Answer:
See explanation
Step-by-step explanation:
Consider the parallelogram ABCD (see attached diagram). First, find the lengths of all sides:
[tex]AB=\sqrt{(-1-4)^2+(2-4)^2}=\sqrt{25+4}=\sqrt{29}\\ \\BC=\sqrt{(4-2)^2+(4-(-1))^2}=\sqrt{4+25}=\sqrt{29}\\ \\CD=\sqrt{(2-(-3))^2+(-1-(-3))^2}=\sqrt{25+4}=\sqrt{29}\\ \\AD=\sqrt{(-1+3)^2+(2-(-3))^2}=\sqrt{4+25}=\sqrt{29}[/tex]
Since all sides are of equal length, this parallelogram is rhombus.
Check if it has right angles:
[tex]\overrightarrow{AB}=(5,2)\\ \\\overrightarrow {DA}=(2,5)\\ \\\cos\angle A=\dfrac{\overrightarrow {AB}\cdot \overrightarrow {DA}}{|\overrightarrow {AB}|\cdot|\overrightarrow{DA}|}=\dfrac{5\cdot2+2\cdot 5}{\sqrt{29}\cdot\sqrt{29}}=\dfrac{20}{29}\neq 0[/tex]
Angle A is not right angle, so this parallelogram is neither rectangle, nor square.
The perimeter is
[tex]P_{ABCD}=AB+BC+CD+DA=\sqrt{29}+\sqrt{29}+\sqrt{29}+\sqrt{29}=4\sqrt{29}\ un.[/tex]
The area is
[tex]A_{ABCD}=2A_{\triangle ABD}=2\cdot \dfrac{1}{2}\cdot AB\cdot AD\cdot \sin \angle A\\ \\\\A_{ABCD}=\sqrt{29}\cdot \sqrt{29}\cdot \sqrt{1-\left(\dfrac{20}{29}\right)^2}=29\cdot\dfrac{\sqrt{29^2-20^2}}{29}=21\ un^2.[/tex]
