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You are building a sand castle and want to use a bucket that holds a volume of 606 in.cubed and has height 10.4 in. What is the...Question
You are building a sand castle and want to use a bucket that holds a volume of 734 in.cubed and has height 11.5 in. What is the radius of the bucket? Use 3.14 for pi. Use pencil and paper. If the height of the bucket is changed, but the volume stays the same, then how will the radius change? Explain.
11.5 in.
Question content area bottom
Part 1
The radius of the bucket is about
enter your response here in. (Round to the nearest tenth as needed.)
11.5 in.
Question content area bottom
Part 1
The radius of the bucket is about
enter your response here in. (Round to the nearest tenth as needed.)
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To find the radius of the bucket, we can use the formula for the volume of a cylinder, which is given by:
\[
V = \pi r^2 h
\]
where:
- \( V \) is the volume,
- \( r \) is the radius,
- \( h \) is the height, and
- \( \pi \) is approximately 3.14 in this case.
You have the following values:
- \( V = 734 \, \text{in}^3 \)
- \( h = 11.5 \, \text{in} \)
We can rearrange the formula to solve for the radius \( r \):
\[
r^2 = \frac{V}{\pi h}
\]
Now substituting the given values:
\[
r^2 = \frac{734}{3.14 \times 11.5}
\]
Calculating the denominator:
\[
3.14 \times 11.5 = 36.11
\]
Now substituting back into the equation:
\[
r^2 = \frac{734}{36.11} \approx 20.34
\]
Now, take the square root to find \( r \):
\[
r \approx \sqrt{20.34} \approx 4.51 \, \text{in}
\]
Rounding to the nearest tenth:
\[
r \approx 4.5 \, \text{in}
\]
So, the radius of the bucket is about **4.5 in**.
### Part 2: How does changing the height affect the radius?
If the height of the bucket is changed but the volume stays the same, the radius will change according to the volume formula.
From the formula \( V = \pi r^2 h \), if \( V \) is constant and \( h \) changes, then \( r^2 \) must change in such a way that \( r^2 \) is inversely proportional to \( h \). This means:
\[
r^2 \propto \frac{1}{h}
\]
- If the height \( h \) increases, the radius \( r \) must decrease in order to keep the volume \( V \) the same.
- If the height \( h \) decreases, the radius \( r \) must increase.
In summary, \( r \) and \( h \) are inversely related when the volume remains constant.
\[
V = \pi r^2 h
\]
where:
- \( V \) is the volume,
- \( r \) is the radius,
- \( h \) is the height, and
- \( \pi \) is approximately 3.14 in this case.
You have the following values:
- \( V = 734 \, \text{in}^3 \)
- \( h = 11.5 \, \text{in} \)
We can rearrange the formula to solve for the radius \( r \):
\[
r^2 = \frac{V}{\pi h}
\]
Now substituting the given values:
\[
r^2 = \frac{734}{3.14 \times 11.5}
\]
Calculating the denominator:
\[
3.14 \times 11.5 = 36.11
\]
Now substituting back into the equation:
\[
r^2 = \frac{734}{36.11} \approx 20.34
\]
Now, take the square root to find \( r \):
\[
r \approx \sqrt{20.34} \approx 4.51 \, \text{in}
\]
Rounding to the nearest tenth:
\[
r \approx 4.5 \, \text{in}
\]
So, the radius of the bucket is about **4.5 in**.
### Part 2: How does changing the height affect the radius?
If the height of the bucket is changed but the volume stays the same, the radius will change according to the volume formula.
From the formula \( V = \pi r^2 h \), if \( V \) is constant and \( h \) changes, then \( r^2 \) must change in such a way that \( r^2 \) is inversely proportional to \( h \). This means:
\[
r^2 \propto \frac{1}{h}
\]
- If the height \( h \) increases, the radius \( r \) must decrease in order to keep the volume \( V \) the same.
- If the height \( h \) decreases, the radius \( r \) must increase.
In summary, \( r \) and \( h \) are inversely related when the volume remains constant.
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