Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: Acreage of uneven lot (or sq ft)
Replies: 4   Last Post: Oct 13, 2009 11:00 AM

 Messages: [ Previous | Next ]
 Lee Crandell Posts: 12 Registered: 2/28/05
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, counter-clockwise.
(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
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 perpendicular--but 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 non-mathematical 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 non-mathematician, I found a fairly accurate non-math 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
>

Date Subject Author
2/19/05 Pat Foss
10/13/09 Bob L Petersen
2/24/05 Westley
2/25/05 Mary Krimmel
2/28/05 Lee Crandell