Area of a Kite Formula

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.Area of a Kite FormulaProperties 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.

The diagonals 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 = \(\frac{1}{2}D_{1}D_{2}\)

D1 = long diagonal of kite

D= 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: Find the area of kite whose long and short diagonals are 22 cm and 12cm respectively.

Solution: Given,

Length of longer diagonal, D1= 22 cm

Length of shorter diagonal, D2= 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: Area of a kite is 126 cm² and one of its diagonal is 21cm long. Find the length of the other diagonal.

Solution: Given,

Area of a kite =126 cm²

Length of one diagonal = 21 cm

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

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

D2 = 12cm

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


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