Question

The radii of the base of a cylinder and a cone are in the ratio 3 : 4 and their heights are in the ratio 2 : 3. What is the ratio of their volumes?

Answer 1

Answer:

The ratio of their volumes is  9 : 8.

Step-by-step explanation:

SOLUTION :

Given:

Let the radius of the base of a cylinder be R & radius of the base of a cone be r.

Let the Height of the base of a cylinder be H & Height of the base of a cone be h

Ratio of radius of the base of a cylinder  & radius of the base of a cone , R : r = 3 : 4 i.e R/r = 3/4

Ratio of Height of the base of a cylinder  & Height of the base of a cone  , H : h = 2 : 3 i.e H/h = 2/3

Volume of Cylinder,V1 / Volume of cone , V2 =  πR²H / ⅓  πr²h

V1/V2 = R²H/ ⅓ r²h

= 3 × (R/r)² × (H/h)

= 3 ×(¾)² × ⅔

= 3 × 9/16 × ⅔

V1/V2= 9/8  

V1 : V2 = 9 : 8

Hence, the ratio of their volumes is  9 : 8.

HOPE THIS ANSWER WILL HELP YOU….

Answer 2

$\huge {\textbf {\underline {\underline {ANSWER}}}}$

Let radius of cylinder be r1 and radius of cone be r2.

Similarly let height of cylinder be h1 and height of cone be h2.

$\frac{r1}{r2} = \frac{3}{4} \\ \\ \\ \frac{h1}{h2} = \frac{2}{3} \\ \\ \\ volume \: of \: cylinder \: = \pi {(r1)}^{2} h1 \\ \\ \\ volume\: of \: cone \: = \frac{1}{3} \pi {(r2)}^{2} h2 \\ \\ \\ \frac{v1}{v2} = \frac{\pi {(r1)}^{2}h1 }{ \frac{1}{3}\pi {(r2)}^{2} h2 } \\ \\ \\ = 3 \times { (\frac{3}{4} )}^{2} \times \frac{2}{3} \\ \\ \\ = \frac{9}{16} \times 2 \\ \\ \\ = \frac{9}{8}$

So the ratio of volumes of cylinder and cone is 9:8