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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

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

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

Commented by ajfour last updated on 12/Dec/17

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

thanks in advance sir ajfour. nice,smart,perfect,beautiful.

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

Commented by ajfour last updated on 12/Dec/17

(x^2 /a^2 )=(1/(1+4)) ⇒ a=x(√5) (((10−x)^2 )/(100))=(3/(3+2)) ⇒ x=10−2(√(15)) so a=10((√5)−(√3)) cos θ+cos (180−θ)=0 ⇒ ((b^2 +a^2 −(12)^2 )/(2ab))+ ((b^2 +a^2 −(10)^2 )/(2ab)) =0 ⇒ b^2 =122−a^2 AE= b =(√(122−100((√5)−(√3))^2 )) =(√(200(√(15))−678)) .

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

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