Question Number 227446 by mr W last updated on 31/Jan/26
Commented by mr W last updated on 31/Jan/26
$${find}\:{the}\:{blue}\:{area}. \\ $$
Commented by Spillover last updated on 31/Jan/26
https://youtu.be/B5LsQI2N1QM?si=J5g-IiwsfY2DelHd
Commented by mr W last updated on 31/Jan/26
$${asking}\:{for}\:{other}\:{approaches}. \\ $$
Answered by mr W last updated on 01/Feb/26
Commented by mr W last updated on 01/Feb/26
$$\frac{{a}}{{a}+{b}}=\frac{{x}+\mathrm{2}}{{x}+\mathrm{2}+\mathrm{8}}=\frac{{x}+\mathrm{2}}{{x}+\mathrm{10}} \\ $$$$\frac{{x}}{{x}+\mathrm{2}+\mathrm{8}+\mathrm{4}}=\left(\frac{{a}}{{a}+{b}}\right)^{\mathrm{2}} =\left(\frac{{x}+\mathrm{2}}{{x}+\mathrm{10}}\right)^{\mathrm{2}} \\ $$$${x}\left({x}+\mathrm{10}\right)^{\mathrm{2}} =\left({x}+\mathrm{14}\right)\left({x}+\mathrm{2}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +\mathrm{20}{x}−\mathrm{28}=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{8}\sqrt{\mathrm{2}}−\mathrm{10}\approx\mathrm{1}.\mathrm{314} \\ $$
Answered by mr W last updated on 01/Feb/26
Commented by mr W last updated on 01/Feb/26
$$\frac{\mathrm{2}}{{y}}=\frac{\mathrm{8}−{y}}{\mathrm{4}} \\ $$$${y}^{\mathrm{2}} −\mathrm{8}{y}+\mathrm{8}=\mathrm{0} \\ $$$$\Rightarrow{y}=\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\frac{{x}}{{x}+\mathrm{14}}=\left(\frac{{DE}}{{BC}}\right)^{\mathrm{2}} =\left(\frac{{y}+\mathrm{2}}{\mathrm{8}−{y}+\mathrm{4}}\right)^{\mathrm{2}} =\left(\frac{\mathrm{3}−\sqrt{\mathrm{2}}}{\mathrm{4}+\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} =\frac{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$${x}=\frac{\mathrm{14}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)}{\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{1}}=\mathrm{8}\sqrt{\mathrm{2}}−\mathrm{10}\approx\mathrm{1}.\mathrm{314} \\ $$