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Example 1: The absolute pressure at a depth h in a liquid open to the atmosphere in increased by the pressure of the atmosphere pushing down on the surface of the liquid.
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Example 2: A hydraulic pump used to lift a car. When a small force f is applied to a small area a of a movable piston it creates a pressure P = f/a. This pressure is transmitted to and acts on a larger movable piston of area A which is then used to lift a car.
Hydrostacic Paradox QT Movie Demonstration
Example: If the height of the fluid's surface above the bottom of the five vessels is the same, in which vessel is the pressure of the fluid on the bottom of the vessel the greatest ? The amount of liquid in each vessel is not necessarily the same.
The pressure P is the same on the bottom of each vessel.
Why the pressure does not depend upon the shape of the vessel or the amount of fluid in the vessel rests upon three things:
a. Pressure is force per unit area and this is not same as the total weight of the liquid in a vessel.
b. A fluid can not support its self without a container. Thus the walls of the container exert a pressure on the fluid equal to the pressure of the fluid at that depth.
c. The pressure at given level is transmitted equally throughout the fluid to be the same value at that level.
Explanations:
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Vessel A: No matter how wide the vessel, the pressure is just the weight of the fluid above unit area on the bottom. Even if you take the whole weight of the fluid in the container mg and divide by the area of the bottom A, you still get the same results since the vessel is equivalent to a column of water.
Vessel B: Vessel B could be divided into three parts. The fluid in parts 1 and 3 is supported by upward force of the vessel on the fluid. Part 2 could be though of a vertical column of liquid similar to vessel A.
One could ask why doesn't the fluid in parts 1 and 3 (which are much bigger) not squeeze column 2 where they meet (along the dashed line) and there by increase the pressure on the bottom of the column ?
The answer is that the fluid in column 2 exerts and equal but opposite pressure outwards on the two other liquids to support itself. From the point of view of column 2, the water outside the dashed lines in sections 1 & 3 could be replace by solid vertical walls (along the dashed lines) and column 2 would still be in equilibrium.
The answer is that top of the container's walls in sections 1 and 3 produce a downward pressure that is equal to the fluid pressure in the middle section at the same level. If you poked a hole in the top of the container in sections 1 or 3, water would fountain upwards from the hole under pressure. From Pascal's principle, this pressure has to be that of the fluid in the middle section at the same level.
Vessel C: Again we could divide the water into three sections. The middle section is similar to that of vessel A or B. Since the height of the fluid in section 1 or 3 is not high enough to produce the same pressure as the height of the fluid in 2, how does the pressure on the bottoms of section 1 and 3 get to be the same as that of 2 ?

Vessel D: Again the center column is similar to vessel A.
The pressure of the vessel's wall creates a pressure that vertically supports the fluid in sections 1 and 3. At the same time the pressure of the walls create a horizontal component of pressure that sustains the fluid in the center column. 
Vessel E: While one can not offer simple arguments like those for the other vessels, the pressure on the bottom is still the same basically because of Pascal's principle.
You go down from the surface to some depth, then move sideways until you can go down again. Repeat the process until you reach the bottom. Since the pressure at the same depth is the same, moving sideways does not change the pressure. Only downwards motion increases the pressure. 
