Given the quadrilateral $ABCD\ :\ A(0,0),B(7,0),C,D$. Point $P(\frac{11}{7},\frac{8\sqrt{5}}{7})$ so that $AP=3$ and $BP=6$. Slope of the line $AP\ :\ m_{AP}=\frac{8\sqrt{5}}{11}$. The circle, midpoint $A$ and radius $5$, intersects the line $AD\ :\ y=-\frac{11}{8\sqrt{5}} \cdot x$ in the point $D(-\frac{40\sqrt{5}}{21},\frac{55}{21})$. Slope of the line $BP\ :\ m_{BP}=-\frac{4\sqrt{5}}{19}$. The circle, midpoint $D$ and radius $10$, intersects the line $CD\ :\ y-y_{D}=\frac{19}{4\sqrt{5}}(x-x_{D})$ in the point $C(0,\frac{35}{3})$. Area $\triangle ABC$ is $\frac{1}{2} \cdot 7 \cdot \frac{35}{3}=\frac{245}{6}$.