Find the area of the shaded region in the figure,where a circular arc of radius $6 \, cm$ has been drawn with vertex $O$ of an equilateral triangle $OAB$ of side $12 \, cm$ as centre. $\left[ \text{Use } \pi = \frac{22}{7} \right]$

(N/A) The shaded region consists of the area of the equilateral triangle $OAB$ and the area of the major circle excluding the sector $OCDE$ that overlaps with the triangle.
$1$. Area of equilateral triangle $OAB = \frac{\sqrt{3}}{4} \times (side)^2 = \frac{\sqrt{3}}{4} \times (12)^2 = 36\sqrt{3} \, cm^2$.
$2$. Area of the circle with radius $r = 6 \, cm = \pi r^2 = \frac{22}{7} \times 6^2 = \frac{792}{7} \, cm^2$.
$3$. The angle of the equilateral triangle is $60^{\circ}$. The area of the sector $OCDE$ (which is the overlapping part) $= \frac{60^{\circ}}{360^{\circ}} \times \pi r^2 = \frac{1}{6} \times \frac{22}{7} \times 36 = \frac{132}{7} \, cm^2$.
$4$. Total area of the shaded region = Area of $\triangle OAB +$ Area of circle $-$ Area of sector $OCDE$.
$= 36\sqrt{3} + \frac{792}{7} - \frac{132}{7} = 36\sqrt{3} + \frac{660}{7} \, cm^2$.