In Fig., OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the 

(i) quadrant OACB, 

(ii) shaded region

We know that,  formula for the area of the sector of a circle

Area of the sector = $\frac{\theta }{{360}^0}$ $\times$ ${\mathrm{\pi r}}^2$

we know that the whole circle is divided into 4 equal quadrant.

Thus, the angle at the center of a quadrant of a circle = $\theta$ = $\frac{{360}^0}{4}$ = 90⁰

and the area of the quadrant OACB = 1/4${\mathrm{\pi r}}^2$

= 1/4 $\times$ 22/7 $\times$ (3.5)²

= 1/5 $\times$ 22/7 $\times$ 7/2 $\times$ 7/2

= 77/8 cm²

Here, triangle BDO is a right angled triangle

Area of triangle = 1/2 $\times$ Base $\times$ Height

= 1/2 $\times$ 7/2 $\times$ 2

= 7/2 cm²

Thus we get 

Area of shaded region = Area of quadrant OACB $-$ Area of triangle

= 77/8 $-$ 7/2

= 49/8 cm²