definition of a cube-faces,edges and vertices?
543718
6 faces, 12 edges, 8 vertices
374445
sphere has no vertices\n\narea of prism is:\n (area of base)*2 + (perimeter of base)*height.
196942
wow, that's pretty amazing.  i was convinced that was wrong until i tried drawing the "counterexample" i had in my head.  other than reading through that proof (props to the guy who posted that link, btw) i guess the best way to realize this is true is to show that for any five closed shapes in two dimensions, there will always be at least two shapes that do not touch one another.  because of this, you can color those two shapes the same color, so you will have a maximum of four colors.\n\nthis problem is reducible to a graph with five nodes, where each node represents a shape and each edge represents a shared boundary.  (showing that this is reducible to this is probably the difficult part of the proof, which is why i'm glossing over it).  you cannot connect each node to every other node without there being at least one set of intersecting edges.  if edges intersect, it indicates that the shapes (or countries) represented by the two nodes connected by one of the intersecting edges do not share a border.\n\nhere's my suggested proof (assuming we accept the reduction to this graph problem):\nstart with a graph G with three nodes, {A, B, C}.  There are also edges {AB, BC, AC}.  So, G is a closed graph with three nodes and three edges where each node is connected to each other node in the graph (i.e. G is a triangle).  If our reduction of the coloring problem is correct, then there exists a map reducible to this graph that requires three colors to be colored in the described way.\n\nNext, construct a graph H by adding a node D to graph G.  There are two possible ways to add D: 1) D can be on the interior of closed graph G; or 2) D can be outside the interior of G.\n\nIn case 1, let the new graph with nodes A,B,C,D be called H.  We need to show that we can make H such that D is connected to A, B, and C without any intersecting edges.  This is simple enough; straight edges to A, B, and C from D do not intersect each other or the existing edges of the graph G.  So, H demonstrates that a map can require four colors.  H now has four separate enclosed interior regions: triangles ABD, ACD, BCD, and ABC.\n\nFrom H, construct a graph I by adding another node E.  Repeating the above procedure, E can be placed in four places: a) in the interior of ABD; b) in the interior of ACD; c) in the interior of BCD; or d) on the exterior of ABC.\n\nCASE a: As with constructing H, edges can be drawn from E to A, B, and D.  However, C is on the exterior of ABD, so any edge from E to C must cross either AB, BD, or AD.  So, in case a, there must be an intersection, so four colors is still an option.\nCASES b and c follow similarly.\nCASE d: E is on the exterior of ABC, but D is on the interior of ABC.  Any edge between E and D must intersect AB, BC, or AC.  So, coloring with four colors is possible in this case.  Since it is possible in all cases, it is possible for any map with five regions with a sub-map reducible to a graph like H.\n\nCase 2) Instead of placing D in the interior of G, make a graph H' by placing D on the exterior of triangle ABC.  Any quick drawing will show that D is easily connected to A, B, and C.  So there is a map with four regions that requires four colors.  To make the rest of this easy, note that H' also produces four enclosed regions: ABD, ABC, BCD, and ADC.  I won't go through the trouble to prove this, but clearly H' is isomorphic to H.  That being said, whatever is true of the general graph H is also true of the general graph H'.  Therefore, a graph I' made by adding a node to H' cannot be complete without there being at least one set of intersecting edges.\n\nSince this is true for all cases, any map reducible to a graph with five nodes is colorable with just four colors.  QED\n\nSorry this is so long.  I hope it's helpful.
274106
Vertex is junction of two or more lines or segments.  Point is any location on a line or segment (or arc), which may or may not also correspond to a vertex.
788320
If it's a cube, all sides have to be squares.\n\nIf the surface area of one side is 1, then the the length of each edge is also 1.\n\nThen, the diagonal of a side would be √(1² + 1²) = √2\n\nAnd, not sure if this is what you're asking, a diagonal connecting one corner of the cube to the opposite corner would have the diagonal of one side as a base, and an edge as the other side, so it would have length √(1+2) = √3
910632
lines, polylines, blocks, circles etc... look in the help files if you have the software\n\nhttp://www.we-r-here.com/cad/tutorials/index.htm
414837
The idea is to build up the corners, you want to get a 2x2x2 corner lined up and then work from there.\n\nHere's a very helpful link\nhttp://lar5.com/cube/
571159
b)hexagon(2d) or cube(3d).\nc)hemisphere\nd)triangle?\ne)net\nf)cube\ng)sphere\nh)decahedron\ni)dodecahedron\nj)parallelogram i think, if wat u mean is opposite equal  angles.
