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.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{Radius}\mathrm{of}\mathrm{the}\mathrm{upper}\mathrm{end}\mathrm{of}\mathrm{the}\mathrm{frustum},R=\frac{45}{2}\mathrm{cm}, \\ \mathrm{Radius}\mathrm{of}\mathrm{the}\mathrm{lower}\mathrm{end}\mathrm{of}\mathrm{the}\mathrm{frustum}=\mathrm{Radius}\mathrm{of}\mathrm{the}\mathrm{cylinder}=r=\frac{25}{2}\mathrm{cm}, \\ \mathrm{Height}\mathrm{of}\mathrm{the}\mathrm{cylinder},h=6\mathrm{cm}\mathrm{and} \\ \mathrm{Total}\mathrm{height}\mathrm{of}\mathrm{the}\mathrm{bucket}=40\mathrm{cm} \\ \mathrm{And},\mathrm{the}\mathrm{height}\mathrm{of}\mathrm{the}\mathrm{frustum},H=40-6=34\mathrm{cm} \\ \mathrm{Also},\mathrm{the}\mathrm{slant}\mathrm{height}\mathrm{of}\mathrm{frustum},l=\sqrt{{\left(R-r\right)}^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} \\ \mathrm{Now}, \\ \mathrm{The}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{metallic}\mathrm{sheet}\mathrm{used}\mathrm{to}\mathrm{make}\mathrm{the}\mathrm{bucket}=\mathrm{CSA}\mathrm{of}\mathrm{the}\mathrm{frustum}+\mathrm{CSA}\mathrm{of}\mathrm{the}\mathrm{cylinder}+\mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{base}\mathrm{of}\mathrm{the}\mathrm{cylinder} \\ =\mathrm{\pi }\left(\mathrm{R}+\mathrm{r}\right)\mathrm{l}+2\mathrm{\pi rh}+{\mathrm{\pi 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 \left(1240.4+150+156.25\right) \\ =\frac{22}{7}\times 1546.65 \\ =4860.9{\mathrm{cm}}^2 \\ \mathrm{Also}, \\ \mathrm{The}\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{water}\mathrm{that}\mathrm{the}\mathrm{bucket}\mathrm{can}\mathrm{hold}=\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{frustum} \\ =\frac{1}{3}\mathrm{\pi }H\left(R^2+r^2+Rr\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}\left(\mathrm{As},1000{\mathrm{cm}}^3=1\mathrm{L}\right) \\ \approx 33.61\mathrm{L} \end{gathered}$$