Question

If the heights of two cylinders are in the ratio of 4:8 and their radii are in the ratio of 3:4 then what is the ratio of the volume??​

Answer 1

Step-by-step explanation:

Given:-

• The ratio of the heights of two cylinders is 4 : 8.

• The radii are in the ratio 3 : 4.

To Find:-

• The ratio of the volumes of the cylinders.

Solution:-

The ratio of the heights of two cylinders is 4 : 8.

$\sf Let\: the \: height\: of\: the\: first\: cylinder\: (h_{1}) = 4h$

$\sf Height\: of\: the\: second \: cylinder\: (h_{2}) = 8h$

The radii are in the ratio 3 : 4.

$\sf Let\: the \: radius \: of\: the\: first\: cylinder\: (r_{1}) = 3r$

$\sf Radius \: of\: the\: second \: cylinder\: (r_{2}) = 4r$

As we know that:-

$\boxed{\sf Volume \: of \: cylinder = \pi r^2 h}$

$\sf \implies \dfrac{V_{1}}{V_{2}} = \dfrac{ \pi {r_{1}}^{2} h_{1}}{\pi {r_{2}}^{2} h_{2}}$

$\sf \implies \dfrac{V_{1}}{V_{2}} = \dfrac{ {r_{1}}^{2} h_{1}}{ {r_{2}}^{2} h_{2}}$

$\sf \implies \dfrac{V_{1}}{V_{2}} = \dfrac{ {(3r)}^{2} }{ {(4r)}^{2} } \times \dfrac{4h}{8h}$

$\sf \implies \dfrac{V_{1}}{V_{2}} = \dfrac{ {9r}^{2} }{ {16r}^{2} } \times \dfrac{1}{2}$

$\sf \implies \dfrac{V_{1}}{V_{2}} = \dfrac{ 9 }{ 32 }$

$\sf \implies \: V_{1} : V_{2} = 9 : 32$

$\large\underline{ \boxed{ \therefore\textsf{ \textbf{Ratio \: of \: the \: volumes = 9 : 32}}}}$