In the given figure, PQRS is 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 cm2) of the circle?

Answer: Option B

Let radius of circle = r
∴ OA = 8 - r and OP = 8 + r
Now ∠OAP is right angle triangle and use pythagoras theorem in ΔOAP
⇒ (8 + r)² = (8 - r)² + 4²
64 + r² + 16r = 64 + r² - 16r + 16
32r = 16
r = $$\frac{1}{2}$$cm
And area of circle = πr²
$$\begin{array}{rl}& =\frac{22}{7}\times \frac{1}{2}\times \frac{1}{2}\\ & =\frac{11}{14}\text{ c}{\text{m}}^2\end{array}$$