Лице на четириъгълник

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Даден е четириъгълник ABCD и AD=[tex]\frac{23}{5 }[/tex] ; AB=10 ; BC=[tex]3\sqrt{5}[/tex] ; cos[tex]\angle[/tex]BAD=[tex]\frac{3}{5 }[/tex] ; cos[tex]\angle[/tex]ABC=[tex]\frac{2\sqrt{5} }{ 5}[/tex] . Да се намери лицето на четириъгълника.
С 2 cos теореми съответно за ▲ABC и ▲ADB намерих дианоналите AC=5 и BD=[tex]\frac{11\sqrt{14} }{5 }[/tex]

[Xixibg]

[tex]H\in AB ; DH\bot AB[/tex]
[tex]cos(\angle BAD)=\frac{AH}{AD} ; =>AH=AD.cos(\angle BAD)=\frac{23}{5}.\frac{3}{5}=\frac{69}{25}[/tex]
[tex]AD^2=AH^2+DH^2 ; =>DH=\sqrt{\frac{23.23.25}{25.25}-\frac{69.69}{25.25}}=\sqrt{\frac{23^2(25-9)}{25^2}}=\frac{92}{25}[/tex]

[tex]M\in AB ; CM\bot AB[/tex]
[tex]cos (\angle ABC)=\frac{BM}{BC} ; =>BM=BC.cos(\angle ABC)=3\sqrt{5}.\frac{2\sqrt{5}}{5}=\frac{3.2.5}{5}=6[/tex]
[tex]BC^2=BM^2+CM^2 ; =>CM=\sqrt{45-36}=\sqrt{9}=3[/tex]

[tex]MH=AB-AH-BM=10-6-\frac{69}{25}=\frac{31}{25}[/tex]
[tex]S_{AHD}=\frac{AH.DH}{2}=\frac{\frac{92}{25}.\frac{69}{25}}{2}=\frac{3174}{625}[/tex]
[tex]S_{BMC}=\frac{BM.CM}{2}=\frac{3.6}{2}=9[/tex]
[tex]S_{CMHD}=\frac{(CM+DH)MH}{2}=\frac{(3+\frac{92}{25}).\frac{31}{25}}{2}=\frac{5177}{1250}[/tex]
[tex]S_{ABCD}=S_{CMHD}+S_{AHD}+S_{BMC}=\frac{5177}{1250}+\frac{3174}{625}+9=\frac{22775}{1250}=\frac{911}{50}[/tex]

[vikc]

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