HELP NEEDED. 37 POINTS
I just need the answers

Answer:
Part 1) [tex]P=[2\sqrt{29}+\sqrt{18}]\ units[/tex] or [tex]P=15.01\ units[/tex]
Part 2) [tex]P=2[\sqrt{20}+\sqrt{45}]\ units[/tex] or [tex]P=22.36\ units[/tex]
Part 3) [tex]P=4[\sqrt{13}]\ units[/tex] or [tex]P=14.42\ units[/tex]
Part 4) [tex]P=[19+\sqrt{17}]\ units[/tex] or [tex]P=23.12\ units[/tex]
Part 5) [tex]P=2[\sqrt{17}+\sqrt{68}]\ units[/tex] or [tex]P=24.74\ units[/tex]
Part 6) [tex]A=36\ units^{2}[/tex]
Part 7) [tex]A=20\ units^{2}[/tex]
Part 8) [tex]A=16\ units^{2}[/tex]
Part 9) [tex]A=10.5\ units^{2}[/tex]
Part 10) [tex]A=6.05\ units^{2}[/tex]
Step-by-step explanation:
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]
Part 1) we have the triangle ABC
[tex]A(0,3),B(5,1),C(2,-2)[/tex]
step 1
Find the distance AB
[tex]A(0,3),B(5,1)[/tex]
substitute in the formula
[tex]AB=\sqrt{(1-3)^{2}+(5-0)^{2}}[/tex]
[tex]AB=\sqrt{(-2)^{2}+(5)^{2}}[/tex]
[tex]AB=\sqrt{29}\ units[/tex]
step 2
Find the distance BC
[tex]B(5,1),C(2,-2)[/tex]
substitute in the formula
[tex]BC=\sqrt{(-2-1)^{2}+(2-5)^{2}}[/tex]
[tex]BC=\sqrt{(-3)^{2}+(-3)^{2}}[/tex]
[tex]BC=\sqrt{18}\ units[/tex]
step 3
Find the distance AC
[tex]A(0,3),C(2,-2)[/tex]
substitute in the formula
[tex]AC=\sqrt{(-2-3)^{2}+(2-0)^{2}}[/tex]
[tex]AC=\sqrt{(-5)^{2}+(2)^{2}}[/tex]
[tex]AC=\sqrt{29}\ units[/tex]
step 4
Find the perimeter
The perimeter is equal to
[tex]P=AB+BC+AC[/tex]
substitute
[tex]P=[\sqrt{29}+\sqrt{18}+\sqrt{29}]\ units[/tex]
[tex]P=[2\sqrt{29}+\sqrt{18}]\ units[/tex]
or
[tex]P=15.01\ units[/tex]
Part 2) we have the rectangle ABCD
[tex]A(-4,-4),B(-2,0),C(4,-3),D(2,-7)[/tex]
Remember that in a rectangle opposite sides are congruent
step 1
Find the distance AB
[tex]A(-4,-4),B(-2,0)[/tex]
substitute in the formula
[tex]AB=\sqrt{(0+4)^{2}+(-2+4)^{2}}[/tex]
[tex]AB=\sqrt{(4)^{2}+(2)^{2}}[/tex]
[tex]AB=\sqrt{20}\ units[/tex]
step 2
Find the distance BC
[tex]B(-2,0),C(4,-3)[/tex]
substitute in the formula
[tex]BC=\sqrt{(-3-0)^{2}+(4+2)^{2}}[/tex]
[tex]BC=\sqrt{(-3)^{2}+(6)^{2}}[/tex]
[tex]BC=\sqrt{45}\ units[/tex]
step 3
Find the perimeter
The perimeter is equal to
[tex]P=2[AB+BC][/tex]
substitute
[tex]P=2[\sqrt{20}+\sqrt{45}]\ units[/tex]
or
[tex]P=22.36\ units[/tex]
Part 3) we have the rhombus ABCD
[tex]A(-3,3),B(0,5),C(3,3),D(0,1)[/tex]
Remember that in a rhombus all sides are congruent
step 1
