How to Find the Area of a Quadrilateral

A quadrilateral is a polygon with four sides, four vertices, and the sum of its interior angles equal to $360^\circ$. The calculation of its area depends on both its type (such as square, rectangle, parallelogram, rhombus, kite, or trapezium) and the information provided (sides, angles, diagonals, or coordinates).

General Formula for the Area of a Quadrilateral Using Diagonals and Perpendiculars

Consider an arbitrary quadrilateral $ABCD$ with diagonal $AC$ dividing it into triangles $ABC$ and $ACD$. Let $h_1$ be the perpendicular distance from $B$ to $AC$ and $h_2$ be the perpendicular distance from $D$ to $AC$.

The area of triangle $ABC$ is given by $A_1 = \dfrac{1}{2} \cdot AC \cdot h_1$.

The area of triangle $ACD$ is given by $A_2 = \dfrac{1}{2} \cdot AC \cdot h_2$.

The total area of quadrilateral $ABCD$ is then the sum of $A_1$ and $A_2$:

\[ \text{Area}_{ABCD} = \dfrac{1}{2} AC \cdot h_1 + \dfrac{1}{2} AC \cdot h_2 \]

\[ = \dfrac{1}{2} AC (h_1 + h_2) \]

This formula holds for any quadrilateral where the length of a diagonal and the lengths of perpendiculars to it from the other two vertices are known.

Area Formulas for Standard Quadrilaterals

For specific classes of quadrilaterals, area formulas depend on their symmetry or the equality of sides and angles. Standard formulas are as follows:

For a square with side $a$, the area is $A = a^2$. Further details can be found at Area Of Square Formula.

For a rectangle with length $l$ and breadth $b$, the area is $A = l \times b$.

For a parallelogram with base $b$ and height $h$, the area is $A = b \times h$.

For a rhombus or kite with diagonals $d_1$ and $d_2$ intersecting at right angles, the area is $A = \frac{1}{2} d_1 d_2$.

For a trapezium (trapezoid) with parallel sides $a$ and $b$ and height $h$ between them, the area is $A = \frac{1}{2}(a + b)h$.

Area of a Quadrilateral in Terms of Sides and Included Angle (Trigonometric Formula)

For a cyclic or tangential quadrilateral with consecutive sides $a$, $b$, $c$, and $d$ and one pair of opposite angles $\theta$, consider dividing it by a diagonal $e$ into triangles with angle $\theta$ at the intersection.

Using the formula for the area of two adjacent triangles sharing a common diagonal:

\[ \text{Area} = \frac{1}{2} ab \sin \theta + \frac{1}{2} cd \sin (180^\circ - \theta) \]

Since $\sin (180^\circ - \theta) = \sin \theta$, this simplifies to:

\[ \text{Area} = \frac{1}{2} \sin \theta (ab + cd) \]

For orthodiagonal quadrilaterals (i.e., quadrilaterals whose diagonals are perpendicular such as rhombus, square, and kite), set $\theta = 90^\circ$, so $\sin \theta = 1$:

\[ \text{Area} = \frac{1}{2} d_1 d_2 \]

Area of a Quadrilateral Using Coordinates (Coordinate Geometry)

For a quadrilateral $ABCD$ with vertices $A(x_1, y_1)$, $B(x_2, y_2)$, $C(x_3, y_3)$, and $D(x_4, y_4)$ taken in order, its area is equal to the sum of the areas of triangles $ABC$ and $CDA$.

Apply the shoelace formula sequentially for quadrilaterals:

\[ \text{Area}_{ABCD} = \frac{1}{2} |(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1) - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1)| \]

Every step in the expansion comes from considering the signed areas of the triangles formed by successive vertices. 

Area of a Cyclic Quadrilateral (Brahmagupta's Formula)

A cyclic quadrilateral has all its vertices lying on a single circle. Let its sides be $a$, $b$, $c$, and $d$. Let $s$ be the semiperimeter given by $s = \dfrac{a + b + c + d}{2}$.

The area is given by Brahmagupta's formula:

\[ \text{Area} = \sqrt{(s-a)(s-b)(s-c)(s-d)} \]

This result is derivable from the properties of inscribed quadrilaterals and generalizes the formula for the area of a triangle to quadrilaterals with all vertices co-cyclic.

Worked Examples: Application of Quadrilateral Area Formulas

Example 1: Calculate the area of a square with side $a = 9\, \mathrm{m}$.

Substituting into the formula $A = a^2$,

\[ A = 9^2 = 81\, \mathrm{m}^2 \]

Example 2: Find the area of a trapezium with parallel sides $a = 200\,\mathrm{m}$, $b = 100\,\mathrm{m}$, and height $h = 50\,\mathrm{m}$.

Use $A = \frac{1}{2}(a + b) h$:

\[ A = \frac{1}{2}(200 + 100) \times 50 \]

\[ A = \frac{1}{2}(300) \times 50 = 150 \times 50 = 7500\, \mathrm{m}^2 \]

Example 3: Determine the area of a parallelogram with base $b = 7\, \mathrm{units}$ and height $h = 9\, \mathrm{units}$.

Apply $A = b \times h$:

\[ A = 7 \times 9 = 63\, \mathrm{units}^2 \]

For further examples linking areas of specific figures, see Area Of A Sector Of A Circle Formula.

Further Notes and Interrelations

For formulas involving surface area and perimeter jointly for quadrilaterals, related reference materials can be consulted at Area And Perimeter Formula.

Each area formula for quadrilaterals is directly dependent on certain geometric properties: parallelism of sides, equality of sides, diagonal intersection, and the possibility to inscribe the quadrilateral within a circle or a circle about it. Thus, the problem's data determines which formula applies and what supplementary constructions or measurements must be made.