A railing of 288 m is required for fencing a semi-circular park. Find the area of the park.

Options

A. `4928 m^(2)`
B. `9865 m^(2)`
C. `8956 m^(2)`
D. `9856 m^(2)`

Answer

To solve the problem of finding the area of a semi-circular park given that a railing of 288 m is required for fencing, we can follow these steps:

### Step 1: Understand the fencing requirement
The fencing required for a semi-circular park includes the curved part (the semi-circular boundary) and the straight part (the diameter). The total length of the fencing is given as 288 m.

### Step 2: Write the formula for the circumference of a semi-circle
The formula for the circumference (perimeter) of a semi-circle is:
\[ \text{Circumference} = \pi r + 2r \]
where \( r \) is the radius of the semi-circle. Here, \( \pi \) can be approximated as \( \frac{22}{7} \).

### Step 3: Set up the equation
We know that the total circumference is 288 m, so we can set up the equation:
\[ \pi r + 2r = 288 \]

### Step 4: Substitute the value of \( \pi \)
Substituting \( \pi \) with \( \frac{22}{7} \):
\[ \frac{22}{7} r + 2r = 288 \]

### Step 5: Combine the terms
To combine the terms, we can express \( 2r \) with a common denominator:
\[ \frac{22}{7} r + \frac{14}{7} r = 288 \]
This simplifies to:
\[ \frac{36}{7} r = 288 \]

### Step 6: Solve for \( r \)
To isolate \( r \), multiply both sides by \( \frac{7}{36} \):
\[ r = 288 \times \frac{7}{36} \]
Calculating this gives:
\[ r = 288 \div 36 \times 7 = 8 \times 7 = 56 \text{ m} \]

### Step 7: Find the area of the semi-circle
The area \( A \) of a semi-circle is given by the formula:
\[ A = \frac{1}{2} \pi r^2 \]
Substituting \( r = 56 \) m:
\[ A = \frac{1}{2} \times \frac{22}{7} \times (56)^2 \]

### Step 8: Calculate \( (56)^2 \)
Calculating \( (56)^2 \):
\[ (56)^2 = 3136 \]

### Step 9: Substitute and simplify
Now substitute back into the area formula:
\[ A = \frac{1}{2} \times \frac{22}{7} \times 3136 \]
Calculating this step-by-step:
\[ A = \frac{22 \times 3136}{14} \]
Calculating \( 22 \times 3136 = 69092 \):
\[ A = \frac{69092}{14} = 4928 \text{ m}^2 \]

### Final Answer
The area of the semi-circular park is:
\[ \boxed{4928 \text{ m}^2} \]