If $r, h, c, v$ are radius curved surface area and volume of a cone respectively, then prove that $3 π v h^{3} - c^{2} h^{2} + 9 v^{2} = 0$

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Solution

The curved surface area of a cone
=C
=πrl
=πr√(h²+r²)
where h, l and r are the height, slant height and radius of the cone.
Volume of the cone
=V
=πr²h/3
∴3πVh³-C²h²+9V²
=3π{(πr²h)/3}h³-{πr√(h²+r²)}²h²+9(πr²h/3)²
=π²r²h⁴-π²r²(h²+r²)h²+9π²r⁴h²/9
=π²r²h⁴-π²r²h⁴-π²r⁴h²+π²r⁴h²
=0 (Proved)

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