Answer:
C. 26 yards
Step-by-step explanation:
Total length of Jamie's fence = sum of all distances from one vertex of the polygon to another = AB + BC + CD + DE + EF + FA
Use the distance formula, [tex] d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} [/tex], to find the distance between vertexes.
Distance between A(-5, 5) and B(0, 5)
Let,
[tex] A(-5, 5) = (x_1, y_1) [/tex]
[tex] B(0, 5) = (x_2, y_2) [/tex]
[tex] AB = \sqrt{(0 -(-5))^2 + (5 - 5)^2} [/tex]
[tex] AB = \sqrt{(5)^2 + (0)^2} [/tex]
[tex] AB = \sqrt{25 + 0} = \sqrt{25} [/tex]
[tex] AB = 5 [/tex]
Distance between B(0, 5) and C(4, 2)
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 [/tex]
Distance between C(4, 2) and D(1, -2)
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 [/tex]
Distance between D(1, -2) and E(-2, -2)
Let,
[tex] D(1, -2) = (x_1, y_1) [/tex]
[tex] E(-2, -2) = (x_2, y_2) [/tex]
[tex] DE = \sqrt{(-2 - 1)^2 + (-2 -(-2))^2} [/tex]
[tex] DE = \sqrt{(-3)^2 + (0)^2} [/tex]
[tex] DE = \sqrt{9 + 0} = \sqrt{9} [/tex]
[tex] DE = 3 [/tex]
Distance between E(-2, -2) and F(-5, 2)
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 [/tex]
Distance between A(-5, 5) and F(-5, 2)
Let,
[tex] A(-5, 5) = (x_1, y_1) [/tex]
[tex] F(-5, 2) = (x_2, y_2) [/tex]
[tex] FA = \sqrt{(-5 -(-5))^2 + (2 - 5)^2} [/tex]
[tex] FA = \sqrt{(0)^2 + (-3)^2} [/tex]
[tex] FA = \sqrt{0 + 9} = \sqrt{9} [/tex]
[tex] FA = 3 [/tex]
Total length of Jamie's fence in yards = 5 + 5 + 5 + 5 + 3 + 5 + 3 = 26 yards
