Answer-
The perimeter and area of the parallelogram are 19.74 units and 15 sq. units respectively.
Solution-
The co-ordinates of the vertices are,
A = (-2, 3)
B = (4, 0)
C = (1, -1)
D = (-5, 2)
E = (-3, 1)
We can get the side length of the parallelogram by calculating the respective distances by applying distance formula,
[tex]\overline{CD}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(-5-1)^2+(2+1)^2}=\sqrt{(-6)^2+(3)^2}=\sqrt{36+9}=\sqrt{45}=3\sqrt5[/tex]
[tex]\overline{AD}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(-2+5)^2+(3-2)^2}=\sqrt{(3)^2+(1)^2}=\sqrt{9+1}=\sqrt{10}[/tex]
[tex]\overline{AE}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(-2+3)^2+(3-1)^2}=\sqrt{(1)^2+(2)^2}=\sqrt{1+4}=\sqrt{5}[/tex]
Perimeter of the parallelogram ABCD is,
[tex]=2(\overline{AD}+\overline{CD})\\\\=2(\sqrt{10}+3\sqrt5)\\\\=19.74\ units[/tex]
Area of the parallelogram ABCD is,
[tex]=\overline{CD}\times \overline{AE}\\\\=3\sqrt5\times \sqrt{5}\\\\=3\times 5\\\\=15\ sq.unit[/tex]