Find the distance AB
[tex]A(-3,3),B(0,5)[/tex]
substitute in the formula
[tex]AB=\sqrt{(5-3)^{2}+(0+3)^{2}}[/tex]
[tex]AB=\sqrt{(2)^{2}+(3)^{2}}[/tex]
[tex]AB=\sqrt{13}\ units[/tex]
step 2
Find the perimeter
The perimeter is equal to
[tex]P=4[AB][/tex]
substitute
[tex]P=4[\sqrt{13}]\ units[/tex]
or
[tex]P=14.42\ units[/tex]
Part 4) we have the quadrilateral ABCD
[tex]A(-2,-3),B(1,1),C(7,1),D(6,-3)[/tex]
step 1
Find the distance AB
[tex]A(-2,-3),B(1,1)[/tex]
substitute in the formula
[tex]AB=\sqrt{(1+3)^{2}+(1+2)^{2}}[/tex]
[tex]AB=\sqrt{(4)^{2}+(3)^{2}}[/tex]
[tex]AB=5\ units[/tex]
step 2
Find the distance BC
[tex]B(1,1),C(7,1)[/tex]
substitute in the formula
[tex]BC=\sqrt{(1-1)^{2}+(7-1)^{2}}[/tex]
[tex]BC=\sqrt{(0)^{2}+(6)^{2}}[/tex]
[tex]BC=6\ units[/tex]
step 3
Find the distance CD
[tex]C(7,1),D(6,-3)[/tex]
substitute in the formula
[tex]CD=\sqrt{(-3-1)^{2}+(6-7)^{2}}[/tex]
[tex]CD=\sqrt{(-4)^{2}+(-1)^{2}}[/tex]
[tex]CD=\sqrt{17}\ units[/tex]
step 4
Find the distance AD
[tex]A(-2,-3),D(6,-3)[/tex]
substitute in the formula
[tex]AD=\sqrt{(-3+3)^{2}+(6+2)^{2}}[/tex]
[tex]AD=\sqrt{(0)^{2}+(8)^{2}}[/tex]
[tex]AD=8\ units[/tex]
step 5
Find the perimeter
The perimeter is equal to
[tex]P=AB+BC+CD+AD[/tex]
substitute
[tex]P=[5+6+\sqrt{17}+8]\ units[/tex]
[tex]P=[19+\sqrt{17}]\ units[/tex]
or
[tex]P=23.12\ units[/tex]
Part 5) we have the quadrilateral ABCD
[tex]A(-1,5),B(3,6),C(5,-2),D(1,-3)[/tex]
step 1
Find the distance AB
[tex]A(-1,5),B(3,6)[/tex]
substitute in the formula
[tex]AB=\sqrt{(6-5)^{2}+(3+1)^{2}}[/tex]
[tex]AB=\sqrt{(1)^{2}+(4)^{2}}[/tex]
[tex]AB=\sqrt{17}\ units[/tex]
step 2
Find the distance BC
[tex]B(3,6),C(5,-2)[/tex]
substitute in the formula
[tex]BC=\sqrt{(-2-6)^{2}+(5-3)^{2}}[/tex]
[tex]BC=\sqrt{(-8)^{2}+(2)^{2}}[/tex]
[tex]BC=\sqrt{68}\ units[/tex]
step 3
Find the distance CD
[tex]C(5,-2),D(1,-3)[/tex]
substitute in the formula
[tex]CD=\sqrt{(-3+2)^{2}+(1-5)^{2}}[/tex]
[tex]CD=\sqrt{(-1)^{2}+(-4)^{2}}[/tex]
[tex]CD=\sqrt{17}\ units[/tex]
step 4
Find the distance AD
[tex]A(-1,5),D(1,-3)[/tex]
substitute in the formula
[tex]AD=\sqrt{(-3-5)^{2}+(1+1)^{2}}[/tex]
[tex]AD=\sqrt{(-8)^{2}+(2)^{2}}[/tex]
[tex]AD=\sqrt{68}\ units[/tex]
step 5
Find the perimeter
The perimeter is equal to
[tex]P=\sqrt{17}+\sqrt{68}+\sqrt{17}+\sqrt{68}[/tex]
substitute
[tex]P=2[\sqrt{17}+\sqrt{68}]\ units[/tex]
or
[tex]P=24.74\ units[/tex]