Question

A right circular cone is divided by a plane parallel to its base in two equal volumes. In what ratio will the plane divide the axis of the cone?​

Answer 1

Answer:

let the radius of the base of the cone (db)=R

height of the base of the cone (ad)=H

slant height of the base of the cone (ab)=L

again, let the radius of the base of the cone (ge)=r

height of the base of the cone(ag)=h

slant height of the base of the cone (ae)= l

now from the figure,

height of the remaining part of the cone (gd)= H - h

given that the right circular cone is divided by a plane parallel to its base in two equal volumes.

so that the volume of the small cone is half of the volume of the whole cone.

=> πr2h/3=(1/2)πR2H/3

=> r2 h= R2 H/2

=> r2 h/R2 H= 1/2

=> r2 /R2 = H/2h............1

again we know that EF parallel to BC

So , ∆AEG- ∆ABD

=> GE/DB = AG/AD

=> r/R =h/H

=> (r/R)2 = (h/H ) 2

=> r2/R2 = H2/H2...........2

from equation 1

=> H /2h = h 2 / H2

=> H/2h = (h/H)2

=> (1/2)(h/H)=(h/H)2

=> 1/2= (h/H) 3

=> h/H= (1/2)1/3

=> h/H= 1/21/3

=> H= h*21/3s

=> Now h/(H - h)= h{h*(21/3-h)}

=> h/ (H - h) = h{ h*( 21/3-1)}

=> h/(H - h)= 1/(21/3-1)

=> h:(H - h) = 1:(21/3-1)

so, the ratio of the line segment into which the axis of the cone is divided by the plane is 1:(21/3-1) .....

hope it helps u aakash............

Answer 2

Step-by-step explanation:

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