To find the sample space of the area in which the cup fell, we need to calculate the area of the square playground and then subtract the area of the semicircular sandpit.
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Area of the square playground: \[ \text{Area} = \text{side}^2 = 22 \text{ ft} \times 22 \text{ ft} = 484 \text{ ft}^2 \]
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Area of the semicircular sandpit: The radius \( r \) of the semicircle is half the diameter: \[ r = \frac{14 \text{ ft}}{2} = 7 \text{ ft} \] The area of a semicircle is given by: \[ \text{Area} = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi (7 \text{ ft})^2 = \frac{1}{2} \pi (49 \text{ ft}^2) = \frac{49\pi}{2} \text{ ft}^2 \approx 76.96 \text{ ft}^2 \]
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Approximate area of the semicircle using \(\pi \approx 3.14\): \[ \text{Area} \approx \frac{49 \times 3.14}{2} \approx \frac{153.86}{2} \approx 76.93 \text{ ft}^2 \]
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Area of the playground minus the area of the sandpit: \[ \text{Available area} = \text{Area of the square} - \text{Area of the semicircle} \] \[ \text{Available area} = 484 \text{ ft}^2 - 76.93 \text{ ft}^2 \approx 407.07 \text{ ft}^2 \]
So, the area in which the cup can fall (the sample space) would be approximately \(407.07 \text{ ft}^2\). Rounding to two decimal places, the best choice from the options presented is:
\[ \boxed{407.03 \text{ ft}^2} \]