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 :
\(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
\((x_1,y_1)=(2,4)\)
\((x_2,y_2)=(5,4)\)
Substitute the values in the formula :
\(AB=\sqrt{(5-2)^2+(4-4)^2}\)
\(AB=\sqrt{(3)^2+(0)^2}\)
\(AB=\sqrt{9}\)
\(AB=3\)
To find BC use distance formula :
\(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
\((x_1,y_1)=(5,4)\)
\((x_2,y_2)=(5,1)\)
Substitute the values in the formula :
\(BC=\sqrt{(5-5)^2+(1-4)^2}\)
\(BC=\sqrt{(0)^2+(-3)^2}\)
\(BC=\sqrt{9}\)
\(BC=3\)
To find CD use distance formula :
\(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
\((x_1,y_1)=(5,1)\)
\((x_2,y_2)=(0,-1)\)
Substitute the values in the formula :
\(CD=\sqrt{(0-5)^2+(-1-1)^2}\)
\(CD=\sqrt{(-5)^2+(-2)^2}\)
\(CD=\sqrt{25+4}\)
\(CD=\sqrt{29}\)
To find AD use distance formula :
\(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
\((x_1,y_1)=(2,4)\)
\((x_2,y_2)=(0,-1)\)
Substitute the values in the formula :
\(CD=\sqrt{(0-2)^2+(-1-4)^2}\)
\(CD=\sqrt{(-2)^2+(-5)^2}\)
\(CD=\sqrt{4+25}\)
\(CD=\sqrt{29}\)
Now perimeter of Kite = Sum of all sides
=AB+BC+CD+AD
= \(3+3+\sqrt{29}+\sqrt{29}\)
= \(16.7\)
Thus the perimeter of the kite is 16.7 units.
