Đáp án:
$\begin{array}{l}
a)Theo\,Pytago:\\
A{B^2} + A{C^2} = B{C^2}\\
\Rightarrow A{C^2} = {10^2} - {6^2} = 64\\
\Rightarrow AC = 8\left( {cm} \right)\\
\Rightarrow AH = \frac{{AB.AC}}{{BC}} = \frac{{6.8}}{{10}} = 4,8\left( {cm} \right)\\
Do:A{B^2} = BH.BC\\
\Rightarrow BH = \frac{{{6^2}}}{{10}} = 3,6\left( {cm} \right)\\
\Rightarrow CH = 6,4\left( {cm} \right)\\
b)\\
A{C^2} = CH.BC\\
\Rightarrow A{C^2} = 6,4.BC\\
Do:A{B^2} + A{C^2} = B{C^2}\\
\Rightarrow 4,{8^2} + 6,4BC = B{C^2}\\
\Rightarrow B{C^2} - 6,4BC - 4,{8^2} = 0\\
\Rightarrow BC = 8,968\left( {cm} \right)\\
\Rightarrow BH = 2,568\left( {cm} \right)\\
\Rightarrow AH = \sqrt {BH.CH} = 4,054\left( {cm} \right)\\
\Rightarrow AC = \sqrt {6,4.8,968} = 4,57\left( {cm} \right)
\end{array}$
