Given:
The points A(-3, 3), B(0, 7), C(4, 10), D(1, 6) create parallelogram ABCD.
To find:
The perimeter of parallelogram ABCD.
Solution:
Distance formula is
[tex]Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
Using distance formula, find the side length of parallelogram.
[tex]AB=\sqrt{(0-(-3))^2+(7-3)^2}[/tex]
[tex]AB=\sqrt{(3)^2+(4)^2}[/tex]
[tex]AB=\sqrt{9+16}[/tex]
[tex]AB=\sqrt{25}[/tex]
[tex]AB=5[/tex]
Similarly,
[tex]BC=\sqrt{\left(4-0\right)^2+\left(10-7\right)^2}=5[/tex]
[tex]CD=\sqrt{\left(1-4\right)^2+\left(6-10\right)^2}=5[/tex]
[tex]AD=\sqrt{\left(1-\left(-3\right)\right)^2+\left(6-3\right)^2}=5[/tex]
Now,
Perimeter of ABCD is
[tex]P=AB+BC+CD+AD[/tex]
[tex]P=5+5+5+5[/tex]
[tex]P=20[/tex]
Therefore, the perimeter of parallelogram ABCD is 20 units.
