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The problem in the image involves determining if a pile of gravel in a gravel pit is enough to meet an order of 275 cubic meters of gravel. Here's a breakdown of the given information:

1. A pile of gravel with a height of 4 meters is in the pit.
2. The circumference of the base of the pile is 48.20 meters.
3. The volume required is 275 cubic meters.

### Step-by-Step Solution:

To find out if the pile is sufficient, we first calculate the volume of the gravel pile. The pile likely has the shape of a cone or something similar, and we'll treat it as such for the volume calculation.

#### 1. Finding the radius of the base:

The circumference \(C\) of the base of a cone is related to the radius \(r\) by the formula:
\[
C = 2\pi r
\]
Given \(C = 48.20 \, \text{m}\), we solve for \(r\):
\[
r = \frac{C}{2\pi} = \frac{48.20}{2\pi} \approx \frac{48.20}{6.2832} \approx 7.67 \, \text{m}
\]
So, the radius of the base is approximately 7.67 meters.

#### 2. Calculating the volume of the cone:

The volume \(V\) of a cone is given by the formula:
\[
V = \frac{1}{3} \pi r^2 h
\]
where:
- \(r\) is the radius of the base,
- \(h\) is the height of the cone.

Substituting the values \(r = 7.67 \, \text{m}\) and \(h = 4 \, \text{m}\):
\[
V = \frac{1}{3} \pi (7.67)^2 (4)
\]
First, calculate the square of the radius:
\[
7.67^2 = 58.85 \, \text{m}^2
\]
Now, calculate the volume:
\[
V = \frac{1}{3} \pi \times 58.85 \times 4 \approx \frac{1}{3} \times 3.1416 \times 235.4 \approx \frac{1}{3} \times 739.56 \approx 246.52 \, \text{m}^3
\]
So, the volume of the gravel pile is approximately 246.52 cubic meters.

#### 3. Comparing with the required amount:

The order is for 275 cubic meters of gravel. Since the available gravel volume is approximately 246.52 cubic meters, the pile does not have enough gravel to meet the order.

#### Conclusion:

**The available gravel is not sufficient to meet the order.** The gravel pile is short by:
\[
275 - 246.52 = 28.48 \, \text{m}^3
\]

### Sketch:

I'll describe how you can create the sketch:
- Draw a cone with a circular base.
- Label the height as 4 meters and the base circumference as 48.20 meters.
- You can add the radius of 7.67 meters as well.

Would you like to proceed with a detailed explanation or need further questions answered? Here are some relative questions you can explore:

1. How does the shape of the pile affect the volume calculation?
2. What would be the volume if the pile had a different height?
3. How do different base shapes (e.g., cylindrical) affect the volume formulas?
4. How can you calculate the amount of gravel that needs to be added?
5. How would the calculation change if the pile was truncated at the top?

**Tip:** Understanding the volume of basic 3D shapes like cones, cylinders, and spheres can be very useful for practical applications such as this one!

Bei einem Kiesgrubenbesitzer wurden 275 m³ Kies bestellt. In der Grube lagert ein 4 m hoher Haufen, dessen Umfang an der Grundfläche 48,20 m beträgt. Reicht der Vorrat?

This problem involves calculating the volume of a gravel pile with a height of 4 meters and a base circumference of 48.20 meters. Learn how to use the volume of a cone formula and compare the result to the required 275 cubic meters. Ideal for students in Grades 9-11.

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