SOLUTION: An inverted regular pyramid whose square base has an area of 16 square units and a height of 6 units contains water with a depth of 3 units. If the said pyramid is set upright, how

Algebra ->  Volume -> SOLUTION: An inverted regular pyramid whose square base has an area of 16 square units and a height of 6 units contains water with a depth of 3 units. If the said pyramid is set upright, how      Log On


   

Question 1044343: An inverted regular pyramid whose square base has an area of 16 square units and a height of 6 units contains water with a depth of 3 units. If the said pyramid is set upright, how high is the water level?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
This problem has numbers and specifications that are not really needed.
If an inverted pyramid of any shape and size
is filled with water to half of its height,
the height of the water after turning it upright is
%282-root%283%2C7%29%29%2F2=about0.0435=%224.35%25%22 of the container pyramid's height.
In the case of a container of height 6 units, originally filled to a depth of 3units=%281%2F2%296units ,
the final water height is
%28%282-root%283%2C7%29%29%2F2%29%2A6units=%282-root%283%2C7%29%29%2A3units=approximatelyhighlight%280.2612units%29

The container is a pyramid that has a height H=6 and a Volume V .
The part originally filled with water is a similar pyramid.
Its height is 3=6%2F2=H%2F2=%281%2F2%29H , half of the height of the container pyramid.
The original water-filled pyramid is a smaller version of the container pyramid,
scaled down by a factor of 1%2F2 .
For scaled up or down objects, if the ratio of corresponding length is r and the ratio of corresponding volume is r%5E3 .
If you know one ratio you can find the other.
So, the volume of that original water-filled pyramid is %281%2F2%29%5E3=1%2F8 of the volume of the container pyramid.
When the container pyramid is set upright, the water still fills 1%2F8 of the volume,
but now the water is against the base of the container pyramid.
Above the water there is an air-filled pyramid with volume V-%281%2F8%29V=%287%2F8%29V .
The height of that pyramid is H%2Aroot%283%2C%287%2F8%29%29H=H%2A%28root%283%2C7%29%2F2%29 .
The height of the water is
H-H%2A%28root%283%2C7%29%2F2%29=H%281-root%283%2C7%29%2F2%29=H%282-root%283%2C7%29%29%2F2