 # Area Of A Kite

Rhombus- A rhombus is a quadrilateral in which all the four sides are of equal length. Unlike square, none of the interior angle of a rhombus are not $90^{\circ}$ in measure. ## Kite-

The diagonals of a kite are perpendicular. ## Area of a Kite-

If we know the diagonals of a Kite, it is possible to calculate the area of a Kite.

Let $D_{1}$ and $D_{2}$ be the long and short diagonals of the Kite respectively.

Then the area of a Kite is given by –

A = $\frac{1}{2}D_{1}D_{2}$

## Proof for Area of a Kite-

Let us consider a Kite ABCD. Let diagonals AB($D_{1}$) and CD($D_{2})$ meet at point E. Thus we see that a diagonal divides a Kite into two triangle. In the figure given above, we see that Diagonal AB divides a Kite in two triangle ACB and ADB.

Thus area of traingles equal to –

$Area (\bigtriangleup ACB) = \frac{1}{2} \times AB \times CE$

and, $Area (\bigtriangleup ADB) = \frac{1}{2} \times AB \times DE$

Area of a Kite $= Area (\bigtriangleup ACB) + Area (\bigtriangleup ADB)= \frac{1}{2} \times AB \times CE + \frac{1}{2} \times AB \times DE$

Area of a Kite $= \frac{1}{2} \times AB \times (CE + DE)$

Area of a Kite $= \frac{1}{2} \times AB \times (CD) = \frac{1}{2} \times D_{1} \times D_{2}$

 Solved Example- Example: Find the area of kite whose diagonals are 20 cm and 15 cm. Solution: We know, Area of a kite $= \frac{1}{2}D_{1}D_{2}$ Area $= \frac{1}{2} \times 20 \times 15 \;\;cm^{2}$ $= 150 cm^{2}$ If lengths of unequal sides are given, using Pythagoras theorem, the length of diagonals can be found.

 Example: The sides of a Kite are given as follows Find the area of a kite.Solution: Given IK = $8\sqrt{2}$KE = 17Construction- Draw in segment KT and segment IE as shown in the figure alongside To find the area of a Kite, first we need to calculate the length of the diagonals KT, EI. In Triangle IKX, it is clear that $\angle XKI = XIK$ So, the length KX = XI Using Pythagoras Theorem, we have $KX^{2}+XI^{2}= KI^{2}$ $XI^{2}+XI^{2}= KI^{2}$ $2 XI^{2}= \left ( 8\sqrt{2} \right )^{2}$ Therefore $XI = KX = 8$ units In Triangle KEX, using Pythagoras theorem we have $KX^{2}+XE^{2}= KE^{2}$ $8^{2} + XE^{2} = 17^{2}$ $XE^{2} = 289 – 64 = 225$ Therefore $XE = 15$ units Thus the length of diagonals are – Diagonal KT = KX + XT = 8 + 8 = 16 units Diagonal IE = IX + XE = 8 + 15 = 23 units Area = $\frac{1}{2} \times D_{1} \times D_{2}$ $= \frac{1}{2} \times 16 \times 23 \;\; unit^{2}$ = $184 \; units^{2}$<

This was all about Kite. Learn various related concepts of topics like Quadrilateral, Trapezoid, Rhombus, Rectangle, Square, etc. in an engaging manner by visiting our site BYJU’S.