Đáp án:
Ta chứng minh được:
$\begin{array}{l}
\Delta ABH \sim \Delta CBA\left( {g - g} \right)\\
\Leftrightarrow \dfrac{{AB}}{{BC}} = \dfrac{{BH}}{{AB}}\\
\Leftrightarrow A{B^2} = BH.BC = 9.\left( {9 + 16} \right)\\
\Leftrightarrow A{B^2} = 9.25 = 225\\
\Leftrightarrow AB = 15\left( {cm} \right)\\
\Leftrightarrow AC = \sqrt {B{C^2} - A{B^2}} \\
= \sqrt {{{25}^2} - {{15}^2}} \\
= 20\\
Do:BM = CM = AM\\
= \dfrac{1}{2}BC = \dfrac{1}{2}.25 = 12,5\left( {cm} \right)\\
HM = BM - BH = 12,5 - 9 = 3,5\left( {cm} \right)\\
{S_{ABC}} = \dfrac{1}{2}.AH.BC = \dfrac{1}{2}.AB.AC\\
\Leftrightarrow AH = \dfrac{{15.20}}{{25}} = 12\\
\Leftrightarrow {S_{AHM}} = \dfrac{1}{2}.AH.HM = \dfrac{1}{2}.12.3,5 = 21\left( {c{m^2}} \right)
\end{array}$
