Question Number 162396 by Mathematification last updated on 29/Dec/21
Commented by Mathematification last updated on 29/Dec/21
$${Thank}\:{you}\:{so}\:{much}.\:{But}\:{how}\:{do}\:{we}\:{know}\: \\ $$$${when}\:{to}\:{apply}\:{the}\:{formula}? \\ $$
Answered by mr W last updated on 29/Dec/21
$${r}_{{red}} =\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\mathrm{2}\sqrt{\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}×\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}×\mathrm{1}}}} \\ $$$${r}_{{red}} =\frac{\mathrm{6}}{\mathrm{23}} \\ $$$${A}_{{red}} =\frac{\mathrm{36}\pi}{\mathrm{529}} \\ $$
Commented by Tawa11 last updated on 29/Dec/21
$$\mathrm{Great}\:\mathrm{sir}. \\ $$
Commented by mr W last updated on 29/Dec/21
Commented by mr W last updated on 29/Dec/21
$${r},{R}=\frac{\mathrm{1}}{\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}\pm\mathrm{2}\sqrt{\frac{\mathrm{1}}{{ab}}+\frac{\mathrm{1}}{{bc}}+\frac{\mathrm{1}}{{ca}}}} \\ $$$$+\:{for}\:{r} \\ $$$$−\:{for}\:{R} \\ $$
Commented by mr W last updated on 29/Dec/21
$${see}\:{Q}\mathrm{77681}\:{for}\:{more}\:{details} \\ $$