Three squares

Consider the diagram. By ratio and proportion, we have

$\dfrac{x}{6}=\dfrac{4}{10}$ $\implies$ $x=\dfrac{12}{5}$

It follows that $y=4-x=4-\dfrac{12}{5}=\dfrac{8}{5}$ .

By ratio and proportion again, we have

$\dfrac{a}{8}=\dfrac{6}{14}$ $\implies$ $a=\dfrac{24}{7}$

It follows that $b=6-a=6-\dfrac{24}{7}=\dfrac{18}{7}$ .

The area of the shaded region is equal to the areas of the three squares minus the areas of the four triangles (unshaded part). We have

$A=4^2+6^2+8^2-\dfrac{1}{2}\left[8\left(\dfrac{8}{5}\right)+6\left(\dfrac{12}{5}\right)+6\left(\dfrac{18}{7}\right)+8\left(\dfrac{24}{7}\right)\right]=116-\dfrac{1114}{35}=\boxed{\dfrac{2946}{35}}$