In the given figure, $$PQRS$$ is a square whose side is $$8 cm$$. $$PQS$$ and $$QPR$$ are two quadrants. A circle is placed touching both the quadrants and the square as shown in the figure. What is the area (in $$cm^2$$) of the circle ? 

We can construct the image in following way :

Let say, $$r_1$$ is radius of smaller circle.

PS=PQ=PM is the radius of PQS sector. 

So, PM=8 cm.

From the above picture we can say that ,

AP=8/2=4 cm.

So, 

$$\left(8-r_1\right)^2+4^2=\left(8+r_1\right)^2\ .$$

or, $$\left(64-16r_1+r_1^2\right)+16=\left(64+16r_1+r_1^2\ \right).$$

or, $$32r_1=16\ .$$

or, $$r_1=\frac{1}{2}\ .$$

So, Area of smaller circle=$$\pi\times\left(\frac{1}{2}\right)^2=\frac{22}{7}\times\frac{1}{2}\times\frac{1}{2}=\frac{11}{14}\ cm^2\ .$$

B is correct choice.