Consider a rectangular park that measures $80ft\times 90ft$ Within this park, there's a circular playground area with a diameter of 50 ft. Near the

playground, there is a semicircular bench area with a radius of 25 ft.
A family loses their car keys somewhere randomly in the park.What is the probability that the keys are not on the playground?
(1 point)
0.27
0.73
0.59
0.86

1 answer

To find the probability that the car keys are not on the playground, we first need to determine the area of the park and the area of the playground.

  1. Calculate the area of the park: The park is rectangular with dimensions 80 ft × 90 ft. Therefore, the area of the park \( A_{\text{park}} \) is: \[ A_{\text{park}} = 80 \text{ ft} \times 90 \text{ ft} = 7200 \text{ ft}^2 \]

  2. Calculate the area of the playground: The playground is circular with a diameter of 50 ft. The radius \( r \) of the playground is half the diameter: \[ r = \frac{50 \text{ ft}}{2} = 25 \text{ ft} \] The area of the playground \( A_{\text{playground}} \) is given by the formula for the area of a circle: \[ A_{\text{playground}} = \pi r^2 = \pi (25 \text{ ft})^2 = 625\pi \text{ ft}^2 \]

  3. Calculate the area of the park not covered by the playground: The area of the region outside the playground within the park is: \[ A_{\text{not playground}} = A_{\text{park}} - A_{\text{playground}} = 7200 \text{ ft}^2 - 625\pi \text{ ft}^2 \]

  4. Calculate the probability that the keys are not on the playground: The probability \( P \) that the keys are not on the playground is the ratio of the area outside the playground to the area of the park: \[ P(\text{not on playground}) = \frac{A_{\text{not playground}}}{A_{\text{park}}} = \frac{7200 \text{ ft}^2 - 625\pi \text{ ft}^2}{7200 \text{ ft}^2} \]

To perform the calculation, we can approximate \( \pi \) as 3.14: \[ 625\pi \approx 625 \times 3.14 \approx 1963.75 \text{ ft}^2 \] So, \[ A_{\text{not playground}} \approx 7200 \text{ ft}^2 - 1963.75 \text{ ft}^2 \approx 5236.25 \text{ ft}^2 \] And now, \[ P(\text{not on playground}) \approx \frac{5236.25 \text{ ft}^2}{7200 \text{ ft}^2} \approx 0.7267 \] This gives us approximately \( 0.73 \).

Thus, the probability that the keys are not on the playground is approximately \( \boxed{0.73} \).