Let A = x , y ∈ ℝ 2 : y ≥ 0 , 2 x ≤ y ≤ 4 - x - 1 2 and B = x , y ∈…

Given,

$2x\le y\le \sqrt{4-{\left(x-1\right)}^2} :y\ge 0$

$\Rightarrow y^2+{\left(x-1\right)}^2=4$ and $y=2x$

Now plotting the diagram we get,

Area of $A$: area of shaded region as shown in the above figure,

Shaded portion $=$ circular $\left(OABC\right)$

$-Ar\left(∆OAB\right)$$=\frac{\pi \left(4\right)}{4}-\frac{1}{2}\left(2\right)\left(1\right)$

Hence, Area of $\mathrm{A}=\left(\pi -1\right)$

Now, plotting the diagram of $0\le y\le \min \left\{2x,\sqrt{4-{\left(x-1\right)}^2}\right\}$ we get,

Area of $B$: area of shaded region as shown in the above figure,

Area of $\mathrm{B}=\mathrm{Ar}\left(∆AOB\right)+$ Area of arc of circle $\left(ABC\right)$

$=\frac{1}{2}\left(1\right)\left(2\right)+\frac{\pi {\left(2\right)}^2}{4}=\pi +1$

Hence, $\frac{A}{B}=\frac{\text{area} of A}{\text{area} of B}=\frac{\pi -1}{\pi +1}$