PQR is a quadrant whose radius is 7 cm.  A circle is inscribed in the quadrant. What is the area of the circle(in cm²)?

154 (3 - 2\(\sqrt{2}\))

QR = 7 cm.,

QT = \(x\) cm,

OT = OU = r cm,

 QU = QO + OU = 2r +  \(x\) = 7cm.   ............(i)

OQ = \(\sqrt {QS^2 + OS^2}\) = \(\sqrt {r^2 + r^2}\)

⇒ \(x\) + r = \(\sqrt {2}\)r

⇒ \(x\) = (\(\sqrt {2}\) - 1) r

From (i)

2r + (\(\sqrt {2}\) - 1) r = 7

r (\(\sqrt {2}\) + 1) = 7

r = \(\frac{7}{\sqrt {2} + 1}\) = 7 (\(\sqrt {2}\) - 1)

⇒ Area of Circle = \(\pi \) r²

                         = \(\frac{22}{7}\) × 7 × 7 (\(\sqrt {2}\) - 1)²

                         = 154 (2 + 1 - 2\(\sqrt {2}\))

                         =154 (3 - 2\(\sqrt {2}\))