Interior angles of an n-sided polygon sum to (n−2)×180°
Split the polygon into triangles from one vertex; each triangle contributes . So the interior-angle sum is — e.g. a pentagon () gives , a hexagon () gives .
The exterior angles of ANY polygon sum to 360°
Walking once around the outside of a convex polygon turns you through one full turn, so the exterior angles always add to — no matter how many sides. For a regular polygon each exterior angle is therefore .
Interior + exterior angle = 180° at every vertex
At each vertex the interior and exterior angles lie on a straight line, so they add to . For a regular polygon: each interior angle .
Drawn from real examiner reports.
Multiplying by n instead of (n−2)
Using for the interior-angle sum (e.g. for a pentagon) instead of . Always subtract 2 from the number of sides first.
Confusing interior and exterior angles when finding n
When given a regular polygon's interior angle, dividing by it directly. You must first find the exterior angle () and then do .
Using the polygon-sum rule on an irregular polygon as if it were regular
Dividing the total interior-angle sum by to get "each angle" only works for a regular polygon. For an irregular polygon you can only use the sum, then subtract the known angles.
Go through the exterior angle — it is usually the quickest route
To find the number of sides, work with the exterior angle: . Given an interior angle, first do to get the exterior angle. This avoids the heavier algebra.
For a polygon with sides: The interior sum comes from cutting the polygon into triangles from a single vertex (each ). The exterior sum is always — one full turn as you walk around the outside — regardless of the number of sides.
| Polygon | Sides | Interior sum |
|---|---|---|
| Triangle | 3 | |
| Quadrilateral | 4 | |
| Pentagon | 5 | |
| Hexagon | 6 | |
| Heptagon | 7 | |
| Octagon | 8 | |
| Decagon | 10 |
Sum of the interior angles of a pentagon?
Work out the sum of the interior angles of a polygon with sides (a nonagon).
