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 Question from kannan: i have a 1.5 acre land having four sides A-B=64.8 meters B-C=94.2 meters C-D=54.4 meters D-A=127.2 meters angle at D is 90 the side AB (frontage)is road facing i want to split the plot in to two halves having same almost area and almost equal frontage, kindly help me, if possible explain with drawing. Regards Kannan

Kannan,

I used our four-sided lot calculator to verify that the area is 1.5 acres. I then put a coordinate system on the plane with D at the origin. A on the x-axis and C on the y-axis. Using the distance formula and solving equation I found that B has the coordinates (94.2, 55.7).

The point M midway between A and B is (110.7, 27.9) so if the frontage is to be divided in half you need a boundary from M to the side CD (the y-axis). I then took a point N on the y-axis with coordinates (0, y) and found the distance from this point to M to be √[110.72 + (27.9 - y)2].

I then experimented with various values of y giving a lot with corners at D, A, M and N and used our four-sided lot calculator to determine the area of this lot. I found that when y = 23.1 meters this lot has an area of 0.75 acres.

Thus my conclusion is to divide the lot by a line from M, the point along AB which is 32.4 meters from A, to N, the point along the side DC which is 23.1 meters from D.

I hope this helps,
Harley

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