Question

If the right circular cone is separated into 3 solids of volume v1,v2,v3 by 2 planes parallel to the base and trisect the altitudes. Then v1:v2:v3 is :

Answer 1

Answer:

1:7:19

Step-by-step explanation:

Volume of cone,OPQ is π*r²*h/3

Since cone is trisected on altitude , OM = MN = NR = h/3, where h is the height of the cone OPQ

Let, v₁ volume of OAB

v₂ be volume of ABCD

v₃ be volume of CDQP.

It is clear that triangle OAB, OCD and OPQ are similar hence AM:CN:PR=1:2:3.

v₁ = 1/3*π*AM²*(OM) = 1/3*π*(r²/9)*(h/3)

v₂ = Volume of OCD - v₁

Volume of OCD = 1/3*π*CN²*(ON) = 1/3*π*(4r²/9)*(2h/3)

On simplification we get,

v₂=1/3*π*(7/27)*r²h

Now, v₃ = Volume of OPQ - Volume of OCD

=1/3*π*(r²)*(h)-1/3*π*(4r²/9)*(2h/3)

=1/3*π*(19/27)*(r²)*(h)

Thus, v₁:v₂:v₃ = 1:7:19.

=

Answer 2

Answer:

1:7:19

Step-by-step explanation:

for video solution search #underoottssc on youtube

have a great day (: