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Question Number 205660 by marie last updated on 26/Mar/24

Answered by Skabetix last updated on 26/Mar/24

$$\mathrm{E}xercicen°4$$$$1)D^{\prime }apresletheoremedePythagore:$$$${BC}^2={AC}^2+{AB}^2$$$$\iff {BC}^2={12}^2+5^2$$$$\iff {BC}^2=144+25$$$$\iff {BC}^2=169$$$$\iff BC=\sqrt{169}$$$$\iff BC=13$$$$2)a)LestrianglesCABetNAMsontdes$$$$trianglesrectangles$$$${\mathbf{D}}^{\prime }\mathbf{apres}\mathbf{le}\mathbf{theoreme}\mathbf{de}\mathbf{Thales}:$$$$\frac{\mathbf{AM}}{\mathbf{AB}}=\frac{\mathbf{MN}}{\mathbf{BC}}$$$$2)\mathrm{b})\mathrm{Produit}\mathrm{en}\mathrm{croix}:$$$$\mathrm{MN}=\frac{BC\times AM}{AB}=\frac{13\times 3}{5}=\frac{39}{5}$$

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