The problem involves a cubical container, partially filled with water, and requires finding the base area of the container.

Step-by-Step Solution:

1. Given Data:

   • The container is $\frac{4}{5}$ filled with water.
   • The container holds $\frac{27}{5} \, \text{liters of water}$.

2. Relating Volume to Base Area:

   • The volume of water in the container when it is $\frac{4}{5}$ full is: $\text{Volume of water} = \text{Base Area} \times \text{Height of Water}$
   • Let the base area be $A$ and the height of the container (when fully filled) be $h$. When $\frac{4}{5}$ full, the height of the water is $\frac{4}{5}h$.

3. Convert Volume to Cubic Units:

   • $\frac{27}{5} \, \text{liters}$ of water equals $\frac{27}{5} \times 1000 = 5400 \, \text{cm}^3$, because $1 \, \text{liter} = 1000 \, \text{cm}^3$.

4. Equation Relating Base Area and Volume:

   • The volume equation becomes: $5400 = A \times \frac{4}{5}h$ Rearrange to find $A \times h$: $A \times h = \frac{5400 \times 5}{4} = 6750 \, \text{cm}^3$

5. Relate Height to the Base Area:

   • In a cubical container, the height equals the side length of the base, meaning $h = \sqrt{A}$. Substituting $h = \sqrt{A}$ into $A \times h = 6750$, we get: $A \times \sqrt{A} = 6750$ Let $A = x^2$. Then $x^2 \times x = x^3 = 6750$, so: $x = \sqrt[3]{6750}$

6. Compute the Base Area:

   • Calculate $x = \sqrt[3]{6750}$ (using approximation or a calculator): $x \approx 18.88 \, \text{cm}$ The base area is: $A = x^2 \approx 18.88^2 = 356.3 \, \text{cm}^2$

Final Answer:

The base area of the container is approximately 356.3 cm².

Would you like me to clarify any step or provide further details?

Related Questions:

1. What is the formula for the volume of a cubical container?
2. How do you convert liters to cubic centimeters?
3. Can this method work for a non-cubical container?
4. What is the significance of $\frac{4}{5}$ in this problem?
5. How does knowing the height simplify the calculation?

Tip: Always ensure consistent units (e.g., cm³, liters) when solving volume problems.