Kite and trapezium

Not every quadrilateral needs two pairs of parallel sides to be interesting. The trapezium guarantees only one such pair, and the kite does not rely on parallel sides at all — its rule is about which sides are equal. Between them, these two shapes round out the classification of quadrilaterals begun with the parallelogram family, and each carries its own small, testable toolkit: for the trapezium, a supplementary-angle rule between the parallel sides; for the kite, a perpendicular-bisecting-diagonal rule tied to its equal adjacent sides. Neither shape is a parallelogram in general, and mixing up their properties with parallelogram properties is the single most common slip students make. This article builds both rules from the ground up, generalises the kite's diagonal relationship into a reusable formula, and works through angle and length problems of the kind these two shapes actually produce on a test.

The core insight

A trapezium is defined by one pair of parallel sides (giving supplementary co-interior angles), while a kite is defined by two pairs of adjacent equal sides (giving one diagonal that perpendicularly bisects the other).

The core idea

Trapezium — defined by parallel sides, not equal sides. A quadrilateral $ABCD$ is a trapezium if at least one pair of sides is parallel. If $AB \parallel CD$ , then $AB$ and $CD$ are called the parallel sides (or bases) and $AD$ , $BC$ are the legs. Unlike a parallelogram, the legs of a general trapezium need not be equal, and they need not be parallel to each other. Only a special case — the isosceles trapezium, where the legs are equal — guarantees equal base angles at each end of a parallel side. Because $AB \parallel CD$ , each leg acts as a transversal cutting two parallel lines, so the co-interior (same-side interior) angles on either leg are supplementary:

\angle A + \angle D = 180°, \qquad \angle B + \angle C = 180°.

where: $\angle A, \angle B, \angle C, \angle D$ are the four interior angles of trapezium $ABCD$ , taken in order, with $AD$ and $BC$ the two legs (transversals) between the parallel sides $AB$ and $CD$ . This single rule, applied to each leg in turn, is enough to find any missing angle once one pair is known.

Kite — defined by adjacent equal sides, not parallel sides. A quadrilateral $ABCD$ is a kite if it has two distinct pairs of adjacent sides equal — say $AB = AD$ and $CB = CD$ , so the two equal pairs meet at $A$ and at $C$ respectively. Three consequences follow directly from this one definition, and all three are testable facts, not extra assumptions:

One diagonal perpendicularly bisects the other. The diagonal joining the two vertices where the different equal-side pairs meet — here, $AC$ — is the perpendicular bisector of the other diagonal, $BD$ . So $AC \perp BD$ and $BD$ is cut exactly in half by $AC$ , though $AC$ itself is generally not bisected by $BD$ .

One pair of opposite angles is equal. The angles between the unequal sides, $\angle B$ and $\angle D$ , are equal to each other. The angles at $A$ and $C$ — where the equal-side pairs meet — are generally not equal to each other.

A kite is generally not a parallelogram. Its equal sides are adjacent, not opposite, so opposite sides are generally not parallel.

\text{Trapezium: exactly the parallel-side rule.} \qquad \text{Kite: exactly the adjacent-equal-side rule.}

where: neither rule implies the other — a trapezium need not have any equal sides, and a kite need not have any parallel sides, which is precisely why they sit outside the parallelogram family.

Figure

A trapezium with one pair of marked parallel sides beside a kite with two pairs of marked adjacent equal sides and its perpendicular diagonals

Derivation ASplitting a kite diagonal using the right-triangle side relation

The goal is to find a general formula for the length of the diagonal $AC$ in a kite $ABCD$ where $AB = AD = a$ , $CB = CD = b$ , and the other diagonal has length $BD = 2h$ , since this single relationship is what turns the kite's perpendicular-bisecting-diagonal property into an actual number whenever a kite problem gives side lengths and one diagonal.

Step 1. Place the known perpendicular bisection. Because $AC \perp BD$ and $AC$ bisects $BD$ at their intersection point $O$ , the half-diagonal on each side of $O$ has the same length: $BO = OD = h$ , and both $\angle AOB$ and $\angle COB$ are right angles.

Step 2. Apply the right-triangle side relation in $\triangle AOB$ . Since $AB = a$ is the hypotenuse and $BO = h$ is one leg, $AO^2 + h^2 = a^2$ , so $AO = \sqrt{a^2 - h^2}$ .

Step 3. Apply the same relation in $\triangle COB$ . Since $CB = b$ is the hypotenuse and $BO = h$ is the same leg, $CO^2 + h^2 = b^2$ , so $CO = \sqrt{b^2 - h^2}$ .

