Question Number 188163 by normans last updated on 26/Feb/23
Commented by mr W last updated on 26/Feb/23
$${is}\:{it}\:{really}\:{a}\:{square}? \\ $$
Commented by normans last updated on 26/Feb/23
$${yes} \\ $$
Answered by mr W last updated on 26/Feb/23
Commented by mr W last updated on 26/Feb/23
$${OA}={OB} \\ $$$$\Rightarrow\mathrm{6}^{\mathrm{2}} +{a}^{\mathrm{2}} =\mathrm{4}^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$$$\Rightarrow{b}^{\mathrm{2}} −{a}^{\mathrm{2}} =\mathrm{20}\:\:\:…\left({i}\right) \\ $$$${OD}={AB} \\ $$$$\Rightarrow\mathrm{5}^{\mathrm{2}} +\left(\frac{{a}+{b}}{\mathrm{2}}\right)^{\mathrm{2}} =\left(\mathrm{6}−\mathrm{4}\right)^{\mathrm{2}} +\left({a}−{b}\right)^{\mathrm{2}} \: \\ $$$$\Rightarrow\mathrm{3}{a}^{\mathrm{2}} −\mathrm{10}{ab}+\mathrm{3}{b}^{\mathrm{2}} =\mathrm{84}\:\:\:…\left({ii}\right) \\ $$$$\Rightarrow\mathrm{3}{a}^{\mathrm{2}} −\mathrm{10}{a}\sqrt{\mathrm{20}+{a}^{\mathrm{2}} }+\mathrm{3}\left(\mathrm{20}+{a}^{\mathrm{2}} \right)=\mathrm{84} \\ $$$$\Rightarrow\mathrm{3}{a}^{\mathrm{2}} −\mathrm{12}=\mathrm{5}{a}\sqrt{\mathrm{20}+{a}^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{9}{a}^{\mathrm{4}} −\mathrm{72}{a}^{\mathrm{2}} +\mathrm{144}=\mathrm{25}{a}^{\mathrm{2}} \left(\mathrm{20}+{a}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\mathrm{4}{a}^{\mathrm{4}} +\mathrm{143}{a}^{\mathrm{2}} −\mathrm{36}=\mathrm{0} \\ $$$$\Rightarrow{a}^{\mathrm{2}} =\frac{−\mathrm{143}+\mathrm{145}}{\mathrm{8}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow{a}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{b}=\sqrt{\mathrm{20}+\frac{\mathrm{1}}{\mathrm{4}}}=\frac{\mathrm{9}}{\mathrm{2}} \\ $$$${eqn}.\:{of}\:{AB}: \\ $$$${y}=−\frac{\mathrm{1}}{\mathrm{2}}−\left({x}−\mathrm{6}\right)\frac{\mathrm{5}}{\mathrm{2}}=−\frac{\mathrm{5}{x}}{\mathrm{2}}+\frac{\mathrm{29}}{\mathrm{2}} \\ $$$$\mathrm{0}=−\frac{\mathrm{5}{x}_{{E}} }{\mathrm{2}}+\frac{\mathrm{29}}{\mathrm{2}} \\ $$$$\Rightarrow{x}_{{E}} =\frac{\mathrm{29}}{\mathrm{5}} \\ $$$${y}_{{D}} =\frac{{a}+{b}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{9}}{\mathrm{2}}\right)=\mathrm{2} \\ $$$${s}={OD}=\sqrt{\mathrm{5}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }=\sqrt{\mathrm{29}} \\ $$$${A}_{{square}} =\left(\sqrt{\mathrm{29}}\right)^{\mathrm{2}} =\mathrm{29} \\ $$$${A}_{{blue}} =\frac{\left(\sqrt{\mathrm{29}}\right)^{\mathrm{2}} }{\mathrm{2}}−\frac{{x}_{{E}} {y}_{{D}} }{\mathrm{2}}=\frac{\mathrm{29}}{\mathrm{2}}−\frac{\mathrm{29}×\mathrm{2}}{\mathrm{2}×\mathrm{5}}=\frac{\mathrm{87}}{\mathrm{10}} \\ $$$$\frac{{A}_{{blue}} }{{A}_{{square}} }=\frac{\mathrm{3}}{\mathrm{10}}=\mathrm{30\%} \\ $$