An interesting geometry problem!

Area of the overlapping region is one half of the product of its diagonals. $A=\frac{1}{2}\times CB \times FE=CB \times DE$ .

We can get the common diagonal from right triangle $ABC$ as $BC=\sqrt{AC^2+CB^2}=\sqrt{3^2+7^2}=\sqrt{58}$ .

Half of the shorter diagonal can be obtained from that result and the similarity of the triangles $\triangle ABC$ and $\triangle DEB$ .

$\frac{DE}{DB}=\frac{3}{7}$ . So $DE=\frac{3}{7}\times DB=\frac{3}{7}\times \frac{\sqrt{58}}{2}$ .

Finally, the area is $A=CB \times DE=\sqrt{58} \times\frac{3}{7}\times \frac{\sqrt{58}}{2}=\frac{3\times 58}{14}\approx12.43$