In the given figure, PQRS is a square whose side is 12 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 ?

$\frac{99}{56}$

Let radius of circle = r

∴ OA = 12 - r, OQ = 12 + r

In ΔOAQ

(OQ)² = (OA)² + (AQ)²

(12 + r)² = (12 - r)² + 6²

144 + r² + 24r = 144 + r² - 24r + 36

       48r = 36

          r = $\frac{36}{48}$

          r = $\frac{3}{4}$

and area of circle = $\pi$r²

= $\frac{22}{7}$ × $\frac{3}{4}$ × $\frac{3}{4}$ = $\frac{99}{56}$ cm²