# A bucket is in the form of a frustum of a cone with

Question:

A bucket is in the form of a frustum of a cone with a capacity of $12308.8 \mathrm{~cm}^{3}$ of water. The radii of the top and bottom circular ends are $20 \mathrm{~cm}$ and $12 \mathrm{~cm}$ respectively. Find the height of the bucket and the area of the metal sheet used in its making. (Use $\pi=3.14$ ).

Solution:

Let the depth of the bucket is cm. The radii of the top and bottom circles of the frustum bucket are r1 =20cm and r2 =12cm respectively.

The volume/capacity of the bucket is

$V=\frac{1}{3} \pi\left(r_{1}^{2}+r_{1} r_{2}+r_{2}^{2}\right) \times h$

$=\frac{1}{3} \pi\left(20^{2}+20 \times 12+12^{2}\right) \times h$

$=\frac{1}{3} \times \frac{22}{7} \times 784 \times h$

$=\frac{1}{3} \times 22 \times 112 \times h \mathrm{~cm}^{3}$

Given that the capacity of the bucket is $12308.8$ Cubic $\mathrm{cm}$. Thus, we have

$\frac{1}{3} \times 22 \times 112 \times h=12308.8$

$\Rightarrow h=\frac{12308.8 \times 3}{22 \times 112}$

$\Rightarrow h=15$

Hence, the height of the bucket is $15 \mathrm{~cm}$

The slant height of the bucket is

$l=\sqrt{\left(r_{1}-r_{2}\right)^{2}+h^{2}}$

$=\sqrt{(20-12)^{2}+15^{2}}$

$=\sqrt{289}$

$=17 \mathrm{~cm}$

The surface area of the used metal sheet to make the bucket is

$S_{1}=\pi\left(r_{1}+r_{2}\right) \times l+\pi r_{2}^{2}$

$=\pi \times(20+12) \times 17+\pi \times 12^{2}$

$=\pi \times 32 \times 17+144 \pi$

$=2160.32 \mathrm{~cm}^{2}$

Hence Surface area of the metal $=2160.32 \mathrm{~cm}^{2}$