Đáp án:
$\begin{array}{l} 2)Theo\,Pytago:\\ A{B^2} + A{C^2} = B{C^2}\\ \Rightarrow B{C^2} = {6^2} + {8^2} = 100\\ \Rightarrow BC = 10\left( {cm} \right)\\ {S_{ABC}} = \dfrac{1}{2}.AB.AC = \dfrac{1}{2}.AH.BC\\ \Rightarrow AH = \dfrac{{6.8}}{{10}} = 4,8\left( {cm} \right)\\ \Rightarrow HB = \sqrt {A{B^2} - A{H^2}} = \sqrt {{6^2} - 4,{8^2}} = 3,6\left( {cm} \right)\\ \Rightarrow HC = 10 - HB = 6,4\left( {cm} \right)\\3)\\ Theo\,Pytago:\\ A{B^2} = A{H^2} + B{H^2}\\ \Rightarrow AH = \sqrt {{{12}^2} - {6^2}} = 6\sqrt 3 \left( {cm} \right)\\ Do:A{B^2} = BH.BC\\ \Rightarrow BC = \dfrac{{{{12}^2}}}{6} = 24\left( {cm} \right)\\ \Rightarrow HC = 24 - 6 = 18\left( {cm} \right)\\ \Rightarrow AC = \sqrt {{{24}^2} - {{12}^2}} = 12\sqrt 3 \left( {cm} \right) \end{array} $
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