Find the area of a quadrant of a circle whose circumference is $22 \, cm$. [use $\pi = \frac{22}{7}$]

(N/A) Let the radius of the circle be $r$.
Circumference $= 2 \pi r = 22 \, cm$.
$r = \frac{22}{2 \pi} = \frac{11}{\pi} = \frac{11}{22/7} = \frac{11 \times 7}{22} = 3.5 \, cm$.
$A$ quadrant of a circle subtends an angle of $90^{\circ}$ at the centre.
Area of the quadrant $= \frac{90^{\circ}}{360^{\circ}} \times \pi r^{2} = \frac{1}{4} \times \frac{22}{7} \times (3.5)^{2}$.
Area $= \frac{1}{4} \times \frac{22}{7} \times 3.5 \times 3.5 = \frac{1}{4} \times 22 \times 0.5 \times 3.5$.
Area $= \frac{11 \times 3.5}{4} = \frac{38.5}{4} = 9.625 \, cm^{2}$.