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# 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 Fig. 12.12. Find :

(i) the total length of the silver wire required.

(ii) the area of each sector of the brooch.

**Solution:**

(i) Diameter of the brooch (d) = 35 mm

Total length of silver wire required = circumference of brooch + 5 × diameter

= πd + 5d

= (π + 5) × 35

= (22/7 + 5) × 35

= (22 + 35)/7 × 35

= 57 × 5

= 285 mm

(ii) Radius of the brooch (r) = 35/2 mm

The wire divides the brooch into 10 equal sectors.

So, angle of the sector (θ) = 360°/10 = 36°

∴ Area of each sector of the brooch = 36°/360° × πr^{2}

= 1/10 × 22/7 × 35/2 mm × 35/2 mm

= 96.25 mm^{2}

**☛ Check: **NCERT Solutions for Class 10 Maths Chapter 12

**Video Solution:**

## 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 Fig. 12.12. Find : (i) the total length of the silver wire required. (ii) the area of each sector of the brooch.

NCERT Solutions Class 10 Maths Chapter 12 Exercise 12.2 Question 9

**Summary:**

A brooch is made with silver wire in the form of a circle with a diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in figure (i) the total length of the silver wire required is 285 mm and (ii) the area of each sector of the brooch is 96.25 mm^{2 }respectively.

**☛ Related Questions:**

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