To solve the problem, we need to determine the volume of the water trough, which appears to be in the shape of a trapezoidal prism.

Given:

• Top width of the trapezoid: 50 cm
• Bottom width of the trapezoid: 30 cm
• Height of the trapezoid: 1.2 m (which is 120 cm)
• Length of the trough: Unknown (not provided)

The problem requires the calculation of the volume, which is typically given by the formula:

$\text{Volume} = \text{Area of the trapezoidal base} \times \text{Length}$

1. Calculate the area of the trapezoid:

The area of a trapezoid is calculated using the formula:

$\text{Area} = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height}$

Where:

• $\text{Base}_1 = 50 \, \text{cm}$
• $\text{Base}_2 = 30 \, \text{cm}$
• $\text{Height} = 120 \, \text{cm}$

Plug in the values:

$\text{Area} = \frac{1}{2} \times (50 + 30) \times 120$

$\text{Area} = \frac{1}{2} \times 80 \times 120$

$\text{Area} = 40 \times 120 = 4800 \, \text{cm}^2$

2. Calculate the volume:

Assuming the length of the trough is in the same unit (meters), let's assume the length is $L$ meters. Convert this to centimeters:

$L = 1.2 \, \text{m} \times 100 = 120 \, \text{cm}$

Now, calculate the volume:

$\text{Volume} = 4800 \, \text{cm}^2 \times 120 \, \text{cm}$

$\text{Volume} = 576000 \, \text{cm}^3$

3. Convert the volume to liters:

Since 1 liter is equal to 1000 cm³, the volume in liters is:

$\text{Volume} = \frac{576000 \, \text{cm}^3}{1000} = 576 \, \text{liters}$

Final Answer:

The water trough can hold 576 liters of water.

Do you want more details or have any questions?

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Related Questions:

1. How would the volume change if the top width was increased to 60 cm?
2. What would be the volume if the height of the trapezoid was only 1 meter?
3. How does changing the shape of the trough affect the volume calculation?
4. What if the bottom width was 20 cm instead of 30 cm? How does it affect the volume?
5. How does the length of the trough influence the total water it can hold?

Tip: Always ensure that all dimensions are in the same units when calculating volume, especially when dealing with mixed units like meters and centimeters.