Giải thích các bước giải:
$\begin{array}{l}
a)\,ap\,dung\,he\,thuc\,luong\,trong\,tam\,giac:\\
+ A{H^2} = BH.CH\\
\Rightarrow CH = \frac{{A{H^2}}}{{BH}} = \frac{{{6^2}}}{2} = 18\left( {cm} \right)\\
+ trong\,tam\,giac\,vuong\,ABH\,co:\\
A{B^2} = A{H^2} + B{H^2} = {6^2} + {2^2} = 40 \Rightarrow AB = \sqrt {40} = 2\sqrt {10} \left( {cm} \right)\\
+ \Delta AHC\,\,vuong:\\
A{C^2} = C{H^2} + A{H^2} = {18^2} + {6^2} = 360\\
\Rightarrow AC = \sqrt {360} = 6\sqrt {10} \left( {cm} \right)\\
b){S_{AHC}} = \frac{1}{2}AH.HC = \frac{1}{2}.6.18 = 54\left( {cm} \right)\\
c)\cot C = \frac{{AC}}{{AB}} = \frac{{6\sqrt {10} }}{{2\sqrt {10} }} = 3
\end{array}$
