Đáp án:
$AH = \dfrac{{12}}{{\sqrt {13} }}cm;HC = \dfrac{{72}}{{13}}cm;BC = \dfrac{{98}}{{13}}cm$
Giải thích các bước giải:
Ta có:
$\Delta ABC;\widehat A = {90^0};AH \bot BC = H;AB = 4cm;AC = 6cm;BH = 2cm$
$\begin{array}{l}
a)\dfrac{1}{{A{H^2}}} = \dfrac{1}{{A{B^2}}} + \dfrac{1}{{A{C^2}}} = \dfrac{1}{{{4^2}}} + \dfrac{1}{{{6^2}}} \Rightarrow AH = \dfrac{{12}}{{\sqrt {13} }}cm\\
b)A{H^2} = HB.HC \Rightarrow HC = \dfrac{{A{H^2}}}{{HB}} = \dfrac{{{{\left( {\dfrac{{12}}{{\sqrt {13} }}} \right)}^2}}}{2} = \dfrac{{72}}{{13}}cm\\
c)BC = BH + CH = 2 + \dfrac{{72}}{{13}} = \dfrac{{98}}{{13}}cm
\end{array}$
Vậy $AH = \dfrac{{12}}{{\sqrt {13} }}cm;HC = \dfrac{{72}}{{13}}cm;BC = \dfrac{{98}}{{13}}cm$
