Question

A kite has vertices at (2, 4), (5, 4), (5, 1), and (0, –1).

Answer 1

The answer is 16.8 units

Answer 2

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.