Question

Find the perimeter AND area of the parallelogram ABCD. Show all work

Answer 1

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,

$\overline{CD}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(-5-1)^2+(2+1)^2}=\sqrt{(-6)^2+(3)^2}=\sqrt{36+9}=\sqrt{45}=3\sqrt5$

$\overline{AD}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(-2+5)^2+(3-2)^2}=\sqrt{(3)^2+(1)^2}=\sqrt{9+1}=\sqrt{10}$

$\overline{AE}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(-2+3)^2+(3-1)^2}=\sqrt{(1)^2+(2)^2}=\sqrt{1+4}=\sqrt{5}$

Perimeter of the parallelogram ABCD is,

$=2(\overline{AD}+\overline{CD})\\\\=2(\sqrt{10}+3\sqrt5)\\\\=19.74\ units$

Area of the parallelogram ABCD is,

$=\overline{CD}\times \overline{AE}\\\\=3\sqrt5\times \sqrt{5}\\\\=3\times 5\\\\=15\ sq.unit$