Geometry problem involving different volumes
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# Geometry problem involving different volumes

[From: ] [author: ] [Date: 12-04-05] [Hit: ]
ft.h = 2 2/3feet.Now,= 2 feet,......
The inner dimensions of a closed rectangular tank are: 4ft (l) x 2ft (w) x 3ft (h). There is 16 cubic feet of water in the tank. The tank will be repositioned on level ground so that it rests on one of the faces that has the least area. What will be the height, in feet, of the water after the tank has been repositioned?

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The areas of the sides of the tank, in square feet, are:
4 x 2 = 8
4 x 3 = 12
2 x 3 = 6.

The tank will be positioned so that a 2x3 face will be on the bottom. Then, a 2x3 face will be on the top, and two side faces will be 4x2, and the other two side faces will be 4x3, with the 2' and 3' sides of the side faces down. It may help to draw a diagram at this point.
What it boils down to is, we have a 4-foot high tank with a 2' x 3' bottom in it.
The area of the bottom is 6 square feet.

Volume = area of the base multiplied by the height of the tank. (A x h)

We need 16 cubic feet. Set up the equation:

16 = A x h.

We know that "A", the area of the base, is 6 sq. ft.

16 = 6 x h

h = 16 / 6

h = 8/3

h = 2 2/3 feet. Now, convert it to feet and inches: 2 2/3 feet = 2 feet + (2/3 foot x 12 inches / foot)

= 2 feet, 8 inches
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