BD=BO+DO=26cm
Xét ΔBAD vuôg tại A có AO là đườg cao
nên \(AO^2=OD\cdot OB\)
hay AO=12(cm)
\(AB=\sqrt{8\cdot26}=4\sqrt{13}\left(cm\right)\)
\(AD=\sqrt{18\cdot26}=6\sqrt{13}\left(cm\right)\)
\(AC=\dfrac{AD^2}{AO}=\dfrac{\left(6\sqrt{13}\right)^2}{12}=39\left(cm\right)\)
\(DC=\sqrt{39^2-\left(6\sqrt{3}\right)^2}=3\sqrt{157}\left(cm\right)\)
\(S_{ABCD}=\dfrac{4\sqrt{13}+3\sqrt{157}}{2}\cdot6\sqrt{13}=3\sqrt{13}\left(4\sqrt{13}+3\sqrt{157}\right)\left(cm^2\right)\)
Đúng 6
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