Question

cylinder and a cone have same base and same height. The ratio their volume is:-​

Answer 1

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

Given that,

• A cylinder and a cone have same base and same height.

Let assume that

• Radius of base of cone and cylinder = r units

• Height of cone and cylinder = h units

Now,

We know,

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

and

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

So,

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

$\rm \: = \: \pi \: {r}^{2} \: h \: : \: \dfrac{1}{3} \: \pi \: {r}^{2} \: h \:$

$\rm \: = \: 1 \: : \: \dfrac{1}{3} \: \:$

$\rm \: = \: 3 \: : \: 1 \: \:$

Hence,

$\rm\implies \:\boxed{ \rm{ \:Volume_{(Cylinder)} : Volume_{(Cone)} = 3 : 1 \: }}$

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

Additional Information :-

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