Question

An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm,
diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, then find the area of the tin sheet required to make the funnel.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{Height}\mathrm{of}\mathrm{the}\mathrm{cylindrical}\mathrm{portion},h=10\mathrm{cm}, \\ \mathrm{Height}\mathrm{of}\mathrm{the}\mathrm{frustum}\mathrm{of}\mathrm{cone}\mathrm{portion},H=22-10=12\mathrm{cm}, \\ \mathrm{Radius}\mathrm{of}\mathrm{the}\mathrm{cylindical}\mathrm{portion}=\mathrm{Radius}\mathrm{of}\mathrm{smaller}\mathrm{end}\mathrm{of}\mathrm{frustum}\mathrm{portion},r=\frac{8}{2}=4\mathrm{cm}\mathrm{and} \\ \mathrm{Radius}\mathrm{of}\mathrm{larger}\mathrm{end}\mathrm{of}\mathrm{frustum}\mathrm{portion},R=\frac{18}{2}=9\mathrm{cm} \\ \mathrm{Also},\mathrm{the}\mathrm{slant}\mathrm{height}\mathrm{of}\mathrm{the}\mathrm{frustum},l=\sqrt{{\left(R-r\right)}^2+H^2} \\ =\sqrt{{\left(9-4\right)}^2+{12}^2} \\ =\sqrt{5^2+{12}^2} \\ =\sqrt{25+144} \\ =\sqrt{169} \\ =13\mathrm{cm} \\ \mathrm{Now}, \\ \mathrm{The}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{tin}\mathrm{sheet}\mathrm{required}=\mathrm{CSA}\mathrm{of}\mathrm{frustum}\mathrm{of}\mathrm{cone}+\mathrm{CSA}\mathrm{of}\mathrm{cylinder} \\ =\mathrm{\pi }\left(R+r\right)l+2\mathrm{\pi }rh \\ =\frac{22}{7}\times \left(9+4\right)\times 13+2\times \frac{22}{7}\times 4\times 10 \\ =\frac{22}{7}\times 13\times 13+\frac{22}{7}\times 80 \\ =\frac{22}{7}\times \left(169+80\right) \\ =\frac{22}{7}\times 249 \\ \approx 782.57{\mathrm{cm}}^2 \end{gathered}$$

So, the area of the tin sheet required to make the funnel is 782.57 cm².