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.

(i) Given,

Diameter = 35 mm

Radius (r) = $\dfrac{\text{Diameter}}{2} = \dfrac{35}{2}$ = 17.5 mm

From figure,

Length of silver required = Circumference of circle + 5 × Diameter

= 2πr + 5 × 35

= $2 \times \dfrac{22}{7} \times 17.5$ + 175

= 2 × 22 × 2.5 + 175

= 110 + 175 = 285 mm.

Hence, total length of silver wire required = 285 mm.

(ii) There are ten sectors.

Angle subtended by each sector = $\dfrac{360°}{10}$ = 36°.

We know that,

Area of sector of angle θ and radius r = $\dfrac{θ}{360°} \times πr^2$

Substituting values we get :

$= \dfrac{36°}{360°} \times \dfrac{22}{7} \times (17.5)^2 \\[1em] = \dfrac{1}{10} \times 22 \times 2.5 \times 17.5 \\[1em] = \dfrac{962.5}{10} \\[1em] = \dfrac{9625}{100} \\[1em] = \dfrac{385}{4} \text{ mm}^2.$

Hence, area of each sector = $\dfrac{385}{4}$ mm².