Square: definition, properties and diagonals

Of all the quadrilaterals in this chapter, the square sits at the very top of the hierarchy — it is the one shape that is simultaneously a rectangle and a rhombus, so it inherits every special property both of those shapes have, all at once. That double inheritance is what makes the square's diagonals so interesting: they are equal in length (a rectangle trait), they cross at right angles (a rhombus trait), and they cut the $90°$ corners exactly in half (also a rhombus trait). No other quadrilateral studied in this chapter combines all three diagonal properties together. This article builds the square up from the rectangle and the rhombus, states its full property list, derives the side–diagonal relationship $d = a\sqrt{2}$ , and works through the same field and diagonal problems that a Class 8 exam favours — so that by the end you can place the square correctly in the quadrilateral family tree and use its diagonal facts confidently and quickly.

The core insight

A square is a parallelogram that is both a rectangle and a rhombus, so its diagonals are simultaneously equal, perpendicular, and bisecting the vertex angles into $45°$ halves, with side $a$ and diagonal related by $d = a\sqrt{2}$ .

The core idea

A square is a quadrilateral that is both a rectangle and a rhombus. Equivalently, it is a parallelogram in which all four sides are equal and all four angles are $90°$ . Because it belongs to both families, a square keeps every property either family has — there is nothing extra to memorise beyond combining the two lists.

From being a parallelogram, a square has opposite sides parallel, and its diagonals bisect each other (each diagonal cuts the other exactly in half at the point where they cross).

From being a rectangle, all four angles of a square equal $90°$ , and its two diagonals are equal in length: $AC = BD$ .

From being a rhombus, all four sides of a square are equal, $AB = BC = CD = DA$ , its diagonals meet at right angles, $AC \perp BD$ , and each diagonal bisects the pair of vertex angles it passes through, splitting every $90°$ corner into two $45°$ parts. Put together, a square's diagonals are equal, perpendicular, and angle-bisecting at the same time — this triple combination is unique to the square among the shapes in this chapter; a plain rectangle has equal diagonals that are not perpendicular, and a plain rhombus has perpendicular diagonals that are not equal. The figure below marks all three facts on one square: the two diagonals $AC$ and $BD$ are drawn equal in length, they cross at a $90°$ angle at the centre $O$ , and each corner angle is split into two $45°$ pieces.

Figure

A square ABCD with both diagonals drawn crossing at the centre point O at a right angle marked there and each ninety degree corner split by a diagonal into two forty five degree parts

If the side of the square is $a$ , the diagonal has length

d = a\sqrt{2}.

where: $a$ is the length of one side of the square and $d$ is the length of either diagonal (both diagonals are equal, so one formula covers both). This relationship comes from applying the Pythagoras result to the right triangle formed by two sides and a diagonal — that derivation belongs to a later chapter, but the relationship itself is fair to use here.

Derivation AWhy the diagonal of a square equals side times root two

The goal is to show, using only the definition of a square and the Pythagoras relation applied to a single right triangle inside it, that a square of side $a$ must have diagonal length exactly $d = a\sqrt{2}$ , and to confirm this matches the two Pass-1 worked numbers of an $8\text{ cm}$ square and a $14\sqrt2\text{ m}$ diagonal.

Step 1. Draw square $ABCD$ with side $a$ , and draw one diagonal, say $AC$ . This diagonal splits the square into two triangles; look at triangle $ABC$ . Since all angles of a square are $90°$ , the angle at $B$ inside this triangle is $\angle ABC = 90°$ , so triangle $ABC$ is right-angled at $B$ .

Step 2. Identify the two legs and the hypotenuse. The two sides of the square meeting at $B$ are $AB$ and $BC$ , both equal to $a$ (all sides of a square are equal), and these are the two legs of the right triangle. The diagonal $AC$ , opposite the right angle, is the hypotenuse.

Step 3. Apply the Pythagoras relation to this right triangle: hypotenuse squared equals the sum of the squares of the two legs, so $AC^2 = AB^2 + BC^2 = a^2 + a^2 = 2a^2$ .

