Finding the Area of an Irregular ShapeDate: 01/03/2007 at 12:42:47 From: Melissa Subject: Area of an irregular shape I am looking to purchase a irregular shaped lot, but can't figure out the square footage. Could you please provide the answer and the formula? The dimensions are clockwise 32.10' (which has two 90 degree angles) x 13.90' x 13.75'x 65.41' x 72.54' (this one has the mentioned above 90 degree angle and a 40 degree angle). Date: 01/03/2007 at 23:27:21 From: Doctor Peterson Subject: Re: Area of an irregular shape Hi, Melissa. As a check, I believe the lot looks something like this, if I'm reading this right: 32 +------------+ |90 90|13 | + | 13\ | + | / 72| / | / | / | / 65 | / | / |40 / | / |/ + If your angle measurements are exact enough, we can find the area by treating it as a trapezoid minus a smaller trapezoid: 32 +------------+.+ | 13| : | + :y | 13\: |..............+ | x / 72| / | / | / | / 65 | / | / |40 / | / |/ + The larger one has altitude x and bases 72.54 and y; the smaller has altitude x-32.10 and bases y and 13.90. All we have to do is find x and y using some trigonometry: x = 65.41 sin(40) = 42.04 y = 72.54 - 65.41 cos(40) = 22.43 So the larger trapezoid has area A1 = (72.54 + y)*x/2 = (72.54 + 22.43)*42.04/2 = 1996 and the smaller is A2 = (13.90 + y)*(x - 32.10)/2 = (13.90 + 22.43)*(42.04 - 32.10)/2 = 180 That gives a net area of A1 - A2 = 1996 - 180 = 1816 square feet As a check, since I didn't use the 13.75' dimension, we can calculate it by the Pythagorean Theorem: sqrt[(y - 13.90)^2 + (x - 32.10)^2] = sqrt[(22.43 - 13.9)^2 + (42.04 - 32.10)^2] = 13.09 That's not quite what you have, so I suspect that at least one of the angles is not exact; but it won't make a huge difference in the area. If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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