Answer:
Part 1) The polygon is a square
Part 2) The perimeter is equal to ![20\ units](https://tex.z-dn.net/?f=20%5C%20units)
Part 3) The area is equal to ![25\ units^{2}](https://tex.z-dn.net/?f=25%5C%20units%5E%7B2%7D)
Step-by-step explanation:
we have
![A(5,0), B(2,4), C(-2,1),D(1,-3)](https://tex.z-dn.net/?f=A%285%2C0%29%2C%20B%282%2C4%29%2C%20C%28-2%2C1%29%2CD%281%2C-3%29)
Plot the points
see the attached figure
we know that
the formula to calculate the distance between two points is equal to
![d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%28y2-y1%29%5E%7B2%7D%2B%28x2-x1%29%5E%7B2%7D%7D)
Find the distance AB
![A(5,0),B(2,4)](https://tex.z-dn.net/?f=A%285%2C0%29%2CB%282%2C4%29)
substitute in the formula
![d=\sqrt{(4-0)^{2}+(2-5)^{2}}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%284-0%29%5E%7B2%7D%2B%282-5%29%5E%7B2%7D%7D)
![d=\sqrt{(4)^{2}+(-3)^{2}}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%284%29%5E%7B2%7D%2B%28-3%29%5E%7B2%7D%7D)
![d=\sqrt{25}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B25%7D)
![AB=5\ units](https://tex.z-dn.net/?f=AB%3D5%5C%20units)
Find the distance BC
![B(2,4), C(-2,1)](https://tex.z-dn.net/?f=B%282%2C4%29%2C%20C%28-2%2C1%29)
substitute in the formula
![d=\sqrt{(1-4)^{2}+(-2-2)^{2}}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%281-4%29%5E%7B2%7D%2B%28-2-2%29%5E%7B2%7D%7D)
![d=\sqrt{(-3)^{2}+(-4)^{2}}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%28-3%29%5E%7B2%7D%2B%28-4%29%5E%7B2%7D%7D)
![d=\sqrt{25}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B25%7D)
![BC=5\ units](https://tex.z-dn.net/?f=BC%3D5%5C%20units)
Find the distance CD
![C(-2,1),D(1,-3)](https://tex.z-dn.net/?f=C%28-2%2C1%29%2CD%281%2C-3%29)
substitute in the formula
![d=\sqrt{(-3-1)^{2}+(1+2)^{2}}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%28-3-1%29%5E%7B2%7D%2B%281%2B2%29%5E%7B2%7D%7D)
![d=\sqrt{(-4)^{2}+(3)^{2}}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%28-4%29%5E%7B2%7D%2B%283%29%5E%7B2%7D%7D)
![d=\sqrt{25}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B25%7D)
![CD=5\ units](https://tex.z-dn.net/?f=CD%3D5%5C%20units)
Find the distance AD
![A(5,0),D(1,-3)](https://tex.z-dn.net/?f=A%285%2C0%29%2CD%281%2C-3%29)
substitute in the formula
![d=\sqrt{(-3-0)^{2}+(1-5)^{2}}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%28-3-0%29%5E%7B2%7D%2B%281-5%29%5E%7B2%7D%7D)
![d=\sqrt{(-3)^{2}+(-4)^{2}}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%28-3%29%5E%7B2%7D%2B%28-4%29%5E%7B2%7D%7D)
![AD=5\ units](https://tex.z-dn.net/?f=AD%3D5%5C%20units)
we have that
AB=BC=CD=AD
Find the distance BD (diagonal)
![B(2,4),D(1,-3)](https://tex.z-dn.net/?f=B%282%2C4%29%2CD%281%2C-3%29)
substitute in the formula
![d=\sqrt{(-3-4)^{2}+(1-2)^{2}}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%28-3-4%29%5E%7B2%7D%2B%281-2%29%5E%7B2%7D%7D)
![d=\sqrt{(-7)^{2}+(-1)^{2}}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%28-7%29%5E%7B2%7D%2B%28-1%29%5E%7B2%7D%7D)
<em>Verify if the polygon is a square</em>
If the triangle BDA is a right triangle, then the polygon is a square
Applying the Pythagoras theorem
![BD^{2}=AD^{2}+AB^{2}](https://tex.z-dn.net/?f=BD%5E%7B2%7D%3DAD%5E%7B2%7D%2BAB%5E%7B2%7D)
substitute
![(\sqrt{50})^{2}=5^{2}+5^{2}](https://tex.z-dn.net/?f=%28%5Csqrt%7B50%7D%29%5E%7B2%7D%3D5%5E%7B2%7D%2B5%5E%7B2%7D)
-----> is true
so
The triangle BDA is a right triangle
therefore
The polygon is a square
<em>Find the Area of the polygon</em>
The area of a square is equal to
![A=b^{2}](https://tex.z-dn.net/?f=A%3Db%5E%7B2%7D)
we have
![b=5\ units](https://tex.z-dn.net/?f=b%3D5%5C%20units)
![A=5^{2}=25\ units^{2}](https://tex.z-dn.net/?f=A%3D5%5E%7B2%7D%3D25%5C%20units%5E%7B2%7D)
<em>Find the perimeter of the polygon</em>
The perimeter of a square is equal to
![P=4b](https://tex.z-dn.net/?f=P%3D4b)
we have
![b=5\ units](https://tex.z-dn.net/?f=b%3D5%5C%20units)
![P=4(5)=20\ units](https://tex.z-dn.net/?f=P%3D4%285%29%3D20%5C%20units)