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Question Number 188163 by normans last updated on 26/Feb/23

Commented by mr W last updated on 26/Feb/23

is it really a square?

$${is}\:{it}\:{really}\:{a}\:{square}? \\ $$

Commented by normans last updated on 26/Feb/23

yes

$${yes} \\ $$

Answered by mr W last updated on 26/Feb/23

Commented by mr W last updated on 26/Feb/23

OA=OB ⇒6^2 +a^2 =4^2 +b^2 ⇒b^2 −a^2 =20 ...(i) OD=AB ⇒5^2 +(((a+b)/2))^2 =(6−4)^2 +(a−b)^2 ⇒3a^2 −10ab+3b^2 =84 ...(ii) ⇒3a^2 −10a(√(20+a^2 ))+3(20+a^2 )=84 ⇒3a^2 −12=5a(√(20+a^2 )) ⇒9a^4 −72a^2 +144=25a^2 (20+a^2 ) ⇒4a^4 +143a^2 −36=0 ⇒a^2 =((−143+145)/8)=(1/4) ⇒a=−(1/2) ⇒b=(√(20+(1/4)))=(9/2) eqn. of AB: y=−(1/2)−(x−6)(5/2)=−((5x)/2)+((29)/2) 0=−((5x_E )/2)+((29)/2) ⇒x_E =((29)/5) y_D =((a+b)/2)=(1/2)(−(1/2)+(9/2))=2 s=OD=(√(5^2 +2^2 ))=(√(29)) A_(square) =((√(29)))^2 =29 A_(blue) =((((√(29)))^2 )/2)−((x_E y_D )/2)=((29)/2)−((29×2)/(2×5))=((87)/(10)) (A_(blue) /A_(square) )=(3/(10))=30%

$${OA}={OB} \\ $$$$\Rightarrow\mathrm{6}^{\mathrm{2}} +{a}^{\mathrm{2}} =\mathrm{4}^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$$$\Rightarrow{b}^{\mathrm{2}} −{a}^{\mathrm{2}} =\mathrm{20}\:\:\:...\left({i}\right) \\ $$$${OD}={AB} \\ $$$$\Rightarrow\mathrm{5}^{\mathrm{2}} +\left(\frac{{a}+{b}}{\mathrm{2}}\right)^{\mathrm{2}} =\left(\mathrm{6}−\mathrm{4}\right)^{\mathrm{2}} +\left({a}−{b}\right)^{\mathrm{2}} \: \\ $$$$\Rightarrow\mathrm{3}{a}^{\mathrm{2}} −\mathrm{10}{ab}+\mathrm{3}{b}^{\mathrm{2}} =\mathrm{84}\:\:\:...\left({ii}\right) \\ $$$$\Rightarrow\mathrm{3}{a}^{\mathrm{2}} −\mathrm{10}{a}\sqrt{\mathrm{20}+{a}^{\mathrm{2}} }+\mathrm{3}\left(\mathrm{20}+{a}^{\mathrm{2}} \right)=\mathrm{84} \\ $$$$\Rightarrow\mathrm{3}{a}^{\mathrm{2}} −\mathrm{12}=\mathrm{5}{a}\sqrt{\mathrm{20}+{a}^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{9}{a}^{\mathrm{4}} −\mathrm{72}{a}^{\mathrm{2}} +\mathrm{144}=\mathrm{25}{a}^{\mathrm{2}} \left(\mathrm{20}+{a}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\mathrm{4}{a}^{\mathrm{4}} +\mathrm{143}{a}^{\mathrm{2}} −\mathrm{36}=\mathrm{0} \\ $$$$\Rightarrow{a}^{\mathrm{2}} =\frac{−\mathrm{143}+\mathrm{145}}{\mathrm{8}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow{a}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{b}=\sqrt{\mathrm{20}+\frac{\mathrm{1}}{\mathrm{4}}}=\frac{\mathrm{9}}{\mathrm{2}} \\ $$$${eqn}.\:{of}\:{AB}: \\ $$$${y}=−\frac{\mathrm{1}}{\mathrm{2}}−\left({x}−\mathrm{6}\right)\frac{\mathrm{5}}{\mathrm{2}}=−\frac{\mathrm{5}{x}}{\mathrm{2}}+\frac{\mathrm{29}}{\mathrm{2}} \\ $$$$\mathrm{0}=−\frac{\mathrm{5}{x}_{{E}} }{\mathrm{2}}+\frac{\mathrm{29}}{\mathrm{2}} \\ $$$$\Rightarrow{x}_{{E}} =\frac{\mathrm{29}}{\mathrm{5}} \\ $$$${y}_{{D}} =\frac{{a}+{b}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{9}}{\mathrm{2}}\right)=\mathrm{2} \\ $$$${s}={OD}=\sqrt{\mathrm{5}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }=\sqrt{\mathrm{29}} \\ $$$${A}_{{square}} =\left(\sqrt{\mathrm{29}}\right)^{\mathrm{2}} =\mathrm{29} \\ $$$${A}_{{blue}} =\frac{\left(\sqrt{\mathrm{29}}\right)^{\mathrm{2}} }{\mathrm{2}}−\frac{{x}_{{E}} {y}_{{D}} }{\mathrm{2}}=\frac{\mathrm{29}}{\mathrm{2}}−\frac{\mathrm{29}×\mathrm{2}}{\mathrm{2}×\mathrm{5}}=\frac{\mathrm{87}}{\mathrm{10}} \\ $$$$\frac{{A}_{{blue}} }{{A}_{{square}} }=\frac{\mathrm{3}}{\mathrm{10}}=\mathrm{30\%} \\ $$

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