The dodecahedron star has 30 edges, 12 faces, and 20 vertices. But, a 3D dodecagon star has 60 edges, 24 faces, and 40 vertices but i'm not sure. Some people still need help on how many edges faces and vertices a dodecagon star has. Thanks for your cooperation by looking at this answer.
A tetrahedron has four faces.
Actually it has ten vertices's you half to count the inside too
An octagon has 8 sides and 8 vertices
To determine the number of triangles in a star inside a hexagon, we need to consider the number of triangles formed by the lines connecting the vertices of the hexagon and the points where the lines of the star intersect. Each intersection point forms a triangle with two adjacent vertices of the hexagon. Therefore, if the star has n points of intersection, the total number of triangles would be n multiplied by 2. Additionally, we need to consider the triangles formed by the lines of the star itself, which would add n triangles to the total count. So, the total number of triangles in a star inside a hexagon would be 3n.
It has 10 vertices, 10 edges, and 0 faces.
The dodecahedron star has 30 edges, 12 faces, and 20 vertices. But, a 3D dodecagon star has 60 edges, 24 faces, and 40 vertices but i'm not sure. Some people still need help on how many edges faces and vertices a dodecagon star has. Thanks for your cooperation by looking at this answer.
A 3D star is not a standard shape. You can easily add another point.
This is probably about Euler's formula V - E + F = 2, where V is the number of vertices, E he number of edges and F the number of faces. Example: a cube has 6 faces (F = 6) and 8 vertices (V = 8). So the formula tells us that 8 - E + 6 = 2, and so E = 12. Yes, a cube has 12 edges. Euler's formula only works for standard polyhedra, not unusual things like star polyhedra.
A star is not a specific shape: it is a generic word for a shape which has an even number of vertices. The interior angles at alternate vertices are usually reflex angles. A star can have six or more vertices.
14
A tetrahedron has four faces.
Actually it has ten vertices's you half to count the inside too
Actually it has ten vertices's you half to count the inside too
A star has 10 sides and 5 vertices but if it's a decagonal star then 20 sides (10 for the star and 10 for the star) and 10 vertices (for the decagon in the middle, hence decagonal star). But, if it's pentagram star then 10 sides and 10 vertices.
A star graph, call it S_k is a complete bipartite graph with one vertex in the center and k vertices around the leaves. To be a tree a graph on n vertices must be connected and have n-1 edges. We could also say it is connected and has no cycles. Now a star graph, say S_4 has 3 edges and 4 vertices and is clearly connected. It is a tree. This would be true for any S_k since they all have k vertices and k-1 edges. And Now think of K_1,k as a complete bipartite graph. We have one internal vertex and k vertices around the leaves. This gives us k+1 vertices and k edges total so it is a tree. So one way is clear. Now we would need to show that any bipartite graph other than S_1,k cannot be a tree. If we look at K_2,k which is a bipartite graph with 2 vertices on one side and k on the other,can this be a tree?
A geometrical star can have various numbers of vertices, and the angles formed at the vertices need not be the same. The number of perpendicular lines depends on exactly how many vertices it has and how their angles are configured.