What is Diagonal — Definition, Formula & Examples in Math

A diagonal is a line segment that connects two corners of a shape without sliding along any edge. A triangle has none — every pair of vertices is already joined by an edge. A square has two. A pentagon has five.

The Formal Definition

A diagonal of a polygon is a line segment joining two non-adjacent (non-consecutive) vertices. In a polyhedron (a 3D solid), a space diagonal joins two non-adjacent vertices that do not share a face; a face diagonal lies inside a single face of the solid.

The smallest polygon that can have a diagonal is the quadrilateral — three-sided polygons (triangles) have only neighbouring vertices.

Quick reference.

• Definition: a segment joining two non-adjacent vertices.

• Notation: $d$ for length; sometimes $AC$ or $BD$ naming the vertices.

• Count for an $n$-sided polygon: $\dfrac{n(n-3)}{2}$.

• Length — square (side $a$): $a\sqrt{2}$.

• Length — rectangle ($l, b$): $\sqrt{l^{2} + b^{2}}$.

• Length — cube (side $a$): $a\sqrt{3}$ (space diagonal).

• Grade introduced: CCSS-M 4.G.A (polygon properties); NCERT Class 8 — Understanding Quadrilaterals.

The Count Formula — Why $\dfrac{n(n-3)}{2}$

From any one vertex, you can draw a line to every other vertex except two — yourself and your two immediate neighbours (those connect by edges, not diagonals). That gives $n - 3$ diagonals per vertex. With $n$ vertices, the total count of vertex-to-diagonal pairs is $n(n - 3)$. But each diagonal has two endpoints, so it's been counted twice — divide by $2$.

$$\text{Number of diagonals} = \frac{n(n-3)}{2}.$$

The count grows quickly because each new side adds connections to roughly all existing vertices.

Length of a Diagonal — The Three Most-Used Formulas

These three come up most often in CBSE and Common Core problems.

Square (side $a$). The diagonal cuts the square into two right triangles whose legs are both $a$. By the Pythagorean theorem:

$$d = \sqrt{a^{2} + a^{2}} = a\sqrt{2}.$$

Rectangle ($l, b$). The diagonal is the hypotenuse of a right triangle with legs $l$ and $b$:

$$d = \sqrt{l^{2} + b^{2}}.$$

Cube (side $a$). The space diagonal runs from one corner of the cube straight through to the opposite corner. It's the hypotenuse of a right triangle whose legs are $a$ (one edge) and $a\sqrt{2}$ (the face diagonal):

$$d = \sqrt{a^{2} + (a\sqrt{2})^{2}} = \sqrt{a^{2} + 2a^{2}} = a\sqrt{3}.$$

A pattern emerges: a square's diagonal grows like $\sqrt{2}$, a cube's like $\sqrt{3}$, a 4D-hypercube's like $\sqrt{4} = 2$, and so on. The $\sqrt{n}$-growth is one of the cleanest results in elementary geometry.

Diagonal Properties Across Common Quadrilaterals

The diagonals of a quadrilateral carry rich information about the shape.

These properties are how a geometry student proves what kind of quadrilateral they're looking at — measure the diagonals, check the three boxes above, identify the shape.

Three Worked Examples — Quick, Standard, Stretch

Quick. How many diagonals does a hexagon have?

Use $\dfrac{n(n-3)}{2}$ with $n = 6$:

$$\frac{6 \times (6 - 3)}{2} = \frac{6 \times 3}{2} = 9.$$

Final answer: $9$ diagonals.

Standard (Wrong Path First — Where Diagonals Trip Students). A rectangle has length $24$ cm and breadth $7$ cm. Find the length of its diagonal.

The wrong path. A student tries to add: $d = l + b = 24 + 7 = 31$ cm.

The flaw: the diagonal is not the sum of the sides. It's the straight-line distance across, which is the hypotenuse of a right triangle. Adding $l$ and $b$ would only work if the rectangle were collapsed flat — which makes no geometric sense.

The rescue. Use the Pythagorean theorem on the right triangle whose legs are $l$ and $b$:

$$d = \sqrt{l^{2} + b^{2}} = \sqrt{24^{2} + 7^{2}} = \sqrt{576 + 49} = \sqrt{625} = 25 \text{ cm}.$$

Final answer: $25$ cm.

