Question

1. In triunghiul ABC se cunosc (∡A)=90° , AB=9cm si AC=12cm. Aflati BC si AD, unde AD
⊥ BC, D ∈ (BC)

2. In triunghiul ABC se cunosc m(∡A)=90° , AB=6 cm si BC=10 cm. Aflati AC si AD, unde AD ⊥ BC, D ∈ (BC)

3. In triunghiul ABC se cunosc m(∡B)=90° , BD ⊥ AD, D ∈ (AC), BC= 16 cm si DC=12,8 cm. Aflati perimetrul triunghiului ABC

Answer 1

$\it 1)\ ABC-dreptunghic,\ m(\hat A)=90^o \stackrel{T.Pitagora}{\Longrightarrow}\ \ BC^2=AB^2+AC^2 \Rightarrow\\ \\ BC^2=9^2+12^2=81+144=225 \Rightarrow BC=\sqrt{225}=15\ cm\\ \\ AD=\dfrac{AB\cdot AC}{BC}=\dfrac{9\cdot12}{15}=\dfrac{\ \ 108^{(3}}{15}=\dfrac{^{2)}36}{\ 5}=\dfrac{72}{10}=7,2\ cm$

$\it 2)\ ABC-dreptunghic,\ m(\hat A)=90^o \stackrel{T.Pitagora}{\Longrightarrow}\ \ AC^2=BC^2-AB^2 \Rightarrow\\ \\ AC^2=10^2-6^2=100-36=64 \Rightarrow AC=\sqrt{64}=8\ cm\\ \\ AD=\dfrac{AB\cdot AC}{BC}=\dfrac{6\cdot8}{10}=\dfrac{\ \ 48}{10}=4,8\ cm$

$\it 3)\ ABC-dreptunghic,\ m(\hat A)=90^o \stackrel{T.catetei}{\Longrightarrow}\ AC^2=BC\cdot DC \Rightarrow\\ \\ \Rightarrow AC^2=16\cdot12,8 =16\cdot\dfrac{\ 128^{(2}}{10}=\dfrac{16\cdot64}{5} \Rightarrow AC=\sqrt{\dfrac{16\cdot64}{5}}=\dfrac{4\cdot8}{\sqrt5} \Rightarrow\\ \\ \\ \Rightarrow AC=\dfrac{^{\sqrt5)}32}{\ \sqrt5} =\dfrac{32\sqrt5}{5}=6,4\sqrt5\ cm$

$\it ABC-dreptunghic,\ m(\hat A)=90^o \stackrel{T.Pitagora}{\Longrightarrow}\ AB^2=BC^2-AC^2 \Rightarrow \\ \\ \Rightarrow AB^2=16^2-(6,4\sqrt5)^2=256-204,8=51,2=51,20=10,24\cdot5 \Rightarrow\\ \\ \Rightarrow AB=\sqrt{10,24\cdot5}=3,2\sqrt5\ cm$

$\it \mathcal{P}_{ABC} =AB+AC+BC=3,2\sqrt5+6,4\sqrt5+16=9,6\sqrt5+16\ cm$