Given:
A figure of a kite.
To find:
The area of the given kite.
Solution:
The area of a kite is the half of the product of its diagonals.
[tex]A=\dfrac{d_1d_2}{2}[/tex] ...(i)
Where, [tex]d_1,d_2[/tex] are two diagonals of the kite.
From the given figure it is clear that the length of one diagonal is the sum of 12 and 8.
[tex]d_1=12+8[/tex]
[tex]d_1=20[/tex]
Let the second diagonal be 2x. The first diagonals bisect the second diagonal. So, the length of one parts of the diagonal is x.
Diagonals of a kite are perpendicular to each other. Using Pythagoras theorem, we get
[tex]Hypotenuse^2=Perpendicular^2+Base^2[/tex]
[tex]13^2=x^2+12^2[/tex]
[tex]169-144=x^2[/tex]
[tex]25=x^2[/tex]
[tex]5=x[/tex]
The length of second diagonal is:
[tex]d_2=2x[/tex]
[tex]d_2=2(5)[/tex]
[tex]d_2=10[/tex]
Substituting [tex]d_1=20,d_2=10[/tex] in (i), we get
[tex]A=\dfrac{20\times 10}{2}[/tex]
[tex]A=\dfrac{200}{2}[/tex]
[tex]A=100[/tex]
Therefore, the area of the kite is 100 square units.
