0 First find the area of the whole rectangle, then subtract from that the area of the small rectangles. You can find the width of the top part by adding 28.5 and 13.0, and its height by subtracting 12.0 from 26.5. The height of the smallest piece is a little tricky, but once you have the height of the top piece, you can subtract. The hard part of these problems is to figure out which sides to add or subtract to get another. If you don't see it quickly, just try filling in all the sides that aren't already labeled, starting with the sides of rectangles opposite labeled sides. That's the key: opposite sides of a rectangle have the same length. Usually once you've labeled everything you can do immediately, you will quickly see how to get the ones you need. For example, here's the second method with everything labeled based on opposite sides: 55.0 +--------------------------+-------+ |? ?| 13.5 | | 28.5 13.0 | | +----------------+---------+13.8 | 26.5| | ?| |26.5 | +---------+ | |12.0 13.0 | | | | | +----------------------------------+ 55.0 I hope this helps. If you need more help, write again and let me know where you're stuck. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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