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Let's solve the given questions step by step.

### Problem Analysis:
The swimming pool in the diagram has:
- A volume of $ 8.57 \, \text{m}^3 $
- Two half-cylindrical ends with a radius of $ 1 \, \text{m} $ and an unknown height $ h $
- A rectangular prism (middle section) with a length of $ 2 \, \text{m} $, width equal to the diameter of the cylinder (so $ 2 \, \text{m} $), and height $ h $.

We need to:
1. **Calculate the depth $ h $ of the swimming pool (5.1).**
2. **Calculate the surface area of the bottom of the swimming pool to repaint it (5.2).**

### Step 1: Calculate the Depth of the Swimming Pool (5.1)

#### Volume Calculation
The total volume of the pool consists of:
- **Volume of the rectangular prism**: $ V_{\text{rectangular prism}} = l \times w \times h $
- **Volume of two half-cylinders** (which combine to form a full cylinder): $ V_{\text{cylinder}} = \pi r^2 h $

Since $ l = 2 \, \text{m} $, $ w = 2 \, \text{m} $, and $ r = 1 \, \text{m} $, the total volume is given by:
\[
8.57 = (2 \times 2 \times h) + (\pi \times 1^2 \times h)
\]

#### Simplify the Equation
Substitute and simplify:
\[
8.57 = (4h) + (\pi h)
\]
\[
8.57 = h(4 + \pi)
\]

Now, solve for $ h $:
\[
h = \frac{8.57}{4 + \pi}
\]

Using $ \pi \approx 3.1416 $:
\[
h = \frac{8.57}{4 + 3.1416} \approx \frac{8.57}{7.1416} \approx 1.2 \, \text{m}
\]

So, the **depth $ h $ of the swimming pool is approximately $ 1.2 \, \text{m} $**.

### Step 2: Calculate the Surface Area of the Bottom of the Swimming Pool (5.2)

The bottom of the swimming pool consists of:
- The rectangular base of the middle section.
- The circular bases of the two half-cylindrical ends.

#### Surface Area Calculation
- **Area of the rectangular base**: $ A_{\text{rectangle}} = l \times w = 2 \times 2 = 4 \, \text{m}^2 $
- **Area of the circular base** (for one circle): $ A_{\text{circle}} = \pi r^2 = \pi \times 1^2 = \pi \, \text{m}^2 $
  - Since there are two half-cylinders, we have one full circle, so total circular area is $ \pi \, \text{m}^2 $.

Adding these together:
\[
\text{Total surface area of the bottom} = 4 + \pi \approx 4 + 3.1416 = 7.1416 \, \text{m}^2
\]

Thus, the **surface area of the bottom of the swimming pool to repaint is approximately $ 7.14 \, \text{m}^2 $**.

---

### Final Answers:
1. **Depth of the swimming pool (5.1):** $ h \approx 1.2 \, \text{m} $
2. **Surface area of the bottom of the swimming pool (5.2):** $ \approx 7.14 \, \text{m}^2 $

Would you like further clarification on any steps, or do you need additional details?

### Related Questions:
1. How would the depth change if the volume of the pool was different?
2. How is the surface area affected if the radius of the half-cylinders increases?
3. How would the calculation change if the pool had full cylindrical sections on each end rather than half-cylinders?
4. What would be the cost to repaint the bottom if the paint costs $5 per square meter?
5. How would we approach this problem if the pool were three-dimensional?

#### Tip:
For geometry problems involving volume and surface area, always break down the structure into simpler shapes (like prisms and cylinders) for easier calculations.

The diagram below shows a swimming pool with a volume 8.57 m^3 and an inner radius 1 m. The swimming pool is made out of a rectangular prism and has two half-cylinders at each end. Ignore the thickness of the swimming tool. Calculate the depth of the swimming pool (5.1) and the surface area of the bottom of the swimming pool to repaint it (5.2).

Find the depth and surface area of a swimming pool with a volume of 8.57 m^3, composed of a rectangular prism and two half-cylindrical ends. Learn to apply volume and surface area formulas for composite shapes, suitable for high school geometry students in Grades 9-11.

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