a) Áp dụng định lý Pytago ta có:
\(BC^2=AB^2+AC^2\Rightarrow AB^2=BC^2-AC^2=20^2-12^2=256\Rightarrow AB=16\left(cm\right)\)
Xét \(\Delta ABH\) và \(\Delta CBA\) có:
\(\widehat{BAC}=\widehat{BHA}=90^0\)
\(\widehat{B}\) chung
\(\Rightarrow\Delta ABH\sim CBA\) (g.g) \(\Rightarrow\dfrac{AB}{AH}=\dfrac{BC}{AC}\Rightarrow AH=\dfrac{AB.AC}{BC}=\dfrac{16.12}{20}=9,6\left(cm\right)\)
Ta cũng có: \(\dfrac{AB}{BH}=\dfrac{BC}{AB}\Rightarrow BH=\dfrac{AB^2}{BC}=\dfrac{16^2}{20}=12,8\left(cm\right)\)
b) \(S_{\Delta ABC}=\dfrac{1}{2}AB.AC=\dfrac{1}{2}.16.12=96\left(cm^2\right)\)