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Question Number 137528 by I want to learn more last updated on 03/Apr/21

Commented by I want to learn more last updated on 03/Apr/21

$$\mathrm{Find}\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{circle}.$$

Answered by mr W last updated on 03/Apr/21

Commented by mr W last updated on 03/Apr/21

$$\sqrt{R^2-2^2}+\sqrt{R^2-4^2}=6$$$$R^2-2^2=6^2+R^2-4^2-12\sqrt{R^2-4^2}$$$$2=\sqrt{R^2-4^2}$$$$R^2=20$$$$R=2\sqrt{5}$$

Commented by I want to learn more last updated on 03/Apr/21

$$\mathrm{Thanks}\mathrm{sir},\mathrm{God}\mathrm{bless}\mathrm{you}.$$

Answered by mr W last updated on 03/Apr/21

Commented by mr W last updated on 03/Apr/21

$$2R=\sqrt{4^2+8^2}=4\sqrt{5}$$$$\Rightarrow R=2\sqrt{5}$$

Commented by I want to learn more last updated on 03/Apr/21

$$\mathrm{I}\mathrm{really}\mathrm{appreciate}\mathrm{sir}.$$

Commented by I want to learn more last updated on 03/Apr/21

$$\mathrm{Sir},\mathrm{most}\mathrm{people}\mathrm{got}\mathrm{radius}=5.$$

Commented by I want to learn more last updated on 03/Apr/21

$$\mathrm{But}\mathrm{i}\mathrm{understand}\mathrm{your}\mathrm{workings}\mathrm{sir}.\mathrm{They}\mathrm{are}\mathrm{judt}\mathrm{getting}5.$$

Commented by mr W last updated on 03/Apr/21

$$ifyoureallyhaveunderstood,you$$$$shouldbeabletoknowbyyourself$$$$whatisright.$$

Commented by I want to learn more last updated on 04/Apr/21

$$\mathrm{Yes}\mathrm{sir}.\mathrm{I}\mathrm{know}.$$

Answered by mr W last updated on 03/Apr/21

$$R=\frac{a}{2\sin A}=\frac{6\sqrt{2}}{2\times \frac{6}{\sqrt{40}}}=2\sqrt{5}$$

Commented by mr W last updated on 04/Apr/21

Commented by I want to learn more last updated on 04/Apr/21

$$\mathrm{Thanks}\mathrm{sir}.\mathrm{I}\mathrm{really}\mathrm{appreciate}.$$

Commented by mr W last updated on 04/Apr/21

$$canyousolvefollowingcaseusing$$$$thismethod:$$

Commented by mr W last updated on 05/Apr/21

Commented by mr W last updated on 05/Apr/21

$$BC=\sqrt{b^2+c^2}$$$$AC=\sqrt{b^2+{(a+c)}^2}$$$$\sin \alpha =\frac{b}{AC}=\frac{b}{\sqrt{b^2+{(a+c)}^2}}$$$$BC=2R\sin \alpha$$$$\Rightarrow R=\frac{\sqrt{(b^2+c^2)[b^2+{(a+c)}^2]}}{2b}$$

Commented by I want to learn more last updated on 04/Apr/21

$$\mathrm{I}\mathrm{appreciate}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{more}.$$

Commented by mr W last updated on 05/Apr/21

$$youshouldbeabletogetR=\frac{\sqrt{65}}{2}for$$$$thiscase.$$

Commented by I want to learn more last updated on 10/Apr/21

$$\mathrm{I}\mathrm{appreciate}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}.$$

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