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Intr-un triunghi dreptunghic ,raportul lungimilor catetelor este egal cu 3/4 ,iar lungimea ipotenuzei este egala cu 40 cm.Calculati :
a) lungimile catetelor triunghiului dat ;
b)lungimea inaltimii corespunzatoare ipotenuzei.


Răspuns :

   
[tex]\displaystyle \\ \texttt{Se da: } \\ \Delta ABC ~~cu~~\ \textless \ A=90^o \\ BC = 40 cm \\ \\ \frac{AB}{AC} = \frac{3}{4} \\ \\ AD \perp BC ~~~D \in BC~~~(inaltime)\\ \texttt{Se cere: } \\ AB=? \\ AC=? \\ AD=? \\ \\ Rezolvare: \\ \\ \frac{AB}{AC} = \frac{3}{4} \\ \\ \Longrightarrow~~~AB =3k~~~si~~~ AC=4k \\ \texttt{Aplicam Pitagora: } \\ AB^2+AC^2 = BC^2 \\ (3k)^2 + (4k)^2=40^2 \\ 9k^2 + 16K^2 = 1600 \\ k^2(9+16) = 1600 \\ 25 k^2 = 1600 \\ \\ k^2 = \frac{1600}{25}[/tex]


[tex]\displaystyle \\ k = \sqrt{\frac{1600}{25}} =\frac{40}{5} = \boxed{8} \\ \\ AB = 3k = 3\times 8 = \boxed{24 ~cm} \\ AC = 4k = 4\times 8 = \boxed{32 ~cm} \\ \\ \texttt{Calculam inaltimea:} \\ BC \times AD =AB \times AC\\ \\ AD = \frac{AB \times AC}{BC}= \frac{24 \times 32}{40}= \frac{24 \times 8}{10}=\frac{192}{10}= \boxed{19,2 ~cm}[/tex]