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?

Question 886692: 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?
Answer by JulietG(1812) (Show Source): You can put this solution on YOUR website!
Area of a circle is Pi R squared.
At full pressure, the sprinkler covers Pi 4 squared (radius = 1/2 diameter) or 16 Pi.
At half pressure, assuming it covers half the distance, the sprinkler area is Pi 2 squared, or 4 Pi.
Therefore, at half pressure, the area of coverage is 12 Pi less. 4 Pi is 25% of 16 Pi, so reducing the pressure by 50% reduces the coverage area by 75%.
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