If the perimeter of a semicircular park is 360\text{ m}, then what is its area? (Take \pi=\frac{22}{7})
- A. 3850\text{ m}^2
- B. 7700\text{ m}^2 ✓
- C. 11550\text{ m}^2
- D. 15400\text{ m}^2
Correct Answer: B. 7700\text{ m}^2
Explanation
Perimeter of a semicircle = \pi r + 2r = r(\frac{22}{7} + 2) = r(\frac{36}{7}). Given r(\frac{36}{7}) = 360 \implies r = 70\text{ m}. Area = \frac{1}{2}\pi r^2 = \frac{1}{2} \times \frac{22}{7} \times 70^2 = 7700\text{ m}^2.
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