Sum of angles of a polygon. What is the sum of angles of a (planar non-self-intersecting) n-gon? We know that the answer is 180° for n = 3 and 360° for n = 4. Find the next few values. Then find a formula in terms of n and prove it.

Step-by-step solution

Idea: Cut the polygon into triangles using diagonals that lie inside it. Each cut by an inside diagonal adds one piece; a polygon with n sides ends up as n − 2 triangles.

1. Values: a pentagon splits into 3 triangles: 540°; a hexagon into 4: 720°; a 7-gon into 5: 900°; an 8-gon into 6: 1080°. Pattern: each extra side adds 180°, so the sum is (n − 2) × 180°.1 mark
2. Key fact: every polygon with n ≥ 4 sides has a diagonal lying inside it. Take a vertex V with internal angle less than 180° (the lowest vertex has one), with neighbours U and W. If no other vertex lies inside or on ∆UVW, then UW is an inside diagonal. Otherwise, among the vertices inside ∆UVW take the one farthest from line UW; joining it to V gives an inside diagonal.1½ marks
3. Proof by building up: an inside diagonal splits an n-gon into a k-gon and an (n − k + 2)-gon, and the angles of the two pieces together make up the angles of the n-gon. If the formula holds for the smaller polygons, the sum is (k − 2)180° + (n − k)180° = (n − 2) × 180°. Starting from triangles (180°), this proves it for n = 4, 5, 6, … in turn.1½ marks

Check: For the non-convex hexagon in the picture, the four triangles give 4 × 180° = 720° = (6 − 2) × 180° ✓, the reflex angle included.

Answer to write in the exam

Common mistakes that cost marks

• Using n × 180° (counting the angles at a central point that are not angles of the polygon).
• Proving it only for convex polygons by drawing all diagonals from one vertex; for a non-convex polygon some of those diagonals lie outside.
• Forgetting the reflex angle of a non-convex polygon when adding.

How this can come in the exam

Try one yourself

Find the sum of the angles of a 12-gon and each angle of a regular 12-gon.

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