Question Number 215445 by BaliramKumar last updated on 07/Jan/25
Answered by A5T last updated on 07/Jan/25
$$\sqrt{\left(\mathrm{x}+\mathrm{r}\right)^{\mathrm{2}} −\mathrm{r}^{\mathrm{2}} }=\sqrt{\left(\mathrm{r}+\mathrm{r}\right)^{\mathrm{2}} −\mathrm{r}^{\mathrm{2}} }−\left(\mathrm{r}+\mathrm{x}\right) \\ $$$$\Rightarrow\sqrt{\mathrm{x}\left(\mathrm{x}+\mathrm{2r}\right)}=\mathrm{r}\sqrt{\mathrm{3}}−\mathrm{r}−\mathrm{x} \\ $$$$\Rightarrow\mathrm{4r}−\mathrm{2r}\sqrt{\mathrm{3}}−\mathrm{2x}\sqrt{\mathrm{3}}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{r}=\frac{\mathrm{2}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\sqrt{\mathrm{3}}}{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}}}=\sqrt{\mathrm{3}} \\ $$
Answered by mr W last updated on 07/Jan/25
$${r}+{x}=\frac{\mathrm{2}\sqrt{\mathrm{3}}{r}}{\mathrm{3}} \\ $$$$\mathrm{2}−\sqrt{\mathrm{3}}=\frac{\mathrm{2}\sqrt{\mathrm{3}}{r}}{\mathrm{3}}−{r} \\ $$$$\Rightarrow{r}=\frac{\mathrm{3}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)}{\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{3}}=\sqrt{\mathrm{3}} \\ $$
Commented by BaliramKumar last updated on 07/Jan/25
$${thanks} \\ $$
Answered by A5T last updated on 07/Jan/25
$$\left(\mathrm{r}+\mathrm{x}\right)^{\mathrm{2}} =\left(\mathrm{2r}\right)^{\mathrm{2}} +\left(\mathrm{r}+\mathrm{x}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{2r}\right)\left(\mathrm{r}+\mathrm{x}\right)\mathrm{cos30} \\ $$$$\Rightarrow\mathrm{2r}=\left(\mathrm{r}+\mathrm{x}\right)\sqrt{\mathrm{3}}\Rightarrow\mathrm{r}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)=\sqrt{\mathrm{3}}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right) \\ $$$$\Rightarrow\mathrm{r}=\sqrt{\mathrm{3}} \\ $$
Commented by BaliramKumar last updated on 07/Jan/25
$${thanks} \\ $$