189752
sphere, cube, pyramid, cilinder
494335
The five named after Plato are\n\nTetrahedron: 4 triangular faces, 4 vertices, 6 edges\nCube: 6 square faces, 8 vertices, 12 edges\nOctahedron: 8 triangular faces, 6 vertices, 12 edges\nDodecahedron: 12 pentagonal faces, 20 vertices, 30 edges\nIcosahedron: 20 triangular faces, 12 vertices, 30 edges\n\nEuler's Rule E + 2 = F + V applies to all five aolids.
765048
Cuboid is a 3D rectangle\n\nExamples in home are:\nComputer midi towers\nwardrobes\nNew fashion coffee tables\nMicrowaves\nOven\nFridge\netc..\n\nAnything that is a cube is also a cuboid as a cube is a special cuboid where all the sides are equal
588234
Polygons are many-sided figures, with sides that are line segments. \n\nPolygons are named according to the number of sides and angles they have. The most familiar polygons are the triangle, the rectangle, and the square. A regular polygon is one that has equal sides. \n\nPolygons also have diagonals, which are segments that join two vertices and are not sides.\n\nFor more details you can visit the maths site given below:
801416
sphere none of the list\ncone one vertex and one face\ncylinder two faces-two circular edges
692594
one and two. and edge is where two faces meet. think about it, you got that ring at the bottom of the cone, and you have two on a cylinder.
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float ComputeEdgeCollapseCost(Vertex *u,Vertex *v)\n{\n   // if we collapse edge uv by moving u to v then how\n   // much different will the model change, i.e. the\n   // "error".\n   float edgelength = magnitude(v->position - u->position);\n   float curvature=0;\n   // find the "sides" triangles that are on the edge uv\n   List<Triangle *> sides;\n   for(i=0;i<u->face.num;i++)\n   {\n      if(u->face[i]->HasVertex(v))\n      {\n         sides.Add(u->face[i]);\n      }\n   }\n   // use the triangle facing most away from the sides\n   // to determine our curvature term\n   for(i=0;i<u->face.num;i++)\n   {\n      float mincurv=1;\n      for(int j=0;j < sides.num;j++)\n      {\n         // use dot product of face normals.\n         float dotprod = \n            u->face[i]->normal ^   sides[j]->normal;\n         mincurv = min(mincurv,(1-dotprod)/2.0f);\n      }\n      curvature = max(curvature,mincurv);\n   }\n   return edgelength * curvature;\n}\n\nvoid ComputeEdgeCostAtVertex(Vertex *v)\n{\n   if(v->neighbor.num==0)\n   {\n      v->collapse=NULL;\n      v->cost=-0.01f;\n      return;\n   }\n   v->cost = 1000000;\n   v->collapse=NULL;\n   // search all neighboring edges for Òleast costÓ edge\n   for(int i=0;i < v->neighbor.num;i++)\n   {\n      float c;\n      c = ComputeEdgeCollapseCost(v,v->neighbor[i]);\n      if(c < v->cost)\n      {\n         v->collapse=v-neighbor[i];\n         v->cost=c;\n      }\n   }\n}\n\nvoid Collapse(Vertex *u,Vertex *v)\n{\n   // Collapse the edge uv by moving vertex u onto v\n   if(!v)\n   {\n      // u is a vertex all by itself so just delete it\n      delete u;\n      return;\n   }\n   int i;\n   List<Vertex *>tmp;\n   // make tmp a list of all the neighbors of u\n   for(i=0;i<u->neighbor.num;i++)\n   {\n      tmp.Add(u->neighbor[i]);\n   }\n   // delete triangles on edge uv:\n   for(i=u->face.num-1;i>=0;i--)\n   {\n      if(u->face[i]->HasVertex(v))\n      {\n         delete(u->face[i]);\n      }\n   }\n   // update remaining triangles to have v instead of u\n   for(i=u->face.num-1;i>=0;i--)\n   {\n      u->face[i]->ReplaceVertex(u,v);\n   }\n   delete u;\n   // recompute the edge collapse costs in neighborhood\n   for(i=0;i<tmp.num;i++)\n   {\n      ComputeEdgeCostAtVertex(tmp[i]);\n   }\n}
462376
20\n\n 12 edges and 8 vertices
254189
cube = a 3  \n\nrectangular prism = a b c  \n\nirregular prism = b h  \n\ncylinder = b h = pi r 2 h  \n\npyramid = (1/3) b h  \n\ncone = (1/3) b h = 1/3 pi r 2 h  \n\nsphere = (4/3) pi r 3  \n\nellipsoid = (4/3) pi r1 r2 r3