A cone of radius 8 cm and height 12 cm is divided into two parts by a plane through the mid-point of its axis parallel to its base. Find the ratio of the volumes of two parts.
NCERT Class X
Mathematics - Exemplar Problems
Chapter _SURFACE AREAS AND VOLUMES
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radius of cone=8cms
height=12cms
since the cone is divided through the mid point which is parellel to its base..
so by mid point theorem we have...2*radius of upper cone=radius of bigger cone or lower radius of frustum {by mid point theorem}
so take radius of smaller cone x and that of lower radius of frustum of radius of bigger cone = 2x
and height of upper cone=height of lower frustum {or 1/2 height of biggest cone}
so their ratio==
(1/3*pi*r1^2*h ) / (1/3*pi*h(r1^2 + r2^2 + r1*r2) {where r1 is radius of upper cone and r2 is of biggest cone or lower radius of frusum)
x^2/x^2 + (2x)^2 + x*2x) (cancelling pi,h,1/3) (as proved that the radius of smaller cone x and radius of bigger one 2x)
x^2/7x^2
2/7
answer=2/7
height=12cms
since the cone is divided through the mid point which is parellel to its base..
so by mid point theorem we have...2*radius of upper cone=radius of bigger cone or lower radius of frustum {by mid point theorem}
so take radius of smaller cone x and that of lower radius of frustum of radius of bigger cone = 2x
and height of upper cone=height of lower frustum {or 1/2 height of biggest cone}
so their ratio==
(1/3*pi*r1^2*h ) / (1/3*pi*h(r1^2 + r2^2 + r1*r2) {where r1 is radius of upper cone and r2 is of biggest cone or lower radius of frusum)
x^2/x^2 + (2x)^2 + x*2x) (cancelling pi,h,1/3) (as proved that the radius of smaller cone x and radius of bigger one 2x)
x^2/7x^2
2/7
answer=2/7
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