Originally posted by ChronicLeakyyeah u r right, its pretty much the same thing i told earlier..i simply expanded the determinant..😀
EDIT What I said before should be done in two steps, one per triangle.
The are of a triangle with vertices (a,b), (c,d), (e,f) is 1/2*det(a b 1, c d 1, e f 1).
cheers!
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Originally posted by ChronicLeakyI also didn't know about it... Cute, isn't it?
Nice! I guess Green's theorem will give this, which is probably what OP should have gone for in the first place.
Geometry is really not my thing, but I liked the method you posted, went looking for a proof and found this. It's funny how we (I) immediately think of integrals when dealing with curved shapes, but we tend to go for 'easier' ways when the lines are straight (obviously forgetting how easy the integrals would be!).
Originally posted by PalynkaThat's a good observation. I think maybe it's because we learn calculus after we already have experience measuring things like polygons (which is exhausting). I often notice how often I don't think to use big technical machinery to do things that are special cases of such machinery. Sometimes it's a case of "Yesssss, saved some work" and sometimes it puts a real hole in the like unified picture of things. I suspect that in the case of measuring polygons, with integration being the big machinery, avoiding same is in the "Yessss" category.
I also didn't know about it... Cute, isn't it?
Geometry is really not my thing, but I liked the method you posted, went looking for a proof and found this. It's funny how we (I) immediately think of integrals when dealing with curved shapes, but we tend to go for 'easier' ways when the lines are straight (obviously forgetting how easy the integrals would be!).