Question
A right circular cone and a right circular cylinder have equal base and equal height. If the radius of the base and height are in the ratio 5 : 12, write the ratio of the total surface area of the cylinder to that of the cone.
✓
Answer
Let r = 5x and h = 12x be the base radius and height of the cone and cylinder respectively.
Slant height of the cone
$\text{l}=\sqrt{\text{r}^2+\text{h}^2}$
$=\sqrt{(5\text{x})^2+(12\text{x}^2)^2}$
$=\sqrt{25\text{x}^2+144\text{x}^2}$
$\Rightarrow\text{l}=13\text{x}$
The total surface area of cylinder
$\text{S}_1=2\pi\text{r}(\text{h}+\text{r})$
The total surface area of cone
$\text{s}_2=2\pi\text{r}(\text{l}+\text{r})$
Now,
$\Rightarrow\frac{\text{S}_1}{\text{S}_2}=\frac{2\pi\text{r}(\text{h}+\text{r})}{\pi\text{r}(\text{l}+\text{r})}$
$=\frac{2(\text{h}+\text{r})}{(\text{l}+\text{r})}$
$\Rightarrow\frac{\text{S}_1}{\text{S}_2}=\frac{2(12\text{x}+5\text{x})}{13\text{x}+5\text{x}}$
$\Rightarrow\frac{\text{S}_1}{\text{S}_2}=\frac{2\times17\text{x}}{18\text{x}}$
$\Rightarrow\text{S}_1:\text{S}_2=17:9$
Hence, the required ratio are 17 : 9
Slant height of the cone
$\text{l}=\sqrt{\text{r}^2+\text{h}^2}$
$=\sqrt{(5\text{x})^2+(12\text{x}^2)^2}$
$=\sqrt{25\text{x}^2+144\text{x}^2}$
$\Rightarrow\text{l}=13\text{x}$
The total surface area of cylinder
$\text{S}_1=2\pi\text{r}(\text{h}+\text{r})$
The total surface area of cone
$\text{s}_2=2\pi\text{r}(\text{l}+\text{r})$
Now,
$\Rightarrow\frac{\text{S}_1}{\text{S}_2}=\frac{2\pi\text{r}(\text{h}+\text{r})}{\pi\text{r}(\text{l}+\text{r})}$
$=\frac{2(\text{h}+\text{r})}{(\text{l}+\text{r})}$
$\Rightarrow\frac{\text{S}_1}{\text{S}_2}=\frac{2(12\text{x}+5\text{x})}{13\text{x}+5\text{x}}$
$\Rightarrow\frac{\text{S}_1}{\text{S}_2}=\frac{2\times17\text{x}}{18\text{x}}$
$\Rightarrow\text{S}_1:\text{S}_2=17:9$
Hence, the required ratio are 17 : 9
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