Quadrilaterals

⚡ Key Takeaways

A quadrilateral is a polygon with four sides, four vertices, and four interior angles that sum to 360°. Key result: every simple quadrilateral can be split by a diagonal into two triangles, which governs its angle sum and area. Applies throughout Euclidean geometry, trigonometry, surveying, and engineering. Studied since Euclid (c. 300 BCE) and systematically organised in the modern era.

💡 The Core Insight

A single diagonal cuts any quadrilateral into two triangles. Since each triangle's angles sum to 180°, the quadrilateral's angles sum to 360°, and its area equals the sum of the two triangle areas. Almost every quadrilateral formula descends from this one decomposition.

DefinitionQuadrilateral

A quadrilateral is a polygon with four edges (sides) and four vertices. Label the vertices $A, B, C, D$ in order, with side lengths

1

\quad a = AB,\quad b = BC,\quad c = CD,\quad d = DA

Derivation hint: definitional — sides are the distances between consecutive vertices. The two diagonals are the segments connecting opposite vertices, $AC$ and $BD$ . A quadrilateral is simple if its sides do not cross, convex if every interior angle is less than 180° (equivalently, both diagonals lie inside), and concave (re-entrant) if one interior angle exceeds 180°.

DefinitionSpecial Quadrilaterals

Trapezoid (UK: trapezium): at least one pair of parallel sides.
Parallelogram: both pairs of opposite sides parallel.
Rectangle: a parallelogram with four right angles.
Rhombus: a parallelogram with four equal sides.
Square: both a rectangle and a rhombus.
Kite: two distinct pairs of adjacent equal sides.
Cyclic quadrilateral: all four vertices lie on a common circle.

TheoremAngle Sum

The interior angles of any simple quadrilateral satisfy

2

\quad \angle A + \angle B + \angle C + \angle D = 360^\circ

Derivation hint: a diagonal splits it into two triangles, each summing to 180°.

TheoremArea by Diagonals

For a quadrilateral with diagonals $p$ and $q$ meeting at angle $\theta$ ,

3

\quad K = \tfrac{1}{2}\, p\, q\, \sin\theta

Derivation hint: each of the four triangles formed by the diagonals has area $\tfrac12 r_i r_j \sin\theta$ ; sum and factor.

TheoremBretschneider's Formula

For a general quadrilateral with sides $a,b,c,d$ , semiperimeter $s = \tfrac{a+b+c+d}{2}$ , and opposite angles $\alpha, \gamma$ ,

4

\quad K = \sqrt{(s-a)(s-b)(s-c)(s-d) - abcd\cos^2\!\left(\tfrac{\alpha+\gamma}{2}\right)}

Derivation hint: apply the law of cosines to both triangles formed by a diagonal, then eliminate the diagonal between area and length relations.

ProofAngle Sum via Triangulation

1. Take a convex quadrilateral $ABCD$ and draw diagonal $AC$ . 2. This produces two triangles, $\triangle ABC$ and $\triangle ACD$ , which together cover the quadrilateral exactly once. 3. By the triangle angle-sum theorem, $\angle ABC + \angle BCA + \angle CAB = 180^\circ$ and $\angle ACD + \angle CDA + \angle DAC = 180^\circ$ . 4. Adding: the six angles total $360^\circ$ . 5. But $\angle CAB + \angle DAC = \angle A$ (full angle at $A$ ) and $\angle BCA + \angle ACD = \angle C$ (full angle at $C$ ), while $\angle ABC = \angle B$ and $\angle CDA = \angle D$ . 6. Therefore $\angle A + \angle B + \angle C + \angle D = 360^\circ$ . $\blacksquare$

∎

ProofVarignon's Theorem

1. Let $P, Q, R, S$ be the midpoints of sides $AB, BC, CD, DA$ . 2. In $\triangle ABC$ , segment $PQ$ joins midpoints of $AB$ and $BC$ ; by the Midsegment Theorem, $PQ \parallel AC$ and $PQ = \tfrac12 AC$ . 3. In $\triangle ACD$ , segment $SR$ joins midpoints of $DA$ and $CD$ ; likewise $SR \parallel AC$ and $SR = \tfrac12 AC$ . 4. Hence $PQ \parallel SR$ and $PQ = SR$ . 5. A quadrilateral with one pair of opposite sides equal and parallel is a parallelogram, so $PQRS$ is a parallelogram. $\blacksquare$

∎

mathematician · Euclid of Alexandria | c. 300 BCE

Euclid did not isolate "quadrilateral" as a major category but treated parallelograms extensively in Elements Book I. Propositions I.34–I.45 establish that a parallelogram's opposite sides and angles are equal, that its diagonal bisects it into congruent triangles, and that parallelograms on the same base and between the same parallels have equal area. These results — proved without trigonometry, using only congruence and parallel-line axioms — remain the logical foundation for every area formula in this article.

🧠 Intuition

Think of a quadrilateral as two triangles glued along a shared diagonal. A triangle is rigid: fix three side lengths and the shape is locked. A quadrilateral is not — fix four side lengths and you can still flex it like a hinged frame, sweeping out a whole family of shapes (this is why a four-bar gate sags but a triangular brace does not). The "extra freedom" is exactly the diagonal angle that Bretschneider's formula captures through the $\cos^2$ term. When you force the opposite angles to be supplementary, that term vanishes — the quadrilateral becomes cyclic and area is maximised.

