Find the area of a quadrant of a circle whose circumference is 22 cm.

Let radius of the circle = r

$\therefore \quad 2 \pi \mathrm{r}=22$

$\Rightarrow 2 \times \frac{22}{7} \times r=22$

$\Rightarrow \mathrm{r}=22 \times \frac{\mathbf{7}}{\mathbf{2 2}} \times \frac{\mathbf{1}}{\mathbf{2}}=\frac{\mathbf{7}}{\mathbf{2}} \mathrm{cm}$

Here, $\theta=90^{\circ}$

$\therefore$ Area of the $\left(\frac{\mathbf{1}}{\mathbf{4}}\right)^{\text {th }}$ quadrant of the circle,

$=\frac{\theta}{360} \times \pi r^{2}=\frac{90^{\circ}}{360^{\circ}} \times \frac{22}{7}\left(\frac{7}{2}\right)^{2} \mathrm{~cm}^{2}$

$=\frac{1}{4} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \mathrm{~cm}^{2}=\frac{77}{8} \mathrm{~cm}^{2}$