Answer:
Part 1) The shape is a trapezoid
Part 2) The perimeter is [tex]25(4+\sqrt{2})\ units[/tex] or approximately [tex]135.4\ units[/tex]
Part 3) The area is [tex]937.5\ units^2[/tex]
Step-by-step explanation:
step 1
Plot the figure to better understand the problem
we have
A(-28,2),B(-21,-22),C(27,-8),D(-4,9)
using a graphing tool
The shape is a trapezoid
see the attached figure
step 2
Find the perimeter
we know that
The perimeter of the trapezoid is equal to
[tex]P=AB+BC+CD+AD[/tex]
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
we have
A(-28,2),B(-21,-22)
substitute in the formula
[tex]d=\sqrt{(-22-2)^{2}+(-21+28)^{2}}[/tex]
[tex]d=\sqrt{(-24)^{2}+(7)^{2}}[/tex]
[tex]d=\sqrt{625}[/tex]
[tex]d_A_B=25\ units[/tex]
Find the distance BC
we have
B(-21,-22),C(27,-8)
substitute in the formula
[tex]d=\sqrt{(-8+22)^{2}+(27+21)^{2}}[/tex]
[tex]d=\sqrt{(14)^{2}+(48)^{2}}[/tex]
[tex]d=\sqrt{2,500}[/tex]
[tex]d_B_C=50\ units[/tex]
Find the distance CD
we have
C(27,-8),D(-4,9)
substitute in the formula
[tex]d=\sqrt{(9+8)^{2}+(-4-27)^{2}}[/tex]
[tex]d=\sqrt{(17)^{2}+(-31)^{2}}[/tex]
[tex]d=\sqrt{1,250}[/tex]
[tex]d_C_D=25\sqrt{2}\ units[/tex]
Find the distance AD
we have
A(-28,2),D(-4,9)
substitute in the formula
[tex]d=\sqrt{(9-2)^{2}+(-4+28)^{2}}[/tex]
[tex]d=\sqrt{(7)^{2}+(24)^{2}}[/tex]
[tex]d=\sqrt{625}[/tex]
[tex]d_A_D=25\ units[/tex]
Find the perimeter
[tex]P=25+50+25\sqrt{2}+25[/tex]
[tex]P=(100+25\sqrt{2})\ units[/tex]
simplify
[tex]P=25(4+\sqrt{2})\ units[/tex] ----> exact value
[tex]P=135.4\ units[/tex]
therefore
The perimeter is [tex]25(4+\sqrt{2})\ units[/tex] or approximately [tex]135.4\ units[/tex]
step 3
Find the area
The area of trapezoid is equal to
[tex]A=\frac{1}{2}[BC+AD]AB[/tex]
substitute the given values
[tex]A=\frac{1}{2}[50+25]25=937.5\ units^2[/tex]
