Diagonals Formula n(n-3)/2 with Worked Examples

Key Points

• The diagonals formula is $D = \frac{n(n-3)}{2}$, giving the number of diagonals in a polygon with $n$ sides.

• The OEIS records that $n(n-3)/2$ for $n \geq 3$ is the number of diagonals of an $n$-gon.

• Each vertex joins to $n-3$ others, excluding itself and its two neighbours, which is where the $n-3$ comes from.

• The division by 2 is needed because counting from every vertex names each diagonal twice, once from each end.

• A triangle has zero diagonals, which the formula returns correctly since $3(3-3)/2 = 0$.

What Is the Diagonals Formula?

The diagonals formula counts how many diagonals a polygon has from its number of sides:

$$D = \frac{n(n-3)}{2}$$

A diagonal is a line segment joining two non-adjacent vertices. The On-Line Encyclopedia of Integer Sequences records that $n(n-3)/2$ for $n \geq 3$ gives the number of diagonals of an $n$-gon.

This article covers only that count. For the length of a diagonal in a specific shape, see the diagonals page, which deals with diagonal segments and their measurements rather than how many exist.

Why the Formula Is n Times n Minus 3, Halved

Pick any vertex of an $n$-sided polygon. It cannot join to itself, and joining to either of its two neighbours produces a side rather than a diagonal, so it has $n-3$ possible diagonals.

Doing that from every vertex gives $n(n-3)$ segments. But each diagonal has two endpoints and gets counted once from each, so the true total is half that figure.

A second route reaches the same place. The number of segments joining any two of the $n$ vertices is $\frac{n(n-1)}{2}$, and removing the $n$ sides leaves $\frac{n(n-1)}{2} - n = \frac{n(n-3)}{2}$.

Variable Key

Examples of the Diagonals Formula

Example 1

How many diagonals does a hexagon have?

$$D = \frac{6(6-3)}{2} = \frac{18}{2} = 9$$

Final answer: 9 diagonals

Example 2

How many diagonals does an octagon have, counting from every vertex without halving?

Each of the 8 vertices reaches $8-3 = 5$ others, so counting from every vertex gives

$$8 \times 5 = 40$$

That 40 is too large, and the reason is visible on a drawing. The diagonal from vertex A to vertex D gets counted once when working from A and again when working from D, so every diagonal appears twice in the total.

Halving corrects it:

$$D = \frac{8(8-3)}{2} = \frac{40}{2} = 20$$

Final answer: 20 diagonals

Example 3

How many diagonals does a triangle have?

$$D = \frac{3(3-3)}{2} = \frac{3 \times 0}{2} = 0$$

Final answer: 0

Every pair of vertices in a triangle is adjacent, so no diagonal exists. The formula returns this correctly rather than breaking down.

Example 4

A polygon has 35 diagonals. How many sides does it have?

Set the formula equal to 35 and solve:

$$\frac{n(n-3)}{2} = 35 \implies n^2 - 3n - 70 = 0 \implies (n-10)(n+7) = 0$$

Final answer: $n = 10$, a decagon. The root $n = -7$ is discarded as a side count cannot be negative.

Example 5

How many diagonals does a 20-sided polygon have?

$$D = \frac{20(20-3)}{2} = \frac{20 \times 17}{2} = 170$$

Final answer: 170 diagonals

The count grows roughly with the square of the number of sides, which is why diagonals multiply so quickly on larger polygons.

Don't Confuse the Diagonal Count With These Three

The last row is a different question entirely. A square's diagonal length is $a\sqrt{2}$, and that has nothing to do with how many diagonals a square has, which is 2.

Where Intuition Breaks on the Diagonals Formula

Mistake 1: Forgetting to halve

Where it slips in: Counting $n-3$ from each of $n$ vertices is the natural first move, and the result looks like a finished total.

Don't do this: $D = n(n-3)$

The correct way: $D = \frac{n(n-3)}{2}$. Each diagonal has two ends and is counted once from each.

Students who draw a pentagon and count its diagonals by hand, getting 5, then test both candidate formulas against that number, never forget the halving. Verifying a counting formula on the smallest case you can draw is the quickest structural check there is.

Mistake 2: Using n minus 2 instead of n minus 3

Where it slips in: Excluding the vertex itself and one neighbour gives $n-2$, and the second neighbour is easy to overlook.

Don't do this: $D = \frac{n(n-2)}{2}$

The correct way: Three vertices are unavailable from any starting vertex: itself and both neighbours. On a hexagon that leaves 3 diagonals per vertex, not 4.

Mistake 3: Applying the formula to a non-polygon or to a shape with fewer than 3 sides

Where it slips in: The formula is algebraic, so it returns values for $n = 1$ and $n = 2$.

Don't do this: Compute $D$ for $n = 2$ and report 1.

The correct way: The formula holds for $n \geq 3$, as the OEIS entry states. Below three sides there is no polygon, so the output is meaningless rather than merely wrong.

Bottom Line

• The diagonals formula is $D = \frac{n(n-3)}{2}$ for a polygon with $n$ sides.

• The $n-3$ comes from excluding the vertex itself and its two neighbours.

• The halving removes the double count, since each diagonal is reached from both of its ends.

• The formula holds from $n = 3$ upwards, correctly returning zero diagonals for a triangle.

• To work through polygon geometry with a teacher, explore a geometry tutor or math tuition.

Where to Go From Here

Find the number of diagonals in a pentagon and a dodecagon. Then work out how many sides a polygon has if it holds 54 diagonals.

Want a live Bhanzu trainer to walk through more geometry problems? Book a free demo class.

Read More

• Convex Polygon

• Area of Polygons

• Diagonals of a Rectangle

• Exterior Angles of a Polygon

• Diagonal of a Rhombus