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.
Kite
A quadrilateral figure is 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 \(\begin{array}{l}D_{1}\end{array} \)
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.
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 \(\begin{array}{l} = \frac{1}{2} \times 20 \times 15 \;\;cm^{2}\end{array} \)
\(\begin{array}{l}= 150 cm^{2}\end{array} \)
|
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 = \(\begin{array}{l}8\sqrt{2}\end{array} \) 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.
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