Finding the area

By similar triangles,

$\dfrac{x}{10}=\dfrac{8}{18}$ $\implies$ $x=\dfrac{40}{9}$

$\dfrac{w}{4}=\dfrac{8}{18}$ $\implies$ $w=\dfrac{16}{9}$

It follows that, $y=8-\dfrac{40}{9}=\dfrac{32}{9}$ .

$A_1=\dfrac{1}{2}\left(\dfrac{32}{9}\right)(8)=\dfrac{128}{9}$

$A_2=\dfrac{1}{2}\left(\dfrac{40}{9}+\dfrac{16}{9}\right)(6)=\dfrac{56}{3}$

$A_3=4^2-\dfrac{1}{2}\left(\dfrac{16}{9}\right)(4)=\dfrac{112}{9}$

Therefore, the desired area is,

$A=\dfrac{128}{9}+\dfrac{56}{3}+\dfrac{112}{9}=\dfrac{136}{3}$ .

Finally,

$a+b=136+3=$ $\color{#D61F06}\boxed{139}$