If h,c and v be the height, curved surface and volume of a cone, show that 3πvh3−c2h2+9v2=0.
Let r and l denote respectively the radius of the base and slant height of the cone.
l=√r2+h2
V=13πr2h
C=πrl
Therefore,
3πVh3−C2h2+9V2
=3π×13πr2h×h3−(πrl)2h2+9×(13πr2h)2
=π2r2h4−π2r2l2h2+π2r4h2
=π2r2h4−π2r2h2(r2+h2)+π2r4h2
=π2r2h4−π2r4h2−π2r2h4+π2r4h2=0