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.

$\text{Area of sector}=\frac{\theta}{360}\times \pi r^2$, where,

$\theta$ = Central angle of sector,

$r$ = Radius.

For our given problem $\theta = 225^{\circ}$ and $r=11$.

$\text{Area of sector}=\frac{225^{\circ}}{360^{\circ}}\times \pi (11)^2$

$\text{Area of sector}=0.625\times 121\pi$

$\text{Area of sector}=237.5829444277281137$

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

Therefore, the sprinkler can water approximately 238 square feet.