Question Number 150305 by Tawa11 last updated on 10/Aug/21
Commented by Tawa11 last updated on 10/Aug/21
$$\mathrm{Find}\:\:\:\:\frac{\mathrm{R}}{\mathrm{r}} \\ $$
Answered by nimnim last updated on 11/Aug/21
$${With}\:{midpoint}\:{M} \\ $$$${ar}\left(\bigtriangleup{MBC}\right)=\mathrm{2}×{ar}\left(\bigtriangleup{MDC}\right) \\ $$$${AB}=\mathrm{2}{a}\:\left({say}\right) \\ $$$${BM}={CM}=\sqrt{\left(\mathrm{2}{a}\right)^{\mathrm{2}} +{a}^{\mathrm{2}} }={a}\sqrt{\mathrm{5}} \\ $$$${In}\:\bigtriangleup{MBC},\:{s}=\frac{{a}\sqrt{\mathrm{5}}+{a}\sqrt{\mathrm{5}}+\mathrm{2}{a}}{\mathrm{2}}=\left(\sqrt{\mathrm{5}}+\mathrm{1}\right){a} \\ $$$${R}=\frac{{ar}\left(\bigtriangleup{MBC}\right)}{\left(\sqrt{\mathrm{5}}+\mathrm{1}\right){a}} \\ $$$${In}\:\bigtriangleup{MDC},\:{s}=\frac{{a}\sqrt{\mathrm{5}}+{a}+\mathrm{2}{a}}{\mathrm{2}}=\frac{\left(\sqrt{\mathrm{5}}+\mathrm{3}\right){a}}{\mathrm{2}} \\ $$$${r}=\frac{{ar}\left(\bigtriangleup{MDC}\right)}{\left(\sqrt{\mathrm{5}}+\mathrm{3}\right){a}/\mathrm{2}}=\frac{\mathrm{2}×{ar}\left(\bigtriangleup{MDC}\right)}{\:\left(\sqrt{\mathrm{5}}+\mathrm{3}\right){a}} \\ $$$$\therefore\:\frac{{R}}{{r}}=\frac{{ar}\left(\bigtriangleup{MBC}\right)}{\left(\sqrt{\mathrm{5}}+\mathrm{1}\right){a}}×\frac{\left(\sqrt{\mathrm{5}}+\mathrm{3}\right){a}}{\mathrm{2}×{ar}\left(\bigtriangleup{MDC}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\sqrt{\mathrm{5}}+\mathrm{3}}{\:\sqrt{\mathrm{5}}+\mathrm{1}}\bigstar \\ $$$$\:\:{or}\:\:\:=\frac{\sqrt{\mathrm{5}}+\mathrm{3}}{\:\sqrt{\mathrm{5}}+\mathrm{1}}×\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\:\sqrt{\mathrm{5}}−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{5}+\mathrm{3}\sqrt{\mathrm{5}}−\sqrt{\mathrm{5}}−\mathrm{3}}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}=\:\varphi\:\left({golden}\:{ratio}\right) \\ $$
Commented by Tawa11 last updated on 11/Aug/21
$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$