Find the area of a quadrant of a circle whose circumference is $\mathbf{44}\mathbf{ }\mathbf{cm}$.

Given: Circumference of the circle = $\mathbf{44}\mathbf{ }\mathbf{cm}$

Suppose r is the circle's radius.

We know that the circumference of a circle = $\mathbf{2}\mathbf{\pi R}$

Substituting the values:

$\Rightarrow 2\mathrm{\pi r} = 44$

$\Rightarrow \mathrm{r} = \frac{22}{\mathrm{\pi }}$

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

$\mathbf{\Rightarrow }\mathbf{r}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{7}\mathbf{ }\mathbf{cm}$

Now, a quadrant is one-fourth of a circle in terms of area.

So, the area of the quadrant = $\frac{\mathbf{Area}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{the}\mathbf{ }\mathbf{circle}}{\mathbf{4}}$

= $\frac{{\mathbf{\pi r}}^{\mathbf{2}}}{\mathbf{4}}$

= $\frac{1}{4}\times \frac{22}{7}\times 7\times 7$ cm²

= $\frac{77}{2}$ cm²

= $\mathbf{38}\mathbf{.}\mathbf{5}\mathbf{ }{\mathbf{cm}}^{\mathbf{2}}$

Thus, the required area is $\mathbf{38}\mathbf{.}\mathbf{5}\mathbf{ }{\mathbf{cm}}^{\mathbf{2}}$.