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Question Number 128947 by ajfour last updated on 11/Jan/21

Commented by ajfour last updated on 11/Jan/21

Find radius of the blue arc.

$${Find}\:{radius}\:{of}\:{the}\:{blue}\:{arc}. \\ $$

Answered by mr W last updated on 11/Jan/21

Commented by mr W last updated on 11/Jan/21

a=GF+FE=1+(1/( (√3))) DE=(2/( (√3))) AE=2(1+(1/( (√3)))) AB=(1+(1/( (√3)))) sin 30°=(1/2)(1+(1/( (√3)))) BC=2(1+(1/( (√3))))−(2/( (√3)))−1−(1/2)(1+(1/( (√3)))) =(1/2)(1−(1/( (√3)))) HB=(1+(1/( (√3))))cos 30°=(1/2)(1+(√3)) HB^2 =(2R−BC)BC (((1+(√3))^2 )/4)=(2R−(1/2)(1−(1/( (√3)))))(1/2)(1−(1/( (√3)))) ⇒R=((3(√3)+4)/2)≈4.598076

$${a}={GF}+{FE}=\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$${DE}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}} \\ $$$${AE}=\mathrm{2}\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$${AB}=\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\:\mathrm{sin}\:\mathrm{30}°=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$${BC}=\mathrm{2}\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)−\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$$\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$${HB}=\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\mathrm{cos}\:\mathrm{30}°=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right) \\ $$$${HB}^{\mathrm{2}} =\left(\mathrm{2}{R}−{BC}\right){BC} \\ $$$$\frac{\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }{\mathrm{4}}=\left(\mathrm{2}{R}−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\right)\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$$\Rightarrow{R}=\frac{\mathrm{3}\sqrt{\mathrm{3}}+\mathrm{4}}{\mathrm{2}}\approx\mathrm{4}.\mathrm{598076} \\ $$

Commented by ajfour last updated on 12/Jan/21

Thanks Sir! (power of point wrt a circle is really helpful).

$${Thanks}\:{Sir}!\:\left({power}\:{of}\:{point}\right. \\ $$$$\left.{wrt}\:{a}\:{circle}\:{is}\:{really}\:{helpful}\right). \\ $$

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