[tex]\it \Delta ABC-dreptunghic, \widehat A=90^o,\ \stackrel{T.P.}{\Longrightarrow } AC^2=BC^2-AB^2=12^2-10^2=\\ \\ =(12-10)(12+10)=2\cdot22=2\cdot2\cdot11=2^2\cdot11 \Rightarrow AC=2\sqrt{11} cm\\ \\ \mathcal{A}=\dfrac{c_1\cdot c_2}{2} =\dfrac{AB\cdot AC}{2}=\dfrac{10\cdot2\sqrt{11}}{2}=10\sqrt{11}\ cm^2[/tex]
