Question
You are given the great circle of a sphere is a length of 25 miles. What is the volume of the sphere?
Question 8 options:
268 mi³
896 mi³
103 mi³
424 mi³
Question 8 options:
268 mi³
896 mi³
103 mi³
424 mi³
Answers
GPT-4o mini
To find the volume of the sphere given the length of its great circle, we first need to determine the radius of the sphere from the circumference of the great circle.
The formula for the circumference \(C\) of a sphere is given by:
\[
C = 2 \pi r
\]
where \(r\) is the radius of the sphere. We know that the circumference (great circle length) is 25 miles:
\[
25 = 2 \pi r
\]
To solve for \(r\), we rearrange the equation:
\[
r = \frac{25}{2\pi}
\]
Now, we will calculate \(r\):
\[
r \approx \frac{25}{6.2832} \approx 3.9789 \text{ miles}
\]
Next, we can find the volume \(V\) of the sphere using the formula:
\[
V = \frac{4}{3} \pi r^3
\]
First, we need to calculate \(r^3\):
\[
r^3 \approx (3.9789)^3 \approx 62.78 \text{ miles}^3
\]
Now we can find the volume:
\[
V \approx \frac{4}{3} \pi (62.78) \approx \frac{4}{3} \cdot 3.1416 \cdot 62.78
\]
\[
V \approx \frac{4}{3} \cdot 197.887 \approx 263.84 \text{ mi}^3
\]
Finally, we compute:
\[
V \approx 263.84 \text{ mi}^3
\]
Since the exact value is not among the options provided, let's double-check calculations and rounding or errors.
Calculating \(r\) and then using its approximate integer value could yield closer results, or minute differences in values might lead to round-off relevant to finding the final options.
Now going through the options given:
1. **268 mi³**
2. **896 mi³**
3. **103 mi³**
4. **424 mi³**
From these calculations, the closest option that makes sense relative to our approximation would be **268 mi³**, which is a close estimate from our calculations and approximation, therefore we choose this option.
Thus, the answer is:
**268 mi³**
The formula for the circumference \(C\) of a sphere is given by:
\[
C = 2 \pi r
\]
where \(r\) is the radius of the sphere. We know that the circumference (great circle length) is 25 miles:
\[
25 = 2 \pi r
\]
To solve for \(r\), we rearrange the equation:
\[
r = \frac{25}{2\pi}
\]
Now, we will calculate \(r\):
\[
r \approx \frac{25}{6.2832} \approx 3.9789 \text{ miles}
\]
Next, we can find the volume \(V\) of the sphere using the formula:
\[
V = \frac{4}{3} \pi r^3
\]
First, we need to calculate \(r^3\):
\[
r^3 \approx (3.9789)^3 \approx 62.78 \text{ miles}^3
\]
Now we can find the volume:
\[
V \approx \frac{4}{3} \pi (62.78) \approx \frac{4}{3} \cdot 3.1416 \cdot 62.78
\]
\[
V \approx \frac{4}{3} \cdot 197.887 \approx 263.84 \text{ mi}^3
\]
Finally, we compute:
\[
V \approx 263.84 \text{ mi}^3
\]
Since the exact value is not among the options provided, let's double-check calculations and rounding or errors.
Calculating \(r\) and then using its approximate integer value could yield closer results, or minute differences in values might lead to round-off relevant to finding the final options.
Now going through the options given:
1. **268 mi³**
2. **896 mi³**
3. **103 mi³**
4. **424 mi³**
From these calculations, the closest option that makes sense relative to our approximation would be **268 mi³**, which is a close estimate from our calculations and approximation, therefore we choose this option.
Thus, the answer is:
**268 mi³**