[tex]\it \Delta BDA\ -\ dreptunghic,\ m(\widehat{D})=90^o,\ \stackrel{TP}{\Longrightarrow} BD^2=AB^2-AD^2=\\ \\ \\ =20^2-12^2=(20-12)(20+12)=8\cdot32=8\cdot8\cdot4=64\cdot4 \Rightarrow \\ \\ \\ \Rightarrow BD=\sqrt{64\cdot4}=8\cdot2=16\ cm[/tex]
[tex]\it AO-median\breve a\ \hat\imath n\ \Delta BDA \Rightarrow \mathcal{A}_{AOB}=\dfrac{\mathcal{A}_{BDA}}{2}=\dfrac{\dfrac{c_1\cdot c_2}{2}}{2}=\\ \\ \\ =\dfrac{16\cdot12}{2\cdot2}=4\cdot12=48\ cm^2[/tex]
[tex]\it R\breve a spunsul\ \ corect\ \ este:\ \ c)\ 48\ cm^2[/tex]
