[tex]\it ABCD-dreptunghi \Rightarrow DC=AB=12\sqrt3\ cm\\ \\ \Delta ABM-echilateral \Rightarrow \hat M=\widehat{(MAB)}=\widehat{(ABM)}=60^o\\ \\ \widehat{(MBC)}=30^o\ (complementul\ lui\ 60^o)\\ \\ tg\widehat{(MBC)}=tg\widehat{(QBC)}=\dfrac{QC}{BC} \Rightarrow tg30^o=\dfrac{QC}{12} \Rightarrow \dfrac{^{4)}\sqrt3}{\ 3}=\dfrac{QC}{12} \Rightarrow\\ \\ \\ \Rightarrow \dfrac{4\sqrt3}{12}=\dfrac{QC}{12} \Rightarrow QC=4\sqrt3\ cm[/tex]
[tex]\it Analog,\ pentru\ \Delta DAP \Rightarrow DP=4\sqrt3\ cm\\ \\ PQ=DC-(DP+CQ)=12\sqrt3-(4\sqrt3+4\sqrt3)=12\sqrt3-8\sqrt3=4\sqrt3\ cm[/tex]
[tex]\it Din\ \Delta QBC \Rightarrow \widehat{(CQB)}=60^o\ (complementul\ lui\ 30^o)\\ \\ \widehat{(PQM)}=\widehat{(CQB)}=60^o\ (opuse\ la\ v\hat arf)\\ \\ Deci, \Delta PQC-echilateral\ (av\hat and\ dou\breve a\ unghiuri\ de\ 60^o)[/tex]
[tex]\it \mathcal{A}_{MPQ}=\dfrac{\ell^2\sqrt3}{4}=\dfrac{PQ^2\sqrt3}{4}=\dfrac{(4\sqrt3)^2\sqrt3}{4}=\dfrac{16\cdot3\sqrt3}{4}=4\cdot3\sqrt3=12\sqrt3cm^2[/tex]
