Find the area of the shaded region. Given that ABCD is a square with side length 'b' units and E is the midpoint of AB.

For Triangle ECD,

$$EC=DE=\sqrt{\ b^2+\left(\frac{b}{2}\right)^2}=\frac{\sqrt{\ 5}b}{2}$$

CD = b

$$Area\ of\ \triangle\ ECD=r\times\ s$$....(1)

$$s=\frac{\left(\frac{\sqrt{\ 5}b}{2}+\frac{\sqrt{\ 5}b}{2}+b\right)}{2}=\frac{b\left(\sqrt{\ 5}+1\right)}{2}$$...(2)

$$Area\ of\ \triangle\ ECD\ =\ \frac{B\times\ H}{2}=\frac{b\times\ b}{2}=\frac{b^2}{2}$$.......(3)
From EQ (1), (2) & (3)

we get, $$\frac{b^2}{2}=\frac{b\left(\sqrt{\ 5}+1\right)}{2}\times\ r$$

$$\ r=\frac{b}{\sqrt{\ 5}+1}=\frac{b\left(\sqrt{\ 5}-1\right)}{4}$$

Area of the incircle of ECD = $$\ \pi\ r^2$$ 

= $$\ \pi\ \times\ \left(\frac{b\left(\sqrt{\ 5}-1\right)}{4}\right)^2$$

$$=\frac{\pi\ b^2\left(5+1-2\sqrt{\ 5}\right)}{16}=\frac{\pi\ b^2\left(3-\sqrt{\ 5}\right)}{8}$$....(4)

For Triangle AED,

Inradius, $$r=\frac{AE+AD-DE}{2}=\frac{b+\frac{b}{2}-\frac{\sqrt{\ 5}b}{2}}{2}=\frac{b\left(3-\sqrt{\ 5}\right)}{4}$$

Area of the incircle of Triangle AED, = $$\frac{\pi\ b^2\ \left(3-\sqrt{\ 5}\right)^2}{4^2}$$

$$=\pi\ b^2\ \frac{\left(9+5-6\sqrt{\ 5}\right)}{16}=\frac{\pi\ b^2\ \left(14-6\sqrt{\ 5}\right)}{16}$$......(5)

As, Triangle AED $$\simeq\ $$ Triangle BEC

Area of the incircle of $$\triangle\ AED$$ = Area of the incircle of $$\triangle\ BEC$$.

Hence, the area of the shaded region $$=\frac{\pi\ b^2\left(14-6\sqrt{\ 5}\right)}{16}+\frac{\pi\ b^2\left(14-6\sqrt{\ 5}\right)}{16}+\frac{\pi b^2\ \left(3-\sqrt{\ 5}\right)}{8}$$

$$=\frac{\pi\ b^2}{16}\times\ \left(14-6\sqrt{\ 5}+14-6\sqrt{\ 5}+6-2\sqrt{\ 5}\right)$$

$$=\frac{\pi\ b^2\left(34-14\sqrt{\ 5}\right)}{16}$$