An open metal bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet.

Solution:

We have,

Radius of the upper end of the frustum, $R=\frac{45}{2} \mathrm{~cm}$,

Radius of the lower end of the frustum = Radius of the cylinder $=r=\frac{25}{2} \mathrm{~cm}$,

Height of the cylinder, $h=6 \mathrm{~cm}$ and

Total height of the bucket $=40 \mathrm{~cm}$

And, the height of the frustum, $H=40-6=34 \mathrm{~cm}$

Also, the slant height of frustum, $l=\sqrt{(R-r)^{2}+H^{2}}$

$=\sqrt{\left(\frac{45}{2}-\frac{25}{2}\right)^{2}+34^{2}}$

$=\sqrt{10^{2}+34^{2}}$

$=\sqrt{100+1156}$

$=\sqrt{1256}$

$\approx 35.44 \mathrm{~cm}$

Now,

The area of the metallic sheet used to make the bucket = CSA of the frustum $+$ CSA of the cylinder $+$ Area of the base

of the cylinder

$=\pi(\mathrm{R}+\mathrm{r}) \mathrm{l}+2 \pi \mathrm{rh}+\pi \mathrm{r}^{2}$

$=\frac{22}{7} \times\left(\frac{45}{2}+\frac{25}{2}\right) \times 35.44+2 \times \frac{22}{7} \times \frac{25}{2} \times 6+\frac{22}{7} \times \frac{25}{2} \times \frac{25}{2}$

$=\frac{22}{7} \times 35 \times 35.44+\frac{22}{7} \times 150+\frac{22}{7} \times \frac{625}{4}$

$=\frac{22}{7} \times(1240.4+150+156.25)$

$=\frac{22}{7} \times 1546.65$

$=4860.9 \mathrm{~cm}^{2}$

Also,

The volume of the water that the bucket can hold = Volume of the frustum

$=\frac{1}{3} \pi H\left(R^{2}+r^{2}+R r\right)$

$=\frac{1}{3} \times \frac{22}{7} \times 34 \times\left[\left(\frac{45}{2}\right)^{2}+\left(\frac{25}{2}\right)^{2}+\left(\frac{45}{2}\right)\left(\frac{25}{2}\right)\right]$

$=\frac{1}{3} \times \frac{22}{7} \times 34 \times\left[\left(\frac{45}{2}\right)^{2}+\left(\frac{25}{2}\right)^{2}+\left(\frac{45}{2}\right)\left(\frac{25}{2}\right)\right]$

$=\frac{748}{21} \times\left(\frac{2025}{4}+\frac{625}{4}+\frac{1125}{4}\right)$

$=\frac{748}{21} \times \frac{3775}{4}$

$=33615.48 \mathrm{~cm}^{3}$

$=33.61548 \mathrm{~L} \quad\left(\right.$ As, $\left.1000 \mathrm{~cm}^{3}=1 \mathrm{~L}\right)$

$\approx 33.61 \mathrm{~L}$