Đáp án:
$\left[\begin{array}{l} AB=20(cm); AC=15(cm); BH=16(cm); CH=9(cm)\\ AB=15(cm); AC=20(cm);BH=9(cm);CH=16(cm) \end{array} \right.$
Giải thích các bước giải:
$\Delta ABC $ vuông tại $A$, đường cao $AH$
$\Rightarrow \left\{\begin{array}{l} AH.BC=AB.AC\\AB^2+AC^2=BC^2 \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} AB.AC=300\\AB^2+AC^2=625\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} AB=\dfrac{300}{AC}\\\dfrac{300^2}{AC^2}+AC^2=625\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} AB=\dfrac{300}{AC}\\300^2+AC^4=625AC^2\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} AB=\dfrac{300}{AC}\\AC^4-625AC^2+90000=0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} AB=\dfrac{300}{AC}\\\left[\begin{array}{l} AC^2=400\\ AC^2=225\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} AC=20; AB=15 \\AC=15;AB=20 \end{array} \right.$
$\Delta ABC$ vuông tại $A$, đường cao $AH$
$\Rightarrow AB^2=BH.BC\\ \Rightarrow BH=\dfrac{AB^2}{BC}\\ \Rightarrow \left[\begin{array}{l} AB=20 \Rightarrow BH=16\\ AB=15 \Rightarrow BH=9 \end{array} \right.\\ \Rightarrow \left[\begin{array}{l} AB=20 \Rightarrow BH=16; CH=BC-BH=9\\ AB=15 \Rightarrow BH=9;CH=BC-BH=16 \end{array} \right.$
