# 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-

Question Number 76974 by Maclaurin Stickker last updated on 02/Jan/20
$${Calculate}\:{the}\:{side}\:{of}\:{an}\:{equilateral} \\$$$${triangle}\:{whose}\:{vertices}\:{are}\:{situated} \\$$$${on}\:{three}\:{parallel}\:{coplanar}\:{lines}, \\$$$${knowing}\:{that}\:\boldsymbol{{a}}\:{and}\:\boldsymbol{{b}}\:{are}\:{the}\:{distances} \\$$$${of}\:{the}\:{parallel}\:{line}\:{to}\:{the}\:{others}. \\$$
Answered by mr W last updated on 02/Jan/20
$$\sqrt{{l}^{\mathrm{2}} −\left({a}+{b}\right)^{\mathrm{2}} }=\sqrt{{l}^{\mathrm{2}} −{b}^{\mathrm{2}} }−\sqrt{{l}^{\mathrm{2}} −{a}^{\mathrm{2}} } \\$$$${l}^{\mathrm{2}} −{a}^{\mathrm{2}} −{b}^{\mathrm{2}} −\mathrm{2}{ab}={l}^{\mathrm{2}} −{b}^{\mathrm{2}} +{l}^{\mathrm{2}} −{a}^{\mathrm{2}} −\mathrm{2}\sqrt{\left({l}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\left({l}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)} \\$$$${l}^{\mathrm{2}} +\mathrm{2}{ab}=\mathrm{2}\sqrt{\left({l}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\left({l}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)} \\$$$${l}^{\mathrm{4}} +\mathrm{4}{abl}^{\mathrm{2}} +\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} =\mathrm{4}\left({l}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\left({l}^{\mathrm{2}} −{a}^{\mathrm{2}} \right) \\$$$$\mathrm{3}{l}^{\mathrm{2}} −\mathrm{4}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{ab}\right)=\mathrm{0} \\$$$$\Rightarrow{l}=\mathrm{2}\sqrt{\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{ab}}{\mathrm{3}}} \\$$
Commented by Maclaurin Stickker last updated on 02/Jan/20
$${how}\:{did}\:{you}\:{get}\:{the}\:{first}\:{expression}? \\$$
Commented by mr W last updated on 02/Jan/20
Commented by mr W last updated on 02/Jan/20
$${AB}=\sqrt{{l}^{\mathrm{2}} −{b}^{\mathrm{2}} } \\$$$${CD}=\sqrt{{l}^{\mathrm{2}} −\left({a}+{b}\right)^{\mathrm{2}} } \\$$$${DE}=\sqrt{{l}^{\mathrm{2}} −{a}^{\mathrm{2}} } \\$$$${CD}+{DE}={AB} \\$$$$\sqrt{{l}^{\mathrm{2}} −\left({a}+{b}\right)^{\mathrm{2}} }+\sqrt{{l}^{\mathrm{2}} −{a}^{\mathrm{2}} }=\sqrt{{l}^{\mathrm{2}} −{b}^{\mathrm{2}} } \\$$