Diagonal of Square Formula — d = s√2

The diagonal of square formula sizes TV and monitor screens, picture frames, floor tiles, and any layout where the corner-to-corner span decides whether something fits.

The Formula

For a square with side length $s$, the diagonal $d$ is:

$$\boxed{;d = s\sqrt{2};}$$

Each symbol points to the figure above. $s$ is the side of the square — and because all four sides are equal, you only need one. $d$ is the diagonal — the line joining opposite corners, which is always longer than a side by the fixed factor $\sqrt{2} \approx 1.414$.

If you do not know the side directly, two related forms reach the diagonal from other measurements:

• From the area $A$: since $s = \sqrt{A}$, the diagonal is $d = \sqrt{A}\cdot\sqrt{2} = \sqrt{2A}$.

• From the perimeter $P$: since $s = \frac{P}{4}$, the diagonal is $d = \frac{P}{4}\sqrt{2}$.

How the Diagonal of Square Formula Is Derived

The factor of $\sqrt{2}$ is not a memorised constant — it comes straight out of the Pythagorean theorem.

A diagonal cuts the square into two identical right triangles. In each triangle, the two legs are sides of the square (both length $s$), and the hypotenuse is the diagonal $d$ we want. Apply the Pythagorean theorem to one of those right triangles:

$$d^2 = s^2 + s^2 = 2s^2.$$

Take the square root of both sides:

$$d = \sqrt{2s^2} = s\sqrt{2}.$$

That is the whole derivation. The diagonal is a side scaled by $\sqrt{2}$ because the two equal legs of the inner right triangle force the hypotenuse to that exact multiple. Each of those right triangles is a 45-45-90 triangle — the same isosceles right triangle that appears in our square and diagonal of parallelogram formula guides, since a square is a special rhombus and a special parallelogram all at once.

Why Is the Diagonal s√2 and Not 2s?

A common first guess is that the diagonal is twice the side, but it is not. Walking along two sides (corner to corner the long way) covers $2s$; the diagonal is the straight shortcut, which must be shorter than that detour. The straight-line distance comes out to $s\sqrt{2} \approx 1.414,s$ — longer than one side, but well short of two. The $\sqrt{2}$ sits exactly between 1 and 2, which is the geometric reason the diagonal is longer than a side but shorter than two of them.

Examples of the Diagonal of Square Formula

Example 1

A square has a side of 7 cm. Find its diagonal.

Apply the formula directly:

$$d = s\sqrt{2} = 7\sqrt{2} \approx 9.9 \text{ cm}.$$

Final answer: $d = 7\sqrt{2} \approx 9.9$ cm.

Example 2

A square has a diagonal of $10\sqrt{2}$ cm. Find its side length.

Reverse the formula — divide the diagonal by $\sqrt{2}$:

$$s = \frac{d}{\sqrt{2}} = \frac{10\sqrt{2}}{\sqrt{2}} = 10 \text{ cm}.$$

Final answer: $s = 10$ cm.

Example 3

A square has a diagonal of 12 cm. Find its side length.

The most common slip is to assume the diagonal is twice the side and halve it.

Wrong attempt. A student writes $s = \frac{d}{2} = \frac{12}{2} = 6$ cm, treating the diagonal as $2s$. But check it: if the side were 6 cm, the true diagonal would be $6\sqrt{2} \approx 8.49$ cm, not 12 — so 6 cm cannot be right. The diagonal is $s\sqrt{2}$, not $2s$.

Correct. Divide by $\sqrt{2}$, not 2:

$$s = \frac{d}{\sqrt{2}} = \frac{12}{\sqrt{2}} = \frac{12}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{12\sqrt{2}}{2} = 6\sqrt{2} \approx 8.49 \text{ cm}.$$

Final answer: $s = 6\sqrt{2} \approx 8.49$ cm.

Example 4

A square has an area of 36 cm². Find its diagonal.

First the side, then the diagonal — or use $d = \sqrt{2A}$ directly:

$$s = \sqrt{36} = 6 \text{ cm}, \qquad d = 6\sqrt{2} \approx 8.49 \text{ cm}.$$

Check with the area form: $d = \sqrt{2 \times 36} = \sqrt{72} = 6\sqrt{2} \approx 8.49$ cm. Same answer.

Final answer: $d = 6\sqrt{2} \approx 8.49$ cm.

Example 5

A square has a perimeter of 32 m. Find its diagonal.

The side is a quarter of the perimeter, then apply the formula:

$$s = \frac{P}{4} = \frac{32}{4} = 8 \text{ m}, \qquad d = 8\sqrt{2} \approx 11.31 \text{ m}.$$

Final answer: $d = 8\sqrt{2} \approx 11.31$ m.

