Question

A container, open at the top and made up of metal sheet, is in the form of a frustum of a cone of height 16 cm with diameters of its lower and upper ends as 16 cm and 40 cm, respectively. Find the cost of metal sheet used to make the container, if it costs ₹10 per 100 cm².
[CBSE 2013]

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{Height},h=16\mathrm{cm}, \\ \mathrm{Radius}\mathrm{of}\mathrm{the}\mathrm{upper}\mathrm{end},R=\frac{40}{2}=20\mathrm{cm}\mathrm{and} \\ \mathrm{Radius}\mathrm{of}\mathrm{the}\mathrm{lower}\mathrm{end},r=\frac{16}{2}=8\mathrm{cm} \\ \mathrm{Also},\mathrm{the}\mathrm{slant}\mathrm{height},l=\sqrt{{\left(R-r\right)}^2+h^2} \\ =\sqrt{{\left(20-8\right)}^2+{16}^2} \\ =\sqrt{{12}^2+{16}^2} \\ =\sqrt{144+256} \\ =\sqrt{400} \\ =20\mathrm{cm} \\ \mathrm{Now}, \\ \mathrm{Total}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{container}=\mathrm{\pi }\left(R+r\right)l+\mathrm{\pi }{r}^2 \\ =\frac{22}{7}\times \left(20+8\right)\times 20+\frac{22}{7}\times 8\times 8 \\ =\frac{22}{7}\times 28\times 20+\frac{22}{7}\times 64 \\ =\frac{22}{7}\times 560+\frac{22}{7}\times 64 \\ =\frac{22}{7}\times \left(560+64\right) \\ =\frac{22}{7}\times 624 \\ =\frac{13728}{7}{\mathrm{cm}}^2 \\ \mathrm{So},\mathrm{the}\mathrm{cost}\mathrm{of}\mathrm{metal}\mathrm{sheet}\mathrm{used}=\frac{13728}{7}\times \frac{10}{100} \\ =\frac{13728}{70} \\ \approx ₹196.11 \end{gathered}$$

Hence, the cost of metal sheet used to make the container is ₹196.11.

Disclaimer: The answer given in the textbook is incorrect. The same has been corrected above.