SOLUTION: A lawn sprinkler sprays water 8 feet at full pressure as it rotates 360 degrees. If the water pressure is reduced by 50%, what is the difference in the area covered? The choices a

Question 887628: A lawn sprinkler sprays water 8 feet at full pressure as it rotates 360 degrees. If the water pressure is reduced by 50%, what is the difference in the area covered?
The choices are:
1)78.5
2)50.2
3)201.0
4)150.8
Found 2 solutions by Theo, Alan3354:
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
i believe your answer is 150.8

at full pressure, it sprays 8 feet out so 8 is your radius and the area covered is 8^2 * pi = 64 * pi.

at half pressure, it sprays 4 feet out so 4 is your radius and the area covered is 4^2 * pi = 16 * pi.

the difference in the area covered is 64 * pi - 16 * pi = 48 * pi.

48 * pi = 150.7964... which rounds to 150.8.

Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
A lawn sprinkler sprays water 8 feet at full pressure as it rotates 360 degrees. If the water pressure is reduced by 50%, what is the difference in the area covered?
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You have to specify the relation between pressure and the distance of the stream. You can't assume it's linear, I don't think it is.
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