View Full Version : Need some geometry help

Trying to figure the acreage and square feet of a piece of property that is odd shaped.

Any math whiz's on here tonight?

http://i1064.photobucket.com/albums/u372/22rimfire22/gt/Project1.jpg

Slice it into triangles and find the area of each triangle, then add the areas together.

fgutie35

11-27-2012, 22:45

Find the perimeter first by adding all sides. Then divide it by 4 to give you a square chunk of land which now you can find the area by multiplying L x L= Asq. So your perimeter will be 863ft. Divide by 4 to get 215.75ft. Multiply it by itself to get 46,548.1sq ft. I think. There is 43,560sq ft in an acre, so that land you showed is a little bit over an acre.

devildog2067

11-27-2012, 22:53

Find the perimeter first by adding all sides. Then divide it by 4 to give you a square chunk of land which now you can find the area by multiplying L x L= Asq

This will not work. If I draw a strip of land that's 2 miles long and 1 inch wide, this method will (wrongly) estimate the area to be ~1 square mile.

For the area shown, the right way to do it is with the Law of Cosines. Hang on, I'll crank through it.

EDITED TO ADD: To actually solve it rigorously will take a while, because actually you need to write a system of equations for all of the unknown angles as well. I'm going to eyeball one of the angles and just do it that way, since the measurements probably aren't accurate down to the inch anyway.

17119jfkioe

11-27-2012, 22:57

This will not work. If I draw a strip of land that's 2 miles long and 1 inch wide, this will estimate the area to be ~1 square mile.

For the area shown, the right way to do it is with the Law of Cosines. Hang on, I'll crank through it.

You are a badass DD:cool::wavey:

CitizenOfDreams

11-27-2012, 22:58

Find the perimeter first by adding all sides. Then divide it by 4 to give you a square chunk of land which now you can find the area by multiplying L x L= Asq. So your perimeter will be 863ft. Divide by 4 to get 215.75ft. Multiply it by itself to get 46,548.1sq ft. I think. There is 43,560sq ft in an acre, so that land you showed is a little bit over an acre.

That is not a valid method. You cannot calculate the area of a random shape from its perimeter.

Example: a 3x3 square and a 2x4 rectangle have the same perimeter but different area.

CitizenOfDreams

11-27-2012, 23:17

Geometry never was my strong suit, but something tells me that there's not enough data to find the area. We need to know at least 2 angles, or 2 more measurements.

devildog2067

11-27-2012, 23:27

http://i1252.photobucket.com/albums/hh568/devildog267/LoS-1.png

I arbitrarily split it into 3 triangles and labeled the sides. Unknown sides labeled in red, unknown angles labeled in green.

Angle 1 is ~100 degrees if I hold a protractor up to the screen. With the two sides (113' and 60') and an estimate of angle 1, we can solve for the area of the bottom triangle:

http://latex.codecogs.com/gif.latex?%5Cfrac%7B1%7D%7B2%7D%28113%29%2860%29%5Csin%7B100%5E%5Ccirc%7D%20%5Capprox%203340%5C%20ft %5E2

We can also use the Law of Cosines to get side B:

http://latex.codecogs.com/gif.latex?B%20%3D%20%5Csqrt%7B113%5E2+60%5E2-2%28113%29%2860%29%5Ccos%20100%5E%5Ccirc%7D%20%5Capprox%20136%20%5C%20ft

You crank through the same deal by estimating the magnitude of angle 6 (~105 degrees) and angle 9 (~100 degrees) and you get the area of the other two triangles (18980 and 12880) for an approx. area of 35,200 square feet or ~0.8 acres.

Feel free to check my math, I did it using the crappy MS calculator app, but the principle is correct.

devildog2067

11-27-2012, 23:29

Geometry never was my strong suit, but something tells me that there's not enough data to find the area. We need to know at least 2 angles, or 2 more measurements.

That's exactly right. I "measured" some of the angles with a protractor; without that information the area can't be found.

CitizenOfDreams

11-27-2012, 23:31

I "measured" some of the angles with a protractor

Cheater. :shakehead:

That's exactly right. I "measured" some of the angles with a protractor; without that information the area can't be found.

