Question:
What is the perimeter of the path, in miles?
Answer:
35 miles
Step-by-step explanation:
Given
Let the corners be ABCD
[tex]A = (-2,2)[/tex]
[tex]B = (1,2)[/tex]
[tex]C = (1,-2)[/tex]
[tex]D = (-2,-2)[/tex]
First, we calculate the perimeter of the corners in units.
This is done by calculating the distance between the corners.
Distance is calculated as:
[tex]D = \sqrt{(x_2 - x_1)^2 +(y_2 - y_1)^2 }[/tex]
For AB:
[tex](x_1,y_1) = (-2,2)[/tex]
[tex](x_2,y_2) = (1,2)[/tex]
[tex]D_1 = \sqrt{(1 - (-2))^2 +(2 - 2)^2 }[/tex]
[tex]D_1 = \sqrt{(1 +2)^2 +(0)^2 }[/tex]
[tex]D_1 = \sqrt{3^2}[/tex]
[tex]D_1 = 3[/tex]
For BC
[tex](x_1,y_1) = (1,2)[/tex]
[tex](x_2,y_2) = (1,-2)[/tex]
[tex]D_2 = \sqrt{(1 - 1)^2 +(-2 - 2)^2 }[/tex]
[tex]D_2 = \sqrt{(0)^2 +(-4)^2 }[/tex]
[tex]D_2 = \sqrt{16 }[/tex]
[tex]D_2 = 4[/tex]
For CD:
[tex](x_1,y_1) = (1,-2)[/tex]
[tex](x_2,y_2) = (-2,-2)[/tex]
[tex]D_3 = \sqrt{(-2 - 1)^2 +(-2 - (-2))^2 }[/tex]
[tex]D_3 = \sqrt{(-3)^2 +(-2 +2)^2 }[/tex]
[tex]D_3 = \sqrt{9 +0^2 }[/tex]
[tex]D_3 = \sqrt{9 }[/tex]
[tex]D_3 = 3[/tex]
For DA
[tex](x_1,y_1) = (-2,-2)[/tex]
[tex](x_2,y_2) = (-2,2)[/tex]
[tex]D_4 = \sqrt{(-2 - (-2))^2 +(2 - (-2))^2 }[/tex]
[tex]D_4 = \sqrt{(-2 +2)^2 +(2 +2))^2 }[/tex]
[tex]D_4 = \sqrt{0^2 +4^2 }[/tex]
[tex]D_4 = \sqrt{4^2 }[/tex]
[tex]D_4 = 4[/tex]
The perimeter P in units is:
[tex]P = D_1 + D_2 + D_3 + D_4[/tex]
[tex]P = 3+4+3+4[/tex]
[tex]P = 14[/tex]
Given that:
[tex]1\ units = 2.5\ miles[/tex]
Multiply both sides by 14
[tex]14 * 1\ units = 14 * 2.5\ miles[/tex]
[tex]14 \ units = 35\ miles[/tex]
Hence, the perimeter is 35 miles
