The park is rectangular with dimensions 60 m by 40 m.
Area of the park = Length × Width
Area_{park} = $60 \text{ m} \times 40 \text{ m} = 2400 \text{ m}^2$
The pond is semicircular with a diameter of 40 m.
The radius (r) is half the diameter:
r = $\frac{40 \text{ m}}{2} = 20 \text{ m}$
The area of a full circle is $\pi r^2$. The area of a semicircle is $\frac{1}{2} \pi r^2$.
Area_{pond} = $\frac{1}{2} \times \pi \times (20 \text{ m})^2$
Area_{pond} = $\frac{1}{2} \times \pi \times 400 \text{ m}^2$
Area_{pond} = $200 \pi \text{ m}^2$
Using $\pi \approx 3.14159$:
Area_{pond} $\approx 200 \times 3.14159 \text{ m}^2 \approx 628.318 \text{ m}^2$
To find the percentage of the park's area the pond occupies, divide the pond's area by the park's area and multiply by 100.
Percentage = $(\frac{\text{Area}_{pond}}{\text{Area}_{park}}) \times 100$
Percentage $\approx (\frac{628.318 \text{ m}^2}{2400 \text{ m}^2}) \times 100$
Percentage $\approx 0.261799 \times 100 \approx 26.18\%$
The semicircular pond occupies approximately 26.18% of the park's area.
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