Find the area of the shaded triangle

Let $O$ be the origin $(0,0)$ , $AF$ be perpendicular to $CD$ , and the $x$ -axis bisects $DE$ at $G$ . Since $OC=7$ is a diameter of the big circle, $\angle ODC = 90^\circ$ . Then $\triangle ODC$ and $\triangle AFC$ are similar. Then $\dfrac {CD}{OC} = \dfrac {CF}{AC} \implies CD = \dfrac {CF}{AC} \times OC = \dfrac {\sqrt{AC^2-AF^2}}{AC}\times OC = \dfrac {14\sqrt{14}}9$ .

We also note that $\triangle GDC$ and $\triangle AFC$ are similar. Then $DG = \dfrac {CD}{AC} \times AF = \dfrac {70 \sqrt{14}}{81}$ , $CG = dfrac {CD}{AC} \times CF = {70 \sqrt{14}}{81}$ , and the area of $\triangle CDE = DG \times CG = \dfrac {70 \sqrt{14}}{81} \times {392}{81} = \dfrac {27440\sqrt{14}}{6561} \approx \boxed{15.6}$ .