The problem you're working with involves calculating the area and perimeter of a regular polygon. Based on the image, the following information is given:

• Side length of the polygon: $8.72 \, \text{m}$
• The polygon is a pentagon (5 sides).
• The apothem (denoted as $L$) is given as $6 \, \text{m}$.

Perimeter Calculation:

For a regular polygon, the perimeter $P$ is calculated as: $P = n \times \text{side length}$ Where $n$ is the number of sides.

Here, $n = 5$ and the side length is $8.72 \, \text{m}$, so: $P = 5 \times 8.72 = 43.6 \, \text{m}$

Area Calculation:

The area $A$ of a regular polygon can be calculated using the formula: $A = \frac{1}{2} \times P \times \text{apothem}$ Where:

• $P$ is the perimeter
• The apothem is the distance from the center of the polygon to the middle of one of its sides.

Given:

• $P = 43.6 \, \text{m}$
• Apothem $= 6 \, \text{m}$

Thus: $A = \frac{1}{2} \times 43.6 \times 6 = 130.8 \, \text{m}^2$

Summary:

• Perimeter: $43.6 \, \text{m}$
• Area: $130.8 \, \text{m}^2$

Let me know if you'd like any further details or clarifications.

Here are 5 questions that expand on this topic:

1. How would the formula for area change if the polygon had more sides, such as a hexagon?
2. Can you derive the formula for the area of a regular polygon from basic geometric principles?
3. What happens to the area of a polygon as the number of sides increases?
4. How is the apothem related to the radius of the circumscribed circle in a regular polygon?
5. How would the area calculation differ for an irregular polygon?

Tip: Always verify the number of sides and apothem values to ensure correct calculations for polygons!