Yes, but not a polygon (or polyhedron).
Consider a quadrilateral with one diagonal. The end points of the diagonal are at odd vertices while the other two are even.
The shape you are describing is a triangular prism. It has an odd number of vertices (5), four of its faces are isosceles (two triangular and three rectangular faces), and it has an even number of edges (9). This configuration meets all the specified criteria.
Break up the odd shape into even shapes and add the areas.
It is traversable if there is an even number of edges at each vertex, or at every vertex except two. In the latter case the traverse must start at one of the "odd vertices" and finish at the other.
To eulerize a graph, you need to ensure that all vertices have even degrees, as this is a requirement for a graph to have an Eulerian circuit. If any vertices have odd degrees, you can add edges between pairs of odd-degree vertices to make their degrees even. The added edges can be chosen carefully to minimize the total length of the resulting Eulerian circuit. Finally, the resulting graph will have all vertices with even degrees, allowing for an Eulerian path or circuit.
All you do is add up all the sides to find the perimeter even though it is a odd shape
Well, honey, let me break it down for you. A pyramid can have an odd or even number of vertices, depending on the base shape. If the base has an odd number of sides, then the pyramid will have an odd number of vertices. But if the base has an even number of sides, then the pyramid will have an even number of vertices. It's as simple as that, darling.
No. Not can it have an odd number of vertices.
In order for a network to be transversable, it either needs to have all of the vertices even, or just 2 odd vertices
No it is not possible to get a even number by adding odd and even number. 1+2=3-odd number.3+4=7-odd number.
Break up the odd shape into even shapes and add the areas.
It is traversable if there is an even number of edges at each vertex, or at every vertex except two. In the latter case the traverse must start at one of the "odd vertices" and finish at the other.
No. even and even or even and odd give even only
All you do is add up all the sides to find the perimeter even though it is a odd shape
No.
no
3 if n is odd 2 if n is even where n is the number of vertices.
I'm gonna guess no, because numbers are even and odd by the last number, so... No.