In geometry, the area of a plane 2D shape is the region covered by the sides of it (region bounded by the perimeter) in a two-dimensional plane. Or we can say that area of any shape is the number of unit squares that can fit into it. Here a unit square refers to a square of side $1$ unit. One good example is graph paper. By counting the number of squares in a region, you can find the area of that region.

Let’s learn about the area of a kite, its formula, and its properties.

## What is the Area of a Kite?

The area of a kite can be defined as the region covered by the perimeter of a kite in a two-dimensional plane. Like a square, and a rhombus, a kite does not have all four sides equal. The area of a kite is always expressed in terms of $units^{2}$ such as $cm^{2}$, $in^{2}, $m^{2}$, $ft^{2}$ etc.

### Area of a Kite Formula

A kite has two unequal diagonals bisecting each other at right angles. If $d_{1}$ and $d_{2}$ are the lengths of the two diagonals of a kite, its area is given by the formula $A = \frac {1}{2} \times d_{1} \times d_{2}$.

### Area of Kite Formula Derivation Using the Diagonals

Consider a kite $ABCD$, such that the adjacent sides $AB = BC$ and $CD = DA$. The diagonals of the kite are $AC = d_{1}$ and $BD = d_{2}$.

Area of kite $ABCD$ = (Area of $\triangle AOD$) + (Area of $\triangle COD$) + (Area of $\triangle AOB$) + (Area of $\triangle BOC$)

= $\frac {1}{2} \times AO \times DO$ + $\frac {1}{2} \times DO \times CO$ + $\frac {1}{2} \times AO \times BO$ + $\frac {1}{2} \times BO \times CO$

= $\frac {1}{2} \times DO \times \left(AO + CO \right) + \frac {1}{2} \times BO \times \left(AO + CO \right)$

= $\frac {1}{2} \times DO \times AC + \frac {1}{2} \times BO \times AC$

= $\frac {1}{2} \times AC \times (DO + BO) = \frac {1}{2} \times AC \times BD = \frac {1}{2} \times d_{1} \times d_{2}$

### Area of a Kite Examples

**Ex 1:** Find the area of a kite whose diagonals are of length $12 cm$ and $8 cm$.

The length of the first diagonal of a kite $d_{1} = 12 cm$

The length of the second diagonal of a kite $d_{2} = 8 cm$

Area of a kite = $\frac {1}{2} \times d_{1} \times d_{2} = \frac {1}{2} \times 12 \times 8 = 48 cm^{2}$

**Ex 2:** The area of a kite is $120 mm^{2}$. If one of the diagonals is $16 mm$, find the length of the second diagonal.

Area of a kite $A = 120 mm^{2}$

Length of one diagonal of a kite $d_{1} = 16 mm$

Area of a kite $A = \frac {1}{2} \times d_{1} \times d_{2} => 120 = \frac {1}{2} \times 16 \times d_{2} => 120 = 8 \times d_{2} => d_{2} = \frac {120}{8} => d_{2} = 15 mm$

## Kite – A 2D Plane Figure

A kite is a quadrilateral in which two pairs of adjacent sides are of equal length. No pair of sides in a kite are parallel but one pair of opposite angles are equal. The two diagonals are unequal and intersect at right angles. The longer diagonal of a kite bisects the shorter one.

### Properties of A Kite

The following are the characteristic features of a kite

- A kite has two pairs of adjacent equal sides. Here, $AC = BC$ and $AD = BD$
- It has one pair of opposite angles (obtuse) that are equal. Here, $\angle A = \angle B$
- The longer diagonal bisects the shorter one. Here, the diagonal $AB$, $AO = OB$
- The shorter diagonal forms two isosceles triangles. Here, diagonal $AB$ forms two isosceles triangles: $\triangle ACB$ and $\triangle ADB$. $AC = BC$ and $AD = BD$ in two isosceles triangles
- The longer diagonal forms two congruent triangles. Here, diagonal $CD$ forms two congruent triangles – $\triangle CAD$ and $\triangle CBD$ by $SSS$ criteria
- The diagonals are perpendicular to each other. Here, $AB \perp CD$
- The longer diagonal bisects the pair of opposite angles. Here, $\angle ACD = \angle DCB$, and $\angle ADC = \angle CDB$

## Conclusion

A kite is a 2D plane figure whose two pairs of adjacent equal sides. The kite has two unequal diagonals intersecting at right angles and you can find the area of a kite by multiplying half by the product of its diagonals.

## Practice Problems

- Find the area of a kite whose diagonals are of length
- $14 in$ and $8 in$
- $11 cm$ and $9 cm$

- Find the length of a diagonal, if the area of a kite is $84 cm^{2}$ and the other diagonal is of length $4 cm$.

## Recommended Reading

- Area of Rectangle – Definition, Formula & Examples
- Area of Square – Definition, Formula & Examples
- Area of a Triangle – Formulas, Methods & Examples
- Area of a Circle – Formula, Derivation & Examples
- Area of Rhombus – Formulas, Methods & Examples
- Perimeter of a Polygon(With Formula & Examples)
- Perimeter of Trapezium – Definition, Formula & Examples
- Perimeter of Kite – Definition, Formula & Examples
- Perimeter of Rhombus – Definition, Formula & Examples
- Circumference (Perimeter) of a Circle – Definition, Formula & Examples
- Perimeter of Square – Definition, Formula & Examples
- Perimeter of Rectangle – Definition, Formula & Examples
- Perimeter of Triangle – Definition, Formula & Examples
- What Are 2D Shapes – Names, Definitions & Properties

## FAQs

### How do you find the area of a kite?

The area of a kite can be calculated using the formula Area = $\frac {1}{2} \times d_{1} \times d_{2}$, , where $d_{1}$ and $d_{2}$ are its diagonals.

### How do you find the diagonals of a kite?

The length of one diagonal of a kite can be found using the Pythagorean theorem. The length of the other diagonal can be found by substituting the length of the first diagonal into the area of a kite formula if the area is known.

### What is the formula for the area of a kite?

The area of a kite is half the product of the lengths of its diagonals. The formula for the area of a kite is given as $\frac {1}{2} \times d_{1} \times d_{2}$, where $d_{1}$ and $d_{2}$ are its diagonals.