To find the length and width of the rectangular plot, we need to use two equations: one equation for the area and another equation for the perimeter (total length of fencing).
Let's assume that the length of the plot is 'L' yards and the width is 'W' yards.
The equation for the area of a rectangle is: A = L * W, where A represents the area.
The equation for the perimeter of a rectangle is: P = 2L + 2W, where P represents the perimeter.
Given that the area is 2000 sq. yards and the total length of fencing is 180 yards, we can set up the following equations:
Equation 1: A = L * W = 2000
Equation 2: P = 2L + 2W = 180
To solve these equations simultaneously, we can rearrange Equation 1 to solve for one variable in terms of the other. Since we need to find the length and width, let's solve for W:
W = 2000 / L
Now substitute this value of W into Equation 2:
2L + 2W = 180
2L + 2(2000/L) = 180
Simplify the equation:
2L + 4000/L = 180
Next, multiply the entire equation by L to eliminate the fraction:
2L^2 + 4000 = 180L
Rearrange the equation:
2L^2 - 180L + 4000 = 0
To solve this quadratic equation, we can either factor it or use the quadratic formula. Let's use the quadratic formula:
L = (-b ± √(b^2 - 4ac)) / 2a
For this equation, a = 2, b = -180, and c = 4000. Now, substitute these values into the formula:
L = (-(-180) ± √((-180)^2 - 4(2)(4000))) / 2(2)
L = (180 ± √(32400 - 32000)) / 4
L = (180 ± √400) / 4
L = (180 ± 20) / 4
L = 200/4 = 50 or L = 160/4 = 40
Since the length cannot be longer than the perimeter, we can discard the solution L = 50. Therefore, L = 40 yards.
Substitute this value back into Equation 1 to find the width:
W = 2000 / L
W = 2000 / 40
W = 50 yards
So, the length of the plot (L) is 40 yards and the width (W) is 50 yards.