We have been given that 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.
We are asked to find the area of the sector to nearest square foot.
We will use area of sector formula to solve our given problem.
[tex]\text{Area of sector}=\frac{\theta}{360}\times \pi r^2[/tex], where,
[tex]\theta[/tex] = Central angle of sector,
[tex]r[/tex] = Radius.
For our given problem [tex]\theta = 225^{\circ}[/tex] and [tex]r=11[/tex].
[tex]\text{Area of sector}=\frac{225^{\circ}}{360^{\circ}}\times \pi (11)^2[/tex]
[tex]\text{Area of sector}=0.625\times 121\pi[/tex]
[tex]\text{Area of sector}=237.5829444277281137[/tex]
[tex]\text{Area of sector}\approx 238[/tex]
Therefore, the sprinkler can water approximately 238 square feet.
