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Question-212760

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Question Number 212760 by Spillover last updated on 23/Oct/24
Answered by mr W last updated on 23/Oct/24
Commented by mr W last updated on 23/Oct/24
AB^2 =R^2 +35^2 AC^2 =R^2 +91^2 AD^2 =R^2 +65^2 BC=CD=DB=2R cos α=((R^2 +91^2 +4R^2 −R^2 −35^2 )/(2×2R(√(R^2 +91^2 ))))=((R^2 +1764)/(R(√(R^2 +91^2 )))) cos β=((R^2 +91^2 +4R^2 −R^2 −65^2 )/(2×2R(√(R^2 +91^2 ))))=((R^2 +1014)/(R(√(R^2 +91^2 )))) β=60°−α cos β=(1/2)(cos α+(√3) sin α) ((R^2 +1014)/(R(√(R^2 +91^2 ))))=(1/2)(((R^2 +1764)/(R(√(R^2 +91^2 ))))+((√(3(4753R^2 −1764^2 )))/(R(√(R^2 +91^2 ))))) 2R^2 +2×1014=R^2 +1764+(√(3(4753R^2 −1764^2 ))) R^2 +264=(√(3(4753R^2 −1764^2 ))) R^4 +2×264R^2 +264^2 =3(4753R^2 −1764^2 ) R^4 −13731R^2 +9404784=0 ⇒R^2 =((13731±12285)/2)=13008 or 723 ⇒R=4(√(813)) or (√(723)) for the case as shown R=(√(723))
$${AB}^{\mathrm{2}} ={R}^{\mathrm{2}} +\mathrm{35}^{\mathrm{2}} \\ $$$${AC}^{\mathrm{2}} ={R}^{\mathrm{2}} +\mathrm{91}^{\mathrm{2}} \\ $$$${AD}^{\mathrm{2}} ={R}^{\mathrm{2}} +\mathrm{65}^{\mathrm{2}} \\ $$$${BC}={CD}={DB}=\mathrm{2}{R} \\ $$$$\mathrm{cos}\:\alpha=\frac{{R}^{\mathrm{2}} +\mathrm{91}^{\mathrm{2}} +\mathrm{4}{R}^{\mathrm{2}} −{R}^{\mathrm{2}} −\mathrm{35}^{\mathrm{2}} }{\mathrm{2}×\mathrm{2}{R}\sqrt{{R}^{\mathrm{2}} +\mathrm{91}^{\mathrm{2}} }}=\frac{{R}^{\mathrm{2}} +\mathrm{1764}}{{R}\sqrt{{R}^{\mathrm{2}} +\mathrm{91}^{\mathrm{2}} }} \\ $$$$\mathrm{cos}\:\beta=\frac{{R}^{\mathrm{2}} +\mathrm{91}^{\mathrm{2}} +\mathrm{4}{R}^{\mathrm{2}} −{R}^{\mathrm{2}} −\mathrm{65}^{\mathrm{2}} }{\mathrm{2}×\mathrm{2}{R}\sqrt{{R}^{\mathrm{2}} +\mathrm{91}^{\mathrm{2}} }}=\frac{{R}^{\mathrm{2}} +\mathrm{1014}}{{R}\sqrt{{R}^{\mathrm{2}} +\mathrm{91}^{\mathrm{2}} }} \\ $$$$\beta=\mathrm{60}°−\alpha \\ $$$$\mathrm{cos}\:\beta=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{cos}\:\alpha+\sqrt{\mathrm{3}}\:\mathrm{sin}\:\alpha\right) \\ $$$$\frac{{R}^{\mathrm{2}} +\mathrm{1014}}{{R}\sqrt{{R}^{\mathrm{2}} +\mathrm{91}^{\mathrm{2}} }}=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{R}^{\mathrm{2}} +\mathrm{1764}}{{R}\sqrt{{R}^{\mathrm{2}} +\mathrm{91}^{\mathrm{2}} }}+\frac{\sqrt{\mathrm{3}\left(\mathrm{4753}{R}^{\mathrm{2}} −\mathrm{1764}^{\mathrm{2}} \right)}}{{R}\sqrt{{R}^{\mathrm{2}} +\mathrm{91}^{\mathrm{2}} }}\right) \\ $$$$\mathrm{2}{R}^{\mathrm{2}} +\mathrm{2}×\mathrm{1014}={R}^{\mathrm{2}} +\mathrm{1764}+\sqrt{\mathrm{3}\left(\mathrm{4753}{R}^{\mathrm{2}} −\mathrm{1764}^{\mathrm{2}} \right)} \\ $$$${R}^{\mathrm{2}} +\mathrm{264}=\sqrt{\mathrm{3}\left(\mathrm{4753}{R}^{\mathrm{2}} −\mathrm{1764}^{\mathrm{2}} \right)} \\ $$$${R}^{\mathrm{4}} +\mathrm{2}×\mathrm{264}{R}^{\mathrm{2}} +\mathrm{264}^{\mathrm{2}} =\mathrm{3}\left(\mathrm{4753}{R}^{\mathrm{2}} −\mathrm{1764}^{\mathrm{2}} \right) \\ $$$${R}^{\mathrm{4}} −\mathrm{13731}{R}^{\mathrm{2}} +\mathrm{9404784}=\mathrm{0} \\ $$$$\Rightarrow{R}^{\mathrm{2}} =\frac{\mathrm{13731}\pm\mathrm{12285}}{\mathrm{2}}=\mathrm{13008}\:{or}\:\mathrm{723} \\ $$$$\Rightarrow{R}=\mathrm{4}\sqrt{\mathrm{813}}\:{or}\:\sqrt{\mathrm{723}} \\ $$$${for}\:{the}\:{case}\:{as}\:{shown}\:{R}=\sqrt{\mathrm{723}} \\ $$
Commented by Spillover last updated on 24/Oct/24
great solution.thank you
$${great}\:{solution}.{thank}\:{you} \\ $$

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