[tex]\it DC=#MC=3\cdot10=30\ cm\\ \\ AB=DC=30\ cm\ (laturi\ \ opuse\ \ \^{i}n\ paralelogram)[/tex]
[tex]\it \Delta MNC-dr,\ \widehat N=90^o,\ \stackrel{T.Pitagora}{\Longrightarrow}\ NC=8\ cm\\ \\ (6,\ \ 8,\ \ 10)\ -\ triplet\ \ pitagoreic[/tex]
[tex]\it CM||AB\ \stackrel{T.f.a.}{\Longrightarrow}\ \Delta NAB\sim\Delta NCM \Rightarrow \dfrac{NA}{NC}=\dfrac{AB}{CM} \Rightarrow \dfrac{NA}{8}=\dfrac{30}{10} \Rightarrow NA=24cm\\ \\ \\ AC=NA+NC=24+8=32\ cm[/tex]
