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A community pool that is shaped like a regular pentagon needs a new cover for the winter months. The radius of the pool is 20.10 ft. The pool is 23.62 ft on each side.

To the nearest square foot, what is the area of the pool that needs to be covered?

A. [tex]\(192 \text{ ft}^2\)[/tex]
B. [tex]\(960 \text{ ft}^2\)[/tex]
C. [tex]\(1,921 \text{ ft}^2\)[/tex]
D. [tex]\(3,842 \text{ ft}^2\)[/tex]


Sagot :

To determine the area of a regular pentagon-shaped pool, follow these steps:

1. Identify the Given Information:
- Side length of the pentagon: [tex]\(23.62\)[/tex] feet.
- Radius (distance from the center to a vertex) of the pentagon: [tex]\(20.10\)[/tex] feet.

2. Recall the Formula for the Area of a Regular Pentagon:
The area [tex]\(A\)[/tex] of a regular pentagon with side length [tex]\(s\)[/tex] can be calculated using the formula:
[tex]\[ A = \left( \frac{5}{4} \right) \cdot s^2 \cdot \left( \frac{1}{\tan(\pi/5)} \right) \][/tex]
This formula uses the tangent of [tex]\(\pi/5\)[/tex], which accounts for the interior angles of the pentagon.

3. Plug in the Given Side Length into the Formula:
[tex]\[ s = 23.62 \text{ feet} \][/tex]
Calculate the area using the formula:
[tex]\[ A = \left( \frac{5}{4} \right) \cdot (23.62)^2 \cdot \left( \frac{1}{\tan(\pi/5)} \right) \][/tex]

4. Computations (Intermediary steps are here for clarity):
[tex]\[ \left( \frac{5}{4} \right) \approx 1.25 \][/tex]
[tex]\[ s^2 = (23.62)^2 \approx 557.2644 \][/tex]
[tex]\[ \frac{1}{\tan(\pi/5)} \approx 1.3763819204711735 \][/tex]
Combining these, we get:
[tex]\[ A = 1.25 \times 557.2644 \times 1.3763819204711735 \approx 960.44 \][/tex]

5. Round to the Nearest Square Foot:
[tex]\[ \text{Area} \approx 960 \text{ square feet} \][/tex]

Therefore, the area of the pool that needs to be covered is approximately [tex]\(960\)[/tex] square feet.

Hence, the correct answer is:
[tex]\[ \boxed{960 \, \text{ft}^2} \][/tex]