https://file.aiquickdraw.com/mathbot/file/1727275264435afqug3en.jpg

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**.

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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!

1. How many small cubes with sides 1/4 m can be placed in a cube with side 2 m?
2. In the given figure, if BD' = 2√3 cm, find the volume of the given cube.
3. If AB = 4 cm, BC = 3 cm, AE = 5 cm, then find the length of EC.
4. A swimming pool is a rectangular prism. The 3/4 of this swimming pool has 48 m^3 of water. If the depth of this pool is 1.6 m, find the area of the base of the swimming pool.

Solve geometry problems involving cubes, space diagonals, and rectangular prisms. Learn how to calculate volumes, find diagonals using the Pythagorean theorem, and determine base areas of 3D shapes. These problems are suitable for students in Grades 9-11.

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