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.

Rhombus

Kite-

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

The diagonals of a kite are perpendicular.

Kite

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.

Kite

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 \( = \frac{1}{2} \times 20 \times 15 \;\;cm^{2}\)

\(= 150 cm^{2}\)

Kite

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

Kite

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

\(= \frac{1}{2} \times 16 \times 23 \;\; unit^{2}\)

= \(184 \; units^{2}\)<

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Practise This Question

If 14.287628762876........ can be represented in pq form then 'p - q ' is