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Old   June 1, 2001, 14:27
Default area of triangle
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clifford bradford
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How do I find the area of a triangle in 2-D if I know the coordinates of the corners?
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Old   June 1, 2001, 14:36
Default Re: area of triangle
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bronze
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use Heron's law:

a^2 = s(s-a)(s-b)(s-c)

a=area a,b,c = side lengths s=0.5*(a+b+c)

(works well for 3-d too)
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Old   June 1, 2001, 14:38
Default Re: area of triangle
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bronze
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http://www.cs.mtu.edu/~shene/COURSES...06/area-3.html

The method and the code
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Old   June 1, 2001, 17:05
Default Re: area of triangle
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Sid
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The three vertices are ordered in counterclockwise order, (x1,y1), (x2,y2), (x3,y3). Then,

S=0.5*(x1*(y2-y3)+x2*(y3-y1)+x3*(y1-y2))

which can be derived easily by computing a cross product of a parir (any pair) of the side vectors.

Bye
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Old   June 1, 2001, 18:38
Default Re: area of triangle
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clifford bradford
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Aha! Thanks.
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Old   June 2, 2001, 09:33
Default Re: area of triangle
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Barry
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Hi

The very best way to do it is to calculate the cross product using two vectors (sharing the same origen)(one of the corner vertex of the triangle) which forms the edges of the triangle. Half the norm of the resulting vector will give ya the surface area AND by doing so you already calculated a normal to the surface which ye want anyhow. Just make sure your convention is correct, otherwise your normal vector might point in the wrong direction.

Cheers Barry
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