Answer:
16.7 units
Step-by-step explanation:
We are given that A kite has vertices at (2, 4), (5, 4), (5, 1), and (0, –1).
So, Let A = (2,4)
B =(5,4)
C =(5,1)
D =(0,-1)
So, Find the sides AB ,BC,CD,AC
To find AB use distance formula :
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
[tex](x_1,y_1)=(2,4)[/tex]
[tex](x_2,y_2)=(5,4)[/tex]
Substitute the values in the formula :
[tex]AB=\sqrt{(5-2)^2+(4-4)^2}[/tex]
[tex]AB=\sqrt{(3)^2+(0)^2}[/tex]
[tex]AB=\sqrt{9}[/tex]
[tex]AB=3[/tex]
To find BC use distance formula :
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
[tex](x_1,y_1)=(5,4)[/tex]
[tex](x_2,y_2)=(5,1)[/tex]
Substitute the values in the formula :
[tex]BC=\sqrt{(5-5)^2+(1-4)^2}[/tex]
[tex]BC=\sqrt{(0)^2+(-3)^2}[/tex]
[tex]BC=\sqrt{9}[/tex]
[tex]BC=3[/tex]
To find CD use distance formula :
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
[tex](x_1,y_1)=(5,1)[/tex]
[tex](x_2,y_2)=(0,-1)[/tex]
Substitute the values in the formula :
[tex]CD=\sqrt{(0-5)^2+(-1-1)^2}[/tex]
[tex]CD=\sqrt{(-5)^2+(-2)^2}[/tex]
[tex]CD=\sqrt{25+4}[/tex]
[tex]CD=\sqrt{29}[/tex]
To find AD use distance formula :
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
[tex](x_1,y_1)=(2,4)[/tex]
[tex](x_2,y_2)=(0,-1)[/tex]
Substitute the values in the formula :
[tex]CD=\sqrt{(0-2)^2+(-1-4)^2}[/tex]
[tex]CD=\sqrt{(-2)^2+(-5)^2}[/tex]
[tex]CD=\sqrt{4+25}[/tex]
[tex]CD=\sqrt{29}[/tex]
Now perimeter of Kite = Sum of all sides
=AB+BC+CD+AD
= [tex]3+3+\sqrt{29}+\sqrt{29}[/tex]
= [tex]16.7[/tex]
Thus the perimeter of the kite is 16.7 units.
