A sprinkler is designed to rotate 360∘ clockwise, and then 360∘ counterclockwise to water a circular region with a radius of 11 feet. The sprinkler is located in the middle of the circular region. The sprinkler begins malfunctioning and is only able to rotate 225∘ in each direction. Find the area of the sector to the nearest square foot. The sprinkler can water ____ square feet.

To determine the area of the sector that the malfunctioning sprinkler can now water, we first need to calculate the area of the entire circle and then find the fraction of the circle that corresponds to the sprinkler's reduced range of motion.

The formula for the area of a circle is given by: \[A = \pi r^2\]

where \(A\) is the area and \(r\) is the radius.

Given that the radius of the circular region is 11 feet, we can calculate the area as follows: \[A = \pi (11)^2\] \[A = \pi (121)\] \[A = 121\pi\]

This is the area of the entire circle that would be watered if the sprinkler were not malfunctioning.

Now, since the sprinkler rotates 225 degrees in each direction (clockwise and counterclockwise), it covers a total of 450 degrees (225 degrees for each direction).

One complete revolution is 360 degrees, so the sprinkler is now covering \(\frac{450}{360}\) of the entire circle.

To find the area of the sector that the sprinkler can water, we multiply the area of the entire circle by this fraction. So, \[Area_{sector} = A \times \frac{450}{360}\] \[Area_{sector} = 121\pi \times \frac{5}{4}\] \[Area_{sector} = 121\pi \times 1.25\] \[Area_{sector} = 151.25\pi\]

To get the area to the nearest square foot, we need to approximate \(\pi\) as approximately 3.14: \[Area_{sector} ≈ 151.25 \times 3.14\] \[Area_{sector} ≈ 475.325\]

Rounded to the nearest square foot, the sprinkler can water approximately 475 square feet.