Area of Polygon
August 25, 2024 7:06 AM Subscribe
Is there an online tool where I can draw this polygon and get a calculation for the area?
Best answer: Do you want a tool so that you can do this again with a different shape, or just an answer to this specific problem?
For this polygon, it's 9,664.529 sq [unit of measurement you used].
I did fudge this a little- the 80' side was 79'-11 7/8" but I think that is probably close enough.
This will admittedly not help if you are looking for an online tool- I drafted this up real quick in a CAD program.
You could probably learn to do this in Sketchup or an online light CAD program, but I'm not sure it would be time-efficient over just calculating with triangles and rectangles.
posted by Adridne at 8:11 AM on August 25 [2 favorites]
For this polygon, it's 9,664.529 sq [unit of measurement you used].
I did fudge this a little- the 80' side was 79'-11 7/8" but I think that is probably close enough.
This will admittedly not help if you are looking for an online tool- I drafted this up real quick in a CAD program.
You could probably learn to do this in Sketchup or an online light CAD program, but I'm not sure it would be time-efficient over just calculating with triangles and rectangles.
posted by Adridne at 8:11 AM on August 25 [2 favorites]
Response by poster: Thanks, you arrived at a similar number to what I did so that is reassuring.
posted by falsedmitri at 8:13 AM on August 25
posted by falsedmitri at 8:13 AM on August 25
Best answer: Here's my calculation, which agrees with Adridne's.
posted by aws17576 at 8:16 AM on August 25 [2 favorites]
posted by aws17576 at 8:16 AM on August 25 [2 favorites]
I got a very slightly different answer (9665.302), also using CAD.
posted by aubilenon at 12:32 PM on August 25
posted by aubilenon at 12:32 PM on August 25
Without taking additional measurements from the drawing, is it possible to uniquely determine the dimensions of the 2 triangles involved? It's not obvious to me that there is enough information there to to determine this with only the length measurements given.
edit: I see that aws17576 has solved it.
posted by DarkForest at 5:15 PM on August 25
edit: I see that aws17576 has solved it.
posted by DarkForest at 5:15 PM on August 25
Yes, if you assume the lines that LOOK horizontal and vertical ARE, then the polygon's shape is uniquely determined from the given distances. This wouldn't be true for a general shape (for instance, if the lower left corner was cut off with a length of 6 then the shape is "not fully constrained").
I can't think of a succinct geometric explanation for this fact, but if you think of this in terms of equations, you can write 4 equations in 4 variables for the legs of the right triangles that the 2 cut-off corners represent; if you add another cut corner, you get 2 more unknowns but only 1 more equation, so it's not a fully solvable system of equations anymore.
FWIW I also got an answer of around 9665.3 after sketching the given shape & going through some additional steps; a friend solved the 4 equations by hand & also got around 9665.3.
posted by the antecedent of that pronoun at 6:07 AM on August 26 [1 favorite]
I can't think of a succinct geometric explanation for this fact, but if you think of this in terms of equations, you can write 4 equations in 4 variables for the legs of the right triangles that the 2 cut-off corners represent; if you add another cut corner, you get 2 more unknowns but only 1 more equation, so it's not a fully solvable system of equations anymore.
FWIW I also got an answer of around 9665.3 after sketching the given shape & going through some additional steps; a friend solved the 4 equations by hand & also got around 9665.3.
posted by the antecedent of that pronoun at 6:07 AM on August 26 [1 favorite]
a succinct geometric explanation for this fact
What worked for me was to realize that there are only two sides of unknown slope, and think about what would have to happen if I "flexed" one of them. If one gets taller vertically, the other also has to get taller so the top and bottom of the polygon stay parallel. But if both those sides get taller, then they must also both lose some horizontal reach, yet their combined horizontal reach is fixed at 128–42. So they can't flex!
(I used the law of cosines to actually determine the angles labeled in my diagram.)
posted by aws17576 at 9:55 PM on August 26
What worked for me was to realize that there are only two sides of unknown slope, and think about what would have to happen if I "flexed" one of them. If one gets taller vertically, the other also has to get taller so the top and bottom of the polygon stay parallel. But if both those sides get taller, then they must also both lose some horizontal reach, yet their combined horizontal reach is fixed at 128–42. So they can't flex!
(I used the law of cosines to actually determine the angles labeled in my diagram.)
posted by aws17576 at 9:55 PM on August 26
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I'm trying to figure out how much concrete I need, fwiw
posted by falsedmitri at 7:35 AM on August 25 [1 favorite]