Answer:
The approximate perimeter of the trapezoid is 31 units
Step-by-step explanation:
step 1
Plot the trapezoid
Let
A(-5, -3), B(-2, 5), C(2, 5), and D(5, -3)
see the attached figure
step 2
Find the perimeter of trapezoid
we know that
The perimeter of 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
[tex]A(-5, -3),B(-2, 5)[/tex]
substitute in the formula
[tex]d=\sqrt{(5+3)^{2}+(-2+5)^{2}}[/tex]
[tex]d=\sqrt{(8)^{2}+(3)^{2}}[/tex]
[tex]d_A_B=\sqrt{73}\ units[/tex]
Find the distance BC
we have
[tex]B(-2, 5),C(2, 5)[/tex]
substitute in the formula
[tex]d=\sqrt{(5-5)^{2}+(2+2)^{2}}[/tex]
[tex]d=\sqrt{(0)^{2}+(4)^{2}}[/tex]
[tex]d_B_C=4\ units[/tex]
Find the distance CD
we have
[tex]C(2, 5),D(5, -3)[/tex]
substitute in the formula
[tex]d=\sqrt{(-3-5)^{2}+(5-2)^{2}}[/tex]
[tex]d=\sqrt{(-8)^{2}+(3)^{2}}[/tex]
[tex]d_C_D=\sqrt{73}\ units[/tex]
Find the distance AD
we have
[tex]A(-5, -3),D(5, -3)[/tex]
substitute in the formula
[tex]d=\sqrt{(-3+3)^{2}+(5+5)^{2}}[/tex]
[tex]d=\sqrt{(0)^{2}+(10)^{2}}[/tex]
[tex]d_A_D=10\ units[/tex]
step 3
Find the perimeter
[tex]P=AB+BC+CD+AD[/tex]
substitute the values
[tex]P=\sqrt{73}+4+\sqrt{73}+10[/tex]
[tex]P=31\ units[/tex]
