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Question-188310




Question Number 188310 by normans last updated on 27/Feb/23
Commented by normans last updated on 27/Feb/23
yellow lenght(x)??
$${yellow}\:{lenght}\left({x}\right)?? \\ $$
Answered by HeferH last updated on 27/Feb/23
15
$$\mathrm{15} \\ $$
Answered by mr W last updated on 27/Feb/23
Commented by mr W last updated on 27/Feb/23
y_1 +z=y_2 +x  ⇒y_2 =y_1 +z−x  ((zh)/y_1 )=2h+((xh)/y_2 )  ⇒(z/y_1 )=2+(x/y_2 )  ⇒zy_2 =2y_1 y_2 +xy_1   ⇒z(y_1 +z−x)=2y_1 (y_1 +z−x)+xy_1   ⇒2y_1 ^2 +(z−x)y_1 −(z−x)z=0  ⇒y_1 =((−(z−x)+(√((z−x)^2 +8(z−x)z)))/4)  ⇒y=y_1 +y_2 =2y_1 +z−x           =(((z−x)+(√((z−x)^2 +8(z−x)z)))/2)  ⇒(√((z−x)^2 +8(z−x)z))=2y−(z−x)  ⇒(z−x)^2 +8(z−x)z=4y^2 −4y(z−x)+(z−x)^2   ⇒(z−x)(2z+y)=y^2   ⇒x=z−(y^2 /(y+2z))=15−(6^2 /(6+2×15))=14
$${y}_{\mathrm{1}} +{z}={y}_{\mathrm{2}} +{x} \\ $$$$\Rightarrow{y}_{\mathrm{2}} ={y}_{\mathrm{1}} +{z}−{x} \\ $$$$\frac{{zh}}{{y}_{\mathrm{1}} }=\mathrm{2}{h}+\frac{{xh}}{{y}_{\mathrm{2}} } \\ $$$$\Rightarrow\frac{{z}}{{y}_{\mathrm{1}} }=\mathrm{2}+\frac{{x}}{{y}_{\mathrm{2}} } \\ $$$$\Rightarrow{zy}_{\mathrm{2}} =\mathrm{2}{y}_{\mathrm{1}} {y}_{\mathrm{2}} +{xy}_{\mathrm{1}} \\ $$$$\Rightarrow{z}\left({y}_{\mathrm{1}} +{z}−{x}\right)=\mathrm{2}{y}_{\mathrm{1}} \left({y}_{\mathrm{1}} +{z}−{x}\right)+{xy}_{\mathrm{1}} \\ $$$$\Rightarrow\mathrm{2}{y}_{\mathrm{1}} ^{\mathrm{2}} +\left({z}−{x}\right){y}_{\mathrm{1}} −\left({z}−{x}\right){z}=\mathrm{0} \\ $$$$\Rightarrow{y}_{\mathrm{1}} =\frac{−\left({z}−{x}\right)+\sqrt{\left({z}−{x}\right)^{\mathrm{2}} +\mathrm{8}\left({z}−{x}\right){z}}}{\mathrm{4}} \\ $$$$\Rightarrow{y}={y}_{\mathrm{1}} +{y}_{\mathrm{2}} =\mathrm{2}{y}_{\mathrm{1}} +{z}−{x} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\left({z}−{x}\right)+\sqrt{\left({z}−{x}\right)^{\mathrm{2}} +\mathrm{8}\left({z}−{x}\right){z}}}{\mathrm{2}} \\ $$$$\Rightarrow\sqrt{\left({z}−{x}\right)^{\mathrm{2}} +\mathrm{8}\left({z}−{x}\right){z}}=\mathrm{2}{y}−\left({z}−{x}\right) \\ $$$$\Rightarrow\left({z}−{x}\right)^{\mathrm{2}} +\mathrm{8}\left({z}−{x}\right){z}=\mathrm{4}{y}^{\mathrm{2}} −\mathrm{4}{y}\left({z}−{x}\right)+\left({z}−{x}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\left({z}−{x}\right)\left(\mathrm{2}{z}+{y}\right)={y}^{\mathrm{2}} \\ $$$$\Rightarrow{x}={z}−\frac{{y}^{\mathrm{2}} }{{y}+\mathrm{2}{z}}=\mathrm{15}−\frac{\mathrm{6}^{\mathrm{2}} }{\mathrm{6}+\mathrm{2}×\mathrm{15}}=\mathrm{14} \\ $$

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