# 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.

Question:

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. The diameters of the two circular ends of the bucket are 45 cm and 25 cm, the total vertical height of the bucket is 40 cm and that of the cylindrical base is 6 cm. Find the area of the metallic sheet used to make the bucket. Also, find the volume of water the bucket can hold, in litres.

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

$\approx 33.61 \mathrm{~L}$