Let G be a connected planar graph with 10 vertices. If the number of edges on…

Concept: For any connected planar graph drawn in the plane, Euler's formula relates the number of vertices (V), edges (E), and faces (F, counting the unbounded outer face) as V − E + F = 2. Separately, if every face of the embedding is bounded by exactly k edges, summing the edge-count over all faces counts each edge exactly twice (since every edge borders exactly two faces), giving kF = 2E.

Application: Apply both relations to this graph.

1. Here V = 10, and every face (including the outer face) has k = 3 bounding edges, so 3F = 2E, i.e. F = 2E/3.

2. Substitute F = 2E/3 into Euler's formula V − E + F = 2: 10 − E + 2E/3 = 2.

3. Multiply throughout by 3 to clear the fraction: 30 − 3E + 2E = 6.

4. Simplify: 30 − E = 6, so E = 24.

Cross-check: A connected planar graph in which every face, including the outer one, is a triangle is a maximal planar graph (a triangulation), which always satisfies E = 3V − 6 for V ≥ 3. With V = 10, E = 3(10) − 6 = 24, matching. Also, F = 2E/3 = 16 faces in total, and 3F = 48 = 2E confirms consistency.

Note: the condition "every face has 3 edges" applies to the outer (unbounded) face as well as the inner ones — this is exactly what makes 3F = 2E hold for the whole graph. Check the student's own attempt against both identities: if a drawing has V = 10 and F = 12 faces, then 3F = 2E would require E = 18, while applying Euler's formula directly with F = 12 gives E = 10 + 12 − 2 = 20. Since these two values disagree, that drawing cannot actually have every face — including the outer one — bounded by exactly 3 edges, so it does not satisfy this problem's condition and cannot be used to check the answer.

Answer: The graph has 24 edges.