The figure shows a quarter circle in a square. Its centre is at one vertex, and it passes through two adjacent vertices. There are two semicircles on two adjacent sides as diameters. They create the shaded regions A and B.

Show that A and B have equal area.

Let the side of the square be $2r$.

Then, the quarter circle has radius $2r$, each semicircle has radius $r$.

The area of the quarter circle is $\frac{1}{4}\pi(2r)^2 = \pi r^2$.

The area of each semicircle is $\frac{1}{2}\pi r^2$.

So, the total area of the two semicircles is

$$2\left(\frac{1}{2}\pi r^2\right) = \pi r^2.$$

Hence, Area of quarter circle = Sum of areas of the two semicircles.

Now observe the figure carefully:

Region A is the overlap (common part) of the two semicircles.

Region B is the part of the quarter circle left after removing the parts covered by the semicircles.

Inside the square:

(sum of semicircle areas) = A + (parts outside A) and

quarter circle area = B + (same parts outside A)

Since the total areas are equal, subtracting the common remaining parts gives

$$A = B$$

Therefore, Area of region A = Area of region B.