**Rhombus-** A quadrilateral in which all the four sides are of equal length.

**Kite-**

A quadrilateral figure having two pairs of equal adjacent sides, symmetrical only about one diagonal.

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)[/latex]

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

Solved Examples-
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.

KE = 17
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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