Step 4. Add the two segments, since $O$ lies between $A$ and $C$ on the diagonal:

AC = AO + CO = \sqrt{a^2 - h^2} + \sqrt{b^2 - h^2}.

where: $a$ and $b$ are the two distinct side lengths of the kite, and $h$ is half the length of the other diagonal $BD$ . This formula recovers the worked example below when $a = 13$ , $b = 20$ , $h = 12$ , and it makes clear why $AC$ is generally not bisected at $O$ : $AO = \sqrt{a^2-h^2}$ and $CO = \sqrt{b^2-h^2}$ are equal only in the special case $a = b$ , which is a rhombus, not a general kite.

Worked example

Question 1 — trapezium angles. In trapezium $ABCD$ , $AB \parallel CD$ , $\angle A = 118°$ and $\angle B = 100°$ . Find $\angle D$ and $\angle C$ .

Step 1. $\angle A$ and $\angle D$ are co-interior angles on leg $AD$ , so $\angle A + \angle D = 180°$ , giving $\angle D = 180° - 118° = 62°$ .

Step 2. $\angle B$ and $\angle C$ are co-interior angles on leg $BC$ , so $\angle B + \angle C = 180°$ , giving $\angle C = 180° - 100° = 80°$ .

Step 3. Check with the angle sum property: $118° + 100° + 80° + 62° = 360°$ , confirming consistency. Answer: $\angle D = 62°$ , $\angle C = 80°$ .

Question 2 — kite diagonal. $ABCD$ is a kite with $AB = AD = 13\text{ cm}$ , $CB = CD = 20\text{ cm}$ , and $BD = 24\text{ cm}$ . Find diagonal $AC$ .

Step 1. Since $AC$ perpendicularly bisects $BD$ , $BO = OD = 24/2 = 12\text{ cm}$ , i.e. $h = 12$ .

Step 2. Using the derived formula with $a = 13$ , $b = 20$ , $h = 12$ : $AO = \sqrt{13^2 - 12^2} = \sqrt{169-144} = \sqrt{25} = 5\text{ cm}$ , and $CO = \sqrt{20^2 - 12^2} = \sqrt{400-144} = \sqrt{256} = 16\text{ cm}$ .

Step 3. Add the parts: $AC = AO + CO = 5 + 16 = 21\text{ cm}$ . Answer: diagonal $AC = 21\text{ cm}$ , and since $AO = 5 \ne CO = 16$ , this confirms $AC$ is not bisected by $O$ , unlike $BD$ .

Competitive stretch

If instead $BD = 10\text{ cm}$ with the same $a = 13$ , $b = 20$ , then $h = 5$ , so $AO = \sqrt{169-25} = 12\text{ cm}$ and $CO = \sqrt{400-25} = \sqrt{375} \approx 19.36\text{ cm}$ , giving $AC \approx 31.36\text{ cm}$ — the same four-step method handles any consistent set of kite measurements.

Common traps

Assuming a trapezium's legs are equal. Only an isosceles trapezium has equal legs; a general trapezium's legs can be any lengths.
Assuming a kite is a parallelogram. A kite's equal sides are adjacent, not opposite, so it is generally not a parallelogram.
Bisecting the wrong diagonal. Only the diagonal joining the vertices where the two different equal-side pairs meet is the one that does the bisecting; the other diagonal is the one getting cut in half, and it is not automatically the bisector itself.
Assuming both pairs of opposite kite angles are equal. Only the angles between the unequal sides are equal; the angles at the two "equal-pair" vertices are generally different from each other.

Ask the tutor: Want more practice finding trapezium angles or kite diagonals with different numbers? Ask your AI tutor for a fresh set of problems on kites and trapeziums.

How is a trapezium different from a parallelogram if both can have parallel sides?

A parallelogram requires both pairs of opposite sides to be parallel, while a trapezium requires only one pair. Every parallelogram technically satisfies "at least one pair parallel," but in classification problems trapezium is used for the case where exactly one pair is parallel and the legs are not. If a question specifically says "trapezium," assume only one guaranteed parallel pair unless it also says "isosceles trapezium," which adds equal legs and equal base angles as a bonus property.

Why does only one diagonal of a kite get bisected, not both?

The kite's symmetry comes from folding it along the diagonal that joins the two vertices where the equal-side pairs meet — that fold line is a line of symmetry, and folding along a line of symmetry always makes the other diagonal fall exactly onto itself, perpendicularly bisected. But that same fold does not make the symmetry diagonal bisect itself, because the two unequal side-lengths $a$ and $b$ generally split it into two unequal parts, $\sqrt{a^2-h^2}$ and $\sqrt{b^2-h^2}$ , as the derivation shows.

Can a shape be both a kite and a trapezium at the same time?

In general, no — a kite's defining feature is adjacent equal sides with no requirement of parallel sides, while a trapezium's defining feature is a parallel pair with no requirement of equal sides. It is possible to construct special quadrilaterals that happen to satisfy both definitions at once, but that is not the everyday case tested in this chapter, and treating "kite" and "trapezium" as generally overlapping categories will lead you to misapply properties from one to the other.

Kite and trapezium

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads