The fountain is depicted by the white circle in the picture. The surrounding walkway is depicted by the grey areas.
From the sketch shown above, the semi-circle inscribed in the rectangle is one half of the fountain. We shall calculate the area of the semi-circle and subtract this from the area of the rectangle.
The area of the rectangle is;
![\begin{gathered} \text{Area}=l* w \\ \text{Area}=30*42.5 \\ \text{Area}=1275ft^2 \\ \text{The area of the semicircle is,} \\ \text{Area=}(1)/(2)(\pi* r^2) \\ \text{The diameter is 18 ft, and therefore the radius is 9 ft} \\ \text{Area}=(1)/(2)(3.14*9^2) \\ \text{Area}=(1)/(2)(3.14*81) \\ \text{Area}=(1)/(2)(254.34) \\ \text{Area}=127.17ft^2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/84i67jvro0fjkmea83d475reytrv0pnht7.png)
Therefore, the area of the shaded region would be,
Area = 1275 - 127.17
Area = 1147.83
Next step is to calculate the other half of the figure (the right side), as follows;
Observe that the outer semi-circle is the shaded region while the inner one is the white portion.
The area is
![\begin{gathered} \text{Shaded region;} \\ \text{Area}=(1)/(2)(\pi* r^2) \\ \text{Area}=(1)/(2)(3.14*15^2) \\ \text{Area}=(1)/(2)(3.14*225) \\ \text{Area}=(1)/(2)(706.5) \\ \text{Area}=353.25ft^2 \\ \text{White region;} \\ \text{Area}=(1)/(2)(\pi*9^2) \\ \text{Area}=(1)/(2)(3.14*81) \\ \text{Area}=127.17ft^2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/wqdmvubjjkdvpq378pwr2noz1s8rpabe1t.png)
The area of the shaded region is;
Area = 353.25 - 127.17
Area = 226.38
Therefore the total area of the walkway surrounding the fountain is;
Area = 1147.83 + 226.38
Area = 1374.21
Area = 1,374 feet squared (rounded to the nearest foot)