[tex]\it \Delta ABC-dr,\ \hat A=90^o\ \stackrel{T.P.}{\Longrightarrow}\ AC^2=BC^2-AB^2=20^2-12^2=\\ \\ =(20-12)(20+12)=8\cdot32=8\cdot8\cdot4=64\cdot4=8^2\cdot4^2 \Rightarrow \\ \\ \Rightarrow AC=8\cdot2=16\ cm[/tex]
[tex]\it AD=\dfrac{AB\cdot AC}{BC}=\dfrac{\ 12\cdot16^{(2}}{20}=\dfrac{12\cdot 8}{10}=\dfrac{96}{10}=9,6\ cm\\ \\ \\ \mathcal{P}=AB+AC+BC=12+16+20=48\ cm[/tex]
