Let \(x\) be the width of the strip. Thus the new length of the rectangle is \(120 + x\) and the new width is \(80+x\).

The area of the old plot: \(A_o = 120x80 = 9600 yd^3\)

The area of the new plot is twice the rea of the old plot. Thus, \(A_n = 2A_o\)

\(A_n = (120 +x)(80+x) \)

\(A_n =9600 + 200x + 80x^2\)

p>\(9600 + 200x + 80x^2 = 2(9600)\)\(x^2 + 200 x -9600 = 0 \)

\(x = \frac{-200+\sqrt{200^2 - 4(1)(-9600)}}{2(1)}\)

\(x = 40~yards\)Answer