Calculate the area of the designed region in the figure,which is common between the two quadrants of circles of radius $8 \, cm$ each. $\left[\text{Use } \pi = \frac{22}{7}\right]$

(N/A) The designed area is the common region between two sectors $BAEC$ and $DAFC$.
Area of sector $BAEC = \frac{90^{\circ}}{360^{\circ}} \times \frac{22}{7} \times (8)^{2}$
$= \frac{1}{4} \times \frac{22}{7} \times 64$
$= \frac{22 \times 16}{7} = \frac{352}{7} \, cm^{2}$
Area of $\triangle BAC = \frac{1}{2} \times BA \times BC$
$= \frac{1}{2} \times 8 \times 8 = 32 \, cm^{2}$
Area of the designed portion $= 2 \times (\text{Area of segment } AEC)$
$= 2 \times (\text{Area of sector } BAEC - \text{Area of } \triangle BAC)$
$= 2 \times \left(\frac{352}{7} - 32\right) = 2 \left(\frac{352 - 224}{7}\right)$
$= \frac{2 \times 128}{7} = \frac{256}{7} \, cm^{2}$