Question

The ratio of volume of two cones is
4:5 and the ratio of the radii of their
base is 2:3 then ratio of their vertical
heights is
​

Answer 1

Given :–

• The ratio of volume of two cones = 4 : 5.
• The ratio of the radio of their base = 2 : 3.

To Find :–

• The ratio of their vertical height.

Solution :–

For first cone,

• Let the radii be $\sf r_{1}$, vertically height be $\sf h_{1}$ and the volume be $\sf v_{1}$

For second cone,

• Let the radii be $\sf r_{2}$, vertically height be $\sf h_{2}$ and the volume be $\sf v_{2}$

According to the question,

$\implies \dfrac{v_{1} }{ v_{2} } = \dfrac{4}{5}$

And,

$\implies \dfrac{r_{1} }{ r_{2} } = \dfrac{2}{3}$

We know that,

$\gray{ \boxed{ \red{volume \: of \: a \: cone = \dfrac{1}{3} \pi {r}^{2} h}}}$

So,

$\implies\frac{ \cancel\dfrac{1}{3} \cancel\pi { r_{1}}^{2}h }{ \cancel\dfrac{1}{3} \cancel\pi { r_{2} }^{2}h } = \dfrac{4}{5}$

$\implies \dfrac{ { (\dfrac{2}{3} )}^{2} h_{1}}{{ (\dfrac{2}{3} )}^{2} h_{2}} = \dfrac{4}{5}$

$\implies {(\dfrac{r_{1}}{r_{2}})^{2}} \times \dfrac{h_{1}}{h_{2}}$

$\implies { (\dfrac{2}{3}) }^{2} \times \dfrac{ h_{1} }{ h_{2}} = \dfrac{4}{5}$

$\implies \dfrac{ h_{1} }{ h_{2} } = \dfrac{4}{5} \times { (\dfrac{3}{2}) }^{2}$

$\implies \dfrac{ h_{1}}{ h_{2} } = \dfrac{ \cancel4}{5} \times \dfrac{9}{ \cancel4}$

$\implies\dfrac{ h_{1}}{ h_{2} } = \dfrac{1 \times 9}{5 \times 1}$

$\implies\dfrac{ h_{1}}{ h_{2} } = \dfrac{9}{5}$

$\implies h_{1} \ratio h_{2} = 9 \ratio5$

Hence,

The ratio of their vertically height is 9 : 5.