Example 6

A square floor tile must fit through a doorway with a clear opening of 50 cm. The tile has a side of 36 cm. Will it pass through diagonally?

The widest the tile can angle is along its diagonal:

$$d = 36\sqrt{2} \approx 50.9 \text{ cm}.$$

The diagonal ($\approx 50.9$ cm) is just larger than the $50$ cm opening, so the tile will not quite pass through corner-first.

Final answer: no — the diagonal is about $50.9$ cm, slightly wider than the $50$ cm doorway.

Where the Diagonal of Square Formula Shows Up

The diagonal is the measurement that decides corner-to-corner fit, so it appears wherever square or square-ish objects must clear a span.

• Screens and displays. A "32-inch monitor" quotes the diagonal; for a square-ish display, the diagonal relates the width and height through the same Pythagorean logic.

• Fitting and moving. Whether a square table, tile, or panel fits through a doorway depends on its diagonal, not its side — exactly the Example 6 problem.

• Construction squaring. Builders check that a frame is truly square by measuring both diagonals; equal diagonals of length $s\sqrt{2}$ confirm right-angled corners.

• Tiling and layout. Diagonal tile patterns and packing problems all lean on the $s\sqrt{2}$ relationship to space cuts evenly.

For a Grade 8 or 9 student, the most-met setting is the mensuration chapter, but the same $s\sqrt{2}$ appears again in the 45-45-90 special right triangle of trigonometry and in the coordinate-geometry distance between two diagonal corners.

Spot the Trap Before You Fall In — Diagonal of Square Formula

Mistake 1: Thinking the diagonal is twice the side

Where it slips in: Going from a known diagonal back to the side, where "diagonal is bigger, so halve it" feels right.

Don't do this: Write $s = \frac{d}{2}$ or $d = 2s$. The diagonal is $s\sqrt{2} \approx 1.414,s$, not $2s$ — walking two sides is the long way around, the diagonal is the shortcut.

The correct way: $d = s\sqrt{2}$, and to reverse it, $s = \frac{d}{\sqrt{2}}$.

Mistake 2: Dividing by 2 instead of √2

Where it slips in: Reversing the formula correctly in words ("divide by root two") but typing $\div 2$ into the calculator.

Don't do this: Compute $s = \frac{12}{2} = 6$ when you meant $\frac{12}{\sqrt{2}}$. Those differ by a factor of about $1.414$.

The correct way: Divide by $\sqrt{2}$ (about $1.414$), then rationalise if needed: $\frac{d}{\sqrt{2}} = \frac{d\sqrt{2}}{2}$. The memorizer who recalls "$\sqrt{2}$ is involved somewhere" but applies it as a plain 2 gets an answer that is too small by $\sqrt{2}$.

Mistake 3: Forgetting to take the square root in the area route

Where it slips in: Finding the diagonal from the area, but skipping the step that recovers the side.

Don't do this: Treat the area $A$ as if it were the side and write $d = A\sqrt{2}$. The area is $s^2$, not $s$.

The correct way: Recover the side first, $s = \sqrt{A}$, then $d = s\sqrt{2}$ — or combine them as $d = \sqrt{2A}$. Because area is a squared length, the square root is what brings you back to a linear measurement before the $\sqrt{2}$ factor applies.

Key Takeaways

• The diagonal of square formula is $d = s\sqrt{2}$ — the side times the square root of 2.

• It comes from the Pythagorean theorem on the two right triangles a diagonal creates: $d^2 = s^2 + s^2$.

• From the area use $d = \sqrt{2A}$; from the perimeter use $d = \frac{P}{4}\sqrt{2}$.

• The diagonal is about $1.414$ times the side — longer than one side, but shorter than two.

• The costliest mistake is treating the diagonal as $2s$ (or dividing by 2 instead of $\sqrt{2}$), which is off by a factor of $\sqrt{2}$.

Practice These Problems to Solidify Your Understanding

1. A square has a side of 9 cm. Find its diagonal, leaving the answer in surd form and as a decimal.

2. A square has a diagonal of 14 cm. Find its side length.

3. A square has an area of 81 cm². Find its diagonal.

Answer to Question 1: $d = 9\sqrt{2} \approx 12.73$ cm. Answer to Question 2: $s = \frac{14}{\sqrt{2}} = 7\sqrt{2} \approx 9.9$ cm. Answer to Question 3: $s = 9$ cm, so $d = 9\sqrt{2} \approx 12.73$ cm. If Question 2 gave you $7$ cm, you divided by 2 instead of $\sqrt{2}$ — return to Mistake 2.

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