Question

va rog frumos dau coronita Pentru acest exercițiu (Geometrie)​

Answer 1

.............................

Answer 2

$\it \Delta ABC\ dreptunghic,\ \widehat A=90^o,\ \ \stackrel{T.Pitagora}{\Longrightarrow}\ BC^2=AB^2+AC^2 \Rightarrow \\ \\ \Rightarrow BC^2=15^2+20^2=225+400=625=25^2 \Rightarrow BC=25\ cm\\ \\ \mathcal{P}=AB+AC+BC=15+20+25=60\ cm\\ \\ \mathcal{A}=\dfrac{c_1\cdot c_2}{2}=\dfrac{AB\cdot AC}{2}=\dfrac{15\cdot20}{2}=15\cdot10=150\ cm^2\\ \\ \\ AD=\dfrac{AB\cdot AC}{BC}=\dfrac{15\cdot20}{25}=\dfrac{300}{25}=12\ cm\\ \\ \\ AM=\dfrac{BC}{2}=\dfrac{25}{2}=12,5\ cm\\ \\ \\ \Delta ABD\ \Rightarrow BD=9 cm,$

deoarece (9,  12,  15) - triplet  pitagoreic