Question Number 207008 by mr W last updated on 03/May/24
Answered by A5T last updated on 03/May/24
Commented by A5T last updated on 03/May/24
$$\frac{\mathrm{5}}{{BO}}=\frac{\mathrm{6}}{{CO}}\Rightarrow\frac{{BO}}{{CO}}=\frac{\mathrm{5}}{\mathrm{6}};\:{BO}=\mathrm{5}{x};{CO}=\mathrm{6}{x} \\ $$$$\Rightarrow\mathrm{11}{x}=\mathrm{7}\Rightarrow{BO}=\frac{\mathrm{35}}{\mathrm{11}};{CO}=\frac{\mathrm{42}}{\mathrm{11}} \\ $$$$\left(\mathrm{6}−{x}\right)^{\mathrm{2}} +{r}^{\mathrm{2}} =\frac{\mathrm{42}^{\mathrm{2}} }{\mathrm{11}^{\mathrm{2}} }…\left({i}\right) \\ $$$$\left(\mathrm{5}−{x}\right)^{\mathrm{2}} +{r}^{\mathrm{2}} =\frac{\mathrm{35}^{\mathrm{2}} }{\mathrm{11}^{\mathrm{2}} }…\left({ii}\right) \\ $$$$\left({i}\right)\&\left({ii}\right)\Rightarrow{r}=\frac{\mathrm{12}\sqrt{\mathrm{6}}}{\mathrm{11}}\approx\mathrm{2}.\mathrm{6722} \\ $$
Commented by mr W last updated on 03/May/24
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Answered by cherokeesay last updated on 03/May/24
Commented by mr W last updated on 03/May/24
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