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Q9(i):

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. 11.9. Find : (i) the total length of the silver wire required. (Unless stated otherwise, use $\pi = \frac{22}{7}$)

Solution :

Given:

1. The brooch is circular in shape.
2. The diameter of the circle ($d$) = $35\text{ mm}$.
3. The number of diameters used to divide the circle = $5$.
4. The value of $\pi = \frac{22}{7}$.

To Find:

The total length of the silver wire required to make the brooch.

Diameter = 35 mm

Step 1: Calculate the circumference of the circle.

The silver wire forms the outer boundary (circumference) of the circle. The formula for the circumference ($C$) of a circle is given by:
$C = \pi \times d$
Substituting the given values:
$C = \frac{22}{7} \times 35\text{ mm}$
$C = 22 \times 5\text{ mm}$ [Since $35 \div 7 = 5$]
$C = 110\text{ mm}$

Step 2: Calculate the length of the 5 diameters.

The wire is also used to make 5 diameters. The length of one diameter is $35\text{ mm}$.
Total length of 5 diameters = $5 \times d$
Total length of 5 diameters = $5 \times 35\text{ mm}$
Total length of 5 diameters = $175\text{ mm}$

Step 3: Calculate the total length of the silver wire.

The total length of the wire required is the sum of the circumference of the circle and the length of the 5 diameters.
Total Length = Circumference + (5 $\times$ Diameter)
Total Length = $110\text{ mm} + 175\text{ mm}$
Total Length = $285\text{ mm}$

Final Answer: The total length of the silver wire required is 285 mm.


More Questions from Class 10 Mathematics Areas Related to Circles EXERCISE 11.1


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