Question
To find the volume of a Cylinder, we would use the formula: V= pir^2h
Farmer Ted bales large round bales of hay to store in his barn for winter feeding. How much hay is in a round bale, in terms of pi? Height is 4ft and diameter is 5ft.
40pi ft^3
10pi ft^3
20pi ft^3
80pi ft^3
Farmer Ted bales large round bales of hay to store in his barn for winter feeding. How much hay is in a round bale, in terms of pi? Height is 4ft and diameter is 5ft.
40pi ft^3
10pi ft^3
20pi ft^3
80pi ft^3
Answers
Answered by
GPT-4o mini
To find the volume of the round bale of hay modeled as a cylinder, we can use the formula:
\[
V = \pi r^2 h
\]
Where:
- \( V \) is the volume,
- \( r \) is the radius,
- \( h \) is the height.
Given that the diameter of the bale is 5 feet, we find the radius:
\[
r = \frac{diameter}{2} = \frac{5 \text{ ft}}{2} = 2.5 \text{ ft}
\]
The height \( h \) is given as 4 ft.
Now we can substitute \( r \) and \( h \) into the volume formula:
\[
V = \pi (2.5)^2 (4)
\]
Calculating \( (2.5)^2 \):
\[
(2.5)^2 = 6.25
\]
Now substituting this back into the volume equation:
\[
V = \pi (6.25) (4)
\]
Calculating \( 6.25 \times 4 \):
\[
6.25 \times 4 = 25
\]
Thus, the volume is:
\[
V = 25\pi \text{ ft}^3
\]
None of the options provided (40π ft³, 10π ft³, 20π ft³, 80π ft³) is correct based on this calculation.
Based on our calculations, the volume of the round bale is \( 25\pi \text{ ft}^3 \).
\[
V = \pi r^2 h
\]
Where:
- \( V \) is the volume,
- \( r \) is the radius,
- \( h \) is the height.
Given that the diameter of the bale is 5 feet, we find the radius:
\[
r = \frac{diameter}{2} = \frac{5 \text{ ft}}{2} = 2.5 \text{ ft}
\]
The height \( h \) is given as 4 ft.
Now we can substitute \( r \) and \( h \) into the volume formula:
\[
V = \pi (2.5)^2 (4)
\]
Calculating \( (2.5)^2 \):
\[
(2.5)^2 = 6.25
\]
Now substituting this back into the volume equation:
\[
V = \pi (6.25) (4)
\]
Calculating \( 6.25 \times 4 \):
\[
6.25 \times 4 = 25
\]
Thus, the volume is:
\[
V = 25\pi \text{ ft}^3
\]
None of the options provided (40π ft³, 10π ft³, 20π ft³, 80π ft³) is correct based on this calculation.
Based on our calculations, the volume of the round bale is \( 25\pi \text{ ft}^3 \).
There are no human answers yet. A form for humans to post answers is coming very soon!