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Re: Acreage of uneven lot (or sq ft)
Posted:
Feb 28, 2005 7:06 PM


We don't know what Mr. Foss's lot looks like so the problem still stands. I did a paper and scissor thing. On Paint program I draw four horizonal lines with each dimensional foot rounded to the nearest pixel length. Stretched the lines horizontally by 200% , then printed the diagram and cut out the four lines. I don't know if he listed the dimensions in sequence so I placed the cut out lines in the respective given dimensions, counterclockwise. (Clockwise would produce the mirror image). The longest side I assigned as the :"base". I then ran one of the sides perpendicular to the base. Then simply placed the other two side from the opposite vertices so they can (with some swivelling adjustments) have the other two endpoints meet at the fourth vertex. With a little bit of adjustments, neither side has to even be perpendicular to the "base" to get the for lengths of paper strips to meet at their ends.
It appears if I could drill holes in the ends of each stick so that the sticks could be hinged together with a bolt and nut for each junctions (vertex) such that the given proportions are maintained; I would now have an adjustable quadrilateral with the sides in the same proportion and sequence as the given dimensions. Then I could swivel together two opposite sides (within certain limitations)
This swivelling would be analogous to flattening a square or rectangle into a rhombus or parallelogram respectively, except that the shorter end would limit the swivel range of the longer end. It seems that this would also change the area, as it would with a parallelogram or rhombus. I'd have to inspect this further to see if this swivelling would change the area.
>From this it would seem essential to list one of the diagonals in order to determine the area.
If you draw two adjacent sides (not necessarily perpendicularbut not too obtuse or acute). Then draw two circles with the radii of the other two given dimensions in the same respective sequence as given, then the two circles should intersect. However since two intersecting circles meet at two points, so in many cases, there could be the fourth vertex in either of two positions, one of which with a concave vertex (external angle < 180 deg.)
Lee Crandell
"Mary Krimmel" <mary@krimmel.net> wrote in message news:5.1.0.14.0.20050224222148.00a14ca0@popmail.ltsp.com... > Congratulations! > > An ingenious nonmathematical solution! I am astonished that you could get > 4 digit accuracy with your method. I intend to try it ASAP. > > Mary Krimmel > > At 11:13 PM 2/24/05 0500, you wrote: >>On 18 Feb 05 22:12:15 0500 (EST), Pat Foss wrote: >> >Demensions are: 189.49 ft, 116.25 ft, 173.51 ft, and 141.96 ft. >> >>Hi Pat, >>I had the same question years ago. I have 1.347 acres. You can draw >>lines to form multiple triangles until you're blue in the face. Being >>a nonmathematician, I found a fairly accurate nonmath solution. I >>drew on a piece of construction paper a perfect rectangular 1 acre >>lot. Then I cut it out and weighed it on a chemistry lab scale. Then I >>cut out (same scale) a diagram of my lot. I sure you get the picture. >>You will be surprized on how close you can get if you're careful. >>Cordially, >>west > >  End of Forwarded Message >



