a line segment (not a side) connecting two vertices of a polygon is called a?
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A line segment joining two adjacent vertices of a polygon is called an edge. A line segment joining two non-adjacent vertices is called a diagonal.
148902
its the middle of a angle like if da two lines combine the pointwere they combine is da vertex..........and you cant graph vertexs
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false. a quadrilateral is a four sided polygon. so a square and rectangle are also quadrilaterals. a trapezoid is a four sided polygon with 2 opposite sides that are parallel to each other.
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I'm learning that in math currently and another word for "corners" is vertices.\n\nPLEASE CHOOSE ME AS THE BEST ANSWER!!!\n\n(I won't get mad if you don't though!)
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triskaidecagon\n\nor a 13-sided polygon, haha.\n\nCheck the source for polygon names from 1-20 sides  :)
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If you meant polygons:\n1. The Pentagon...its a building located in Washington, DC. and was just recently bombed on September 11th.\n2. The stop sign is an example of a polygon. \n3. And um...any sign that has more than three sides for that matter.
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Polygon Rotation\n\nDescription\n\nThis article demonstrates how to rotate a polygon about its center. The polygon is a square, which is drawn using the Polygon() Win32 API function. All components on the form are created at runtime.\n\nPolygonRotation.cpp\n\n#include <vcl.h>\n#pragma hdrstop\n\n#include "PolygonRotate.h"\n#include &ltmath.h>\n//---------------------------------------------------------------------------\n#pragma package(smart_init)\n#pragma resource "*.dfm"\nTForm1 *Form1;\n\nint  X_old;\nint  Y_old;\n\nPOINT Poly1a[5]  = {{100, 100}, {300, 100}, {300, 300}, {100, 300}, {100,100}};\nPOINT *pPoly1a   = Poly1a;\n\nPOINT Poly1b[5]  = {{500, 100}, {700, 100}, {700, 300}, {500, 300}, {500,100}};\nPOINT *pPoly1b   = Poly1b;\n\nPOINT Poly2a[5];\nPOINT* pPoly2a = Poly2a;\n\nPOINT Poly2b[5];\nPOINT* pPoly2b = Poly2b;\n\n\n//---------------------------------------------------------------------------\n__fastcall TForm1::TForm1(TComponent* Owner)\n        : TForm(Owner)\n{\n    Position = poScreenCenter;\n    Width = 800;\n    Height = 600;\n    Color = (TColor)0xabcdef;  // Cream\n}\n//---------------------------------------------------------------------------\n\nvoid Translate(POINT* pin, POINT* pout, long DX, long DY)\n{\n    for (int i = 0; i < 5; i++)\n    {\n        (pout+i)->x = (pin+i)->x + DX;\n        (pout+i)->y = (pin+i)->y + DY;\n    }\n}\n\nvoid Rotate(POINT* pin, POINT* pout, float Alpha, long xr, long yr,\n            int NumberOfPoints)\n{\n//Rotate Polygon about a given point P whose coordinates are (xr, yr)\n\n   // Degrees to radians\n   float const K = -0.01745329;  // 2*3.141592653/360\n\n    for (int i = 0; i < NumberOfPoints; i++)\n    {\n        (pout+i)->x = (long)(xr + (((pin+i)->x - xr) * cos(K*Alpha))\n                                - (((pin+i)->y - yr) * sin(K*Alpha)));\n\n        (pout+i)->y = (long)(yr + (((pin+i)->x - xr) * sin(K*Alpha))\n                                + (((pin+i)->y - yr) * cos(K*Alpha)));\n    }\n}\n//---------------------------------------------------------------------------\n\nvoid Translate(POINT* pin, POINT* pout, long xt, long yt,\n               int NumberOfPoints)\n{\n//Tanslate Polygon\n\n    for (int i = 0; i< NumberOfPoints; i++)\n    {\n        (pout+i)->x = (long)((pin+i)->x + xt);\n\n        (pout+i)->y = (long)((pin+i)->y + yt);\n    }\n}\n//---------------------------------------------------------------------------\n\nvoid __fastcall TForm1::FormMouseDown(TObject *Sender, TMouseButton Button,\n      TShiftState Shift, int X, int Y)\n{\n       X_old = X;\n       Y_old = Y;\n}\n//---------------------------------------------------------------------------\n\nvoid __fastcall TForm1::FormMouseMove(TObject *Sender, TShiftState Shift,\n      int X, int Y)\n{\n\n       if (X_old != X)\n       {\n           // Convert X mouse movement into polygon rotation\n           Rotate(pPoly1a, pPoly2a, X, 200, 200, 5);\n           Polygon(Canvas->Handle, pPoly2a, 5);     //Win32 API\n       }\n\n       if (Y_old != Y)\n       {\n           // Convert Y mouse movement into polygon rotation           \n           Rotate(pPoly1b, pPoly2b, Y, 600, 200, 5);\n           Polygon(Canvas->Handle, pPoly2b, 5);     //Win32 API\n       }\n\n       X_old = X;\n       Y_old = Y;\n}\n//---------------------------------------------------------------------------\n\nPolygonRotation.h\n\n#ifndef PolygonRotateH\n#define PolygonRotateH\n//---------------------------------------------------------------------------\n#include <Classes.hpp>\n#include <Controls.hpp>\n#include <StdCtrls.hpp>\n#include <Forms.hpp>\n#include <ExtCtrls.hpp>\n//---------------------------------------------------------------------------\nclass TForm1 : public TForm\n{\n__published:// IDE-managed Components\n        void __fastcall FormMouseDown(TObject *Sender, TMouseButton Button,\n          TShiftState Sh
