Question

the volume of a right circular cone, whose radius of the base is half of its allitute, and the volume of a hemisphere are equal. The ratio of the radius of the cone to the hemisphere is?

Answer 1

Dear Student:

  Volume of cone(V) =$\frac{1}{3}\pi r^{2}h$

Where,   r is the radius of cone

              h is the altitude

                           r=h/2

                           h=2r

                           V=$\frac{1}{3}\pi r^{2}*2r$

                           V=$\frac{2}{3} \pi r^{3}$

Volume of hemisphere(U)= $\frac{2}{3} \pi a^{3}$

Where,a is the radius of hemisphere.

                            V=U

$r^{3}= a^{3}$

Hence,       r=a

So,  r/a=1

HOPE IT HELPS

THANKS

WITH REGARDS

Answer 2

Hi ,

i ) Dimensions of the right circular

cone :

radius = r units

altitude ( h ) = 2r units

Volume of

the cone = (π×radius² ×altitude)/3

V1 = ( πr² × 2r )/3

V1 = (2πr³ )/3 ---( 1 )

ii ) Let the radius of the Sphere = R units

Volume of the Sphere = (2/3 ) πR³ --( 2 )

according to the problem given ,

( 1 ) = ( 2 )

( 2πr³ )/3 = ( 2/3 )πR³

=> r³/R³ = ( 2/3 )/( 2/3 )

=> ( r/R )³ = 1/1

r : R = 1 : 1

I hope this helps you.

: )