Menu Close

Question-205660




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

Leave a Reply

Your email address will not be published. Required fields are marked *