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Let's address the math problems step by step.

### Question 6
**Problem**: 
- You are asked to determine how many cans of each type of paint (roof and wall) are needed to paint the house shown in the diagram. The roof paint can cover 24 m² per 4 L can, and the wall paint can cover 28 m² per 4 L can.
  
**Approach**:
1. **Surface Area of the Roof**: The roof appears to be composed of two identical triangular faces and two rectangular sloping sides. To calculate the area of these faces, we need to figure out the dimensions of the roof sections based on the provided diagram (slopes and height).

2. **Surface Area of the Walls**: The walls consist of two rectangular side walls and two end walls (one triangular and one rectangular).

I’ll calculate the areas of each face of the house in detail to find out how many cans of each type are needed.

### Question 7
This is about doubling the height of a pyramid without changing its base.

a) **Surface Area**:
- The total surface area includes both the base and the lateral (slanted) faces. Doubling the height increases the area of the lateral faces but does not change the base area.
- The slant height changes as a result of the height doubling, so the lateral surface area does not simply double.

b) **Volume**:
- The volume of a pyramid is given by:
  \[
  V = \frac{1}{3} \times \text{Base Area} \times \text{Height}
  \]
- If only the height is doubled while keeping the base the same, the volume doubles because the volume is directly proportional to the height.

### Question 8
**Explanation**: 
- The lateral faces are triangular, and they share a base with the base of the pyramid. Since the lateral faces rise upwards to a point, their area depends on both the base and the slant height, which typically exceeds the height of the base.
  
Now, let's break down the computations for question 6 and move on to question 7 after that.### Question 6 Solution:

1. **Wall Paint**:
   - The total area of the walls is approximately **37.68 m²**.
   - Since one can of wall paint covers **28 m²**, you will need:
     \[
     \lceil \frac{37.68}{28} \rceil = 2 \text{ cans}
     \]

2. **Roof Paint**:
   - The total area of the roof is approximately **37.64 m²**.
   - Since one can of roof paint covers **24 m²**, you will need:
     \[
     \lceil \frac{37.64}{24} \rceil = 2 \text{ cans}
     \]

So, Gembo will need **2 cans of wall paint** and **2 cans of roof paint**.

---

### Question 7:
- a) The total surface area **does not double**. The base area remains constant, while the lateral area increases but does not double because the slant height increases at a different rate compared to the vertical height.
- b) The volume **does double**, since the volume formula \( V = \frac{1}{3} \times \text{Base Area} \times \text{Height} \) is directly proportional to the height.

---

### Question 8:
The combined area of the lateral faces of the pyramid must be greater than the base because the lateral faces rise to a point, forming triangles. These triangles have a larger area due to the slant height, which is longer than the perpendicular height of the base.

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Would you like more details or clarifications? Here are some related questions:

1. How does slant height affect the surface area of a pyramid?
2. Can we generalize the relationship between height and surface area for any pyramid?
3. What happens to the surface area if both base and height double?
4. How can you calculate the lateral surface area of a regular pyramid?
5. What are other examples where volume increases at a different rate than surface area?

**Tip**: For pyramids and cones, always consider how slant height affects surface area differently than vertical height!

Gembo was asked to paint a house with the given dimensions, needing to calculate how many cans of paint are required for the roof and walls. If you double the height of a pyramid without changing the base, does the total surface area or volume double? Also, explain why the combined area of the lateral faces of a pyramid must be greater than the area of the base.

Solve a practical geometry problem involving calculating the amount of paint needed for a house, and analyze the effects of doubling the height of a pyramid on surface area and volume. Learn key geometry concepts related to 3D shapes, surface area, and volume.

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