Answer:
Part 1) The polygon is a square
Part 2) The perimeter is equal to [tex]20\ units[/tex]
Part 3) The area is equal to [tex]25\ units^{2}[/tex]
Step-by-step explanation:
we have
[tex]A(5,0), B(2,4), C(-2,1),D(1,-3)[/tex]
Plot the points
see the attached figure
we know that
the formula to calculate the distance between two points is equal to
[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]
Find the distance AB
[tex]A(5,0),B(2,4)[/tex]
substitute in the formula
[tex]d=\sqrt{(4-0)^{2}+(2-5)^{2}}[/tex]
[tex]d=\sqrt{(4)^{2}+(-3)^{2}}[/tex]
[tex]d=\sqrt{25}[/tex]
[tex]AB=5\ units[/tex]
Find the distance BC
[tex]B(2,4), C(-2,1)[/tex]
substitute in the formula
[tex]d=\sqrt{(1-4)^{2}+(-2-2)^{2}}[/tex]
[tex]d=\sqrt{(-3)^{2}+(-4)^{2}}[/tex]
[tex]d=\sqrt{25}[/tex]
[tex]BC=5\ units[/tex]
Find the distance CD
[tex]C(-2,1),D(1,-3)[/tex]
substitute in the formula
[tex]d=\sqrt{(-3-1)^{2}+(1+2)^{2}}[/tex]
[tex]d=\sqrt{(-4)^{2}+(3)^{2}}[/tex]
[tex]d=\sqrt{25}[/tex]
[tex]CD=5\ units[/tex]
Find the distance AD
[tex]A(5,0),D(1,-3)[/tex]
substitute in the formula
[tex]d=\sqrt{(-3-0)^{2}+(1-5)^{2}}[/tex]
[tex]d=\sqrt{(-3)^{2}+(-4)^{2}}[/tex]
[tex]d=\sqrt{25}[/tex]
[tex]AD=5\ units[/tex]
we have that
AB=BC=CD=AD
Find the distance BD (diagonal)
[tex]B(2,4),D(1,-3)[/tex]
substitute in the formula
[tex]d=\sqrt{(-3-4)^{2}+(1-2)^{2}}[/tex]
[tex]d=\sqrt{(-7)^{2}+(-1)^{2}}[/tex]
[tex]BD=\sqrt{50}\ units[/tex]
Verify if the polygon is a square
If the triangle BDA is a right triangle, then the polygon is a square
Applying the Pythagoras theorem
[tex]BD^{2}=AD^{2}+AB^{2}[/tex]
substitute
[tex](\sqrt{50})^{2}=5^{2}+5^{2}[/tex]
[tex]50=50[/tex] -----> is true
so
The triangle BDA is a right triangle
therefore
The polygon is a square
Find the Area of the polygon
The area of a square is equal to
[tex]A=b^{2}[/tex]
we have
[tex]b=5\ units[/tex]
[tex]A=5^{2}=25\ units^{2}[/tex]
Find the perimeter of the polygon
The perimeter of a square is equal to
[tex]P=4b[/tex]
we have
[tex]b=5\ units[/tex]
[tex]P=4(5)=20\ units[/tex]
