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Question-176946

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Question Number 176946 by Ar Brandon last updated on 28/Sep/22
Answered by mr W last updated on 29/Sep/22
Commented by mr W last updated on 29/Sep/22
R=radius, K=diameter of circle sin α=(a/(2R))=(a/K) sin β=(b/(2R))=(b/K) α+β=θ=60°=(π/3) cos (α+β)=cos θ=(1/2) (√((1−(a^2 /K^2 ))(1−(b^2 /K^2 ))))−(a/K)×(b/K)=cos θ (1−(a^2 /K^2 ))(1−(b^2 /K^2 ))=(((ab)/K^2 )+cos θ)^2 1−(a^2 /K^2 )−(b^2 /K^2 )=2cos θ((ab)/K^2 )+cos^2 θ K^2 =4R^2 =((a^2 +b^2 +2ab cos θ)/(sin^2 θ)) ⇒R=((√(a^2 +b^2 +2 ab cos θ))/(2 sin θ)) =((√(2^2 +5^2 +2×2×5×(1/2)))/(2×((√3)/2)))=(√(13)) A_(segment,a) =αR^2 −(a/2)(√(R^2 −(a^2 /4))) A_(segment,b) =βR^2 −(b/2)(√(R^2 −(b^2 /4))) A_(shaded) =R^2 θ−(a/2)(√(R^2 −(a^2 /4)))−(b/2)(√(R^2 −(b^2 /4))) =13×(π/3)−(5/2)(√(13−(5^2 /4)))−(2/2)(√(13−(2^2 /4))) =((13π)/3)−((23(√3))/4) ≈3.654
$${R}={radius},\:{K}={diameter}\:{of}\:{circle} \\ $$$$\mathrm{sin}\:\alpha=\frac{{a}}{\mathrm{2}{R}}=\frac{{a}}{{K}} \\ $$$$\mathrm{sin}\:\beta=\frac{{b}}{\mathrm{2}{R}}=\frac{{b}}{{K}} \\ $$$$\alpha+\beta=\theta=\mathrm{60}°=\frac{\pi}{\mathrm{3}} \\ $$$$\mathrm{cos}\:\left(\alpha+\beta\right)=\mathrm{cos}\:\theta=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\sqrt{\left(\mathrm{1}−\frac{{a}^{\mathrm{2}} }{{K}^{\mathrm{2}} }\right)\left(\mathrm{1}−\frac{{b}^{\mathrm{2}} }{{K}^{\mathrm{2}} }\right)}−\frac{{a}}{{K}}×\frac{{b}}{{K}}=\mathrm{cos}\:\theta \\ $$$$\left(\mathrm{1}−\frac{{a}^{\mathrm{2}} }{{K}^{\mathrm{2}} }\right)\left(\mathrm{1}−\frac{{b}^{\mathrm{2}} }{{K}^{\mathrm{2}} }\right)=\left(\frac{{ab}}{{K}^{\mathrm{2}} }+\mathrm{cos}\:\theta\right)^{\mathrm{2}} \\ $$$$\mathrm{1}−\frac{{a}^{\mathrm{2}} }{{K}^{\mathrm{2}} }−\frac{{b}^{\mathrm{2}} }{{K}^{\mathrm{2}} }=\mathrm{2cos}\:\theta\frac{{ab}}{{K}^{\mathrm{2}} }+\mathrm{cos}^{\mathrm{2}} \:\theta \\ $$$${K}^{\mathrm{2}} =\mathrm{4}{R}^{\mathrm{2}} =\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{ab}\:\mathrm{cos}\:\theta}{\mathrm{sin}^{\mathrm{2}} \:\theta} \\ $$$$\Rightarrow{R}=\frac{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}\:{ab}\:\mathrm{cos}\:\theta}}{\mathrm{2}\:\mathrm{sin}\:\theta} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\sqrt{\mathrm{2}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} +\mathrm{2}×\mathrm{2}×\mathrm{5}×\frac{\mathrm{1}}{\mathrm{2}}}}{\mathrm{2}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}=\sqrt{\mathrm{13}} \\ $$$${A}_{{segment},{a}} =\alpha{R}^{\mathrm{2}} −\frac{{a}}{\mathrm{2}}\sqrt{{R}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}} \\ $$$${A}_{{segment},{b}} =\beta{R}^{\mathrm{2}} −\frac{{b}}{\mathrm{2}}\sqrt{{R}^{\mathrm{2}} −\frac{{b}^{\mathrm{2}} }{\mathrm{4}}} \\ $$$${A}_{{shaded}} ={R}^{\mathrm{2}} \theta−\frac{{a}}{\mathrm{2}}\sqrt{{R}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}}−\frac{{b}}{\mathrm{2}}\sqrt{{R}^{\mathrm{2}} −\frac{{b}^{\mathrm{2}} }{\mathrm{4}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{13}×\frac{\pi}{\mathrm{3}}−\frac{\mathrm{5}}{\mathrm{2}}\sqrt{\mathrm{13}−\frac{\mathrm{5}^{\mathrm{2}} }{\mathrm{4}}}−\frac{\mathrm{2}}{\mathrm{2}}\sqrt{\mathrm{13}−\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{4}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{13}\pi}{\mathrm{3}}−\frac{\mathrm{23}\sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\approx\mathrm{3}.\mathrm{654} \\ $$
Commented by Ar Brandon last updated on 29/Sep/22
Thank you, Sir!
Commented by Tawa11 last updated on 02/Oct/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$

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