If a cone is cut into two parts by a horizontal plane passing through the mid-point of its axis, then the ratio of their volumes is:-
a) 1:2
b) 2:1
c) 1:7
d) 1:8
Please explain the answer also.
Answers
use similarly concept .
let r is radius of above part and R is the radius of base of cone .
R/r=h/(h/2)=2/1
R=2r
now
volume of cone =πR^2h
=π(2r)^2h=4πr^2h
volume of small cone =πr^2h/2
now,
volume of below part =4πr^2h-πr^2h/2
=7/2 πr^2h
now ratio = volume of above part/volume of below part
=(πr^2h/2)/(7/2πr^2h)=1/7
hence ratio =1:7
option (c) is correct
Answer:
Let the height of the cone be H and base radius be R units.
Then its volume V=
3
1
πR²H cu. units
As the horizontal plane cuts the cone into two parts through the mid point of its axis, the height of the cone is divided into two equal parts, forming a top cone of height
2
H
units. If the base radius of the top cone is r units, then
2
H
r
=
H
R
[Since the vertical angle of both are same].
Solving, r=
2
R
units.
Hence volume of the small cone at the top is:
V=
3
1
π(
2
R
)
2
(
2
H
)=
24
πR
2
H
cu. units
So volume of the bottom part (frustum) is:
V−v=
3
πR
2
H
−
24
πR
2
H
=
24
7
πR
2
H cu. units
Therefore, ratio of their volumes, Top part : Bottom part that is:
24
πR
2
H
:
24
7
πR
2
H=1:7
Step-by-step explanation: