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Question Number 191714 by Mingma last updated on 29/Apr/23

Answered by mr W last updated on 29/Apr/23

Commented by Shlock last updated on 30/Apr/23

Very detailed, sir!

Commented by mr W last updated on 29/Apr/23

$$sayAG=x$$$$x(x+4)=1\times (1+3)$$$$\Rightarrow x=2\sqrt{2}-2$$$$EF=\sqrt{4^2+{(2\sqrt{2}+2)}^2-2\times 4\times (2\sqrt{2}+2)\times \frac{1}{2}}=2\sqrt{5}$$$$\frac{\sin \alpha }{4}=\frac{\sin (\alpha -\frac{\pi }{3})}{2\sqrt{2}-2}$$$$\Rightarrow \tan \alpha =\frac{2\sqrt{3}}{3-\sqrt{2}}$$$$\Rightarrow \sin \alpha =\frac{2\sqrt{3}}{\sqrt{{\left(2\sqrt{3}\right)}^2+{(3-\sqrt{2})}^2}}=\frac{2\sqrt{3}}{\sqrt{23-6\sqrt{2}}}$$$$\Rightarrow R=\frac{EF}{2\sin \alpha }=\frac{2\sqrt{5}\times \sqrt{23-6\sqrt{2}}}{2\times 2\sqrt{3}}=\frac{\sqrt{15(23-6\sqrt{2})}}{6}\approx 2.459$$

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