Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :

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

\(\begin{array}{l}90^{\circ}\end{array} \)
in measure.

Rhombus

Kite

A quadrilateral figure is 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

\(\begin{array}{l}D_{1}\end{array} \)
and
\(\begin{array}{l}D_{2}\end{array} \)
be the long and short diagonals of the kite, respectively.

Then the area of a kite is given by –

A =

\(\begin{array}{l}\frac{1}{2}D_{1}D_{2}\end{array} \)

Proof for Area of a Kite

Let us consider a kite ABCD. Let diagonals AB(

\(\begin{array}{l}D_{1}\end{array} \)
) and CD(
\(\begin{array}{l}D_{2}) \end{array} \)
meet at point E. Thus we see that a diagonal divides a kite into two triangles.

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 –

\(\begin{array}{l}Area (\bigtriangleup ACB) = \frac{1}{2} \times AB \times CE\end{array} \)

and,

\(\begin{array}{l}Area (\bigtriangleup ADB) = \frac{1}{2} \times AB \times DE\end{array} \)

Area of a Kite

\(\begin{array}{l} = Area (\bigtriangleup ACB) + Area (\bigtriangleup ADB)= \frac{1}{2} \times AB \times CE + \frac{1}{2} \times AB \times DE\end{array} \)

Area of a Kite

\(\begin{array}{l} = \frac{1}{2} \times AB \times (CE + DE)\end{array} \)

Area of a Kite

\(\begin{array}{l}= \frac{1}{2} \times AB \times (CD) = \frac{1}{2} \times D_{1} \times D_{2}\end{array} \)

Solved Example-

Example: Find the area of kite whose diagonals are 20 cm and 15 cm.

Solution: We know, Area of a kite

\(\begin{array}{l} = \frac{1}{2}D_{1}D_{2}\end{array} \)

Area

\(\begin{array}{l} = \frac{1}{2} \times 20 \times 15 \;\;cm^{2}\end{array} \)

\(\begin{array}{l}= 150 cm^{2}\end{array} \)

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 =

\(\begin{array}{l}8\sqrt{2}\end{array} \)
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

\(\begin{array}{l}\angle XKI = XIK\end{array} \)

So, the length KX = XI

Using Pythagoras Theorem, we have

\(\begin{array}{l}KX^{2}+XI^{2}= KI^{2}\end{array} \)

\(\begin{array}{l}XI^{2}+XI^{2}= KI^{2}\end{array} \)

\(\begin{array}{l}2 XI^{2}= \left ( 8\sqrt{2} \right )^{2}\end{array} \)

Therefore

\(\begin{array}{l}XI = KX = 8\end{array} \)
units

In Triangle KEX, using Pythagoras theorem we have

\(\begin{array}{l}KX^{2}+XE^{2}= KE^{2}\end{array} \)

\(\begin{array}{l}8^{2} + XE^{2} = 17^{2}\end{array} \)

\(\begin{array}{l}XE^{2} = 289 – 64 = 225 \end{array} \)

Therefore

\(\begin{array}{l}XE = 15\end{array} \)
units

Thus the length of diagonals are –

Diagonal KT = KX + XT = 8 + 8 = 16 units

Diagonal IE = IX + XE = 8 + 15 = 23 units

Area =

\(\begin{array}{l}\frac{1}{2} \times D_{1} \times D_{2}\end{array} \)

\(\begin{array}{l}= \frac{1}{2} \times 16 \times 23 \;\; unit^{2}\end{array} \)

=

\(\begin{array}{l}184 \; units^{2}\end{array} \)

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.

Leave a Comment

Your Mobile number and Email id will not be published. Required fields are marked *

*

*

DOWNLOAD

App NOW