Find the perimeter of quadrilateral ABCD with vertices

Ответ:

Odog9458

07.08.2021

Given:

The vertices of a quadrilateral ABCD are A(0, 4), B(4, 1), C(1, -3), and D(-3, 0).

To find:

The perimeter of quadrilateral ABCD.

Solution:

Distance formula:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Using the distance formula, we get

$AB=\sqrt{(4-0)^2+(1-4)^2}$

$AB=\sqrt{(4)^2+(-3)^2}$

$AB=\sqrt{16+9}$

$AB=\sqrt{25}$

AB=5

Similarly,

$BC=\sqrt{(1-4)^2+(-3-1)^2}$

BC=5

$CD=\sqrt{(-3-1)^2+(0-(-3))^2}$

CD=5

And,

$AD=\sqrt{(-3-0)^2+(0-4)^2}$

AD=5

Now, the perimeter of the quadrilateral ABCD is:

P=AB+BC+CD+AD

P=5+5+5+5

P=20

Therefore, the perimeter of the quadrilateral ABCD is 20 units.

0,0(0 оценок)