There is enough information on there to find the angles without having to measure them. Give me a second and I'll double check what you have going on there. What you've done looks right though, I just wouldn't trust a protractor as a good measurement off this picture alone.

CitizenOfDreams

11-28-2012, 00:01

There is enough information on there to find the angles without having to measure them.

No, there is not. Without constraints such as 2 known angles or 2 known diagonals, the 5 sides can form an infinite number of different figures, each having a different area.

devildog2067

11-28-2012, 00:11

There is enough information on there to find the angles without having to measure them.

No, there is not. Without constraints such as 2 known angles or 2 known diagonals, the 5 sides can form an infinite number of different figures, each having a different area.

Yep. CitizenOfDreams is right.

Imagine if you took 5 strips of wood, the lengths of the 5 measured sides, and joined them at the ends with swivels. You could squish it into an infinite number of shapes.

... an approx. area of 35,200 square feet or ~0.8 acres.

Feel free to check my math, I did it using the crappy MS calculator app, but the principle is correct.

I think you're correct sir and might I say that you did it in a much fancier fashion than I.

I used a pencil, ruler and a business card and came up with .8 acres also. I just wanted someone to check my math before I make an offer on this tomorrow morning.

Thanks again.

No, there is not. Without constraints such as 2 known angles or 2 known diagonals, the 5 sides can form an infinite number of different figures, each having a different area.

But wouldn't it be possible to get the tangent of a few of the angles and go from there?

For instance, using DD's picture, angle 2 would be:

arctan(60/113)

Hypotenuse of B would be 128.

kateean2

11-28-2012, 01:56

Find the perimeter first by adding all sides.

http://bookstoday.info/go/smile.jpg

http://bookstoday.info/lo/sad.jpg

But wouldn't it be possible to get the tangent of a few of the angles and go from there?

For instance, using DD's picture, angle 2 would be:

arctan(60/113)

Hypotenuse of B would be 128.

No right angles. That's what got me too. There is a way to solve this without measuring the angles. I'll work on it tomorrow. Maybe I'm wrong.

No right angles. That's what got me too. There is a way to solve this without measuring the angles. I'll work on it tomorrow. Maybe I'm wrong.

Ah, forgot about them needing to be right angles. Nevermind.

686Owner

11-28-2012, 04:22

No right angles. That's what got me too. There is a way to solve this without measuring the angles. I'll work on it tomorrow. Maybe I'm wrong.

No, because the perimeter of a shape does not determine it's volume. All we really know is the perimeter. I can draw a bunch of shapes (even concave) with the sides having those measurements.

SouthpawG26

11-28-2012, 05:27

Easiest way IMO is to get an engineering student to import the image into Autocad and to trace the outlines on the image and then scale to correct size. Autocad will give you the area of the scaled shape.

Q. What did the acorn say 20 years later?

A. Geometry. :supergrin:

CitizenOfDreams

11-28-2012, 06:56

Easiest way IMO is to get an engineering student to import the image into Autocad and to trace the outlines on the image and then scale to correct size. Autocad will give you the area of the scaled shape.

A solution for a reloader: draw the figure on a piece of thick paper, cut it out and weigh it. Then compare it with a square piece of paper (you probably know how to calculate its area).

glockrod

11-28-2012, 07:29

Or, 0.81 acre.

There is a slight bit of rounding, but that is its size within 0.01+/-.

Ok, here how to do it:

You dont need right angles.

Find the Hypotenuse of the 60x113 triangle - 127.94, or 128'

This gives you the main area with measures of 82'(front), 128'(rear), and 289 & 319' for the sides. Avg the F & R sides for 105'. Avg the sides for 304'. Multiply 105x304 = 31,920

Multiply 60x113 and divide by 2=3,390

31920 + 3390 = 35310

devildog2067

11-28-2012, 08:18

You dont need right angles.

Find the Hypotenuse of the 60x113 triangle - 127.94, or 128'

This is wrong.

By definition, only a right triangle has a hypotenuse.

