1. **Problem Statement:** We have a graph with vertices A, B, C, D, E, F and edges connecting them. We need to verify if the graph is planar using Euler's formula, state why it is not Eulerian, and find a path visiting every stage exactly once starting at F. 2. **Euler's Formula for Planar Graphs:** Euler's formula states that for a connected planar graph: $$ V - E + F = 2 $$ where $V$ is the number of vertices, $E$ is the number of edges, and $F$ is the number of faces (including the outer face). 3. **Count vertices and edges:** - Vertices $V = 6$ (A, B, C, D, E, F) - Edges $E = 9$ (A-B, A-C, A-F, B-E, B-F, C-D, D-E, D-F, E-F) 4. **Calculate faces $F$ using Euler's formula:** Rearranged: $$ F = 2 - V + E = 2 - 6 + 9 = 5 $$ So, the graph has 5 faces. 5. **Planarity verification:** Since $V - E + F = 2$ holds with integer values and the graph can be redrawn without edges crossing (as per part (a)), the graph is planar. 6. **Why the graph is not Eulerian:** A graph is Eulerian if every vertex has an even degree (even number of edges). Degrees: - A: 3 edges (odd) - B: 3 edges (odd) - C: 2 edges (even) - D: 3 edges (odd) - E: 3 edges (odd) - F: 4 edges (even) Since vertices A, B, D, and E have odd degrees, the graph is not Eulerian. 7. **Path visiting every stage exactly once starting at F:** This is a Hamiltonian path (visiting each vertex once). One such path starting at F is: $$ F \to B \to A \to C \to D \to E $$ 8. **Name of this path:** This path is called a **Hamiltonian path**. **Final answers:** - The graph is planar by Euler's formula with $V=6$, $E=9$, $F=5$. - It is not Eulerian because some vertices have odd degree. - A Hamiltonian path starting at F is $F \to B \to A \to C \to D \to E$. - This path is called a Hamiltonian path.