$A$ brooch is made with silver wire in the form of a circle with diameter $35 \, mm$. The wire is also used in making $5$ diameters which divide the circle into $10$ equal sectors as shown in the figure. Find:
$(i)$ The total length of the silver wire required.
$(ii)$ The area of each sector of the brooch. [Use $\pi = \frac{22}{7}$]

(N/A) The total length of wire required is the sum of the lengths of $5$ diameters and the circumference of the brooch.
Radius of the circle $r = \frac{35}{2} \, mm$.
Circumference of the brooch $= 2 \pi r = 2 \times \frac{22}{7} \times \frac{35}{2} = 110 \, mm$.
Length of $5$ diameters $= 5 \times 35 = 175 \, mm$.
Total length of wire required $= 110 + 175 = 285 \, mm$.
It can be observed from the figure that each of the $10$ sectors of the circle subtends an angle of $\frac{360^{\circ}}{10} = 36^{\circ}$ at the centre of the circle.
Therefore,the area of each sector $= \frac{36^{\circ}}{360^{\circ}} \times \pi r^{2} = \frac{1}{10} \times \frac{22}{7} \times \left(\frac{35}{2}\right) \times \left(\frac{35}{2}\right) = \frac{385}{4} \, mm^{2} = 96.25 \, mm^{2}$.