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:
[tex]\begin{gathered} A=(40+2x)(20+2x) \\ =40\cdot20+40\cdot2x+2x\cdot20+2x\cdot2x \\ =800+80x+40x+4x^2 \\ =800+120x+4x^2 \end{gathered}[/tex]
On the other hand, the area of the outer rectangle (in square meters) is equal to 1000. Then:
[tex]A=1000[/tex]
Replace the expression for A in terms of x to obtain a quadratic equation:
[tex]4x^2+120x+800=1000[/tex]
Bring all the terms to the left member of the equation to write it in standard form:
[tex]\begin{gathered} 4x^2+120x+800-1000=0 \\ \Rightarrow4x^2+120x-200=0 \end{gathered}[/tex]
Use the quadratic formula to find the solutions to this quadratic equation. Use a=4, b=120 and c=-200:
[tex]\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}[/tex]
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:
[tex]x=\frac{-120+\sqrt[]{17600}}{8}=1.583123952\ldots\approx1.58[/tex]
Notice that, in fact, a rectangle with sides (40+2*1.58) and (20+2*1.58) has an area of 1000:
[tex](40+2\cdot1.583)(20+2\cdot1.583)=43.166\times23.166=999.98\approx1000[/tex]
Therefore, the exact value of the width of a strip of the sidewalk is:
[tex]x=\frac{-120+\sqrt[]{17600}}{8}=5\cdot\sqrt[]{11}-15[/tex]
And the approximate value of the width of a strip of the sidewalk is 1.58 meters.
