A closed rectangular tank contains a certain amount of water. When the tank is placed on its 3 ft. by 4 ft. side, the height of the water in the tank is 5 ft. when the tank is placed on another side of dimension 4 ft. by 5 ft., what is the height, in feet, of the surface of the water above the ground?

Options

A. 2
B. 3
C. 4
D. 5

Answer

To solve the problem, we need to find the height of the water in a rectangular tank when it is placed on a different side. We will use the principle that the volume of water remains constant regardless of the orientation of the tank.

### Step-by-Step Solution:

1. **Identify the dimensions and height in the first orientation:**
- When the tank is placed on its 3 ft by 4 ft side, the height of the water is 5 ft.
- The dimensions of the base are 3 ft and 4 ft.

2. **Calculate the volume of water in the first orientation:**
- Volume (V) = Base Area × Height
- Base Area = Length × Width = 3 ft × 4 ft = 12 ft²
- Therefore, Volume = Base Area × Height = 12 ft² × 5 ft = 60 ft³.

3. **Identify the dimensions in the second orientation:**
- When the tank is placed on its 4 ft by 5 ft side, we need to find the new height (H) of the water.
- The dimensions of the base in this orientation are 4 ft and 5 ft.

4. **Set up the equation for the volume in the second orientation:**
- Volume in the second orientation = Base Area × Height
- Base Area = Length × Width = 4 ft × 5 ft = 20 ft²
- Therefore, Volume = Base Area × Height = 20 ft² × H.

5. **Set the volumes equal to each other:**
- Since the volume of water remains constant, we can set the two volumes equal:
- 60 ft³ = 20 ft² × H.

6. **Solve for H:**
- H = 60 ft³ / 20 ft² = 3 ft.

### Conclusion:
The height of the surface of the water above the ground when the tank is placed on its 4 ft by 5 ft side is **3 ft**.

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