Step 4. Take the square root of both sides to get the diagonal itself: $AC = \sqrt{2a^2} = a\sqrt{2}$ . Since a square's two diagonals are equal (a rectangle property), the same relationship $d = a\sqrt{2}$ holds for either diagonal, and it holds only for a square, because it depends on both legs of the triangle being equal — a plain rectangle has unequal legs and no such simple formula.

Worked example

Problem. $WXYZ$ is a square with side $WX = 8\text{ cm}$ . Its diagonals meet at $O$ . Find (a) $\angle WOX$ , (b) each half-angle that diagonal $WY$ makes at vertex $W$ , and (c) the length of diagonal $WY$ .

Step 1. Find $\angle WOX$ . Since $WXYZ$ is a square, it is also a rhombus, so its diagonals bisect each other at right angles. The angle between the diagonals at $O$ is therefore

\angle WOX = 90°.

Step 2. Find the half-angle at vertex $W$ . Every interior angle of a square is $90°$ (rectangle property), and the diagonal $WY$ bisects the vertex angle it passes through (rhombus property), splitting it into two equal parts:

\text{Each half-angle} = \frac{90°}{2} = 45°.

Step 3. Find the diagonal length using $d = a\sqrt{2}$ with side $a = 8\text{ cm}$ :

WY = 8\sqrt{2}\text{ cm} \approx 11.3\text{ cm}.

Answer: $\angle WOX = 90°$ , the diagonal splits each $90°$ vertex angle into two $45°$ parts, and $WY = 8\sqrt{2}\text{ cm} \approx 11.3\text{ cm}$ .

Competitive stretch

A square field has a diagonal walking path of length $14\sqrt{2}\text{ m}$ . Find the side of the field. Using $a\sqrt{2} = 14\sqrt{2}$ and dividing both sides by $\sqrt{2}$ gives $a = 14\text{ m}$ directly — no calculator needed, because the $\sqrt2$ factors cancel exactly. Each side of the field is $14\text{ m}$ .

Common traps

Thinking "rhombus" and "square" are the same. Every square is a rhombus, but a rhombus need only have equal sides — a square additionally needs all four angles equal to $90°$ .
Forgetting one of the three diagonal properties. Students often recall that square diagonals are equal (from the rectangle) but forget they are also perpendicular and angle-bisecting (from the rhombus) — a square diagonal does all three simultaneously.
Misapplying $d = a\sqrt2$ to non-square shapes. This exact relationship holds only for a square, because it needs both legs of the right triangle to be equal; it does not hold for a general rectangle or a general rhombus.
Mixing up "diagonals bisect each other" with "diagonals bisect the angles". These are two separate true facts about a square — one is about the diagonals' own lengths, the other about the $90°$ corners — and a problem may need either one or both.

Ask the tutor: Want to see why $d = a\sqrt2$ follows from Pythagoras step by step, or get more practice combining the rectangle and rhombus properties on one square diagram? Ask your AI tutor for a walkthrough.

Why does a square's diagonal split the corner into exactly 45 degrees, but a rectangle's diagonal usually does not

A square's diagonal bisects the vertex angle because a square is a rhombus, and every rhombus diagonal bisects the angles it passes through — this follows from the two triangles the diagonal creates being congruent (equal sides all round). A general rectangle is not a rhombus unless its sides happen to be equal, so its diagonal does not generally split the $90°$ corner into two equal parts; the two pieces depend on the rectangle's length and breadth instead.

Do I need to prove d equals a root 2 using Pythagoras, or can I just use the formula

For this chapter, you can use $d = a\sqrt2$ directly once you know it comes from the Pythagoras relation applied to the right triangle formed by two sides and a diagonal. The full Pythagoras chapter, where the theorem itself is proved, comes later in the syllabus — here it is enough to know the relationship, understand roughly where it comes from, and apply it correctly to find a side from a diagonal or a diagonal from a side.

If all four sides of a quadrilateral are equal, is it automatically a square

No — that only guarantees a rhombus. A rhombus with equal sides can still be "squashed" into a non-right-angled shape, like a diamond, so its angles need not be $90°$ . To be a square, a quadrilateral needs equal sides and all four angles equal to $90°$ ; either condition alone is not enough.

Square: definition, properties and diagonals

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads