My dad needs to be able to find the area of several irregularly shaped fields. He can determine the perimeter. Is there any mathematical way to find the area?

Thanks,
Teresa

Unfortunately, knowing just the perimeter of a shape doesn't mean that you can find its unique area. For instance, take a square with side length 4 and a rectangle with dimensions of 5x3. The perimeters are both 16, but the areas are different (16, 15). However, if you know other things about the shape, then it might be possible to figure it out. For instance, you can determine an irregular shape's area if:

1) The shape is a polygon (meaning no curved sides) where you know the lengths of all the sides AND the measures of all the angles.
2) The shape is a polygon and you know the coordinates of all the vertices (it's even easier if the coordinates are all whole numbers).
3) You know an algebraic equation whose graph is identical to the shape of the field in question.

If you could provide some more information about these irregular shapes, then maybe we could help you out.

Usually the easiest way is to mark off the shape with lines that create common geometric forms: squares, rectangles, and triangles. Then measure the areas for each of these shapes and add them up to find the total area. If the shape has an edge or edges that are not straight the solution is a bit more involved.

Draw the shape on a sheet of graph paper. Count 1 for any squares completely within the shape. Count .5 for any that are cut by the perimeter (even if they’re almost completely inside or outside).
As a check, draw it again at double the scale. If you don’t get an answer pretty close to four times as big, then either you counted wrong first time, or the scale was too coarse first time round.

I do remember a theorem for finding the area of a given piece of land if you know the perimeter and the number of sides, as well as the lengths of the sides...

I think for a triangle it was something like
s=(a+b+c)/2
A=sqrt(s+(s-a)+(s-b)+(s-c))
or something like that... sorry i just can't remember

WoK, the formula you're remembering is called Heron's Formula, and it's used to find the area of any given triangle assuming you know the lengths of the sides. It is:

Area = sqrt[s*(s-a)*(s-b)*(s-c)] where s is 1/2 the perimeter.

This formula can be extended to certain quadrilaterals, the only difference in the formula being:

Area = sqrt[(s-d)*(s-a)*(s-b)*(s-c)]

This formula only works for cyclic quadrilaterals (quadrilaterals that can be inscribed in a circle) and all triangles (every triangle is cyclic).

If you access to a balance that's good down to 0.1 grams, or better yet, 0.01 grams there is another very quick and very accurate method of determining the area... Before GLC's (gas/liquid chromatagraphs) were equippped with integrators we used two methods to determine the area under curves. One of which has already been mentioned by several of the respondents - was to count whole squares and or triangles and estimate partials.

However another method which I preferred since it was both quicker and more accurate was to use a pair of scissors and a Mettler balance. You cut out the area under the curve and weigh it on a very accurate balance. You could use this same technique as follows: Cut a piece of paper in half. On one half of the paper draw (to scale) the area you're trying to determine. Cut it out and weigh it. On the other half of the paper using the same scale, draw a square that's 209 feet by 209 feet. Cut it out and weigh it. This weight is your standard and is almost exactly equal to one acre. So your area in acres equals:

Area = (wt unknown area)/(wt of square)

Thank you all for your responses.

I apologize for being so slow. I had requested email notification and didn't get any, so I assumed there were no answers.

Unfortunately the fields are very oddly shaped and don't divide easily into common geometric shapes. Some sides are curved and some are straight.

You guys have provided some helpful ideas that I can pass on to my dad. I do appreciate your time and effort.

Thanks,
Teresa