Đáp án:
$\begin{array}{l}
a)Theo\,Pytago:\\
B{C^2} = A{B^2} + A{C^2} = {9^2} + {12^2} = 225\\
\Rightarrow BC = 15\left( {cm} \right)\\
{S_{ABC}} = \dfrac{1}{2}.AH.BC = \dfrac{1}{2}.AB.AC\\
\Rightarrow AH = \dfrac{{AB.AC}}{{BC}} = \dfrac{{9.12}}{{15}} = 7,2\left( {cm} \right)\\
b)Xet:\Delta ABH;\Delta AHE:\\
+ \widehat {BAH}\,chung\\
+ \widehat {AHB} = \widehat {AEH} = {90^0}\\
\Rightarrow \Delta ABH \sim \Delta AHE\left( {g - g} \right)\\
\Rightarrow \dfrac{{AB}}{{AH}} = \dfrac{{AH}}{{AE}}\\
\Rightarrow AB.AE = A{H^2}\\
TT:AC.AF = A{H^2}\\
\Rightarrow AB.AE = AC.AF\\
c)Do:\left\{ \begin{array}{l}
AB.AE = A{H^2}\\
AC.AF = A{H^2}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
AE = \dfrac{{7,{2^2}}}{9} = 5,76\left( {cm} \right)\\
AF = 4,32\left( {cm} \right)
\end{array} \right.\\
\Rightarrow {S_{BECF}} = {S_{ABC}} - {S_{AFE}}\\
= \dfrac{1}{2}.9.12 - \dfrac{1}{2}.5,76.4,32\\
= 41,5584\left( {c{m^2}} \right)
\end{array}$
