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Question Number 132433 by john_santu last updated on 14/Feb/21

Answered by liberty last updated on 14/Feb/21

$$\mathrm{by}\mathrm{using}\mathrm{Cosine}\mathrm{of}\mathrm{law}$$$${(\mathrm{r}-0.5)}^2=0.5^2+{(\sqrt{2}-\mathrm{r})}^2-2\times 0.5\times (\sqrt{2}-\mathrm{r})\cos 45°$$$$\cancel{{\mathrm{r}}^2}-\mathrm{r}+\cancel{\frac{1}{4}}=\cancel{\frac{1}{4}}+2-2\sqrt{2}\mathrm{r}+\cancel{{\mathrm{r}}^2}-\frac{1}{\sqrt{2}}(\sqrt{2}-\mathrm{r})$$$$(2\sqrt{2}-1)\mathrm{r}=2-1+\frac{1}{2}\sqrt{2}\mathrm{r}$$$$(4\sqrt{2}-2-\sqrt{2})\mathrm{r}=2$$$$\mathrm{r}=\frac{2}{3\sqrt{2}-2}=\frac{3\sqrt{2}+2}{7}$$

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