Question

The ratio of volumes of two cones is 4.5and the ratio of their bases is 2.3 find the ratio of their vertical heights

Answer 1

Answer:-

Given:

Ratio of volumes of two cones = 4 : 5

Ratio of their bases = 2 : 3

Let their base radius be 2y , 3y.

We have to find the ratio of their heights.

So,

Let their heights be h1 and h2.

We know that,

Volume of a cone = 1/3 * πr²h

Hence,

Volume of 1st cone/Volume of 2nd cone = 4/5

→ (1/3 * π(2y)² * h1) / (1/3 * π(3y)² * h2) = 4/5

(1/3 , π are cancelled in LHS)

→ 4y² * h1 / 9y² * h2 = 4/5

→ (h1 / h2) * (4y² / 9y²) = 4/5

→ h1 / h2 = 4/5 * 9y²/4y²

→ h1 / h2 = 9 / 5

→ h1 : h2 = 9 : 5

Hence, their heights are in the ratio 9 : 5.

Answer 2

Step-by-step explanation:

$\bf { \underline{ \underline \red{Given:-}}}$

• The ratio of volumes of two cones = 4 : 5

• Ratio of the bases of the cones = 2 : 3

$\bf { \underline{ \underline \blue{To\:Find:-}}}$

• The ratio of the vertical heights of the two cones.

$\bf { \underline{ \underline \green{Solution:-}}}$

Let the common ratio of the radius be ' r '

$\rm \: Let \: the \: common \: ratio \: of \: the \: radius \: be \: ( r )$

$\rm \: Radius \: of \: the \: 1st \: cone \: (r_{1})= \: 4r</p><p>$

$\rm \: Radius \: of \: the \: 2nd \: cone \: ( r_{2}) = \: 5r$

Similarly

$\rm \: Let \: the \: common \: ratio \: of \: the \: volume \: be \: ( v)</p><p>$

$\rm \: Volume \: of \: the \: 1st \: cone (V_{1}) = 2v$

$\rm \: Volume \: of \: 2nd \: cone \: (V_{2}) = 3v$

As we know that:-

$\boxed{ \rm \: Volume \: of \: cone = \frac{1}{3} \pi {r}^{2} h}$

$\rm \: \dfrac{V _{1} }{ V_{2} } = \dfrac{ \frac{1}{3} \pi { r_{1}}^{2} h_{1} }{\frac{1}{3} \pi { r_{2}}^{2} h_{2}}$

$\rm \: \dfrac{4v}{5v} = \dfrac{ h_{1} {(2r)}^{2} }{ h_{2}( {3r)}^{2} }$

$\rm \longrightarrow\dfrac{4}{5} = \dfrac{ h_{1}}{ h_{2} } \times \dfrac{4{r}^{2} }{9 {r}^{2} }$

$\rm \longrightarrow\dfrac{4}{5} = \dfrac{ h_{1}}{ h_{2} } \times \dfrac{4}{9 }$

$\rm \longrightarrow\dfrac{4}{5} \times \dfrac{9}{4} = \dfrac{ h_{1}}{ h_{2} }$

$\rm \longrightarrow\dfrac{36}{20} = \dfrac{ h_{1}}{ h_{2} }$

$\rm \longrightarrow\dfrac{9}{5} = \dfrac{ h_{1}}{ h_{2} }$

$\rm \therefore \: \: h_{1} : h_{2} = 9 : 5$

Therefore:-

$\large\boxed{ \sf \: Ratio \: of \: the \: vertical \: heights = 9 :5 }$