Area of an Irregular Polygon Given Side LengthDate: 8/2/96 at 12:45:20 From: Paul Lemieux Subject: Area of an Irregular Polygon Given Side Length I have a question: I am trying to determine the square footage of a piece of property. The problem is that it is not rectangular. The dimensions of the 4 sides in a clockwise order from the top are 43.61, 133.64, 146.96, and 110.85. I have tried dividing it into 3 parts (2 right triangles and a rectangle) and have tried several different formulas I found on your page but it seems I don't have enough information without knowing at least one of the angles. It's been quite a while since school so please forgive my ignorance if this is actually a simple problem or if there's not enough info to solve it. Thanks in advance, Paul Lemieux Date: 8/2/96 at 13:13:21 From: Doctor Tom Subject: Re: Area of an Irregular Polygon Given Side Length Hi Paul, You're right - you can't determine the area of a quadrilateral from the lengths of the sides alone. Imagine that you have rods of the lengths above which are attached together at their tips by pivoting joints. It should be clear that the whole thing can flex in such a way that wildly different shapes can be formed. If you have the length of either diagonal, or any of the angles and you know it's concave or convex, you can work it out. With just the sides, you're out of luck. -Doctor Tom, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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