Question
Sheldon randomly chooses a point in the square below, what is the probability that point is not in the circle? Assume that pi = 3.14 and round to the nearest tenth of a percent.
A circle inscribed in a square. All four sides of the circle touch the edges of the square. The radius of the circle is 3 units. The side of the square is 6 units.
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A circle inscribed in a square. All four sides of the circle touch the edges of the square. The radius of the circle is 3 units. The side of the square is 6 units.
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Answers
GPT-4o mini
To determine the probability that a randomly chosen point in the square is not within the inscribed circle, we first need to calculate the areas of both the square and the circle.
1. **Area of the square**:
The side length of the square is 6 units. Therefore, the area \( A_{square} \) is given by:
\[
A_{square} = \text{side}^2 = 6^2 = 36 \text{ square units}
\]
2. **Area of the circle**:
The radius of the circle is 3 units. The area \( A_{circle} \) is given by the formula:
\[
A_{circle} = \pi r^2
\]
Plugging in \( r = 3 \) and \( \pi = 3.14\):
\[
A_{circle} = 3.14 \times (3^2) = 3.14 \times 9 = 28.26 \text{ square units}
\]
3. **Area outside the circle**:
To find the area that is not in the circle, we subtract the area of the circle from the area of the square:
\[
A_{outside} = A_{square} - A_{circle} = 36 - 28.26 = 7.74 \text{ square units}
\]
4. **Probability that a point is not in the circle**:
The probability \( P \) that a randomly chosen point in the square is not in the circle is the ratio of the area outside the circle to the area of the square:
\[
P = \frac{A_{outside}}{A_{square}} = \frac{7.74}{36}
\]
5. **Calculating the probability**:
To find the decimal value:
\[
P \approx 0.215
\]
6. **Converting to percentage**:
Convert to percentage by multiplying by 100:
\[
P \times 100 \approx 21.5\%
\]
Rounding to the nearest tenth of a percent, we find that the probability that the point is not in the circle is:
\[
\boxed{21.5\%}
\]
1. **Area of the square**:
The side length of the square is 6 units. Therefore, the area \( A_{square} \) is given by:
\[
A_{square} = \text{side}^2 = 6^2 = 36 \text{ square units}
\]
2. **Area of the circle**:
The radius of the circle is 3 units. The area \( A_{circle} \) is given by the formula:
\[
A_{circle} = \pi r^2
\]
Plugging in \( r = 3 \) and \( \pi = 3.14\):
\[
A_{circle} = 3.14 \times (3^2) = 3.14 \times 9 = 28.26 \text{ square units}
\]
3. **Area outside the circle**:
To find the area that is not in the circle, we subtract the area of the circle from the area of the square:
\[
A_{outside} = A_{square} - A_{circle} = 36 - 28.26 = 7.74 \text{ square units}
\]
4. **Probability that a point is not in the circle**:
The probability \( P \) that a randomly chosen point in the square is not in the circle is the ratio of the area outside the circle to the area of the square:
\[
P = \frac{A_{outside}}{A_{square}} = \frac{7.74}{36}
\]
5. **Calculating the probability**:
To find the decimal value:
\[
P \approx 0.215
\]
6. **Converting to percentage**:
Convert to percentage by multiplying by 100:
\[
P \times 100 \approx 21.5\%
\]
Rounding to the nearest tenth of a percent, we find that the probability that the point is not in the circle is:
\[
\boxed{21.5\%}
\]