Question

if the radius of a cone is half of the radius of cylinder and its height is double to that of cylinder .show that its volume is 1 by 6th Volume of Cylinder​

Answer 1

Radius of cone (r) = Radius of cylinder(R)/2

height of cone(h) = 2 ×height of cylinder ( H)

Volume of cone = 1/3 pie ( R/2)^2 ( 2H)

volume of cylinder = pie R^2 H


volume of cone/volume of cylinder = 1/3 × 1/4 × 2

= 1/6

Answer 2

$\sf\huge{\underline{\mathbb{Answer}}}$

$\sf\bold{\boxed{\underline{Question}}}$

$<b><h>If the radius of a cone is half of the radius of cylinder and its height is double to that of cylinder. Show that its volume is [tex]\frac{1}{6}$th Volume of Cylinder..[/tex]

$\sf\huge{\underline{\mathfrak{Solution}}}$

Given:

▶radius of cone is $\frac{1}{2}$ of radius of cylinder.i. e, r =$\frac{1}{2}$ r'.

where r is radius of cone, r' is radius of cylinder.

▶height of cone is double of cylinder, i.e, h = 2h', where h= height of cone, h' is height of cylinder.

we have!.

volume of cone = $\frac{1}{3}πr^{2}h$==(1)

volume of cylinder = $πr'^{2}h'$==(2)

substituting the value of r and h in (1)

volume of cone = $\frac{1}{3}πr^{2}h$==(1)

V(cone) =$\frac{1}{3}π\frac{r'^{2}}{2^{2}}2h'$

V(cone) = $\frac{πr'^{2}2h'}{3\times 4}$

V(cone) = $\frac{πr'^{2}h'}{3\times 2}$

V(cone) = $\frac{πr'^{2}h'}{6}$

V(cone) = $\frac{1}{6}(πr'^{2}h')$

▶ from equation 2, we have..

$πr'^{2}h'$ = volume of cylinder..

hence..

V(cone)= $\frac{1}{6}$ V(cylinder).......(from equation 2)..

$\sf\mathscr{HENCE\: THE\: PROOF}$