A quadrilateral is a polygon which has 4 vertices and 4 sides enclosing 4 angles. Kite is a special quadrilateral in which each pair of the consecutive sides is congruent, but the opposite sides are not congruent. Rhombus is a kite with all the four sides congruent.**Properties of a Kite:**

- Angles between unequal sides are equal.
- A kite can be viewed as a pair of congruent triangles with a common base.
- Diagonals of a kite intersect each other at right angles.
- The longer diagonal is the perpendicular bisector of the shorter diagonal.
- A kite is symmetrical about its main diagonal.
- The shorter diagonal divides the kite into two isosceles triangles.

TheÂ diagonalsÂ of a kite are perpendicular. Area of a kite is given as half of the product of the diagonals which is same as that of a rhombus. Area of a kite can be expressed by the formula:

Area of Kite = \(\frac{1}{2}D_{1}D_{2}\)

D_{1}Â = long diagonal of kite

D_{2Â }= short diagonal of kite

**Derivation for Area of a Kite:**

Consider the area of the following kite PQRS.

Here the diagonals are PR and QS

Let diagonal PR =a and diagonal QS = b

Diagonals of a kite cut one another at right angles as shown by diagonal PR bisecting diagonal QS.

OQ = OS = \( \frac{OS}{2}=\frac{b}{2}\)

Area of the kite = Area of triangle PQR + Area of triangle PSR

Area of Triangle = \(\frac{1}{2}\;base \times height\)

Here, base = a and height = OQ = OS= b/2

Area of triangle PQR = \(\frac{1}{2}\times a\times \frac{b}{2}\)

Area of triangle PSR= \(\frac{1}{2}\times a\times \frac{b}{2}\)

Area of the kite = \(\frac{1}{2}\times a\times \frac{b}{2}\) + \(\frac{1}{2}\times a\times \frac{b}{2}\)

= \(\frac{ab}{4}+\frac{ab}{4}\)

= \(\frac{2ab}{4}=\frac{1}{2}ab\)

Hence,

Area of the kite = \(\frac{1}{2}PR*QS\)= Half of the product of the diagonals

**Note:**

- If lengths of unequal sides are given, using Pythagoras theorem, the length of diagonals can be found. The area of a kite can be calculated by using the lengths of its diagonals.

**Solved Examples:**

**Example 1:Â **Find the area of kite whose long and short diagonals are 22 cm and 12cm respectively.

**Solution: **Given,

Length of longer diagonal, D_{1}= 22 cm

Length of shorter diagonal, D_{2}= 12 cm

**Area of Kite =\(\frac{1}{2}D_{1}D_{2}\)**

**Area of kite = \(\frac{1}{2}\) x 22 x 12 = 132\(cm^{2}\)**

**Example 2:Â **Area of a kite is 126 cmÂ² and one of its diagonal is 21cm long. Find the length of the other diagonal.

**Solution: **Given,

Area of a kite =126 cmÂ²

Length of one diagonal = 21 cm

**Area of Kite =\(\frac{1}{2}D_{1}D_{2}\)**

\(126 = \frac{1}{2}\times 21\times D_{2}\)

D_{2} = 12cm

To solve more problems on the topic and for video lessons on kite and other quadrilaterals, downloadÂ Byju’s -The Learning App.