In the given figure,$OACB$ 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. $\left[ \text{Use } \pi = \frac{22}{7} \right]$

(N/A) $(i)$ Since $OACB$ is a quadrant,it subtends a $90^{\circ}$ angle at the centre $O$.
Area of quadrant $OACB = \frac{90^{\circ}}{360^{\circ}} \times \pi r^{2}$
$= \frac{1}{4} \times \frac{22}{7} \times (3.5)^{2} = \frac{1}{4} \times \frac{22}{7} \times \left( \frac{7}{2} \right)^{2}$
$= \frac{1}{4} \times \frac{22}{7} \times \frac{49}{4} = \frac{11 \times 7}{8} = \frac{77}{8} \, cm^{2} = 9.625 \, cm^{2}$.
$(ii)$ Area of $\triangle BOD = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times OB \times OD$
$= \frac{1}{2} \times 3.5 \times 2 = 3.5 \, cm^{2} = \frac{7}{2} \, cm^{2}$.
Area of the shaded region = Area of quadrant $OACB - \text{Area of } \triangle BOD$
$= \frac{77}{8} - \frac{7}{2} = \frac{77 - 28}{8} = \frac{49}{8} \, cm^{2} = 6.125 \, cm^{2}$.