Figure

The diagonal decomposition and the family tree of special quadrilaterals

ExampleArea of a Parallelogram from Diagonals

Given: A parallelogram has diagonals of length $p = 10\text{ cm}$ and $q = 8\text{ cm}$ crossing at an angle of $\theta = 60^\circ$ .

Steps: 1. Use equation [3]: $K = \tfrac12 p q \sin\theta$ . 2. Substitute: $K = \tfrac12 (10)(8)\sin 60^\circ$ . 3. $\sin 60^\circ = \tfrac{\sqrt3}{2} \approx 0.8660$ . 4. $K = \tfrac12 (80)(0.8660) = 40 \times 0.8660 = 34.64$ .

\boxed{K \approx 34.6\ \text{cm}^2}

ExampleArea of a General Quadrilateral via Bretschneider

Given: A quadrilateral has sides $a=5, b=6, c=4, d=3$ (metres) with opposite angles $\alpha = 80^\circ$ and $\gamma = 100^\circ$ .

Steps: 1. Semiperimeter: $s = \tfrac{5+6+4+3}{2} = 9$ . 2. Factors: $(s-a)=4,\ (s-b)=3,\ (s-c)=5,\ (s-d)=6$ ; product $= 4\cdot3\cdot5\cdot6 = 360$ . 3. Correction term: $\tfrac{\alpha+\gamma}{2} = \tfrac{180^\circ}{2} = 90^\circ$ , so $\cos^2 90^\circ = 0$ — the quadrilateral is cyclic. 4. Apply [4]: $K = \sqrt{360 - abcd\cdot 0} = \sqrt{360}$ . 5. $\sqrt{360} = 6\sqrt{10} \approx 18.97$ .

\boxed{K \approx 18.97\ \text{m}^2}

Since opposite angles are supplementary, this reduces to Brahmagupta's formula $K=\sqrt{(s-a)(s-b)(s-c)(s-d)}$ .

⚠️ Common mistake

"A trapezoid has exactly one pair of parallel sides." Under the inclusive definition (standard in most modern texts), a trapezoid has at least one pair, so every parallelogram is also a trapezoid. Under the exclusive definition, parallelograms are excluded. Always check which convention your course uses — theorems about trapezoids can flip depending on it.

⚠️ Common mistake

Bretschneider's formula uses $\cos^2\!\left(\tfrac{\alpha+\gamma}{2}\right)$ , not $\cos\left(\tfrac{\alpha+\gamma}{2}\right)$ . Because it is squared, you can use either pair of opposite angles, $(\alpha,\gamma)$ or $(\beta,\delta)$ — they give the same result since $\alpha+\gamma$ and $\beta+\delta$ sum to $360^\circ$ and have cosines of equal magnitude.

Counterexample

Claim: "If a quadrilateral has all four sides equal, it must be a square." Refutation: A rhombus with $60^\circ$ and $120^\circ$ angles has four equal sides but no right angles, so it is not a square. Equal sides force a rhombus; you additionally need a right angle (or equal diagonals) to pin down a square.

connection

Quadrilaterals sit between Triangles (rigid, fully determined by three sides) and general Polygons ( $n$ -gons whose angles sum to $(n-2)\cdot 180^\circ$ ). The diagonal decomposition links directly to Area of a Triangle and to Trigonometry through the law of cosines used in Bretschneider's derivation. Cyclic quadrilaterals connect to Circle Theorems via Ptolemy's theorem ( $AC\cdot BD = AB\cdot CD + BC\cdot DA$ ) and to the inscribed-angle relation that makes opposite angles supplementary. In coordinate form, the Shoelace Formula computes any polygon's area from vertex coordinates, generalising equation [3].

🔬 Where this appears

⏳ History

- c. 1800 BCE — Babylonian tablets compute areas of trapezoidal fields using averaging of parallel sides.

- c. 300 BCE — Euclid's Elements (Book I) proves the fundamental parallelogram theorems.

- c. 628 CE — Brahmagupta gives his formula for the area of a cyclic quadrilateral in the Brāhmasphuṭasiddhānta.

- c. 1000 CE — Ptolemy's theorem on cyclic quadrilaterals (recorded earlier in the Almagest, c. 150 CE) is refined and applied in Islamic astronomy.

- 1842 — Carl Anton Bretschneider (and independently Karl Wilhelm Feuerbach) publishes the general quadrilateral area formula.

- 1731 — Pierre Varignon's midpoint-parallelogram theorem appears posthumously in Éléments de mathématique.

- 1899 — Hilbert's Foundations of Geometry re-axiomatises Euclidean geometry, clarifying the parallel postulate underlying parallelograms.

- 1974 — Roger Penrose introduces aperiodic tilings using kite and dart quadrilaterals.

- 20th–21st c. — Quad meshing becomes a core problem in computational geometry and computer graphics.

open-problem

The clean classification breaks down once we relax "simple." A crossed (self-intersecting) quadrilateral like the "butterfly" or antiparallelogram has a well-defined but ambiguous "area," and its angle sum is no longer 360°. Defining area consistently for non-simple polygons leads to signed area (the shoelace formula handles this with orientation). More broadly, optimal quad-mesh generation — partitioning an arbitrary surface into well-shaped quadrilaterals with minimal distortion — remains an active research problem in computational geometry, with no fully automatic, guaranteed-quality algorithm for all inputs.

Formal Statement

Why It Holds

Building Intuition

Worked Examples

Common Misconceptions

Connections

Real-World Applications

Historical Development

Subtleties & Limits

See Also