Question

Problema asta va rog repede!!!!​

Answer 1

 

$\displaystyle\bf\\Se~d\u a:\\\\\Delta ABC~dreptunghc~cu~m(\angle BAC)=90^o\\AM=mediana~ipotenuzei~BC~cu~M\in~BC\\AD\perp BC, D\in BC=inaltimea~dusa~din~A~pe~ipotenuza~BC\\m(\angle ACB)=15^o\\BC=40~cm\\\\Se~cere:\\\\a)~~AD=?~cm\\b)~~A_{ABC}=?~cm^2\\\\Rezolvare:\\\\AM=\frac{BC}{2}=\frac{40}{2}=20~cm\\\\In~\Delta AMC~avem:\\\\AM=MC=20~cm~~\implies \Delta AMC~este~isoscel\\\\\implies~m(\angle MAC)=m(\angle ACM)=15^o\\\\\angle AMD~este~unghi~exterior~\Delta AMC$

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$\displaystyle\bf\\\implies~m(\angle AMD)=m(\angle MAC)+m(\angle ACM)=15^o+15^o=30^o\\\\\Delta dreptunghic~DMA~cu~m(\angle ADM)=90^o~avem:\\\\AM=20~cm=ipotenuzam(\angle AMD)=30^o\\\\Din~teorema~unghiului~de~30^o~rezulta:AD=\frac{AM}{2}=\frac{20}{2}=10~cm\\\\\boxed{\bf AD=10~cm}\\\\Aria_{\Delta ABC}=\frac{BC\times AD}{2}=\frac{40\times 10}{2}=\frac{400}{2}=200~cm^2\\\\\boxed{\bf Aria_{\Delta ABC}=200~cm^2}$