A circle with a radius of 2 inches and a square with a side length of 3 inches are both inscribed in a rectangle with a width of 7 inches and a length of 8 inches. The circle and square are beside each other in the rectangle. Determine the probability that a point placed randomly in the figure is inside the circle. Round your answer to the nearest hundredth.(1 point)

Responses

0.84
0.84

0.16
0.16

0.22
0.22

0.78

1 answer

To find the probability that a point placed randomly in the rectangle is inside the circle, we need to calculate the area of the circle and the area of the rectangle, and then divide the area of the circle by the area of the rectangle.

  1. Calculate the area of the circle: The formula for the area of a circle is given by: \[ A = \pi r^2 \] where \( r \) is the radius. In this case, the radius of the circle is 2 inches. \[ A_{\text{circle}} = \pi (2^2) = \pi (4) = 4\pi \approx 12.57 \text{ square inches (using } \pi \approx 3.14\text{)} \]

  2. Calculate the area of the rectangle: The area of the rectangle is given by: \[ A = \text{length} \times \text{width} \] Here, the length is 8 inches and the width is 7 inches. \[ A_{\text{rectangle}} = 8 \times 7 = 56 \text{ square inches} \]

  3. Calculate the probability: The probability \( P \) that a randomly placed point in the rectangle is inside the circle is given by: \[ P = \frac{A_{\text{circle}}}{A_{\text{rectangle}}} = \frac{4\pi}{56} \] Substitute \( \pi \approx 3.14 \): \[ P \approx \frac{4 \times 3.14}{56} \approx \frac{12.56}{56} \approx 0.224 \]

  4. Round the result: Rounding \( 0.224 \) to the nearest hundredth, we get: \[ P \approx 0.22 \]

The probability that a point placed randomly in the figure is inside the circle is thus \( \boxed{0.22} \).