Question Number 184718 by sonukgindia last updated on 10/Jan/23
Answered by HeferH last updated on 10/Jan/23
$${Let}\:\mathrm{2}{a}\:{be}\:{the}\:{side}\:{of}\:{those}\:{squares}\:{and}\:{x}\:{the}\: \\ $$$${larger}\:{side}\:{of}\:{the}\:{yellow}\:{triangle}.\: \\ $$$$\:{A}_{{rect}} =\:\mathrm{2}{a}\:\centerdot\:\mathrm{4}{a}\:=\mathrm{8}{a}^{\mathrm{2}} \\ $$$$\:{by}\:{product}\:{of}\:{chords}: \\ $$$$\:{a}\sqrt{\mathrm{5}}\centerdot{x}\:=\:{a}^{\mathrm{2}} \\ $$$$\:{x}\:=\:\frac{{a}}{\:\sqrt{\mathrm{5}}} \\ $$$$\:{the}\:{yellow}\:{triangle}\:{is}\:{a}\:{special}\:{one}\:{k},\:\mathrm{2}{k},\:{k}\sqrt{\mathrm{5}} \\ $$$$\:{from}\:{this}\:{we}\:{get}\:{the}\:{height}\:{and}\:{the}\:{base}. \\ $$$$\:{A}_{\bigtriangleup} \:=\:\left(\frac{\mathrm{2}{a}}{\mathrm{5}}\right)^{\mathrm{2}} \centerdot\frac{\mathrm{1}}{\mathrm{2}}\:=\:\frac{\mathrm{2}{a}^{\mathrm{2}} }{\mathrm{25}}\:\Rightarrow \\ $$$$\:\frac{{A}_{\bigtriangleup} }{{A}_{{rect}} }\:=\:\frac{\mathrm{2}{a}^{\mathrm{2}} }{\mathrm{25}}\centerdot\frac{\mathrm{1}}{\mathrm{8}{a}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{100}}\:\: \\ $$
Answered by mr W last updated on 10/Jan/23
Commented by mr W last updated on 10/Jan/23
$$\frac{{CE}}{{CD}}=\frac{{BC}}{{AC}}\:\Rightarrow{CE}=\frac{\mathrm{1}×\mathrm{1}}{\:\sqrt{\mathrm{5}}}=\frac{\sqrt{\mathrm{5}}}{\:\mathrm{5}}=\frac{{AC}}{\mathrm{5}} \\ $$$$\frac{{CE}}{{AC}}=\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$\frac{{A}_{{yellow}} }{\Delta_{{ACF}} }=\left(\frac{\mathrm{1}}{\mathrm{5}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{25}} \\ $$$$\Delta_{{ACF}} =\frac{{A}_{{rectangle}} }{\mathrm{4}} \\ $$$$\Rightarrow\frac{{A}_{{yellow}} }{{A}_{{rectangle}} }=\frac{\mathrm{1}}{\mathrm{25}}×\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{100}} \\ $$