Đáp án:
${S_{ACH}} = 49,96c{m^2}$
Giải thích các bước giải:
Đặt $BC=x(x>0)$
Ta có:
$9CA = 5BC \Rightarrow CA = \dfrac{5}{9}BC = \dfrac{5}{9}x$
Lại có:
$\begin{array}{l}
\Delta ABC;\widehat A = {90^0};BC = x;CA = \dfrac{5}{9}x\\
\Rightarrow AB = \sqrt {B{C^2} - A{C^2}} = \dfrac{{2x\sqrt {14} }}{9}
\end{array}$
Mặt khác:
$\begin{array}{l}
{S_{ABC}} = \dfrac{1}{2}AB.AC\\
\Leftrightarrow 162 = \dfrac{1}{2}.\dfrac{{2x\sqrt {14} }}{9}.\dfrac{5}{9}x\\
\Leftrightarrow {x^2} = \dfrac{{13122}}{{5\sqrt {14} }}\\
\Leftrightarrow x = \dfrac{{81\sqrt 2 }}{{\sqrt {5\sqrt {14} } }}
\end{array}$
Ta có:
$\begin{array}{l}
{S_{ABC}} = \dfrac{1}{2}AH.BC\\
\Rightarrow AH = \dfrac{{2{S_{ABC}}}}{{BC}} = \dfrac{{2.162}}{{\dfrac{{81\sqrt 2 }}{{\sqrt {5\sqrt {14} } }}}} = 2\sqrt {10\sqrt {14} } cm
\end{array}$
Lại có:
$\begin{array}{l}
\Delta ACH;\widehat {AHC} = {90^0};AC = \dfrac{5}{9}BC = \dfrac{{9\sqrt {10} }}{{\sqrt[4]{{14}}}};AH = 2\sqrt {10\sqrt {14} } \\
\Rightarrow HC = \sqrt {A{C^2} - A{H^2}} = 8,17\\
\Rightarrow {S_{ACH}} = \dfrac{1}{2}AH.CH = \dfrac{1}{2}.2\sqrt {10\sqrt {14} } .8,17 = 49,96cm^2
\end{array}$
Vậy ${S_{ACH}} = 49,96c{m^2}$
