Consider the following diagram,
Let 'y' be the side of the garden. And 'x' be the side of the square formed by the walkway.
Given that the garden is a square with perimeter 260 feet,
![\begin{gathered} \text{Perimeter}=260 \\ 4y=260 \\ y=65 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/p6rldkw2abvfurdxuurertn05q3b0p51ni.png)
From the above diagram, it can be observed that,
![\begin{gathered} x=4+y+4 \\ x=8+y \\ x=8+65 \\ x=73 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/q5ycww0wgthi9jade1rl1m4g4mj19qhz11.png)
Consider that the area of a square is given by,
![\text{Area of square}=Side^2](https://img.qammunity.org/2023/formulas/mathematics/college/qlp60xn6cnk9scajkz5xww615lhq79d4fa.png)
The area of the inner square i.e. garden will be,
![\begin{gathered} A_i=y^2 \\ A_i=65^2 \\ A_i=4225 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/lua05qx1yqb7t1l3qv6dck46p35we3q5br.png)
The area of the outer square will be,
![\begin{gathered} A_o=x^2 \\ A_o=73^2 \\ A_o=5329 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/88u4jmzt5070em6pinyubkps1ldfjg65vk.png)
The difference between the area will give the area of the walkway (A), that can be calculated as,
![\begin{gathered} A=A_o-A_i \\ A=5329-4225 \\ A=1104 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/z62r8dxvczq769e1kn9yc31ufiwdwsvi45.png)
Thus, the area of the walkway is 1104 square feet.