Answer:
C. 26 yards
Step-by-step explanation:
Jamie's fence total length = perimeter of the polygon
Perimeter of the polygon = AB + BC + CD + DE + EF + FA
AB, FA and DE can be worked accordingly as shown below:
AB = |-5 - 0| = 5 units
FA = |5 - 2| = 3 units
DE = |1 -(-2)| = 3 units
BC, CD, and EF can be calculated using the formula [tex] d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} [/tex]
Distance between B(0, 5) and C(4, 2):
[tex] BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} [/tex]
Let,
[tex] B(0, 5) = (x_1, y_1) [/tex]
[tex] C(4, 2) = (x_2, y_2) [/tex]
[tex] BC = \sqrt{(4 - 0)^2 + (2 - 5)^2} [/tex]
[tex] BC = \sqrt{(4)^2 + (-3)^2} [/tex]
[tex] BC = \sqrt{16 + 9} = \sqrt{25} [/tex]
[tex] BC = 5 units [/tex]
Distance between C(4, 2) and D(1, -2)
[tex] CD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} [/tex]
Let,
[tex] C(4, 2) = (x_1, y_1) [/tex]
[tex] D(1, -2) = (x_2, y_2) [/tex]
[tex] CD = \sqrt{(1 - 4)^2 + (-2 - 2)^2} [/tex]
[tex] CD = \sqrt{(-3)^2 + (-4)^2} [/tex]
[tex] CD = \sqrt{9 + 16} = \sqrt{25} [/tex]
[tex] CD = 5 units [/tex]
Distance between E(-2, -2) and F(-5, 2):
[tex] EF = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} [/tex]
Let,
[tex] E(-2, -2) = (x_1, y_1) [/tex]
[tex] F(-5, 2) = (x_2, y_2) [/tex]
[tex] EF = \sqrt{(-5 -(-2))^2 + (2 -(-2))^2} [/tex]
[tex] EF = \sqrt{(-3)^2 + (4)^2} [/tex]
[tex] EF = \sqrt{9 + 16} = \sqrt{25} [/tex]
[tex] EF = 5 units [/tex]
Total length of the wall in yards = 5 + 5 + 5 + 3 + 5 + 3 = 26 yards