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Thread: Perimeter of a triangle...

20050908, 02:07 #1
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Perimeter of a triangle...
Not really a puzzle...
Just need help finding the perimeter of a triangle with coordinates: (3,6), (3,1) and (2,1). Ricky

20050908, 02:46 #2
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Re: Perimeter of a triangle...
<span style="backgroundcolor: #FFFF00; color: #FFFF00; fontweight: bold">The distance from (3,6) to (3,1) = sqrt((33)^2 + (6 1)^2) = sqrt(49) =7
The distance from (3,1) and (2,1) = sqrt((3 2)^2 + (1 1)^2) = sqrt(29) = 5.385
The distance from (2,1) to (3,6) = sqrt((23)^2 + (16)^2) = sqrt(50) = 7.071
The sum = 19.456</span hide>
Steve

20050908, 03:27 #3
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Re: Perimeter of a triangle...
Another way to potentially simplify the problem is to realize that any triangle can be moved around the XY plane, such that one vertex will always be at (0,0) and one side can always lie along one axis. In your case, it's only a simple translation needed to achieve this.
By adding (3,1) to each vertex, and an optional reflection across the Yaxis, its new coordinates become (0,0), (0,7) and (5,2). The formulas Steve used still apply  it just makes the numbers easier to work with.
Alan

20050908, 13:35 #4
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Re: Perimeter of a triangle...
Steve and Alan,
Thanks for the responses. I was helping my 9th grader with some geometry homework. Got through all of the problems just fine until this one popped up. Actually, we came to the same conclusion but thought the answer was a bit "ugly" and figured we must've done something wrong. The other solutions [to the other problems] were cleaner. We backed into the solution by drawing the triangle on graph paper, then dividing the triangle into two separate right triangles using the zero horizontal. Then did the ol' a^2+b^2=c^2 trick on each...
<img src=/S/cheers.gif border=0 alt=cheers width=30 height=16> Ricky

20050908, 14:08 #5
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Re: Perimeter of a triangle...
The calculation I did is general for all coord points on a plane (it can be extended to 3 dimensions by just adding other "squares")
=sqrt( (x1x2)^2 + (y1y2)^2 +(z1z2)^2 )
Steve