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Area of an Irregular Polygon Given Side Length

Date: 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!   
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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