Let's go through the problems one by one and solve them:

1. Small cubes in a large cube

We are asked how many small cubes with side $\frac{1}{4} \, m$ can fit inside a larger cube with side $2 \, m$.

Solution:

• The volume of the large cube is $V_{large} = (2)^3 = 8 \, m^3$.
• The volume of a small cube is $V_{small} = \left( \frac{1}{4} \right)^3 = \frac{1}{64} \, m^3$.
• The number of small cubes that fit inside the large cube is the ratio of their volumes: $N = \frac{V_{large}}{V_{small}} = \frac{8}{\frac{1}{64}} = 8 \times 64 = 512.$ So, the answer is E) 512.

2. Volume of the cube with given diagonal

We are given that $BD' = 2\sqrt{3} \, cm$, where $BD'$ is the space diagonal of the cube. We need to find the volume of the cube.

Solution:

• The space diagonal $BD'$ in a cube is related to the side $s$ of the cube by the formula: $BD' = s\sqrt{3}.$
• Given $BD' = 2\sqrt{3} \, cm$, we can equate it: $s\sqrt{3} = 2\sqrt{3}.$ Solving for $s$, we get: $s = 2 \, cm.$
• The volume of the cube is then: $V = s^3 = 2^3 = 8 \, cm^3.$ So, the answer is A) 8.

3. Finding the length of $EC$

We are given a right triangle formed in 3D, where $AB = 4 \, cm$, $BC = 3 \, cm$, and $AE = 5 \, cm$. We are tasked with finding $EC$.

Solution:

• Using the Pythagorean theorem in 3D, we know that $EC$ can be found by: $EC^2 = AE^2 + AB^2 + BC^2.$
• Substituting the given values: $EC^2 = 5^2 + 4^2 + 3^2 = 25 + 16 + 9 = 50.$
• Therefore: $EC = \sqrt{50} = 5\sqrt{2}.$ So, the answer is E) 5\sqrt{2}.

4. Swimming pool volume and base area

We are given a swimming pool with a total volume of $48 \, m^3$, where $\frac{3}{4}$ of it is filled. The depth of the pool is $1.6 \, m$, and we are asked to find the area of the base of the pool.

Solution:

• The volume filled is $\frac{3}{4} \times 48 = 36 \, m^3$.
• The total volume of the pool is given by the formula: $V = \text{Base Area} \times \text{Depth}.$ Therefore, the base area is: $\text{Base Area} = \frac{V}{\text{Depth}} = \frac{36}{1.6} = 22.5 \, m^2.$
• However, since the total pool's volume is 48 m$^3$, the base area should be: $\frac{48}{1.6} = 30 \, m^2.$ So, the answer is D) 30.

Let me know if you'd like further clarifications or details! Here are 5 additional practice problems:

1. If the side of the cube is doubled, how many more small cubes can fit?
2. How is the space diagonal of a cube calculated in terms of its side length?
3. Given a right triangle in 3D, what other methods can be used to find diagonal distances?
4. How does the volume of a rectangular prism change with respect to changes in depth?
5. If the swimming pool had a trapezoidal base, how would the calculation differ?

Tip: Always check if the units match before performing calculations, especially in volume and area problems!