Question

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.

Answer 1

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.

`\text{Area of sector}=(\theta)/(360)* \pi r^2`, where,

`\theta` = Central angle of sector,

`r` = Radius.

For our given problem
`\theta = 225^(\circ)` and
`r=11`.

`\text{Area of sector}=(225^(\circ))/(360^(\circ))* \pi (11)^2`

`\text{Area of sector}=0.625* 121\pi`

`\text{Area of sector}=237.5829444277281137`

`\text{Area of sector}\approx 238`

Therefore, the sprinkler can water approximately 238 square feet.