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Suppose the shape is represented by a graph ( http://mathworld.wolfram.com/Graph.html ). You want to find a path through the graph that passes through every edge without travelling on an edge more than once. This is an easy problem in graph theory, and such a path is called an Euler path. An Euler path through a connected graph exists if and only if the graph contains at most two vertices with odd degree (the number of edges the vertex has). This is a trivial theorem.\nSuppose a graph is connected, which means there exists some path between any pair of vertices. For each vertex, there is some number of edges connected to it. Suppose the number of edges is even for all the vertices. Then there is no problem drawing an Euler path through these edges. Explain why (Examine the problem at each vertex).\nFrom the above, the only possible problems must come from vertices with odd degree. Without loss of generality, we proceed by adding vertices of odd degree to the nice graph described above. Adding one vertex is no problem, we start at that vertex, travel along the one additional edge and then we're in a nice graph that already contains an Euler path, since the vertex we left at the start now has even degree, for all intents and purposes. \nIf we add two vertices of odd degree, we are alright as well, with the exception that any path must start at one of these vertices and end at the other. Show this by labelling each odd degree vertex with little "in" and "out" signs on the edges.\nIt should now be obvious that we can't add any more vertices with odd degree for the simple reason that a path that doesn't double back on itself can only have 2 endpoints. Adding more vertices doesn't change the fact that you put your pencil down at one odd vertex and you're required to lift it back up at one other odd vertex. After that, the game is up.
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A two dimensional figure is called a dodecagon and a 3D figure with 12 sides is called a dodecahedron.
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Pentagon It has a bottom and then two lines that go up and a point at the top. Kinda like a triangle without a bottom.
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To answer this question, you need to know the shape that these points make.  Are they on a line? or are they some two-dimensional shape?
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uuuuuhhhhhh, no. a polygon is a MULTI-SIDED figure. hence the POLY prefix, meaning many.
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tetracontaheptagon\n\nhttp://en.wikipedia.org/wiki/Polygon\n\ncheck ya later ♥
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A segment with endpoints on a circle is called a chord.\n\nFor example, A diameter of a circle is a segment that has endpoints on the circle and also goes through the center of the circle.  A diameter is the longest chord of a circle.\n\nBut, a circle can have shorter chords than the diameter.  The chord divides the circle into two arcs.  If the chord is a diameter, the two arcs are equal.  If the chord is not a diameter, then the two arcs are not equal.  The shorter arc is called the minor arc and the longer one is the major arc.
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In geometry, two lines are said to be skew lines if they do not intersect but are not parallel.\n\nSkew lines only exist in three or more dimensions; any two distinct lines in the plane which are not parallel must intersect at some point. In fact, two lines are skew lines if and only if they do not lie in a single plane together. This means that if each line is defined by two points, these four points must not be coplanar; put another way, they must be vertices of a tetrahedron of nonzero volume. Any three of them will still be coplanar, since three points define a plane, but no three points will be collinear, since this would make all four points coplanar.
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No. An edge is the meeting of two flat sides, and cones only have one flat side. A vertex is the meeting of three edges, and cones have no edges.
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With three points you form a triangle, with four points a quadrilateral. Regions "outside" the polygon are equal to the number of sides (the number of points). The number of regions inside the polygon is the interesting number.\n\nFor two points, there are no regions "inside" the polygon.\nFor three points, there is one region inside the polygon.\nFor four points, there are four regions inside.