`(D =H)`
Xét \(\Delta ABC\) vuông tại `A`, đường cao `AH`
Ta có: \(BC^2=AB^2+AC^2\) (đl Pytago)
\(\Rightarrow BC^2=5^2+12^2\\ \Rightarrow BC^2=25+144=169\\ \Rightarrow BC=\sqrt{169}=13\left(cm\right)\)
Ta có: \(AH\cdot BC=AB\cdot AC\left(htl\right)\)
\(\Rightarrow AH\cdot13=5\cdot12\\ \Rightarrow AH\cdot13=60\\ \Rightarrow AH=\dfrac{60}{13}\left(cm\right)\)
Ta có: \(AB^2=BH\cdot BC\left(htl\right)\)
\(\Rightarrow5^2=BH\cdot13\\ \Rightarrow25=BH\cdot13\\ \Rightarrow BH=\dfrac{25}{13}\left(cm\right)\)
Ta có: \(AC^2=CH\cdot BC\left(htl\right)\)
\(\Rightarrow12^2=CH\cdot13\\ \Rightarrow144=CH\cdot13\\ \Rightarrow CH=\dfrac{144}{13}\left(cm\right)\)
Vậy: \(BC=13cm;\\ AH=\dfrac{60}{13}\approx4,6cm;\\ BH=\dfrac{25}{13}\approx1,92cm;\\ CH=\dfrac{144}{13}\approx11,1cm.\)
