\begin{tabular}{l}\( \left\lvert\, \begin{array}{l}\text { A water feature can be modeled as a hemisphere } \\ \text { with its base set onto a square patio. If the diameter } \\ \text { of the hemisphere is } 4 \text { feet and the patio has a side } \\ \text { of length } 6 \text { feet,what would the visible area of the } \\ \text { patio be? Give your answer accurately to two } \\ \text { decimal places. }\end{array}\right. \) \\ \hline\end{tabular}

To find the visible area of the patio, first, calculate the area of the patio and then subtract the area covered by the hemisphere. 1. Area of the patio: Since the patio is square with a side length of 6 feet, the area is: \[ \text{Area}_{\text{patio}} = \text{side}^2 = 6^2 = 36 \text{ square feet} \] 2. Area of the visible part of the patio: The hemisphere is placed on the square patio, covering a circle on the patio whose diameter is equal to the diameter of the hemisphere (4 feet). The radius \(r\) of the hemisphere is: \[ r = \frac{\text{diameter}}{2} = \frac{4}{2} = 2 \text{ feet} \] The area of the circular base of the hemisphere is: \[ \text{Area}_{\text{hemisphere base}} = \pi r^2 = \pi (2^2) = 4\pi \text{ square feet} \] 3. Now, calculate the visible area of the patio: \[ \text{Visible Area} = \text{Area}_{\text{patio}} - \text{Area}_{\text{hemisphere base}} = 36 - 4\pi \] 4. Numerically evaluate \(4\pi\) (use approximately 3.14 for \(\pi\)): \[ 4\pi \approx 4 \times 3.14 \approx 12.56 \] So, \[ \text{Visible Area} \approx 36 - 12.56 \approx 23.44 \text{ square feet} \] Thus, the visible area of the patio is approximately \( \boxed{23.44} \) square feet.