Question
Roderick wants to draw a circle for which the numerical value of the circumference is greater than the value of the area.
Which lengths could he use for the radius? Select three options.
One-half
1
1.5
2
2 and one-half
Which lengths could he use for the radius? Select three options.
One-half
1
1.5
2
2 and one-half
Answers
GPT-4o mini
To determine which radii will yield a circumference greater than the area of a circle, we can use the following formulas:
- Circumference \(C\) of a circle:
\[
C = 2\pi r
\]
- Area \(A\) of a circle:
\[
A = \pi r^2
\]
We need to find values of \(r\) such that:
\[
C > A \quad \text{or} \quad 2\pi r > \pi r^2
\]
We can simplify this inequality by dividing both sides by \(\pi\) (noting that \(\pi > 0\)):
\[
2r > r^2
\]
Rearranging gives us:
\[
r^2 - 2r < 0
\]
Factoring:
\[
r(r - 2) < 0
\]
This inequality holds true when \(0 < r < 2\).
Now we can evaluate each of the proposed radii:
1. **One-half \( (0.5) \)**: This is in the range \(0 < r < 2\).
2. **1**: This is also in the range \(0 < r < 2\).
3. **1.5**: This is in the range \(0 < r < 2\).
4. **2**: This is \(r = 2\), which does not satisfy \(r < 2\).
5. **Two and one-half \( (2.5) \)**: This is above the upper limit of \(r < 2\).
Thus, the radii Roderick can choose for which the circumference is greater than the area are:
- One-half
- 1
- 1.5
So the three options are: **One-half, 1, and 1.5**.
- Circumference \(C\) of a circle:
\[
C = 2\pi r
\]
- Area \(A\) of a circle:
\[
A = \pi r^2
\]
We need to find values of \(r\) such that:
\[
C > A \quad \text{or} \quad 2\pi r > \pi r^2
\]
We can simplify this inequality by dividing both sides by \(\pi\) (noting that \(\pi > 0\)):
\[
2r > r^2
\]
Rearranging gives us:
\[
r^2 - 2r < 0
\]
Factoring:
\[
r(r - 2) < 0
\]
This inequality holds true when \(0 < r < 2\).
Now we can evaluate each of the proposed radii:
1. **One-half \( (0.5) \)**: This is in the range \(0 < r < 2\).
2. **1**: This is also in the range \(0 < r < 2\).
3. **1.5**: This is in the range \(0 < r < 2\).
4. **2**: This is \(r = 2\), which does not satisfy \(r < 2\).
5. **Two and one-half \( (2.5) \)**: This is above the upper limit of \(r < 2\).
Thus, the radii Roderick can choose for which the circumference is greater than the area are:
- One-half
- 1
- 1.5
So the three options are: **One-half, 1, and 1.5**.