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$\sqrt {2}$)