In a circular table cover of radius $32 \, cm$,a design is formed by leaving an equilateral triangle $ABC$ in the middle as shown in the figure. Find the area of the design. [Use $\pi = \frac{22}{7}$]

(N/A) Radius $(r)$ of the circle $= 32 \, cm$.
$AD$ is the median of $\triangle ABC$.
Since $O$ is the centroid of the equilateral triangle,$AO = \frac{2}{3} AD = 32 \, cm$.
Therefore,$AD = 32 \times \frac{3}{2} = 48 \, cm$.
In $\triangle ABD$,by Pythagoras theorem:
$AB^2 = AD^2 + BD^2$
$AB^2 = (48)^2 + \left(\frac{AB}{2}\right)^2$
$AB^2 - \frac{AB^2}{4} = 2304$
$\frac{3 AB^2}{4} = 2304$
$AB^2 = \frac{2304 \times 4}{3} = 3072$
$AB = \sqrt{3072} = 32 \sqrt{3} \, cm$.
Area of equilateral triangle $\triangle ABC = \frac{\sqrt{3}}{4} \times (side)^2 = \frac{\sqrt{3}}{4} \times (32 \sqrt{3})^2$
$= \frac{\sqrt{3}}{4} \times 1024 \times 3 = 768 \sqrt{3} \, cm^2$.
Area of circle $= \pi r^2 = \frac{22}{7} \times (32)^2 = \frac{22}{7} \times 1024 = \frac{22528}{7} \, cm^2$.
Area of the design = Area of circle $-$ Area of $\triangle ABC$
$= \left(\frac{22528}{7} - 768 \sqrt{3}\right) \, cm^2$.