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A rotating sprinkler head sprays water as far as 20 feet. The head is set to cover a central angle of [tex]$80^{\circ}$[/tex]. What area of grass will be watered?

A. [tex]\frac{760}{9} \pi \, \text{ft}^2[/tex]
B. [tex]\frac{200}{9} \pi \, \text{ft}^2[/tex]
C. [tex]\frac{800}{9} \pi \, \text{ft}^2[/tex]
D. [tex]\frac{80}{9} \pi \, \text{ft}^2[/tex]


Sagot :

To solve this problem, we need to determine the area of a sector of a circle. The formula for the area ([tex]\(A\)[/tex]) of a sector of a circle with radius [tex]\(r\)[/tex] and central angle [tex]\(\theta\)[/tex] in radians is:

[tex]\[ A = \frac{1}{2} r^2 \theta \][/tex]

Step-by-step solution:

1. Convert the angle from degrees to radians:

The angle given is [tex]\(80^\circ\)[/tex]. To convert degrees to radians, we use the following conversion factor:

[tex]\[ \theta_radians = \theta_{degrees} \times \frac{\pi}{180} \][/tex]

Therefore,

[tex]\[ \theta_radians = 80 \times \frac{\pi}{180} = \frac{80\pi}{180} = \frac{4\pi}{9} \text{ radians} \][/tex]

2. Plug the values into the sector area formula:

The radius ([tex]\(r\)[/tex]) is given as 20 feet, and the central angle ([tex]\(\theta\)[/tex]) in radians is [tex]\(\frac{4\pi}{9}\)[/tex].

[tex]\[ A = \frac{1}{2} \times r^2 \times \theta \][/tex]

Substituting the given values:

[tex]\[ A = \frac{1}{2} \times 20^2 \times \frac{4\pi}{9} \][/tex]

Simplifying inside the multiplication:

[tex]\[ A = \frac{1}{2} \times 400 \times \frac{4\pi}{9} = 200 \times \frac{4\pi}{9} \][/tex]

[tex]\[ A = \frac{800\pi}{9} \][/tex]

Therefore, the area of grass that will be watered by the rotating sprinkler head is [tex]\(\frac{800\pi}{9}\)[/tex] square feet.

The correct answer is:

C. [tex]\(\frac{800}{9} \pi \, \text{ft}^2\)[/tex]
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