Question

For a soccer practice, the coach set up four cones, labeled W, X, Y, and Z, to mark the edges of the rectangular field.Suppose the field is drawn on a coordinate plane, where the x- and y-values represent the position, in yards, from the center of the field, which is located at (0, 0). The location of the cones are as follows.Cone W is located at (-38, -53).Cone X is located at (-38, 53).Cone Y is located at (38, 53).Cone Z is located at (38, -53).What is the perimeter of the field?

Answer 1

The perimeter of the field is $364$ yards.

The distance between two points $A(x_1, y_1)$ and $B(x_2, y_2)$ is $AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.$

$WX=\sqrt{(-38-(-38))^2+(53-(-53))^2}=106$ units

$XY=\sqrt{(38-(-38))^2+(53-(53))^2}=76$ units

Perimeter $= 2 (l+w)$, where $l$ is length and $w$ is width.

Perimeter $= 2(106+76)$

$=364$ yards

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