Question Number 213745 by efronzo1 last updated on 15/Nov/24
Answered by golsendro last updated on 15/Nov/24
$$\:\angle\:\mathrm{AOC}\:=\:\alpha \\ $$$$\:\:\left(\mathrm{1}\right)\:\mathrm{cos}\:\alpha\:=\:\frac{\mathrm{2R}^{\mathrm{2}} −\mathrm{14}}{\mathrm{2R}^{\mathrm{2}} } \\ $$$$\:\left(\mathrm{2}\right)\:\mathrm{cos}\:\alpha\:=\:\frac{\mathrm{5R}^{\mathrm{2}} −\mathrm{49}}{\mathrm{4R}^{\mathrm{2}} } \\ $$$$\:\:\Rightarrow\:\frac{\mathrm{5R}^{\mathrm{2}} −\mathrm{49}}{\mathrm{2}}\:=\:\mathrm{2R}^{\mathrm{2}} −\mathrm{14} \\ $$$$\:\:\:\:\:\:\:\mathrm{R}^{\mathrm{2}} \:=\:\mathrm{21}\Rightarrow\mathrm{R}=\sqrt{\mathrm{21}} \\ $$
Answered by mr W last updated on 15/Nov/24
$$\mathrm{4}×\left(\sqrt{\mathrm{14}}\right)^{\mathrm{2}} =\mathrm{2}×\mathrm{7}^{\mathrm{2}} +\mathrm{2}×{R}^{\mathrm{2}} −\left(\mathrm{2}{R}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{R}=\sqrt{\mathrm{7}^{\mathrm{2}} −\mathrm{2}×\mathrm{14}}=\sqrt{\mathrm{21}} \\ $$