You are getting close only because the interior angle indicated (angle 1, which I estimated to be about 100 degrees) isn't too far off from 90 degrees. By calculating the "hypotenuse" you are assuming this angle is equal to 90 degrees.

glockrod

11-28-2012, 08:19

I dont use perimeter, so unless there is some other equation for using perimeter, it wont work. Perimeter does not equal area due to the infinity of shapes that any particular permiter measurement could equal.

The example of a 2 mile piece that is one inch wide has 880' sq. ft., or .02 acre. 5280' x12 inches x 2(miles) x 1 inch wide =126720 sq in. Converted to sq.ft = 880'

If you use the method mentioned above about adding all the sides together and dividing by 4 to get equal sides, you would have a box 1320' square which yields much more than 0.02 acre.

If I am wrong, I dont mind being called out on it.

devildog2067

11-28-2012, 08:21

Oops hit reply too soon

The second part of your answer relies on the approximation that the side you found (the one I labeled "B") is parallel to the far side. Again, it's close, but it's not.

Try using the same method on the 319' and 60' sides and you willl get a very different answer.

glockrod

11-28-2012, 08:28

You are right about the 90 degree deal.

But, my figure is mighty close. Closer than the over 1 acre estimate.

There is a way to figure the 3rd side when you have the measures on the 2 sides and the degree, but I am 20 years beyond using that college math class to remember or care.

If the OP wants exact size as well as the measurements, he should have a plat done by a licensed surveyor. I know they have the software to accurately figure size. I work in Real Estate and I rarely come across a property with exactly even foot measures.

In essence, this whole problem is all an estimate anyway!!

devildog2067

11-28-2012, 08:46

You are right about the 90 degree deal.

I know.

But, my figure is mighty close.

Yup. Nothing wrong with using approximations, as you point out we are all estimating anywy. But its important to spell out what those approximations are so you can tell when they're not good enough any more.

There is a way to figure the 3rd side when you have the measures on the 2 sides and the degree

Yes there is... That's the way I did it.

mossytxn

11-28-2012, 09:01

traced and scaled in ACAD, 37751 sqft. depending on the accuracy of the endpoints traced, that could be +/- about 500 sqft.

37751/43560 = .86 acres give or take a few hundredths

edit to add:

using the tracing, and scaling based on the 319' length, all of the sides i traced were within a foot or so of the lengths given except for the 289' side which measured 264' so all bets are off. either your info is wrong on the measurements or the picture is grossly out of proportion (on just one side which is odd)

glockrod

11-28-2012, 09:04

You aggravating perfessor!!!:rofl:

I certainly cant argue with you on those points.

Plus, I did mention rounding and gave an approximate +/-range.

I'll stick with a disclaimer that someone who actually knows math witll come along. And then you came along!!

Why is it that no matter how obscure the topic, it can always be discussed on a gun site?

I love it!

Offer was made and accepted. Survey and title search has been ordered. Should be closing next week.

Thanks!

http://i1252.photobucket.com/albums/hh568/devildog267/LoS-1.png

I arbitrarily split it into 3 triangles and labeled the sides. Unknown sides labeled in red, unknown angles labeled in green.

Angle 1 is ~100 degrees if I hold a protractor up to the screen. With the two sides (113' and 60') and an estimate of angle 1, we can solve for the area of the bottom triangle:

http://latex.codecogs.com/gif.latex?%5Cfrac%7B1%7D%7B2%7D%28113%29%2860%29%5Csin%7B100%5E%5Ccirc%7D%20%5Capprox%203340%5C%20ft %5E2

We can also use the Law of Cosines to get side B:

http://latex.codecogs.com/gif.latex?B%20%3D%20%5Csqrt%7B113%5E2+60%5E2-2%28113%29%2860%29%5Ccos%20100%5E%5Ccirc%7D%20%5Capprox%20136%20%5C%20ft

You crank through the same deal by estimating the magnitude of angle 6 (~105 degrees) and angle 9 (~100 degrees) and you get the area of the other two triangles (18980 and 12880) for an approx. area of 35,200 square feet or ~0.8 acres.

Feel free to check my math, I did it using the crappy MS calculator app, but the principle is correct.

Rabbi, would you double check that? :supergrin:

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