$$\mathrm{answer}\:\mathrm{is}\:\mathrm{unknown}\:\mathrm{to}\:\mathrm{me} \\$$
$${let}\angle{BAD}=\angle{BCD}=\theta,\:\angle{BDA}=\emptyset \\$$$$\frac{\mathrm{5}}{{sin}\emptyset}=\frac{\mathrm{6}{sin}\theta}{{sin}\theta}=\mathrm{6} \\$$$$\Rightarrow{sin}\emptyset=\frac{\mathrm{5}}{\mathrm{6}},\:{cos}\emptyset=−\frac{\sqrt{\mathrm{11}}}{\mathrm{6}} \\$$$$\mathrm{2}\left({DM}^{\mathrm{2}} +{AM}^{\mathrm{2}} \right)={DC}^{\mathrm{2}} +{AD}^{\mathrm{2}} \\$$$${DC}^{\mathrm{2}} =\left(\mathrm{6}{cos}\theta\right)^{\mathrm{2}} =\mathrm{36}{cos}^{\mathrm{2}} \theta \\$$$$\frac{{AD}}{{sin}\left(\mathrm{180}°−\theta−\emptyset\right)}=\frac{\mathrm{6}{sin}\theta}{{sin}\theta}=\mathrm{6} \\$$$${AD}=\mathrm{6}{sin}\left(\theta+\emptyset\right)=\mathrm{5}{cos}\theta−\sqrt{\mathrm{11}}{sin}\theta \\$$$${AD}^{\mathrm{2}} =\mathrm{25}{cos}^{\mathrm{2}} \theta−\mathrm{10}\sqrt{\mathrm{11}}{sin}\theta{cos}\theta+\mathrm{11}{sin}^{\mathrm{2}} \theta \\$$$${AC}^{\mathrm{2}} =\mathrm{5}^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} −\mathrm{2}×\mathrm{5}×\mathrm{6}{cos}\left(\mathrm{270}°−\mathrm{2}\theta−\emptyset\right) \\$$$$=\mathrm{61}+\mathrm{60}{sin}\left(\mathrm{2}\theta+\emptyset\right) \\$$$$=\mathrm{61}+\mathrm{60}\left[\mathrm{2}{sin}\theta{cos}\theta{cos}\emptyset+\left(\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \theta\right){sin}\emptyset\right] \\$$$$=\mathrm{61}−\mathrm{20}\sqrt{\mathrm{11}}{sin}\theta{cos}\theta+\mathrm{50}−\mathrm{100}{sin}^{\mathrm{2}} \theta \\$$$${AM}^{\mathrm{2}} =\frac{\mathrm{61}}{\mathrm{4}}−\mathrm{5}\sqrt{\mathrm{11}}{sin}\theta{cos}\theta+\frac{\mathrm{25}}{\mathrm{2}}−\mathrm{25}{sin}^{\mathrm{2}} \theta \\$$$$\mathrm{2}\left({DM}^{\mathrm{2}} +{AM}^{\mathrm{2}} \right)={DC}^{\mathrm{2}} +{AD}^{\mathrm{2}} \\$$$$\mathrm{2}{DM}^{\mathrm{2}} +\frac{\mathrm{61}}{\mathrm{2}}−\mathrm{10}\sqrt{\mathrm{11}}{sin}\theta{cos}\theta+\mathrm{25}−\mathrm{50}{sin}^{\mathrm{2}} \theta \\$$$$=\mathrm{36}{cos}^{\mathrm{2}} \theta+\mathrm{25}{cos}^{\mathrm{2}} \theta−\mathrm{10}\sqrt{\mathrm{11}}{sin}\theta{cos}\theta+\mathrm{11}{sin}^{\mathrm{2}} \theta \\$$$$\mathrm{2}{DM}^{\mathrm{2}} +\frac{\mathrm{61}}{\mathrm{2}}+\mathrm{25}=\mathrm{61}\left({sin}^{\mathrm{2}} \theta+{cos}^{\mathrm{2}} \theta\right) \\$$$${DM}^{\mathrm{2}} =\frac{\mathrm{11}}{\mathrm{4}}\Rightarrow{DM}=\frac{\sqrt{\mathrm{11}}}{\mathrm{2}}=\frac{\sqrt{{a}}}{\mathrm{2}} \\$$$$\Rightarrow{a}=\mathrm{11} \\$$
$$\boldsymbol{\mathrm{sir}}\:\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{question}}\:\boldsymbol{\mathrm{number}}\:\mathrm{65886} \\$$