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Finding the Area of an Irregular Shape

Date: 01/03/2007 at 12:42:47
From: Melissa 
Subject: Area of an irregular shape

I am looking to purchase a irregular shaped lot, but can't figure out 
the square footage.  Could you please provide the answer and the formula? 

The dimensions are clockwise 32.10' (which has two 90 degree angles) x
13.90' x 13.75'x 65.41' x 72.54' (this one has the mentioned above 90
degree angle and a 40 degree angle).



Date: 01/03/2007 at 23:27:21
From: Doctor Peterson
Subject: Re: Area of an irregular shape

Hi, Melissa.

As a check, I believe the lot looks something like this, if I'm
reading this right:

         32
  +------------+
  |90        90|13
  |            +
  |           13\
  |              +
  |             /
72|            /
  |          /
  |         /
  |       / 65
  |      /
  |    /
  |40 /
  | /
  |/
  +

If your angle measurements are exact enough, we can find the area by
treating it as a trapezoid minus a smaller trapezoid:

         32
  +------------+.+
  |          13| :
  |            + :y
  |           13\:
  |..............+
  |      x      /
72|            /
  |          /
  |         /
  |       / 65
  |      /
  |    /
  |40 /
  | /
  |/
  +

The larger one has altitude x and bases 72.54 and y; the smaller has
altitude x-32.10 and bases y and 13.90.  All we have to do is find x
and y using some trigonometry:

  x = 65.41 sin(40) = 42.04

  y = 72.54 - 65.41 cos(40) = 22.43

So the larger trapezoid has area

  A1 = (72.54 + y)*x/2 = (72.54 + 22.43)*42.04/2 = 1996

and the smaller is

  A2 = (13.90 + y)*(x - 32.10)/2
     = (13.90 + 22.43)*(42.04 - 32.10)/2 = 180

That gives a net area of

  A1 - A2 = 1996 - 180 = 1816 square feet

As a check, since I didn't use the 13.75' dimension, we can calculate
it by the Pythagorean Theorem:

  sqrt[(y - 13.90)^2 + (x - 32.10)^2]
  = sqrt[(22.43 - 13.9)^2 + (42.04 - 32.10)^2] = 13.09

That's not quite what you have, so I suspect that at least one of the
angles is not exact; but it won't make a huge difference in the area.

If you have any further questions, feel free to write back.


- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Geometry
High School Practical Geometry
High School Triangles and Other Polygons

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