Ta có : \(\dfrac{BC}{CD}=\dfrac{3}{4}\Rightarrow\) \(\dfrac{AD}{AB}=\dfrac{3}{4}\)
Xét \(\Delta DHA\) và \(\Delta DAB\) ta có :
\(\left\{{}\begin{matrix}\widehat{DHA}=\widehat{DAB}\left(=90^0\right)\\\widehat{D}:chung\end{matrix}\right.\)
\(\Rightarrow\Delta DHA\sim\Delta DAB\left(g-g\right)\)
\(\Rightarrow\) \(\dfrac{HD}{AD}=\dfrac{AH}{AB}\) \(\Rightarrow\) \(\dfrac{HD}{AH}=\dfrac{AD}{AB}\)\(\Rightarrow\) \(\dfrac{HD}{12}=\dfrac{3}{4}\Rightarrow HD=\dfrac{12.3}{4}=9cm\)
Theo định lý py-ta-go cho \(\Delta AHD\) ta có :
\(AD=\sqrt{AH^2+HD^2}=\sqrt{12^2+9^2}=15cm\)
\(\Rightarrow AB=\dfrac{AH.AD}{DH}=\dfrac{12.15}{9}=20cm\)
\(\Rightarrow\left\{{}\begin{matrix}C_{ABC}=2\left(AB+AD\right)=2\left(15+20\right)=70cm\\S_{ABC}=AB.CD=15.20=300cm^2\end{matrix}\right.\)
