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