close

Privacy guaranteed - Your email is not shared with anyone.

# Need some geometry help

Discussion in 'The Okie Corral' started by spotco2, Nov 28, 2012.

1. ### spotco2

Joined:
Sep 11, 2005
Messages:
1,340
3
Location:
GA
Trying to figure the acreage and square feet of a piece of property that is odd shaped.

Any math whiz's on here tonight?

2. ### WoofieDisirregardlessCLM

Joined:
Apr 10, 2007
Messages:
9,993
22
Location:
Here
Slice it into triangles and find the area of each triangle, then add the areas together.

Last edited: Nov 28, 2012

3. ### fgutie35

Joined:
Jul 23, 2007
Messages:
2,906
282
Location:
deep southeast Texas
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.

4. ### devildog2067

Joined:
Apr 20, 2005
Messages:
16,322
1,116
Location:
Chicago
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.

Last edited: Nov 28, 2012
5. ### 17119jfkioe

Joined:
Apr 23, 2010
Messages:
1,336
4

Last edited: Nov 28, 2012
6. ### CitizenOfDreams

Joined:
Feb 23, 2009
Messages:
8,111
5
Location:
Orlando, FL
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.

7. ### CitizenOfDreams

Joined:
Feb 23, 2009
Messages:
8,111
5
Location:
Orlando, FL
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.

8. ### devildog2067

Joined:
Apr 20, 2005
Messages:
16,322
1,116
Location:
Chicago

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:

$\frac{1}{2}(113)(60)\sin{100^\circ} \approx 3340\ ft^2$

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

$B = \sqrt{113^2+60^2-2(113)(60)\cos 100^\circ} \approx 136 \ ft$

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.

9. ### devildog2067

Joined:
Apr 20, 2005
Messages:
16,322
1,116
Location:
Chicago
That's exactly right. I "measured" some of the angles with a protractor; without that information the area can't be found.

Joined:
Feb 23, 2009
Messages:
8,111
5
Location:
Orlando, FL
Cheater.

11. ### DrMaxitDirtbag Airman

Joined:
Sep 15, 2005
Messages:
5,229
0
Location:
Oahu HI
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.

12. ### CitizenOfDreams

Joined:
Feb 23, 2009
Messages:
8,111
5
Location:
Orlando, FL
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.

13. ### devildog2067

Joined:
Apr 20, 2005
Messages:
16,322
1,116
Location:
Chicago
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.

14. ### spotco2

Joined:
Sep 11, 2005
Messages:
1,340
3
Location:
GA
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.

15. ### cgwahlSheriffs a near

Joined:
Feb 15, 2002
Messages:
6,568
512
Location:
CA

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.

16. ### kateean2

Joined:
Nov 13, 2012
Messages:
25
0
Find the perimeter first by adding all sides.

17. ### DrMaxitDirtbag Airman

Joined:
Sep 15, 2005
Messages:
5,229
0
Location:
Oahu HI
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.

18. ### cgwahlSheriffs a near

Joined:
Feb 15, 2002
Messages:
6,568
512
Location:
CA

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

19. ### 686OwnerNRA Life Member

Joined:
Mar 10, 2007
Messages:
12,883
1,828
Location:
KY
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.

Joined:
Sep 4, 2004
Messages:
985