Question

A cylinder and a cone have equal radii of their bases and equal heights. If their curved surface areas are in the ratio 8 : 5., show that the radius and height of each has the ratio 3 : 4.

Answer 1

Suppose that the respective radii and height of the cone and the cylinder are r and h.
Then ratio of curved surface areas = 8 : 5
Let the curved surfaces areas be 8x and 5x.

$$\begin{gathered} \mathrm{i}.\mathrm{e}.,2\mathrm{\pi }rh=8\mathrm{x}\mathrm{and}\mathrm{\pi }rl=5\mathrm{x}\Rightarrow \mathrm{\pi }r\sqrt{h^{\mathit{2}}\mathit{+}{r}^2}=5x \\ \mathrm{Hence}4{\mathrm{\pi }}^{\mathit{2}}{r}^{\mathit{2}}{h}^2=64x^2\mathrm{and}{\mathrm{\pi }}^2{r}^2(h^{\mathit{2}}\mathit{+}{r}^{\mathit{2}}\mathit{)}=25x^2 \\ \therefore \mathrm{Ratio}\mathrm{of}\mathrm{curved}\mathrm{surface}\mathrm{areas}=\frac{4{\mathrm{\pi }}^2{r}^{\mathit{2}}{h}^2}{{\mathrm{\pi }}^{\mathit{2}}{r}^2(h^{\mathit{2}}\mathit{+}{r}^{\mathit{2}})}=\frac{64}{25} \\ \Rightarrow \frac{4{\mathrm{\pi }}^2{r}^{\mathit{2}}{h}^2}{{\mathrm{\pi }}^{\mathit{2}}{r}^2(h^{\mathit{2}}\mathit{+}{r}^{\mathit{2}})}=\frac{64}{25} \\ \Rightarrow \frac{4h^2}{(h^2+r^2)}=\frac{64}{25} \\ \Rightarrow 25h^2=16h^2+16r^2 \\ \Rightarrow 9h^2=16r^2 \\ \Rightarrow \frac{r^2}{h^2}=\frac{9}{16} \\ \Rightarrow \frac{r}{h}=\frac{3}{4} \end{gathered}$$

∴ The ratio of the radius and height of the cone is 3 : 4.