First, draw a diagram of the situation to visualize the problem:
Let x be the width of a strip of the sidewalk.
The dimensions of the outer rectangle are (40+2x) and (20+2x).
Then, the area of the rectangle in terms of x is:

On the other hand, the area of the outer rectangle (in square meters) is equal to 1000. Then:

Replace the expression for A in terms of x to obtain a quadratic equation:

Bring all the terms to the left member of the equation to write it in standard form:

Use the quadratic formula to find the solutions to this quadratic equation. Use a=4, b=120 and c=-200:
![\begin{gathered} ax^2+bx+c=0 \\ \Rightarrow x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \\ \Rightarrow x=\frac{-120\pm\sqrt[]{120^2-4(4)(-200)}}{2(4)} \\ =\frac{-120\pm\sqrt[]{14400+3200}}{8} \\ =\frac{-120\pm\sqrt[]{17600}}{8} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/9ey0fyf6cgrp91srhlbnjo4rx4ommzl6hm.png)
Since x represents the width of the sidewalk (in meters), then, it does not make sense to take the negative solution because x would be negative after subtracting 120 from the value of the square root. Then, the value of x must be:
![x=\frac{-120+\sqrt[]{17600}}{8}=1.583123952\ldots\approx1.58](https://img.qammunity.org/2023/formulas/mathematics/college/u2kzmc0tzk3tzk281x0vty2jauortio0ah.png)
Notice that, in fact, a rectangle with sides (40+2*1.58) and (20+2*1.58) has an area of 1000:

Therefore, the exact value of the width of a strip of the sidewalk is:
![x=\frac{-120+\sqrt[]{17600}}{8}=5\cdot\sqrt[]{11}-15](https://img.qammunity.org/2023/formulas/mathematics/college/bq2iztw1ysi6m7b6a9l745ndvh7rz2jwpu.png)
And the approximate value of the width of a strip of the sidewalk is 1.58 meters.