Circle Quadrant area and ratio I came across a question while doing past papers for a math competition, with such a question:

I have completed it, and gotten the correct answer by showing how the two semicircles add up (not including overlap) to the exact same area as the circle quadrant, and therefore, the overlap must equal the remaining area (1:1), but think that this approach seems a little sketchy or unjustified. Is there any other more "better" way to approach such problem?
 A: Let the area of the quadrant of $2$cm be $Q$ that of semicircle on wall be $S_1$ and that of semicircle on floor be $S_2$.
$Q=\pi(2^2)/4=\pi$
$S_1=S_2=\pi(1^2)/2=\pi/2$
We know: $A\cup B=A+B-A\cap B$
So, $Q=S_1+S_2+A_2-A_1$
$\Rightarrow \pi=\pi/2+\pi/2+A_2-A_1$
$\Rightarrow A_1=A_2$
Thus, $A_1:A_2=1:1$
The "sketchy" thing about this is that what if $A_1=A_2=0$ in which case the ratio won't be defined.
Let's look at this:

Finding area $A_1$:
Consider the square $ADEF$ of side $1$cm. It contains $2$ circle quadrants, $AED$ and $AEF$ of radius $1$cm.
Again using: $A\cup B=A+B-A\cap B$
$Ar(ADEF)=Ar(AED)+Ar(AED)-A_1$
$\Rightarrow 1^2=S_1/2+S_2/2-A_1=\pi/2-A_1$
$\Rightarrow A_1=\pi/2-1$
Finding area $A_2$:
Consider the big quadrant $ABC$ of radius $2$cm. It contains $2$ circle quadrants, $CDE$ and $BFE$ of radius $1$cm each and the unit sqaure $ADEF$ .
$Q=Ar(CDE)+Ar(BFE)+Ar(ADEF)+A_2$
$\Rightarrow \pi=S_1/2+S_2/2+1-A_2=\pi/2+1+A_2$
$\Rightarrow A_2=\pi/2-1$
So, $A_1=A_2=\pi/2-1\neq0$.
Thus, $\bf A_1:A_2=1:1$
