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Determining Area of a Lot of Land

Date: 08/20/2003 at 14:43:29
From: Gary Kearney
Subject: Determining square footage of a lot of land

I own a lot that we are trying to figure out the square footage for. 
There are various records that show various results, but the 
following is the best record I have of the dimensions. Could you 
please provide the answer and the formula? The dimensions are 
98.20' x 100.89' x 140.79' x .89'


Date: 08/20/2003 at 16:17:11
From: Doctor Rob
Subject: Re: Determining square footage of a lot of land

Thanks for writing to Ask Dr. Math, Gary.

You have not given enough information to solve the problem. The best 
that I could tell you is that the area is *at most* 5016.35 square 
feet. In order to answer your question exactly, you will have to give 
at least one of the angles, or a diagonal distance, or some additional 
data. For example, are two of the sides parallel? Is the area convex?

To show you why you need more information, consider a lot whose edges 
have lengths 30, 10, 30, and 10 feet. Then it could be a rectangle, 
whose area would be 300 square feet:

              30
    o---------------------o
    |                     |
  10|                     |10
    |                     |
    o---------------------o
              30

But it also could be a parallelogram, shaped like this:

              30
   o---------------------o
    `.                   :`.
    10`.                h:  `.10
        `.               :    `.
          o--------------o------o
                     30

The area of this would be 30 feet times the height h, which must be
less than 10 feet, so the area would be less than 300 square feet.
The smaller the acute angles in the above figure, the smaller the
area. One can make the area of this parallelogram as close to zero
as you might like. That means that one could only say that the area 
was greater than zero and at most 300 square feet.

It turns out the maximum area occurs when the four corners of the lot 
lie on the circumference of a single circle. That's how I could 
compute the maximum area in your problem.

Feel free to write again if I can help further.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/ 


Date: 08/21/2003 at 07:12:11
From: Gary Kearney
Subject: Determining square footage of a lot of land

Dr. Rob,

I appreciate your taking the time to respond. I do have one 90-degree 
angle at the intersection of the sides that measure 98.20' x 100.89'.

Thanks,
Gary Kearney


Date: 08/21/2003 at 09:08:11
From: Doctor Rob
Subject: Re: Determining square footage of a lot of land

Thanks for writing back to Ask Dr. Math, Gary.

Okay! The right triangle formed at that corner by drawing the other 
diagonal has hypotenuse of length

   L = sqrt(98.20^2 + 100.89^2) = 140.79073... feet.

Then the area of the right triangle is

   A1 = (1/2)*b*h,
      = 0.5*100.89*98.20,
      = 4953.699 square feet,

and the area of the other triangle, using Heron's Formula, is

   A2 = sqrt[s*(s-a)*(s-b)*(s-c)],

where

   a = 140.79073 feet,
   b = 140.79 feet,
   c = 0.89 feet,
   s = (a+b+c)/2 = 141.235365 feet,

so

   A2 = sqrt(141.235365*0.444635*0.445365*140.345365),
      = sqrt(3925.195215...),
      = 62.651378... square feet.

Then the total is

   A = A1 + A2 = 5016.350378... square feet,

which equals the maximum possible value 5016.35054787... to three
decimal places (but not exactly).

Feel free to write again if I can help further.

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

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