Question Number 167280 by puissant last updated on 11/Mar/22
Commented by puissant last updated on 11/Mar/22
$${Area}=\:??? \\ $$
Answered by som(math1967) last updated on 12/Mar/22
![△NAD cos∠NAD=((2^2 +x^2 −5^2 )/(4x)) [let side of square=x] =((x^2 −21)/(4x)) △NBA cos∠NAB=((2^2 +x^2 −17)/(4x)) cos(90−∠NAD)=((x^2 −13)/(4x)) sin∠NAD=((x^2 −13)/(4x)) (((x^2 −13)/(4x)))^2 +(((x^2 −21)/(4x)))^2 =sin^2 ∠NAD +cos^2 ∠NAD x^4 −26x^2 +169+x^4 −42x^2 +441=16x^2 2x^4 −84x^2 +610=0 x^4 −42x^2 +305=0 x^2 =((42±(√(544)))/2)hm^2](https://www.tinkutara.com/question/Q167284.png)
$$\bigtriangleup{NAD} \\ $$$${cos}\angle{NAD}=\frac{\mathrm{2}^{\mathrm{2}} +{x}^{\mathrm{2}} −\mathrm{5}^{\mathrm{2}} }{\mathrm{4}{x}}\:\:\left[{let}\:{side}\:{of}\:{square}={x}\right] \\ $$$$\:\:\:\:=\frac{{x}^{\mathrm{2}} −\mathrm{21}}{\mathrm{4}{x}} \\ $$$$\bigtriangleup{NBA} \\ $$$${cos}\angle{NAB}=\frac{\mathrm{2}^{\mathrm{2}} +{x}^{\mathrm{2}} −\mathrm{17}}{\mathrm{4}{x}} \\ $$$${cos}\left(\mathrm{90}−\angle{NAD}\right)=\frac{{x}^{\mathrm{2}} −\mathrm{13}}{\mathrm{4}{x}} \\ $$$${sin}\angle{NAD}=\frac{{x}^{\mathrm{2}} −\mathrm{13}}{\mathrm{4}{x}} \\ $$$$\left(\frac{{x}^{\mathrm{2}} −\mathrm{13}}{\mathrm{4}{x}}\right)^{\mathrm{2}} +\left(\frac{{x}^{\mathrm{2}} −\mathrm{21}}{\mathrm{4}{x}}\right)^{\mathrm{2}} ={sin}^{\mathrm{2}} \angle{NAD} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{cos}^{\mathrm{2}} \angle{NAD} \\ $$$$\:{x}^{\mathrm{4}} −\mathrm{26}{x}^{\mathrm{2}} +\mathrm{169}+{x}^{\mathrm{4}} −\mathrm{42}{x}^{\mathrm{2}} +\mathrm{441}=\mathrm{16}{x}^{\mathrm{2}} \\ $$$$\mathrm{2}{x}^{\mathrm{4}} −\mathrm{84}{x}^{\mathrm{2}} +\mathrm{610}=\mathrm{0} \\ $$$${x}^{\mathrm{4}} −\mathrm{42}{x}^{\mathrm{2}} +\mathrm{305}=\mathrm{0} \\ $$$$\:{x}^{\mathrm{2}} =\frac{\mathrm{42}\pm\sqrt{\mathrm{544}}}{\mathrm{2}}{hm}^{\mathrm{2}} \\ $$
Commented by puissant last updated on 12/Mar/22
$${Thanks}\:{sir} \\ $$