Question

Dau 50 de puncte + coroană

Answer 1

   
\begin{lgathered}\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}\end{lgathered}​​Se da: ​ΔABC  cu   < A=90​o​​​BC=40cm​​​AC​​AB​​=​4​​3​​​​AD⊥BC   D∈BC   (inaltime)​Se cere: ​AB=?​AC=?​AD=?​​Rezolvare:​​​AC​​AB​​=​4​​3​​​​⟹   AB=3k   si   AC=4k​Aplicam Pitagora: ​AB​2​​+AC​2​​=BC​2​​​(3k)​2​​+(4k)​2​​=40​2​​​9k​2​​+16K​2​​=1600​k​2​​(9+16)=1600​25k​2​​=1600​​k​2​​=​25​​1600​​​​ 


\begin{lgathered}\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}\end{lgathered}​​k=√​​25​​1600​​​​​=​5​​40​​=​8​​​​​AB=3k=3×8=​24 cm​​​​AC=4k=4×8=​32 cm​​​​​Calculam inaltimea:​BC×AD=AB×AC​​AD=​BC​​AB×AC​​=​40​​24×32​​=​10​​24×8​​=​10​​192​​=​19,2 cm​​​​​