6" alt="CH^2=AH\cdot BH
\\\
CH= \sqrt{AH\cdot BH} = \sqrt{3\cdot 9} = \sqrt{27}
\\\
AC= \sqrt{AH^2+HC^2} = \sqrt{9+27} =6
\\\
BC= \sqrt{BH^2+HC^2} = \sqrt{81+27} =\sqrt{108}
\\\
\sqrt{108} >6" align="absmiddle" class="latex-formula">
Ответ: корень 108