Question

A right triangle whose sides are 15 cm and 20 cm (other than hypotenuse), is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of $\mathrm{\pi }$ as found appropriate) [HOTS]

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{In}∆\mathrm{ABC},\angle \mathrm{B}=90°,\mathrm{AB}=l_1=15\mathrm{cm}\mathrm{and}\mathrm{BC}=l_2=20\mathrm{cm} \\ \mathrm{Let}\mathrm{OD}=\mathrm{OB}=r,\mathrm{AO}=h_1\mathrm{and}\mathrm{CO}=h_2 \\ \mathrm{Using}\mathrm{Pythagoras}\mathrm{theorem}, \\ \mathrm{AC}=\sqrt{{\mathrm{AB}}^2+{\mathrm{BC}}^2} \\ =\sqrt{{15}^2+{20}^2} \\ =\sqrt{225+400} \\ =\sqrt{625} \\ \Rightarrow h=25\mathrm{cm} \\ \mathrm{As},\mathrm{ar}\left(∆\mathrm{ABC}\right)=\frac{1}{2}\times \mathrm{AC}\times \mathrm{BO}=\frac{1}{2}\times \mathrm{AB}\times \mathrm{BC} \\ \Rightarrow \mathrm{AC}\times \mathrm{BO}=\mathrm{AB}\times \mathrm{BC} \\ \Rightarrow 25r=15\times 20 \\ \Rightarrow r=\frac{15\times 20}{25} \\ \Rightarrow r=12\mathrm{cm} \\ \mathrm{Now}, \\ \mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{double}\mathrm{cone}\mathrm{so}\mathrm{formed}=\mathrm{Volume}\mathrm{of}\mathrm{cone}1+\mathrm{Volume}\mathrm{of}\mathrm{cone}2 \\ =\frac{1}{3}\mathrm{\pi }{r}^2{h}_1+\frac{1}{3}\mathrm{\pi }{r}^2{h}_2 \\ =\frac{1}{3}\mathrm{\pi }{r}^2\left(h_1+h_2\right) \\ =\frac{1}{3}\mathrm{\pi }{r}^{2}h \\ =\frac{1}{3}\times 3.14\times 12\times 12\times 25 \\ =3768{\mathrm{cm}}^3 \\ \mathrm{Also}, \\ \mathrm{Surace}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{solid}\mathrm{so}\mathrm{formed}=\mathrm{CAS}\mathrm{of}\mathrm{cone}1+\mathrm{CSA}\mathrm{of}\mathrm{cone}2 \\ =\mathrm{\pi }r{l}_1+\mathrm{\pi }r{l}_2 \\ =\mathrm{\pi r}\left(l_1+l_2\right) \\ =\frac{22}{7}\times 12\times \left(15+20\right) \\ =\frac{22}{7}\times 12\times 35 \\ =1320{\mathrm{cm}}^2 \end{gathered}$$