Question Number 43796 by gunawan last updated on 15/Sep/18
Answered by MrW3 last updated on 15/Sep/18
$$\sqrt{\mathrm{2}}{R}+{R}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}×\sqrt{\mathrm{2}} \\ $$$$\mathrm{2}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right){R}=\left(\sqrt{\mathrm{2}}−\mathrm{1}\right) \\ $$$${R}=\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\mathrm{2}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)}=\frac{\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{2}−\mathrm{1}\right)}=\frac{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}}−\sqrt{\mathrm{2}} \\ $$
Commented by Cheyboy last updated on 15/Sep/18
$$\mathrm{Sir}\:\mathrm{Tanmay}\:\mathrm{and}\:\mathrm{Mr}\:\mathrm{W}_{\mathrm{3}} \:\mathrm{will}\:\mathrm{you} \\ $$$$\mathrm{please}\:\mathrm{throw}\:\mathrm{more}\:\mathrm{light}\:\mathrm{on}\:\mathrm{with} \\ $$$$\:\mathrm{diagram}\:\mathrm{so}\:\mathrm{that}\:\mathrm{I}\:\mathrm{can}\:\mathrm{have}\:\mathrm{better} \\ $$$$\mathrm{understanding}.\mathrm{cuz}\:\mathrm{am}\:\mathrm{not}\:\mathrm{getting} \\ $$$$\mathrm{it}.\:\mathrm{Thank}\:\mathrm{you} \\ $$
Commented by MrW3 last updated on 15/Sep/18
Commented by MrW3 last updated on 15/Sep/18
$${I}\:{hope}\:{this}\:{helps}\:{you}. \\ $$
Commented by Cheyboy last updated on 15/Sep/18
$$\mathrm{waaw}!!\:\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much} \\ $$$$\mathrm{sir}\:\mathrm{am}\:\mathrm{so}\:\mathrm{grateful}.\:\mathrm{Now}\:\mathrm{i}\:\mathrm{understand}. \\ $$