Question

if two right circular cylinder of equal volume have their radius in the ratio1:3, the ratio of there height is​

Answer 1

Given :

Volume of two right circular cylinders is equal.

Radius in the ratio = 1:3

To find : Ratio of the heights

Let the radius of cylinder one be R and radius of the cylinder two be r.

Assuming, the height of cylinder one to be H and cylinder two to be h,

Volume of cylinder = πr²h

Let the radius be 1x and 3x respectively. By forming the equation,

πR²H = πr²h

π × (1x)² × H = π × (3x)² × h

π × x² × H = π × 9x² × h

by transposing π × 9x² to the LHS (Left Hand Side) and H to th RHS (Right Hand Side),

$\dfrac{\pi \times {x}^{2} }{\pi \times {9x}^{2} } = \dfrac{h}{H}$

$\dfrac{1}{9} = \dfrac{h}{H}$

(by cancelling π and x² as it's common)

Hence,

H : h = 9 : 1

Answer 2

Question :

if two right circular cylinder of equal volume have their radius in the ratio1:3, the ratio of there height is-

Given:

Volume of two right circular cylinders is equal.

Radius in the ratio = 1:3

To find:

Ratio of the heights.

Solution:

Let the radius of one cylinder be= R

And radius of the other cylinder be = r

Let height of one cylinder be = H

and other be = h

☆Volume of cylinder = πr²h.

Let the radius be 1x and 3x respectively.

By forming the equation= πR²H = πr²h.

solving~~

π × (1x)² × H = π × (3x)² × h

π × x² × H = π × 9x² × h

by transposing π × 9x² to the LHS (Left Hand Side) and H to th RHS (Right Hand Side).

$\frac{\pi \times x {}^{2} }{\pi \times 9 {}^{2} } = \frac{h}{H}$

$\frac{1}{9} = \frac{h}{H}$

➪Here,π and x² will be cancel out as it's common.

$\bold{\huge{\fbox{\color{maroon}{H:h=9:1}}}}$

$\sf\red{hope\:it\:helps\:you}$