Question Number 59975 by vajpaithegrate@gmail.com last updated on 16/May/19
Answered by tanmay last updated on 16/May/19
![T.S.A of cone=πrl+πr^2 T.S.A=πr(r+(√(h^2 +r^2 )) ) ]]when [l^2 =r^2 +h^2 ] (dS/dh)=πr(0+(1/(2(√(h^2 +r^2 )) ))×2h) dS=((h/(√(h^2 +r^2 )))×dh)×πr dS=((πrh)/(√(h^2 +r^2 )))×α [given dh=α]](Q59982.png)
$${T}.{S}.{A}\:{of}\:{cone}=\pi{rl}+\pi{r}^{\mathrm{2}} \\ $$$$\left.{T}\left..{S}.{A}=\pi{r}\left({r}+\sqrt{{h}^{\mathrm{2}} +{r}^{\mathrm{2}} }\:\right)\:\:\:\:\:\right]\right]{when}\:\left[{l}^{\mathrm{2}} ={r}^{\mathrm{2}} +{h}^{\mathrm{2}} \right] \\ $$$$\frac{{dS}}{{dh}}=\pi{r}\left(\mathrm{0}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{{h}^{\mathrm{2}} +{r}^{\mathrm{2}} }\:}×\mathrm{2}{h}\right) \\ $$$${dS}=\left(\frac{{h}}{\sqrt{{h}^{\mathrm{2}} +{r}^{\mathrm{2}} }}×{dh}\right)×\pi{r} \\ $$$${dS}=\frac{\pi{rh}}{\sqrt{{h}^{\mathrm{2}} +{r}^{\mathrm{2}} }}×\alpha\:\left[{given}\:{dh}=\alpha\right] \\ $$$$ \\ $$
Commented by vajpaithegrate@gmail.com last updated on 16/May/19
$$\mathrm{thank}\:\mathrm{u}\:\mathrm{sir} \\ $$