A sector with an angle of 45 degrees is cut out from a circle of radius 3.5cm, find the area of the sector. Take pi = 22/7

The angle of the sector is $\theta = 45^\circ$.
The radius of the circle is $r = 3.5 \text{ cm}$.
We are asked to use $\pi = \frac{22}{7}$.
The formula for the area of a sector is:

$$\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2$$

First, simplify the fraction $\frac{45}{360}$:

$$\frac{45}{360} = \frac{1}{8}$$

Next, calculate $(3.5)^2$:

$$(3.5)^2 = \left(\frac{7}{2}\right)^2 = \frac{49}{4}$$

Now substitute these back into the area formula:

$$\text{Area} = \frac{1}{8} \times \frac{22}{7} \times \frac{49}{4} \text{ cm}^2$$

$$\text{Area} = \frac{1 \times 22 \times 49}{8 \times 7 \times 4} \text{ cm}^2$$ $$\text{Area} = \frac{1 \times (2 \times 11) \times (7 \times 7)}{(2 \times 4) \times 7 \times 4} \text{ cm}^2$$

Cancel out common factors:

$$\text{Area} = \frac{11 \times 7}{4 \times 4} \text{ cm}^2$$ $$\text{Area} = \frac{77}{16} \text{ cm}^2$$

$$\text{Area} = 4.8125 \text{ cm}^2$$

The area of the sector is $\text{4.8125 cm}^2$.