A bucket made up of a metal sheet is in the form of a frustum of a cone of height 16 cm

Solution:

Greater radius of the frustum = R = 20 cm
Smaller radius of the frustum = r = 8 cm
Height of the frustum = h = 16 cm

Slant height, l, of the frustum

$=\sqrt{h^{2}+\left(R^{2}-r^{2}\right)}$

$=\sqrt{16^{2}+(20-8)^{2}}$

$=\sqrt{256+(12)^{2}}$

$=\sqrt{256+144}$

 

$=\sqrt{400}=20 \mathrm{~cm}$

Surface area of the frustum

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

$=\pi\left[r^{2}+l(R+r)\right]$

$=\frac{22}{7}\left[8^{2}+20(20+8)\right]$

$=\frac{22}{7}[64+20(28)]$

 

$=\frac{22 \times 624}{7}=1961.14 \mathrm{~cm}^{2}$

100 cm2 of metal sheet costs Rs 15.
So, total cost of the metal sheet used for fabrication

$=\frac{1961.14}{100} \times 15$

 

$=\operatorname{Rs} 294.171$