Well, for me anyway. I haven't done geo for like 10 years since I learned it in high-school. Anyway, I am trying to figure out the area of a VERY irregular hexagon. Now, knowing only one side (the hypotenuse) of a right triangle, can I infer the other sides and or angles?
stumper of a geometry question
- Thread starter DuffmanOhYeah
- Start date
you really want me to draw this irregular hexagon?
All I need to know is how to get the other sides and or angles of a triangle knowing the hyp length only.
All I need to know is how to get the other sides and or angles of a triangle knowing the hyp length only.
Do you have any angles? If so, surely you remember your basic trig functions?
sin, cos, tan?
SOHCAHTOA
sin = opp/hyp
cos = adj/hyp
tan = opp/adj
If all you have is the hyp, & no other info there's nothing you can do. If you're expected to solve it odds are they've given you a way to determine another piece of info (another side, an angle, etc.).
That's all the help I'm giving you.
Viper GTS
Well for starters you need to know more than one thing about your hexagon. Knowing but one side of a triangle is not good enough. You need at least 2 other pieces of information(angles or sides) to do anything.
You're on the right track. Parting the problem into defineable problems depends on the hexagon. You know how to figure out the area of triangles, squares, and regular shapes.
It'll take brute force to break an irregular shape into more manageable shapes, but it's doable. It'll just take time.
ok, here is the link
pisser aint it. I know I want to break off the corners into triangles so I am left with one or several rectangles to make up the bulk of the area. But breaking them off seems to be the trouble
pisser aint it. I know I want to break off the corners into triangles so I am left with one or several rectangles to make up the bulk of the area. But breaking them off seems to be the trouble
Are the top & bottom lines assumed to be parallel?
What about the right hand endpoints of the top & bottom lines? Are they vertically aligned?
If so, it's easy.
Viper GTS
What about the right hand endpoints of the top & bottom lines? Are they vertically aligned?
If so, it's easy.
Viper GTS
BlueWeasel
Lifer
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I printed it out and am looking at it now.
Are the top & bottom lines assumed to be parallel? Yep
What about the right hand endpoints of the top & bottom lines? Are they vertically aligned? Close, but they don't appear to be aligned.
However, for the most part, I think you could consider the lines to be vertically aligned on the right side. The area would be a little off, but you'd be pretty close.
Are the top & bottom lines assumed to be parallel? Yep
What about the right hand endpoints of the top & bottom lines? Are they vertically aligned? Close, but they don't appear to be aligned.
However, for the most part, I think you could consider the lines to be vertically aligned on the right side. The area would be a little off, but you'd be pretty close.
BlueWeasel
Lifer
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just a shot in the dark here, but maybe the triangle formula of Area = 1/2 (ab sin C) would be helpful.
As someone else said, there are an infinite number of solutions when not a single angle is specified. You can still solve this thing in terms of one independently varied angle, however, where all other angles and the area would be a function of your independant angle. You would need to set up a vector loop and then solve for the angles.
If you know anything about mechanics, this problem is essentially a simple six bar linkage. You could solve it with the Newton-Raphson method, as well, but once again you'd need one independant angle. And nobody likes iterations so you'd be better off with the vector loop anyway.
That said, why not just guess? You'll be close enough, I'm sure.
If you know anything about mechanics, this problem is essentially a simple six bar linkage. You could solve it with the Newton-Raphson method, as well, but once again you'd need one independant angle. And nobody likes iterations so you'd be better off with the vector loop anyway.
That said, why not just guess? You'll be close enough, I'm sure.
Originally posted by: BlueWeasel
Alternatively, I can draw it up really quick in AutoCAD, and calculate the area if you want it.
thats the best idea I've heard all day.
Dont do it if its too much trouble though. Although I certainly appreciate the thought.
Originally posted by: DuffmanOhYeah
Originally posted by: BlueWeasel
Alternatively, I can draw it up really quick in AutoCAD, and calculate the area if you want it.
thats the best idea I've heard all day.
Dont do it if its too much trouble though. Although I certainly appreciate the thought.
That will give you a solution, yes, but it won't give you ALL the solutions. Others have stated this, as well.
BlueWeasel
Lifer
- Jun 2, 2000
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Originally posted by: DuffmanOhYeah
Originally posted by: BlueWeasel
Alternatively, I can draw it up really quick in AutoCAD, and calculate the area if you want it.
thats the best idea I've heard all day.
Dont do it if its too much trouble though. Although I certainly appreciate the thought.
Well, I just realized that I have to know the distance between the parallel horizontal lines to determine the area. I can scale it from the drawing, IF it's drawn to scale.
BlueWeasel
Lifer
- Jun 2, 2000
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Well, it's definately not drawn to scale, as my CAD drawing looks nothing like the picture.
Using my drawing, I got an area of 5042 x^2 (where x is your unit of measurement, ie, inches, feet, meters, etc.)
Using my drawing, I got an area of 5042 x^2 (where x is your unit of measurement, ie, inches, feet, meters, etc.)
Well to be fair, my picture was drawn in about 30 seconds using MS paint. I think it clearly evident that your scale is off due to your use of inferior software. In the future, please try to keep up. Thank you.
Seriously though, thanks. I trust way more than my drawing.
Seriously though, thanks. I trust way more than my drawing.
BlueWeasel
Lifer
- Jun 2, 2000
- 15,944
- 475
- 126
Originally posted by: DuffmanOhYeah
Well to be fair, my picture was drawn in about 30 seconds using MS paint. I think it clearly evident that your scale is off due to your use of inferior software. In the future, please try to keep up. Thank you.
Seriously though, thanks. I trust way more than my drawing.
LOL, I thought you were using a scan of something from a book.
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