Elitmus
Exam
Numerical Ability
Geometry
A right angled cone has its height 'h'. The cone is being cut parallel to its base at h/5 distance from its vertex. What will be the ratio of the cone and the cutting portion
Option: 1:24, 1:125, 1:25, not remember
Read Solution (Total 5)
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- volume of smaller cone / (volume of bigger cone - volume of smaller cone)
also by similarity r/R = h/H
=> r/R=1/5
Substituting and solving we get 1/124 ans :) - 9 years agoHelpfull: Yes(15) No(4)
- let radius of smaller cone=R
radius of bigger cone=r
height of smaller cone=h/5
height of bigger cone=h
volume of smaller cone=1/3*pi*R^2*h
volume of reamaining portion=volume of bigger cone - volume of smaller cone
i.e 1/3*pi*r^2h-1/3*pi*R^2(h/5)
from the similarity of triangle
r/R=5:1
equating value
we get 1:124
- 9 years agoHelpfull: Yes(7) No(1)
- the formula of cone is 1/3 * pi* sqr(r) * h
as the new cone's height will be h/5 so the volume will be 1/5( 1/3 * pi* sqr(r) * h)
=1/5(original cone)
so ans is 1:5 - 9 years agoHelpfull: Yes(3) No(9)
- 1:124 is answer . we use triangle similarity rule by which r/R=h/5/h=h/5h
now ratio of volume of smaller to cutiing portion (1/3*pi*(h/5)^2*h/5) / (1/3*pi(h*h^2-(h/5)^2*h/5)
by solving this we get 1/124 - 9 years agoHelpfull: Yes(3) No(1)
- volume of cone=3.14*r*r*h/3, here after cutting the hight =h/5, height of frustm=h-h/5=4h/5, ratio of volume=1:125
- 9 years agoHelpfull: Yes(1) No(4)
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