Question-137528

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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

(√(R^2 −2^2 ))+(√(R^2 −4^2 ))=6 R^2 −2^2 =6^2 +R^2 −4^2 −12(√(R^2 −4^2 )) 2=(√(R^2 −4^2 )) R^2 =20 R=2(√5)

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

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

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

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

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

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

But i understand your workings sir. They are judt getting 5.

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

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

$${if}\:{you}\:{really}\:{have}\:{understood},\:{you} \\ $$$${should}\:{be}\:{able}\:{to}\:{know}\:{by}\:{yourself} \\ $$$${what}\:{is}\:{right}. \\ $$

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=(a/(2 sin A))=((6(√2))/(2×(6/( (√(40))))))=2(√5)

$${R}=\frac{{a}}{\mathrm{2}\:\mathrm{sin}\:{A}}=\frac{\mathrm{6}\sqrt{\mathrm{2}}}{\mathrm{2}×\frac{\mathrm{6}}{\:\sqrt{\mathrm{40}}}}=\mathrm{2}\sqrt{\mathrm{5}} \\ $$

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

$${can}\:{you}\:{solve}\:{following}\:{case}\:{using} \\ $$$${this}\:{method}: \\ $$

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

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

BC=(√(b^2 +c^2 )) AC=(√(b^2 +(a+c)^2 )) sin α=(b/(AC))=(b/( (√(b^2 +(a+c)^2 )))) BC=2R sin α ⇒R=((√((b^2 +c^2 )[b^2 +(a+c)^2 ]))/(2b))

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

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