If $A_i$  is the area bounded by $|x-a_i|+\left|y\right|=b_i$    $i\in N$    where  $a_{i+1}=a_i+\frac{3}{2}{b}_i$  and     $b_{i+1}=\frac{b_i}{2}{a}_1=0,b_1=32$ then  $A_3=\_\_\_\_$  

$$\begin{gathered} \begin{array}{l}x-a_i+y=b_i\\ x-a_i+y=-b_i\end{array}\}\Rightarrow \begin{array}{l}x+y=a_i+b_i\\ x+y=a_i-b_i\end{array}\} \\ \begin{array}{l}x-a_i-y=b_i\\ x-a_i-y=-b_i\end{array}\}\Rightarrow \begin{array}{l}x-y=a_i+b_i\\ x-y=a_i-b_i\end{array}\} \end{gathered}$$

Required area = $=A_i=\frac{4{b_i}^2}{2}=2{b_i}^2$

  $\begin{array}{l}{A}_1=2{b_1}^2{A}_2=2{b_2}^2,A_3=2{b_3}^2\\ b_2=\frac{b_1}{2}=16,b_3=\frac{b_2}{2}=8\\ \therefore A_3=2{\left(8\right)}^2=2\times 64=128\end{array}$