Sanity check: the diagonal should be less than $l + b = 31$ (triangle inequality) and greater than $l = 24$ (the diagonal is the longest side of the right triangle). $24 < 25 < 31$ ✓.

The lesson — a diagonal is a Pythagorean hypotenuse, not a sum. The right angle at the corner of the rectangle is the whole geometric reason behind the $\sqrt{l^{2} + b^{2}}$ formula.

Stretch. A cube has edge length $6$ cm. Find (a) the face diagonal and (b) the space diagonal.

(a) Face diagonal — diagonal of one square face of side $6$:

$$d_{\text{face}} = 6\sqrt{2} \approx 8.49 \text{ cm}.$$

(b) Space diagonal — corner to opposite corner through the interior:

$$d_{\text{space}} = 6\sqrt{3} \approx 10.39 \text{ cm}.$$

Final answer: face diagonal $\approx 8.49$ cm; space diagonal $\approx 10.39$ cm.

This is the version of diagonal that shows up in NCERT Class 9–10 mensuration problems and in 3D coordinate geometry (the distance from $(0,0,0)$ to $(a, a, a)$ is exactly $a\sqrt{3}$).

Where Diagonals Appear — Beyond the Polygon

A few places the idea is the foundation under a larger result:

• Pythagorean triples in carpentry. The "$3$-$4$-$5$ triangle" comes up because a rectangle of sides $3$ and $4$ has a diagonal of $5$ — exactly the trick builders use to square up a foundation.

• Television screen sizes. A "$55$-inch TV" measures the diagonal of the screen, not the side. A $16:9$ aspect-ratio screen with a $55$-inch diagonal is about $47.9$ inches wide and $26.9$ inches tall.

• Triangulation in maps. Every modern GPS uses diagonals — straight-line distances from satellites to a receiver — to solve for position.

• Graph theory. A diagonal in a Latin square or a magic square is a special row that the puzzle uses. The 1948 Latin Square connection by Sir Ronald A. Fisher extended diagonals into agricultural-trial design.

The earliest known proof that a square's diagonal is $a\sqrt{2}$ — and that $\sqrt{2}$ is not a rational number — is attributed to the Pythagorean school around 500 BCE. The proof is short, four lines, and changed mathematics: it forced the existence of irrational numbers for the first time.

Diagonal: Tripping Points to Avoid

Mistake 1: Counting an edge as a diagonal

Where it slips in: A student lists "the four sides plus two diagonals" of a square as $6$ diagonals.

Don't do this: Treat any line segment between two corners as a diagonal.

The correct way: A diagonal joins non-adjacent vertices only. The sides of the polygon are not diagonals. A square has $2$ diagonals, not $6$.

Mistake 2: Adding the sides to find a diagonal

Where it slips in: Rectangle of $5$ cm × $12$ cm — student writes diagonal $= 5 + 12 = 17$.

Don't do this: Sum the sides.

The correct way: Apply Pythagoras: $d = \sqrt{5^{2} + 12^{2}} = \sqrt{169} = 13$ cm.

Mistake 3: Confusing the face diagonal with the space diagonal of a cube

Where it slips in: A problem asks for the longest diagonal of a cube of side $a$; student gives $a\sqrt{2}$.

Don't do this: Stop at the face diagonal.

The correct way: The longest diagonal of a cube is the space diagonal, $a\sqrt{3}$ — corner to opposite corner through the interior of the cube, not across one face.

Conclusion

• A diagonal joins two non-adjacent vertices of a polygon or polyhedron.

• An $n$-sided polygon has $\dfrac{n(n-3)}{2}$ diagonals.

• A square's diagonal is $a\sqrt{2}$; a rectangle's is $\sqrt{l^{2} + b^{2}}$; a cube's space diagonal is $a\sqrt{3}$.

• The diagonal of every rectangle and square is the hypotenuse of a right triangle — Pythagoras decides the length.

• The properties of the two diagonals (equal, bisect, perpendicular) identify the type of quadrilateral.

Quick Self-Check — Three Problems

1. How many diagonals does a decagon ($n = 10$) have?

2. Find the diagonal of a rectangle of length $9$ cm and breadth $12$ cm.

3. Find the space diagonal of a cube of side $4$ cm.

If you got $21$ for problem 1, recount — the formula gives $35$.

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