Question

If h, c and v respectively are the height the curved surface area and volume of the cone . Prove that 3πvh³-c²h²+9v²=0

Answer 1

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Answer 2

AnswEr:

Let r and l denote respectively the radius of the base and slant height of the cone.

Then,

$\sf \: l = \sqrt{ {r}^{2} + {h}^{2} } \\ \\ \sf \: V = \frac{1}{3} \pi {r}^{2} h \\ \\ \sf C = \pi \: rl \\ \\ \therefore \sf \: 3\pi \: Vh {}^{3} - {c}^{2} {h}^{2} + 9 {V}^{2} \\ \\ = \tt \: 3\pi \times \frac{1}{3} \pi {r}^{2} h \times {h}^{3} - {(\pi \: rl)}^{2} {h}^{2} + \\ \tt \: 9 \times ( \frac{1}{3} \pi {r}^{2} h) {}^{2} \\ \\ = \tt \: {\pi}^{2} {r}^{2} {h}^{4} - {\pi}^{2} {r}^{2} {l}^{2} {h}^{2} + {\pi}^{2} {r}^{4} {h}^{2} \\ \\ = \tt {\pi}^{2} {r}^{2} {h}^{4} - {\pi}^{2} {r}^{2} {h}^{2} ( {r}^{2} + {h}^{2} ) + {\pi}^{2} {r}^{4} {h}^{2} \\ \\ = \tt {\pi}^{2} {r}^{2} {h}^{2} - {\pi}^{2} {r}^{4} {h}^{2} - {\pi}^{2} {r}^{2} {h}^{2} + {\pi}^{2} {r}^{4} {h}^{2} \\ \\ \tt = 0$

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