To find the volume of a cylindrical bucket, we need to use the formula for the volume of a cylinder, which is:
[tex]\[ V = \pi r^2 h \][/tex]
where:
- [tex]\( V \)[/tex] is the volume,
- [tex]\( \pi \)[/tex] (pi) is a constant approximated as [tex]\(\frac{22}{7}\)[/tex],
- [tex]\( r \)[/tex] is the radius of the base,
- [tex]\( h \)[/tex] is the height of the cylinder.
Given the height ([tex]\( h \)[/tex]) is 35 cm and the diameter of the base is 21 cm, we first need to find the radius of the base. The radius ([tex]\( r \)[/tex]) is half of the diameter:
[tex]\[ r = \frac{\text{diameter}}{2} = \frac{21 \, \text{cm}}{2} = 10.5 \, \text{cm} \][/tex]
Next, we substitute the values into the volume formula:
[tex]\[ V = \left( \frac{22}{7} \right) \times (10.5 \, \text{cm})^2 \times 35 \, \text{cm} \][/tex]
Let's calculate [tex]\( (10.5 \, \text{cm})^2 \)[/tex]:
[tex]\[ (10.5 \, \text{cm})^2 = 110.25 \, \text{cm}^2 \][/tex]
Now, substitute back into the volume formula:
[tex]\[ V = \left( \frac{22}{7} \right) \times 110.25 \, \text{cm}^2 \times 35 \, \text{cm} \][/tex]
To simplify the multiplication:
[tex]\[ \left( \frac{22}{7} \right) \times 110.25 = 22 \times \frac{110.25}{7} \][/tex]
[tex]\[ \frac{110.25}{7} = 15.75 \][/tex]
[tex]\[ 22 \times 15.75 = 346.5 \][/tex]
Thus:
[tex]\[ V = 346.5 \times 35 = 12127.5 \, \text{cm}^3 \][/tex]
Finally, after rounding to the nearest tenth:
The volume of the cylindrical bucket is [tex]\(\boxed{12127.5} \, \text{cm}^3\)[/tex].
