The number of diagonals in a hexagon is:

Correct — B. A diagonal joins two vertices that are not neighbours.

From each vertex: 6 − 3 = 3 diagonals (not to itself or its two neighbours)
From all 6 vertices: 6 × 3 = 18, but each diagonal is counted from both ends
Halve the double count: 18 ÷ 2 = 9

Formula check: n(n − 3)⁄2 = 6 × 3 ÷ 2 = 9 → option (b)

Why the others are wrong

• (a)
  6 — 6 is the number of sides of a hexagon, not its diagonals. Each vertex on its own already sends out 3 diagonals.
• (c)
  12 — 12 is not the diagonal count of any polygon: n(n − 3)⁄2 gives 9 for a hexagon and 14 for a heptagon, skipping 12.
• (d)
  15 — 15 is C(6, 2), every pair of vertices joined, which includes the 6 sides. Remove them: 15 − 6 = 9.

Any two vertices of an n-gon define a segment, C(n, 2) = n(n − 1)⁄2 of them. n of those segments are sides.

Diagonals = n(n − 1)⁄2 − n = n(n − 3)⁄2.

The vertex-by-vertex count agrees: each of the n corners sends out n − 3 diagonals, and halving removes the double count.

Key facts

• Number of diagonals of an n-sided polygon = n(n − 3)⁄2.
• In a convex polygon, the n − 3 diagonals from one vertex split it into n − 2 triangles.
• Diagonal counts: pentagon 5, hexagon 9, heptagon 14, octagon 20.

Common traps

• Counting 3 diagonals at each of the 6 vertices and forgetting to halve, which gives 18.
• Using C(6, 2) = 15 without subtracting the 6 sides.

No directly related past PYQ was found.