suppose the polygon is closed, find its surface area given its 100 vertices in x,y ,z coordiantes? please then find the general formula for n-polygon with n given vertices?
2007-08-08 07:09:18 · 5 answers · asked by BenL 2 in Science & Mathematics ➔ Mathematics
okay, polyhedron, just find it.
2007-08-08 07:21:35 · update #1
the geometry can be in any shape, a bottle, a cup or ball or any object.
2007-08-08 08:47:04 · update #2
You can only find a surface area if you have a two-dimensional object that means, the points on the 3-dimensional space are coplanar.
If they are coplanar, then there is a 1-1 correspondence between (x,y,z) and (x',y') ... [not the derivative]
To get the area, arrange the (x', y') points in a column. You now have an n x 2 matrix. Multiply a(i,1) with a(i+1,2), for i from 1 to n-1, add a(n,1)*a(1,2) ... call this A.
Multiply a(i+1,1) with a(i,2) , for i from 1 to n-1, add a(1,1)*a(n,2) ... call this B.
The area of the region is 1/2(A-B).
All of these will work only with a convex polygon.
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Otherwise, if the points are forming a polyhedron, one must first determine the type of polyhedron (or stellation of these polyhedron). Note that there are only 5 regular polyhedra.
2007-08-08 08:06:16 · answer #1 · answered by Alam Ko Iyan 7 · 1⤊ 0⤋
The answer is: You can not!
Imagine a cube on a table and put over it a tetragonal pyramid with a base the upper face of the cube (like a house, covered with a roof). This convex polyhedron has 9 vertexes and 9 faces. Now remove the lower face of the cube (the base of the house) and connect its 4 vertexes with the top of the roof, thus making 4 more triangular faces inside the former solid (turn the 'house' upside down and dig a pyramidal hole). You have now another closed concave polyhedron with the SAME 9 VERTEXES (and 12 faces) whose surface area is DIFFERENT from the area of the former! You can easily master other similar examples.
The point is that the data provided are not sufficient. A set of points does not define unambiguously a polyhedron (as other answers correctly mention), unless they are 4, defining a unique tetrahedron, and the best that can be done is to find the surface area of the so called 'convex envelope' - the smallest convex set, containing the points given (it is a convex polyhedron in our case). Finding the convex envelope is generally a difficult problem, there are algorithms for that but no formulas (except tetrahedron of course).
Sorry if now You feel somewhat disappointed!
2007-08-09 03:07:19 · answer #2 · answered by Duke 7 · 0⤊ 0⤋
Similar shapes have corresponding sides that are related by a fixed ratio k, and areas related by k^2. So (comparing the smaller area to the larger area), k^2 = 64/100 giving k = 8/10. Hence the corresponding side of the smaller polygon = k*36 = 8*36/10 = 8*3.6 = 28.8 units.
2016-05-17 06:56:02 · answer #3 · answered by ? 3 · 0⤊ 0⤋
I don't really see that the coordinates of the vertices is enough information. Which subsets of the vertices are vertices of faces? Do we assume the figure is convex, or do we want something more general than that?
The problem with doing a search on this is wading through all the high school math lessons online. Good luck.
2007-08-08 08:04:11 · answer #4 · answered by Philo 7 · 0⤊ 0⤋
it seems like you're talking about three-dimensional figures. These are called polyhedrons, not polygons. Polygons are all two-dimensional.
2007-08-08 07:12:56 · answer #5 · answered by jibba.jabba 5 · 0⤊ 0⤋