Every polygon is a pile of triangles — count them and the formulas follow.
The problem
The interior angle sum of a decagon
(n − 2) × 180 with n = 10 gives 8 × 180 = 1440°. D uses n instead of n − 2. The −2 is there because a polygon cuts into n − 2 triangles, and each triangle contributes 180°.
Pick one corner and draw to every other corner you can reach.
Always two fewer triangles than sides.
sum = (n − 2) × 180°
The two neighbours of your chosen corner cannot be joined to it — they already share a side. That is where the “− 2” comes from.
n = 10
deca means ten
(10 − 2) × 180
eight triangles
= 1440°
8 × 180
| Polygon | n | Interior sum |
|---|---|---|
| triangle | 3 | 180° |
| quadrilateral | 4 | 360° |
| pentagon | 5 | 540° |
| hexagon | 6 | 720° |
| octagon | 8 | 1080° |
| decagon | 10 | 1440° |
Only a regular polygon has all its angles the same size.
octagon sum = (8 − 2)180 = 1080
six triangles
each = 1080 ÷ 8
split evenly
= 135°
per corner
An ordinary eight-sided figure still totals 1080°, but its corners can be any sizes at all. Dividing by 8 assumes they are equal.
However many sides. This is often the faster route.
each exterior = 360 ÷ 8 = 45°
regular octagon
interior = 180 − 45
they form a linear pair
= 135°
the same answer, two lines
Walk right round the polygon and you turn through one full circle. That is the 360.
Working backwards: “each exterior angle is 24°, how many sides?” gives 360 ÷ 24 = 15 sides — far quicker than the interior formula.
diagonals = n(n − 3)2
each corner joins to n − 3 others
not itself, not its two neighbours
hexagon: 6 × 3 = 18
but each counted twice
18 ÷ 2 = 9
the diagonals
The ÷ 2 is the same idea as in combinations — a diagonal from A to D and one from D to A are the same line.
Can a regular polygon have each interior angle 130°?
exterior = 180 − 130 = 50°
linear pair
n = 360 ÷ 50 = 7.2
not a whole number
no such polygon
sides come in whole numbers
Different: the method can answer “impossible”. The same happens when an angle total is not a multiple of 180, such as 1000°.
A polygon has 35 diagonals. How many sides?
n(n − 3) ÷ 2 = 35
the diagonal formula, backwards
n² − 3n − 70 = 0
× 2, expand
(n − 10)(n + 7) = 0
factor
n = 10
n = −7 is rejected
Different: running the diagonal formula backwards gives a quadratic. Keep only the positive root. A quick test also works: two numbers 3 apart whose product is 70 — 10 and 7.
A pentagon’s angles are x, x, 2x, 2x and 90°. Find the largest angle.
(5 − 2) × 180 = 540
pentagon total
6x + 90 = 540, x = 75
− 90, ÷ 6
2x = 150°
75 + 75 + 150 + 150 + 90 = 540 ✓
Different: the polygon is not regular, so we may not divide by 5. Use the total, not the per-corner formula.
The exam often hands you a total or a ratio and asks for something two steps away. Find n first — every polygon formula runs through it.
Example 1: the interior angles total 1260°. How many diagonals?
(n − 2) × 180 = 1260
the sum formula, run backwards
n − 2 = 7, n = 9
÷ 180, + 2
9 × 6 ÷ 2 = 27
n(n − 3) ÷ 2
Example 2: in a regular polygon each interior angle is 3 times each exterior angle. How many diagonals?
e + 3e = 180
interior and exterior are a linear pair
e = 45°, n = 360 ÷ 45 = 8
exterior angles total 360
8 × 5 ÷ 2 = 20
n(n − 3) ÷ 2
The number of sides is a stop on the way. Choices always include n itself and the diagonal count without the ÷ 2 — read the last line of the question again.
Typical items: find n from one interior or exterior angle, solve for x when an irregular polygon’s angles are expressions (set their total to (n − 2) × 180), or chain to the diagonals. For regular polygons go through the exterior angle — 360 ÷ n — it is the fastest route. Ten-second check: 180 − exterior should give back the interior angle you were told.
How many sides?
A regular polygon has each interior angle 150°. How many sides?
Go through the exterior angle: 180 − 150 = 30°, and exterior angles always total 360°, so n = 360 ÷ 30 = 12. That is far faster than solving (n − 2)180 ÷ n = 150, and it works for every regular polygon.
