Question

The radii of the bases of a cylinder and a cone are in the ratio 3:5 and their heights are in the ratio 3 4. What is the ratio of their volumes?​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

• The radii of the bases of a cylinder and a cone are in the ratio 3:5.

Let assume that

• Radius of cylinder = 3r

• Radius of cone = 5r

Further, given that

• The height of a cylinder and a cone are in the ratio 3:4

Let assume that

• Height of cylinder = 3h

• Height of cone = 4h

So, Volume of cylinder of radius 3r and height 4h is given by

$\rm \: Volume_{(Cylinder)} = \pi \: {(3r)}^{2}(3h)$

$\rm\implies \:Volume_{(Cylinder)} = 27 \: \pi \: {r}^{2} \: h$

Now, Volume of cone of radius 5r and height 4h is given by

$\rm \: Volume_{(Cone)} =\dfrac{1}{3} \pi \: {(5r)}^{2}(4h)$

$\rm\implies \:Volume_{(Cone)} = \dfrac{100}{3} \: \pi \: {r}^{2} \: h$

Now,

$\rm \: Volume_{(Cylinder)} : Volume_{(Cone)}$

$\rm \: = \: 27 \: \pi \: {r}^{2} \: h \: : \:\dfrac{100}{3} \: \pi \: {r}^{2} \: h \:$

$\rm \: = \: 27 \: \: : \:\dfrac{100}{3} \: \:$

$\rm \: = \: 81 \: \: : \:100 \: \:$

Hence,

$\boxed{ \rm{ \:\rm \: Volume_{(Cylinder)} : Volume_{(Cone)} = \: 81: \:100 \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{CSA_{(cylinder)} = 2\pi \: rh}\\ \\ \bigstar \: \bf{Volume_{(cylinder)} = \pi {r}^{2} h}\\ \\ \bigstar \: \bf{TSA_{(cylinder)} = 2\pi \: r(r + h)}\\ \\ \bigstar \: \bf{CSA_{(cone)} = \pi \: r \: l}\\ \\ \bigstar \: \bf{TSA_{(cone)} = \pi \: r \: (l + r)}\\ \\ \bigstar \: \bf{Volume_{(sphere)} = \dfrac{4}{3}\pi {r}^{3} }\\ \\ \bigstar \: \bf{Volume_{(cube)} = {(side)}^{3} }\\ \\ \bigstar \: \bf{CSA_{(cube)} = 4 {(side)}^{2} }\\ \\ \bigstar \: \bf{TSA_{(cube)} = 6 {(side)}^{2} }\\ \\ \bigstar \: \bf{Volume_{(cuboid)} = lbh}\\ \\ \bigstar \: \bf{CSA_{(cuboid)} = 2(l + b)h}\\ \\ \bigstar \: \bf{TSA_{(cuboid)} = 2(lb +bh+hl )}\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

$\small\colorbox{lightyellow} {\text{ \bf♕ Brainliest answer }}$

$\rule{300pt}{0.1pt}$

$\mathbb\red{ \tiny A \scriptsize \: N \small \:S \large \: W \Large \:E \huge \: R}$

$\rule{300pt}{0.1pt}$

Let the ratio of radii be x.

$\Rightarrow$ radii of the bases are 3x and 5x and let ratio of heights be y.

heights of the cylinder and cone will be 3y and 4y .

Ratio of the volumes

$\[ \begin{array}{l} \rm=\dfrac{\text { volume of cylinder }}{\text { volume of cone }} \\ \\ \\ \rm =\dfrac{\pi \times(3 x)^{2} \times 3 y}{\dfrac{1}{3} \pi \times(5 x)^{2} \times 4 y} \\ \\ \\ \\ \rm=\dfrac{3 \times 9 \times 3}{25 \times 4} \\ \\ \\ \rm=\dfrac{81}{100} \\ \\ \\ \rm=81: 100 \end{array} \]$