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:
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
Using the distance formula, we get
[tex]AB=\sqrt{(4-0)^2+(1-4)^2}[/tex]
[tex]AB=\sqrt{(4)^2+(-3)^2}[/tex]
[tex]AB=\sqrt{16+9}[/tex]
[tex]AB=\sqrt{25}[/tex]
[tex]AB=5[/tex]
Similarly,
[tex]BC=\sqrt{(1-4)^2+(-3-1)^2}[/tex]
[tex]BC=5[/tex]
[tex]CD=\sqrt{(-3-1)^2+(0-(-3))^2}[/tex]
[tex]CD=5[/tex]
And,
[tex]AD=\sqrt{(-3-0)^2+(0-4)^2}[/tex]
[tex]AD=5[/tex]
Now, the perimeter of the quadrilateral ABCD is:
[tex]P=AB+BC+CD+AD[/tex]
[tex]P=5+5+5+5[/tex]
[tex]P=20[/tex]
Therefore, the perimeter of the quadrilateral ABCD is 20 units.
