# Area Of A Kite

Rhombus- A rhombus is a quadrilateral in which all the four sides are of equal length. Unlike a square, none of the interior angles of a rhombus is 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 triangles.

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 = 17 Construction- 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 = $$\frac{1}{2} \times D_{1} \times D_{2}$$ $$= \frac{1}{2} \times 16 \times 23 \;\; unit^{2}$$ = $$184 \; units^{2}$$

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