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 (-31, -52). cone x is located at (-31, 52). cone y is located at (31, 52). cone z is located at (31, -52). what is the perimeter of the field? a. 166 yards b. 332 yards c. 36 yards d. 6,448 yards

Explanation:

Step1: Calculate the length of the rectangle

The length of the rectangle can be found by the difference in the \(x -\)coordinates of two opposite - side cones (since the \(y -\)coordinates are the same for \(W(-31,-52)\) and \(Z(31,-52)\) or \(X(-31,52)\) and \(Y(31,52)\)).
Using the formula for the distance between two points \(d=\vert x_2 - x_1\vert\) (when \(y_1 = y_2\)), for points \(W(-31,-52)\) and \(Z(31,-52)\), \(l=\vert31-(-31)\vert=\vert31 + 31\vert=62\) yards.

Step2: Calculate the width of the rectangle

The width of the rectangle can be found by the difference in the \(y -\)coordinates of two opposite - side cones (since the \(x -\)coordinates are the same for \(W(-31,-52)\) and \(X(-31,52)\) or \(Y(31,52)\) and \(Z(31,-52)\)).
Using the formula for the distance between two points \(d = \vert y_2-y_1\vert\) (when \(x_1=x_2\)), for points \(W(-31,-52)\) and \(X(-31,52)\), \(w=\vert52-(-52)\vert=\vert52 + 52\vert = 104\) yards.

Step3: Calculate the perimeter of the rectangle

The formula for the perimeter of a rectangle is \(P=2(l + w)\).
Substitute \(l = 62\) and \(w = 104\) into the formula: \(P=2(62+104)\).
First, calculate the sum inside the parentheses: \(62 + 104=166\).
Then, multiply by 2: \(P=2\times166 = 332\) yards.

Answer:

B. 332 yards