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Question Number 76974 by Maclaurin Stickker last updated on 02/Jan/20

Calculate the side of an equilateral triangle whose vertices are situated on three parallel coplanar lines, knowing that a and b are the distances of the parallel line to the others.

$${Calculate}\:{the}\:{side}\:{of}\:{an}\:{equilateral} \\ $$$${triangle}\:{whose}\:{vertices}\:{are}\:{situated} \\ $$$${on}\:{three}\:{parallel}\:{coplanar}\:{lines}, \\ $$$${knowing}\:{that}\:\boldsymbol{{a}}\:{and}\:\boldsymbol{{b}}\:{are}\:{the}\:{distances} \\ $$$${of}\:{the}\:{parallel}\:{line}\:{to}\:{the}\:{others}. \\ $$

Answered by mr W last updated on 02/Jan/20

(√(l^2 −(a+b)^2 ))=(√(l^2 −b^2 ))−(√(l^2 −a^2 )) l^2 −a^2 −b^2 −2ab=l^2 −b^2 +l^2 −a^2 −2(√((l^2 −b^2 )(l^2 −a^2 ))) l^2 +2ab=2(√((l^2 −b^2 )(l^2 −a^2 ))) l^4 +4abl^2 +4a^2 b^2 =4(l^2 −b^2 )(l^2 −a^2 ) 3l^2 −4(a^2 +b^2 +ab)=0 ⇒l=2(√((a^2 +b^2 +ab)/3))

$$\sqrt{{l}^{\mathrm{2}} −\left({a}+{b}\right)^{\mathrm{2}} }=\sqrt{{l}^{\mathrm{2}} −{b}^{\mathrm{2}} }−\sqrt{{l}^{\mathrm{2}} −{a}^{\mathrm{2}} } \\ $$$${l}^{\mathrm{2}} −{a}^{\mathrm{2}} −{b}^{\mathrm{2}} −\mathrm{2}{ab}={l}^{\mathrm{2}} −{b}^{\mathrm{2}} +{l}^{\mathrm{2}} −{a}^{\mathrm{2}} −\mathrm{2}\sqrt{\left({l}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\left({l}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)} \\ $$$${l}^{\mathrm{2}} +\mathrm{2}{ab}=\mathrm{2}\sqrt{\left({l}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\left({l}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)} \\ $$$${l}^{\mathrm{4}} +\mathrm{4}{abl}^{\mathrm{2}} +\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} =\mathrm{4}\left({l}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\left({l}^{\mathrm{2}} −{a}^{\mathrm{2}} \right) \\ $$$$\mathrm{3}{l}^{\mathrm{2}} −\mathrm{4}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{ab}\right)=\mathrm{0} \\ $$$$\Rightarrow{l}=\mathrm{2}\sqrt{\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{ab}}{\mathrm{3}}} \\ $$

Commented by Maclaurin Stickker last updated on 02/Jan/20

how did you get the first expression?

$${how}\:{did}\:{you}\:{get}\:{the}\:{first}\:{expression}? \\ $$

Commented by mr W last updated on 02/Jan/20

Commented by mr W last updated on 02/Jan/20

AB=(√(l^2 −b^2 )) CD=(√(l^2 −(a+b)^2 )) DE=(√(l^2 −a^2 )) CD+DE=AB (√(l^2 −(a+b)^2 ))+(√(l^2 −a^2 ))=(√(l^2 −b^2 ))

$${AB}=\sqrt{{l}^{\mathrm{2}} −{b}^{\mathrm{2}} } \\ $$$${CD}=\sqrt{{l}^{\mathrm{2}} −\left({a}+{b}\right)^{\mathrm{2}} } \\ $$$${DE}=\sqrt{{l}^{\mathrm{2}} −{a}^{\mathrm{2}} } \\ $$$${CD}+{DE}={AB} \\ $$$$\sqrt{{l}^{\mathrm{2}} −\left({a}+{b}\right)^{\mathrm{2}} }+\sqrt{{l}^{\mathrm{2}} −{a}^{\mathrm{2}} }=\sqrt{{l}^{\mathrm{2}} −{b}^{\mathrm{2}} } \\ $$

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