Area bounded by curves $\sqrt{\left|x\right|}+\sqrt{\left|y\right|}=\sqrt{a}$ and $x^2+y^2=a^2,a>0$ is divided by line $\left|x\right|+\left|y\right|=a$ into two parts and ratio of area of these two parts (larger area to smaller area) in first quadrant is $\frac{A}{3(B\pi -2)}$ . Then the value of A + B is

Given curves are $\sqrt{\left|x\right|}+\sqrt{\left|y\right|}=\sqrt{a} \dots .\left(1\right)$

$x^2+y^2=a^2,a>0 \dots ..\left(2\right)$

$\left|x\right|+\left|y\right|=a \dots .\left(3\right)$

$$\begin{gathered} A_1=\underset{0}{\overset{a}{\int }}[\sqrt{a^2-x^2}-(a-x)]dx={[\frac{x\sqrt{a^2-x^2}}{2}+\frac{a^2}{2}{\sin }^{-1}\left(\frac{x}{a}\right)]}_0^4-a^2+\frac{1}{2}\left(a^2\right)=\frac{a^2}{2}\left(\frac{\pi }{2}\right)-\frac{a^2}{2} \\ =\frac{a^2}{2}(\frac{\pi }{2}-1)=\frac{\left(\mathrm{\pi }-2\right)a^2}{4} \end{gathered}$$

$A_2=\underset{0}{\overset{a}{\int }}[(a-x)-{(\sqrt{a}-\sqrt{x})}^2]dx=\underset{0}{\overset{a}{\int }}(a-x-a-x+2\sqrt{a}\sqrt{x})dx=\underset{0}{\overset{a}{\int }}(2\sqrt{a}\sqrt{x}-2x)dx$

$=2\sqrt{a}\frac{2}{3}{\left[{\left(x\right)}^{3/2}\right]}_0^a-[a^2-0]=\frac{4}{3}\sqrt{a}.a^{3/2}-a^2=(\frac{4}{3}-1)a^2=\frac{1}{3}{a}^2$

$=\frac{A_1}{A_2}=\frac{\frac{a^2}{2}\left(\frac{\pi -2}{2}\right)}{\frac{1}{3}{a}^2}$ and $\frac{A_2}{A_1}=\frac{\frac{1}{3}{a}^2}{\frac{1}{4}{a}^2(\pi -2)}=\frac{4}{3(\pi -2)}=\frac{A}{3(B\pi -2)}\Rightarrow A=4,B=1$

So, $A+B=5.$