Question

Matematica clasa a 7 a probleme geometrie 50 puncte​

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

1.

$\it a)\ MN||BC \stackrel{T.f.a.}{\Longrightarrow}\ \Delta ABC\sim\Delta ANM \Rightarrow\dfrac{AB}{AN}=\dfrac{AC}{AM} \Rightarrow \dfrac{\ 6^{(2}}{14}=\dfrac{x}{12} \Rightarrow\\ \\ \\ \Rightarrow \dfrac{3}{7}=\dfrac{x}{12} \Rightarrow x=\dfrac{3\cdot12}{7}=\dfrac{36}{7}\approx5,14\ cm$

$\it b)\ MN||BC\ \stackrel{T.Thales}{\Longrightarrow}\ \dfrac{AM}{MB}=\dfrac{AN}{NC} \Rightarrow \dfrac{y}{3}=\dfrac{6}{2} \Rightarrow \dfrac{y}{3}=3 \Rightarrow y=3\cdot3=9\ cm$

2.

$\it a)\ a=BC=2\cdot MN=2\cdot34=68\\ \\ AC=2\cdot NC=2\cdot13=26\\ \\ \mathcal{P}_{ABC}=AB+BC+AC=24+68+26=118\\ \\ \mathcal{P}_{AMN}=\dfrac{\mathcal{P}_{ABC}}{2}=\dfrac{118}{2}=59$

$\it b)\ MN=b=\dfrac{BC}{2}=\dfrac{72}{2}=36\\ \\ AB=2\cdotMB=2\cdot16=32\\ \\ \mathcal{P}_{ABC}=AB+BC+AC=32+72+30=134\\ \\ \mathcal{P}_{AMN}=\dfrac{\mathcal{P}_{ABC}}{2}=\dfrac{134}{2}=67$

3.

$\it \mathcal{P}_{ABCD}=AB+BC+CD+AD=10+30+18+16=74\ cm\\ \\ EA=EB-AB=25-10=15\ cm\\ \\ AD||BC\ \stackrel{T.Thales}{\Longrightarrow}\ \ \dfrac{ED}{DC}=\dfrac{EA}{AB} \Rightarrow \dfrac{ED}{18}=\dfrac{15}{10} \Rightarrow ED=\dfrac{18\cdot15}{10}=\dfrac{270}{10}=27\ cm$

$\it \mathcal{P}_{EAD}=EA+AD+ED=15+16+27=58\ cm$