Question Number 205660 by marie last updated on 26/Mar/24
Answered by Skabetix last updated on 26/Mar/24
$$\mathrm{E}{xercice}\:{n}°\mathrm{4} \\ $$$$\left.\mathrm{1}\right)\:{D}'{apres}\:{le}\:{theoreme}\:{de}\:{Pythagore}\:: \\ $$$${BC}^{\mathrm{2}} ={AC}^{\mathrm{2}} +{AB}^{\mathrm{2}} \\ $$$$\Leftrightarrow{BC}^{\mathrm{2}} =\mathrm{12}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} \\ $$$$\Leftrightarrow{BC}^{\mathrm{2}} =\mathrm{144}+\mathrm{25} \\ $$$$\Leftrightarrow{BC}^{\mathrm{2}} =\mathrm{169} \\ $$$$\Leftrightarrow{BC}=\sqrt{\mathrm{169}} \\ $$$$\Leftrightarrow{BC}=\mathrm{13} \\ $$$$\left.\mathrm{2}\left.\right){a}\right)\:{Les}\:{triangles}\:{CAB}\:{et}\:{NAM}\:{sont}\:{des} \\ $$$${triangles}\:{rectangles} \\ $$$$\boldsymbol{\mathrm{D}}'\boldsymbol{\mathrm{apres}}\:\boldsymbol{\mathrm{le}}\:\boldsymbol{\mathrm{theoreme}}\:\boldsymbol{\mathrm{de}}\:\boldsymbol{\mathrm{Thales}}\:: \\ $$$$\frac{\boldsymbol{\mathrm{AM}}}{\boldsymbol{\mathrm{AB}}}=\frac{\boldsymbol{\mathrm{MN}}}{\boldsymbol{\mathrm{BC}}} \\ $$$$\left.\mathrm{2}\left.\right)\mathrm{b}\right)\:\mathrm{Produit}\:\mathrm{en}\:\mathrm{croix}\:: \\ $$$$\mathrm{MN}=\frac{{BC}×{AM}}{{AB}}=\frac{\mathrm{13}×\mathrm{3}}{\mathrm{5}}=\frac{\mathrm{39}}{\mathrm{5}} \\ $$