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 of each is to the height of each as $3 : 4$.

 Given:

A cylinder and a cone have equal radii of their bases and equal heights. 

Their curved surface areas are in the ratio $8 : 5$.

To do:

We have to show that the radius of each is to the height of each as $3 : 4$.

Solution:

Let the radius of the cylinder be $r$ and the radius of the cone be $r$.

Let the height of the cylinder be $h$ and the height of the cone be $h$.

This implies,

Slant height of the cone $(l)=\sqrt{r^{2}+h^{2}}$

Curved surface of the cone $=\pi r l$

$=\pi r \sqrt{r^{2}+h^{2}}$

Curved surface area of the cylinder $=2 \pi r h$

Therefore,

Ratio of the curved surface areas of cone and cylinder $=8: 5$

$\frac{2 \pi r h}{\pi r(\sqrt{r^{2}+h^{2}})}=\frac{8}{5}$

$\frac{2 h}{\sqrt{r^{2}+h^{2}}}=\frac{8}{5}$

Squaring both sides, we get,

$\frac{4 h^{2}}{r^{2}+h^{2}}=\frac{64}{25}$

$100 h^{2}=64 r^{2}+64 h^{2}$

$100 h^{2}-64 h^{2}=64 r^{2}$

$36 h^{2}=64 r^{2}$

$\frac{r^{2}}{h^{2}}=\frac{36}{64}$

$\Rightarrow \frac{r}{h}=\frac{6}{8}$

$\frac{r}{h}=\frac{3}{4}$

The radius of each to the height of each is $3 : 4$.

Hence proved.

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Updated on: 10-Oct-2022

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