Câu 2:
Ta có: \({S_{ABCD}} = {6^2} = 36,{S_{ABM}} = \dfrac{1}{2}BA.BM = \dfrac{1}{2}.6.3 = 9\)
\({S_{ADN}} = \dfrac{1}{2}AD.DN = \dfrac{1}{2}.6.4 = 12,{S_{MCN}} = \dfrac{1}{2}CM.CN = \dfrac{1}{2}.3.2 = 3\)
Vậy \({S_{AMN}} = {S_{ABCD}} - {S_{ABM}} - {S_{ADN}} - {S_{CMN}} = 36 - 9 - 12 - 3 = 12\)
\(MN = \sqrt {M{C^2} + C{N^2}} = \sqrt {9 + 4} = \sqrt {13} \)
\( \Rightarrow d\left( {A,MN} \right) = \dfrac{{2{S_{AMN}}}}{{MN}} = \dfrac{{2.12}}{{\sqrt {13} }} = \dfrac{{24\sqrt {13} }}{{13}}\)