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Question Number 25609 by behi.8.3.4.17@gmail.com last updated on 12/Dec/17

Commented by behi.8.3.4.17@gmail.com last updated on 12/Dec/17

$$\mathbf{BE}=\mathbf{EC},\mathbf{AB}=12,\mathbf{AC}=10$$$$\mathbf{parallel}\mathbf{lines}\mathbf{to}:\mathbf{AE},\mathbf{with}\mathbf{equal}\mathbf{distance}$$$$\mathbf{from}:\mathbf{A}\mathbf{and}\mathbf{B},\mathbf{toward}\mathbf{C},\mathbf{divide}\mathbf{the}\mathbf{area}$$$$\mathbf{of}\mathbf{A}\stackrel{△}{\mathbf{B}\mathit{C}}\mathbf{at}\mathbf{ratio}:1:4:2:3.$$$$....................\mathbf{AE}=?..............$$

Commented by ajfour last updated on 12/Dec/17

Commented by behi.8.3.4.17@gmail.com last updated on 12/Dec/17

$$\mathrm{thanks}\mathrm{in}\mathrm{advance}\mathrm{sir}\mathrm{ajfour}.$$$$\mathrm{nice},\mathrm{smart},\mathrm{perfect},\mathrm{beautiful}.$$

Commented by ajfour last updated on 12/Dec/17

$$\frac{x^2}{a^2}=\frac{1}{1+4}\Rightarrow a=x\sqrt{5}$$$$\frac{{(10-x)}^2}{100}=\frac{3}{3+2}$$$$\Rightarrow x=10-2\sqrt{15}$$$$soa=10(\sqrt{5}-\sqrt{3})$$$$\cos \theta +\cos (180-\theta )=0$$$$\Rightarrow \frac{b^2+a^2-{\left(12\right)}^2}{2ab}+\frac{b^2+a^2-{\left(10\right)}^2}{2ab}=0$$$$\Rightarrow b^2=122-a^2$$$$AE=b=\sqrt{122-100{(\sqrt{5}-\sqrt{3})}^2}$$$$=\sqrt{200\sqrt{15}-678}.$$

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