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If `h , c ,V` are respectively the height, the curved surface and the volume of a cone, prove that `3piV h^3-C^2h^2+9V^2=0.`

Text Solution

Verified by Experts

We know that : `C = pi rl, V=(1)/(3)pi r^(2)h, l^(2)=h^(2)+r^(2)`
`L.H.S. =3pi Vh^(3)-C^(2)h^(2)+9V^(2)`
`=3pi((1)/(3)pir^(2)h)h^(3)-(pi rl)^(2)h^(2)+9 ((1)/(3)pi r^(2)h)^(2)`
`pi^(2)r^(2)h^(4)-pi^(2)r^(2)h^(2)(h^(2)+r^(2))+9xx(1)/(9)pi^(2)r^(4)h^(2) " " (because l^(2)=h^(2)+r^(2))`
`= pi^(2)r^(2)h^(4)-pi^(2)r^(2)h^(4)-pi^(2)r^(4)h^(2)+pi^(2)r^(4)h^(2)`
= 0 = R.H.S.
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