Call \(AD\perp BC=\left\{E\right\}\)

We have: \(a^2+b^2=13^2=169\)(Pythagorean theorem in triangle EAB)

We also have:

\(\left(a+10\right)^2+\left(b+24\right)^2=39^2=1521\)

\(\Rightarrow a^2+20a+100+b^2+48b+576=1521\)

\(\Rightarrow a^2+20a+b^2+48b=1521-576-100=845\)

Because \(a^2+b^2=169\)

\(\Rightarrow20a+48b=845-169=676\)

\(\Rightarrow5a+12b=169\)

\(a^2+b^2=169\)

=> \(\left\{{}\begin{matrix}a^2=5a\\b^2=12b\end{matrix}\right.\)

\(\Rightarrow a=5;b=12\)

=> DE = 15

BE = 12

=> \(S_{DBE}=\dfrac{15.12}{2}=90\left(units^2\right)\)

\(S_{ABE}=\dfrac{5.12}{2}=30\left(units^2\right)\)

\(\Rightarrow S_{ABCD}=90-30=60\left(units^2\right)\)

Lê Quốc Trần Anh selected this answer.