Question

If the volume of 2 cones are in ratio 1:4 and their diameters are in ratio 4:5 then find the ratios of their heights​

Answer 1

Given:

• Volume of 2 cones are in ratio 1:4

$\sf{\dfrac{Voume\:of\:first\:cone}{Voume\:of\:second\:cone}=\dfrac{1}{4}}$

• Diameters of cones are in ratio 4:5

$\sf{\dfrac{Diameter\:of\:first\:cone}{Diameter\:of\:second\:cone}=\dfrac{4}{5}}$

To Find:

• Ratio of heights of the cones

Solution:

We know that,

$\sf{Volume\:of\:cone=\dfrac{1}{3}\pi r^{2}h}$

where,

• r is radius of cone

• h is height of cone

Also,

Diameter(D)=2×Radius(r)

Let the radius of first and second cone be r₁ and r₂ respectively.

Also, let the height of first and second cone be h₁ and h₂ respectively.  

Now,

$\sf{\dfrac{Voume\:of\:first\:cone}{Voume\:of\:second\:cone}=\dfrac{\dfrac{1}{3}\pi r_{1}^{2}h_{1}}{\dfrac{1}{3}\pi r_{2}^{2}h_{2}}}$

$\longrightarrow\sf{\dfrac{\dfrac{1}{3}\pi r_{1}^{2}h_{1}}{\dfrac{1}{3}\pi r_{2}^{2}h_{2}}=\dfrac{1}{4}}$  

On cancelling 1/3 and π from numerator and denominator, we get

$\longrightarrow\sf{\dfrac{r_{1}^{2}h_{1}}{r_{2}^{2}h_{2}}=\dfrac{1}{4}}$

$\longrightarrow\sf{\bigg(\dfrac{r_{1}^{2}}{r_{2}^{2}}\bigg)\times\dfrac{h_{1}}{h_{2}} =\dfrac{1}{4}}$

$\longrightarrow\sf{\bigg(\dfrac{r_{1}}{r_{2}}\bigg)^{2}\times\dfrac{h_{1}}{h_{2}} =\dfrac{1}{4}}$

On multiplying and dividing 2 in the fraction of r₁/r₂, we get

$\longrightarrow\sf{\bigg(\dfrac{2r_{1}}{2r_{2}}\bigg)^{2}\times\dfrac{h_{1}}{h_{2}} =\dfrac{1}{4}}$

$\longrightarrow\sf{\bigg(\dfrac{D_{1}}{D_{2}}\bigg)^{2}\times\dfrac{h_{1}}{h_{2}} =\dfrac{1}{4}}$

where, D₁ and D₂ are diameters of first and second cones respectively

$\longrightarrow\sf{\bigg(\dfrac{4}{5}\bigg)^{2}\times\dfrac{h_{1}}{h_{2}} =\dfrac{1}{4}}$

$\longrightarrow\sf{\dfrac{16}{25}\times\dfrac{h_{1}}{h_{2}} =\dfrac{1}{4}}$

$\longrightarrow\sf{\dfrac{h_{1}}{h_{2}} =\dfrac{1}{4}\times\dfrac{16}{25}}$

$\longrightarrow\sf\pink{\dfrac{h_{1}}{h_{2}} =\dfrac{4}{25}}$

Hence, the required ratio is $\sf\green{\dfrac{4}{25}}$.